[ { "contents": "Turn a Word Doc into a PDF\n\nPDF stands for portable document format. It is a file type (.pdf) just as a Microsoft Word document is a text document (.doc).\n\nPDF is the preferred file type for online publishing because unlike a Word doc, which can be modified, PDFs preserve text and formatting and are easily downloaded to look exactly as it does online.\n\nHere’s how to quickly make a PDF from a Word file:\n\n1. Use a file name that’s all lowercase, inserting hyphens for spaces: vista-community-college.doc\n\n2. Open each Word doc and then for each doc select Print under File (in the main menu bar) just like you were going to print the page.\n\n3. Notice the PDF button on the bottom far left side of the window (see diagram). Select PDF and a drop-down menu appears with Save as PDF as the first option. Select it.\n\n4. Word will now create a PDF file where you want it on your computer (either on your desktop, in a selected folder, or on an external device). Notice now that the file name has changed its extension (vista-community-college.pdf). You now have a PDF, as well as your original Word doc.\n\nFor multiple Word docs, repeat steps 1 through 4 for each doc. Attach the PDF to an email just as you would a Word doc or other file attachment.\n\nViewers can download the free Adobe Reader software to view PDFs or use another image viewer, like Apple’s Preview or Microsoft Reader.", "id": "./materials/6.pdf" }, { "contents": "Investigação Operacional\nOrigem, Metodologias e Aplicações\n\n\"My business card says Data Mining, Performance Measures, and Decision Support. Those are my attempts to translate operations research into English.\"\n\n– Um consultor de IO independente\n\nElisa Correia de Barros\nDepartamento de Gestão Industrial – Instituto Politécnico de Bragança\nA origem da Investigação Operacional\n\nA Investigação Operacional (IO) é uma disciplina recente que utiliza modelos matemáticos e estatísticos para resolver problemas complexos na procura de uma solução ótima que possibilite a melhor tomada de decisão. De vasto campo aplicação, a IO abrange atualmente diversos saberes e técnicas, tais como a Programação Linear e Não-Linear, a Programação Dinâmica, a Simulação, a Teoria de Jogos, a Previsão, a Gestão de Projeto, etc. Identificar e tomar a melhor decisão que conduzisse à vitória foi sempre uma preocupação constante de quem fazia a guerra. Para o conseguirem, os militares recorreram frequentemente aos detentores do conhecimento em cada momento histórico.\n\nÉ por isso que muitos especialistas consideram que a IO data do século III A.C., quando, durante a Segunda Guerra Púnica, Siracusa, sitiada pelos romanos, se defendeu recorrendo à solução proposta por Arquimedes, com um sistema de espelhos que orientava a luz solar, assim conseguindo incendiar os navios inimigos. Em 1503, Leonardo da Vinci participou como engenheiro na guerra que opôs Pisa a Florença, pondo ao serviço desta cidade os seus conhecimentos de construção de navios e outros veículos, canhões, catapultas e outras máquinas de guerra.\n\nA IO sempre se apoiou na matemática, tendo tido enorme impacto os trabalhos que, nos séculos XVII e XVII, Newton, Leibniz, Bernoulli e Lagrange desenvolveram trabalhos ligados à obtenção de máximos e mínimos condicionados a determinadas funções. Ainda na mesma altura, o Fourier delineou os métodos da atual Programação Linear e, nos últimos anos do século XVIII, Monge estabeleceu os precedentes do Método Gráfico a partir dos seus estudos de geometria descritiva.\n\nNo final do século XIX, Taylor, considerado um dos pioneiros da moderna gestão, realizou um estudo que permitiu maximizar o rendimento dos mineiros ao determinar que a variável realmente significativa era o peso combinado entre pá e a sua carga. Deste modo, foram concebidas pás de acordo com os diferentes materiais usados.\n\nNo entanto, o início da IO é normalmente considerado como tendo ocorrido durante a II Guerra Mundial, quando os Aliados se viram confrontados com problemas (de natureza estratégia, tática e logística) de grande dimensão e complexidade. Para apoiar os comandos militares na resolução desses problemas, foram criados grupos multidisciplinares de cientistas em que se incluíam matemáticos, físicos e engenheiros. Como exemplos mais relevantes do trabalho desenvolvido por estes grupos de cientistas, são de referir, no Reino Unido, em 1939, o aumento da eficácia dos radares e o desempenho ótimo do sistema de defesa aérea britânico, fundamentais para a vitória na Batalha de Inglaterra, bem como, nos Estados Unidos da América, em 1942, a utilização de modelos matemáticos na movimentação de navios mercantes para romper o bloqueio que a marinha alemã impunha ao Reino Unido, tendo em conta restrições e condições reais tais como a carga a transportar, a velocidade máxima e o combustível necessário.\n\nNo fim da guerra, incentivado pelo sucesso da IO a nível militar, o mundo empresarial, nomeadamente o sector industrial, começou gradualmente a interessar-se por esta disciplina.\nAs equipas de IO tinham mostrado, no decurso do conflito anterior que eram capazes de resolver problemas complexos, envolvendo muitas variáveis, recorrendo a métodos que tinham permitido obter maior eficiência na utilização do armamento e valiosa economia em vidas humanas e material, sendo suscetíveis de aplicação no âmbito civil.\n\nOs problemas eram basicamente os mesmos que tinham sido tratados pelos militares, mas agora em diferentes contextos. Assim, embora a IO militar não tenha parado de se desenvolver, assistiu-se no período pós-guerra ao rápido crescimento da IO civil, na indústria, nos serviços e no Estado, com o intuito de estabelecer métodos de gestão mais racionais, quer no sector público quer no privado.\n\nPodem identificar-se, pelo menos dois fatores, que tiveram um papel essencial no rápido crescimento da IO durante este período:\n\n- Substancial progresso das técnicas matemáticas disponíveis na IO.\n Depois da guerra, os cientistas sentiam-se motivados para uma investigação profunda nesta nova disciplina, daqui resultando avanços muito importantes como o Método de Simplex para resolver problemas de programação linear, desenvolvido por George Dantzig em 1947.\n\n- Evolução/revolução informática.\n Normalmente é exigida uma grande quantidade de cálculos para tratar, mais eficientemente, os típicos problemas que caracterizam a IO. Contudo, o desenvolvimento da informática, materializado nos computadores, com capacidade para realizar cálculos aritméticos mil vezes, ou mesmo milhões de vezes, mais rapidamente que o Homem, bem como para processar enormes volumes de dados sobre as atividades das empresas, criou condições para que esses problemas de enorme complexidade pudessem ser eficaz e eficientemente resolvidos, assim evidenciando os benefícios decorrentes da utilização da IO.\n\n**Metodologia da Investigação Operacional - Ideias básicas**\n\nA abordagem da IO aplicada aos modelos matemáticos é própria do método científico, o qual é composto pelas seguintes fases:\n\n- **Observar**\n Definir o problema e recolher dados\n\n- **Fazer modelos matemáticos**\n Se possível reduzir o problema a um modelo bem conhecido (é importante ter um “catálogo” de problemas bem conhecidos)\n\n- **Obter soluções a partir do modelo**\n Otimizar resultados, baseados nesses modelos\n\n- **Testar o modelo**\n Verificar se os resultados fazem sentido\n Confirmar/rejeitar hipóteses\n• Preparação e implementação prática\n• Acompanhamento e verificação de resultados práticos\n\nNa prática, as diferentes fases do método científico raramente se sucedem na ordem indicada. Muitas podem ser simultâneas e, em vários estudos, por exemplo, a fase que consiste em formular o problema só fica completa quando a investigação está virtualmente terminada. O processo de investigação é normalmente cíclico. Por exemplo, se ao testar-se o modelo se conclui que ele é deficiente, a formulação do problema e a construção do modelo podem ser revistas e modificadas. Ou seja, as diferentes fases influenciam-se mutuamente durante o trabalho de investigação.\n\nAplicações da Investigação Operacional\n\nComo dissemos acima, os domínios de aplicação da Investigação Operacional são vastos, podendo citar-se fundamentalmente os seguintes:\n\n- Economia e especialmente Economia de Empresa, onde se situam as aplicações mais férteis e os estímulos mais fortes para os desenvolvimentos teóricos da Programação Linear;\n- Matemática, onde a Programação Linear tem impulsionado a obtenção de importantes resultados teóricos e o aperfeiçoamento das técnicas de análise numérica;\n- Militar, onde as aplicações são numerosas mas normalmente pouco divulgadas por razões de segurança.\n\nComo exemplos destas áreas de aplicação podemos referir mais explicitamente:\n\n• Gestão de empresas (determinação das quantidades a produzir dos diferentes produtos da empresa de acordo com os recursos disponíveis, as condições tecnológicas existentes e a situação do mercado.);\n• Problemas de Transportes (conhecido o custo de transporte de uma unidade de produto de cada origem para cada destino, procede-se à determinação do plano de distribuição que torna mínimo o custo total de transporte.);\n• «Trim-Loss» (determinação do número de unidades a cortar com determinadas dimensões de modo a minimizar os desperdícios envolvidos face às dimensões da produção. Exemplos: indústria do papel e do cartão, siderúrgica, têxtil, confeções, vidreira, ...);\n• Estrutura financeira dos bancos (o banco pretende estabelecer a estrutura do ativo que maximiza o seu lucro global, sabendo que devem ser respeitados os condicionalismos legais e de gestão que asseguram o equilíbrio financeiro.);\n• Problemas de Mistura (pretende-se obter, com custo mínimo ou lucro máximo, um ou vários produtos, a satisfazer certos requisitos técnicos, através de vários ingredientes possuidores em grau diferente dessas características técnicas. Exemplo: rações para animais, adubos, produtos alimentares e farmacêuticos, ligas metálicas, tintas, gasolinas.);\n• Planeamento Agrícola (o problema consiste em afectar recursos escassos, tais como superfície arável, mão-de-obra, água, etc..., à produção de diversos bens de modo a maximizar o resultado de exploração.).", "id": "./materials/65.pdf" }, { "contents": "Operational Research\nOrigins, Methods, and Applications\n\n\"My business card says Data Mining, Performance Measures, and Decision Support. Those are my attempts to translate operations research into English.\"\n\n– An independent OR consultant\n\nElisa Correia de Barros\nDepartment of Industrial Management - Polytechnic Institute of Bragança\nThe origins of Operational Research\n\nOperational Research (OR) is a discipline whose origins can be traced back to the second half of the 20th century that uses mathematical and statistical models to solve complex problems in the search for an optimal solution that enables the best decision making.\n\nWith a wide application field, OR currently covers several types of knowledge and techniques, such as Linear and Non-Linear Programming, Dynamic Programming, Simulation, Game Theory, Forecasting, Project Management, etc.\n\nIdentifying and making the best decision that would lead to victory has always been a constant concern of those who waged war. To achieve this, the military often turned to knowledge holders at every historic moment. That is why many experts consider that OR dates from the 3rd century BC, when, during the Second Punic War, Syracuse, besieged by the Romans, defended themselves using the solution proposed by Archimedes, with a system of mirrors that guided sunlight, thus managing to set fire to enemy ships.\n\nIn 1503, Leonardo da Vinci participated as an engineer in the war that opposed Pisa to Florence, putting his knowledge of building ships and other vehicles, cannons, catapults and other war machines at the service of this city.\n\nOR has always relied on mathematics, having had a huge impact on the works that, in the 17th and 17th centuries, Newton, Leibniz, Bernoulli and Lagrange developed works related to obtaining maximum and minimum conditions conditioned to certain functions.\n\nAt the same time, the Fourier outlined the methods of the current Linear Programming and, in the last years of the 18th century, Monge established the precedents of the Graphic Method from his studies of descriptive geometry.\n\nAt the end of the 19th century, Taylor, considered one of the pioneers of modern management, carried out a study that allowed to maximize the miners' income by determining that the really significant variable was the combined weight between shovel and its load. In this way, blades were designed according to the different materials used.\n\nAs most relevant examples of the work carried out by these groups of scientists, in the United Kingdom, in 1939, mention should be made of the increase in the efficiency of radars and the optimal performance of the British air defense system, fundamental to the victory at the Battle of England, as well as, in the United States of America, in 1942, the use of mathematical models in the movement of merchant ships to break the blockade that the German Navy imposed on the United Kingdom, taking into account restrictions and real conditions such as the cargo to be carried, the maximum speed and the necessary fuel.\n\nAt the end of the war, encouraged by the success of the OR at the military level, the business world, namely the industrial sector, gradually began to take an interest in this discipline.\nThe OR teams had shown, in the course of the previous conflict, that they were able to solve complex problems, involving many variables, using methods that had allowed to obtain greater efficiency in the use of armament and valuable savings in human and material lives, being susceptible of application in the civil sphere. The problems were basically the same as those that had been addressed by the military, but now in different contexts. Thus, although the military OR has not stopped developing, there has been a rapid growth in the post-war period of civilian OR, in industry, services and in the State, with the aim of establishing more rational management methods, both in public and private sectors.\n\nAt least two factors can be identified that played an essential role in the rapid growth of OR during this period:\n\n- **Substantial progress in mathematical techniques available at OR**\n After the war, scientists were motivated to further investigate this new discipline, resulting in very important advances such as the Simplex Method to solve linear programming problems, developed by George Dantzig in 1947.\n\n- **Computer evolution**\n Usually, a large number of calculations are required to treat, more efficiently, the typical problems that characterize OR. However, the development of information technology, materialized in computers, with the capacity to perform arithmetic calculations a thousand times, or even millions of times, faster than man, as well as to process huge volumes of data on the activities of companies, created conditions for these extremely complex problems could be effectively and efficiently solved, thus showing the benefits of using OR.\n\n**The methodology of Operational Research**\n\nThe IO approach applied to mathematical models is specific to the scientific method, which is composed of the following phases:\n\n- Define the problem of interest and gather relevant data\n- Formulate a mathematical model to represent the problem\n- Develop a computer-based procedure for deriving solutions to the problem from the model;\n- Test the model and refine it as needed;\n- Prepare for the ongoing application of the model;\n- Implement the model.\n\nHowever, in practice the different phases of the scientific method rarely succeed in the order indicated. Many can be simultaneous and, in several studies, for example, the phase of formulating the problem is only complete when the investigation is virtually finished. The investigation process is usually cyclical. For example, if the model is found to be defective when testing, the problem formulation and model construction can be reviewed and modified. That is, the different phases influence each other during the research work.\nApplications of Operational Research\n\nAs mentioned above, the fields of application of Operational Research are vast, with the following being the main ones:\n\n- Economy and especially Business Economy, where the most rewarding and inspiring applications and the strongest stimuli for theoretical developments in Linear Programming are located;\n- Mathematics, where Linear Programming has driven the achievement of important theoretical results and the improvement of numerical analysis techniques;\n- Military, where applications are numerous but usually under-publicized for security reasons.\n\nAs examples of these application areas we can refer more explicitly:\n\n- Management of companies (determination of the quantities to be produced of the company's different products according to available resources, existing technological conditions and the market situation.);\n- Transportation problems (the cost of transporting a unit of product from each origin to each destination is known, the distribution plan is determined, which minimizes the total transport cost.);\n- «Trim-Loss» (determination of the number of units to be cut with certain dimensions in order to minimize the waste involved in relation to the dimensions of production. Examples: paper and cardboard industry, steel, textile, clothing, glass, ...);\n- Banks' financial structure (the bank intends to establish the asset structure that maximizes its overall profit, knowing that the legal and management constraints that ensure financial balance must be respected);\n- Mixing problems (it is intended to obtain, with minimum cost or maximum profit, one or more products, to satisfy certain technical requirements, through various ingredients possessing a different degree of these technical characteristics. Example: animal feed, fertilizers, food and pharmaceutical products, metal alloys, paints, gasoline.);\n- Agricultural Planning (the problem is to allocate scarce resources, such as arable land, labor, water, etc, to the production of various goods in order to maximize the exploitation result.).", "id": "./materials/66.pdf" }, { "contents": "Investigação Operacional\nOrigem, Metodologias e Aplicações\n\n\"My business card says Data Mining, Performance Measures, and Decision Support. Those are my attempts to translate operations research into English.\"\n\n– Um consultor de IO independente\n\nElisa Correia de Barros\nDepartamento de Gestão Industrial – Instituto Politécnico de Bragança\nA origem da Investigação Operacional\n\nA Investigação Operacional (IO) é uma disciplina recente que utiliza modelos matemáticos e estatísticos para resolver problemas complexos na procura de uma solução ótima que possibilite a melhor tomada de decisão. De vasto campo aplicação, a IO abrange atualmente diversos saberes e técnicas, tais como a Programação Linear e Não-Linear, a Programação Dinâmica, a Simulação, a Teoria de Jogos, a Previsão, a Gestão de Projeto, etc. Identificar e tomar a melhor decisão que conduzisse à vitória foi sempre uma preocupação constante de quem fazia a guerra. Para o conseguirem, os militares recorreram frequentemente aos detentores do conhecimento em cada momento histórico.\n\nÉ por isso que muitos especialistas consideram que a IO data do século III A.C., quando, durante a Segunda Guerra Púnica, Siracusa, sitiada pelos romanos, se defendeu recorrendo à solução proposta por Arquimedes, com um sistema de espelhos que orientava a luz solar, assim conseguindo incendiar os navios inimigos. Em 1503, Leonardo da Vinci participou como engenheiro na guerra que opôs Pisa a Florença, pondo ao serviço desta cidade os seus conhecimentos de construção de navios e outros veículos, canhões, catapultas e outras máquinas de guerra.\n\nA IO sempre se apoiou na matemática, tendo tido enorme impacto os trabalhos que, nos séculos XVII e XVII, Newton, Leibniz, Bernoulli e Lagrange desenvolveram trabalhos ligados à obtenção de máximos e mínimos condicionados a determinadas funções. Ainda na mesma altura, o Fourier delineou os métodos da atual Programação Linear e, nos últimos anos do século XVIII, Monge estabeleceu os precedentes do Método Gráfico a partir dos seus estudos de geometria descritiva.\n\nNo final do século XIX, Taylor, considerado um dos pioneiros da moderna gestão, realizou um estudo que permitiu maximizar o rendimento dos mineiros ao determinar que a variável realmente significativa era o peso combinado entre pá e a sua carga. Deste modo, foram concebidas pás de acordo com os diferentes materiais usados.\n\nNo entanto, o início da IO é normalmente considerado como tendo ocorrido durante a II Guerra Mundial, quando os Aliados se viram confrontados com problemas (de natureza estratégia, tática e logística) de grande dimensão e complexidade. Para apoiar os comandos militares na resolução desses problemas, foram criados grupos multidisciplinares de cientistas em que se incluíam matemáticos, físicos e engenheiros. Como exemplos mais relevantes do trabalho desenvolvido por estes grupos de cientistas, são de referir, no Reino Unido, em 1939, o aumento da eficácia dos radares e o desempenho ótimo do sistema de defesa aérea britânico, fundamentais para a vitória na Batalha de Inglaterra, bem como, nos Estados Unidos da América, em 1942, a utilização de modelos matemáticos na movimentação de navios mercantes para romper o bloqueio que a marinha alemã impunha ao Reino Unido, tendo em conta restrições e condições reais tais como a carga a transportar, a velocidade máxima e o combustível necessário.\n\nNo fim da guerra, incentivado pelo sucesso da IO a nível militar, o mundo empresarial, nomeadamente o sector industrial, começou gradualmente a interessar-se por esta disciplina.\nAs equipas de IO tinham mostrado, no decurso do conflito anterior que eram capazes de resolver problemas complexos, envolvendo muitas variáveis, recorrendo a métodos que tinham permitido obter maior eficiência na utilização do armamento e valiosa economia em vidas humanas e material, sendo suscetíveis de aplicação no âmbito civil.\n\nOs problemas eram basicamente os mesmos que tinham sido tratados pelos militares, mas agora em diferentes contextos. Assim, embora a IO militar não tenha parado de se desenvolver, assistiu-se no período pós-guerra ao rápido crescimento da IO civil, na indústria, nos serviços e no Estado, com o intuito de estabelecer métodos de gestão mais racionais, quer no sector público quer no privado.\n\nPodem identificar-se, pelo menos dois fatores, que tiveram um papel essencial no rápido crescimento da IO durante este período:\n\n- Substancial progresso das técnicas matemáticas disponíveis na IO.\n Depois da guerra, os cientistas sentiam-se motivados para uma investigação profunda nesta nova disciplina, daqui resultando avanços muito importantes como o Método de Simplex para resolver problemas de programação linear, desenvolvido por George Dantzig em 1947.\n\n- Evolução/revolução informática.\n Normalmente é exigida uma grande quantidade de cálculos para tratar, mais eficientemente, os típicos problemas que caracterizam a IO. Contudo, o desenvolvimento da informática, materializado nos computadores, com capacidade para realizar cálculos aritméticos mil vezes, ou mesmo milhões de vezes, mais rapidamente que o Homem, bem como para processar enormes volumes de dados sobre as atividades das empresas, criou condições para que esses problemas de enorme complexidade pudessem ser eficaz e eficientemente resolvidos, assim evidenciando os benefícios decorrentes da utilização da IO.\n\n**Metodologia da Investigação Operacional - Ideias básicas**\n\nA abordagem da IO aplicada aos modelos matemáticos é própria do método científico, o qual é composto pelas seguintes fases:\n\n- Observar\n Definir o problema e recolher dados\n\n- Fazer modelos matemáticos\n Se possível reduzir o problema a um modelo bem conhecido (é importante ter um “catálogo” de problemas bem conhecidos)\n\n- Obter soluções a partir do modelo\n Otimizar resultados, baseados nesses modelos\n\n- Testar o modelo\n Verificar se os resultados fazem sentido\n Confirmar/rejeitar hipóteses\n• Preparação e implementação prática\n• Acompanhamento e verificação de resultados práticos\n\nNa prática, as diferentes fases do método científico raramente se sucedem na ordem indicada. Muitas podem ser simultâneas e, em vários estudos, por exemplo, a fase que consiste em formular o problema só fica completa quando a investigação está virtualmente terminada. O processo de investigação é normalmente cíclico. Por exemplo, se ao testar-se o modelo se conclui que ele é deficiente, a formulação do problema e a construção do modelo podem ser revistas e modificadas. Ou seja, as diferentes fases influenciam-se mutuamente durante o trabalho de investigação.\n\nAplicações da Investigação Operacional\n\nComo dissemos acima, os domínios de aplicação da Investigação Operacional são vastos, podendo citar-se fundamentalmente os seguintes:\n\n- Economia e especialmente Economia de Empresa, onde se situam as aplicações mais férteis e os estímulos mais fortes para os desenvolvimentos teóricos da Programação Linear;\n- Matemática, onde a Programação Linear tem impulsionado a obtenção de importantes resultados teóricos e o aperfeiçoamento das técnicas de análise numérica;\n- Militar, onde as aplicações são numerosas mas normalmente pouco divulgadas por razões de segurança.\n\nComo exemplos destas áreas de aplicação podemos referir mais explicitamente:\n\n• Gestão de empresas (determinação das quantidades a produzir dos diferentes produtos da empresa de acordo com os recursos disponíveis, as condições tecnológicas existentes e a situação do mercado.);\n• Problemas de Transportes (conhecido o custo de transporte de uma unidade de produto de cada origem para cada destino, procede-se à determinação do plano de distribuição que torna mínimo o custo total de transporte.);\n• «Trim-Loss» (determinação do número de unidades a cortar com determinadas dimensões de modo a minimizar os desperdícios envolvidos face às dimensões da produção. Exemplos: indústria do papel e do cartão, siderúrgica, têxtil, confeções, vidreira, ...);\n• Estrutura financeira dos bancos (o banco pretende estabelecer a estrutura do ativo que maximiza o seu lucro global, sabendo que devem ser respeitados os condicionalismos legais e de gestão que asseguram o equilíbrio financeiro.);\n• Problemas de Mistura (pretende-se obter, com custo mínimo ou lucro máximo, um ou vários produtos, a satisfazer certos requisitos técnicos, através de vários ingredientes possuidores em grau diferente dessas características técnicas. Exemplo: rações para animais, adubos, produtos alimentares e farmacêuticos, ligas metálicas, tintas, gasolinas.);\n• Planeamento Agrícola (o problema consiste em afectar recursos escassos, tais como superfície arável, mão-de-obra, água, etc..., à produção de diversos bens de modo a maximizar o resultado de exploração.).", "id": "./materials/67.pdf" }, { "contents": "Turn a Word Doc into a PDF\n\nPDF stands for portable document format. It is a file type (.pdf) just as a Microsoft Word document is a text document (.doc).\n\nPDF is the preferred file type for online publishing because unlike a Word doc, which can be modified, PDFs preserve text and formatting and are easily downloaded to look exactly as it does online.\n\nHere’s how to quickly make a PDF from a Word file:\n\n1. Use a file name that’s all lowercase, inserting hyphens for spaces: vista-community-college.doc\n\n2. Open each Word doc and then for each doc select Print under File (in the main menu bar) just like you were going to print the page.\n\n3. Notice the PDF button on the bottom far left side of the window (see diagram). Select PDF and a drop-down menu appears with Save as PDF as the first option. Select it.\n\n4. Word will now create a PDF file where you want it on your computer (either on your desktop, in a selected folder, or on an external device). Notice now that the file name has changed its extension (vista-community-college.pdf). You now have a PDF, as well as your original Word doc.\n\nFor multiple Word docs, repeat steps 1 through 4 for each doc. Attach the PDF to an email just as you would a Word doc or other file attachment.\n\nViewers can download the free Adobe Reader software to view PDFs or use another image viewer, like Apple’s Preview or Microsoft Reader.", "id": "./materials/7.pdf" }, { "contents": "Is \\( \\{\\emptyset\\} \\) an empty set?\nA collection of objects known as elements is called set. An element can be almost anything, such as numbers, functions, or lines. A set can be finite or infinite as well.\n\n**Example**\n\n1. \\( A = \\{1, 3, 5, 7, 9\\} \\) is an example of finite set.\n2. \\( \\mathbb{Z} \\) is a set of integers which is an example of finite set.\n3. The elements of set is not only limited to numbers, the elements of set can be anything, \\( B = \\{\\text{cow, donkey, rat, horse}\\} \\) is also a valid set.\n4. Sets can also be written in set builder notation. Like \\( A = \\{x \\in \\mathbb{N} \\mid x \\geq 4 \\text{ and } x \\leq 10\\} \\) which is same as \\( A = \\{4, 5, 6, 7, 8, 9, 10\\} \\)\nProperties of Set\n\nProperties\n\n1. Order of elements in a set doesn’t matter.\n2. If one or many elements of a set are repeated, the set remains the same.\n For example \\{1, 2, 3, 1, 2, 3, 1, 2, 3\\} is same as just \\{1, 2, 3\\}.\n3. Two sets are considered equal if and only if every element of each set is an element of the other.\n\nSymbol \\(\\in\\) is used to denote a element belongs to a set. For example: \\(X = \\{a, e, i, o, u\\}\\) Then, \\(a \\in X\\) but \\(b \\notin X\\) or \\(\\{a\\} \\notin X\\).\nSome important Sets\n\n| Symbol | Name |\n|--------|-----------------------------|\n| Z | The set of Integers. |\n| N | The set of Natural numbers. |\n| Q | The set of Rational numbers.|\n| R | The set of Real numbers. |\n| C | The set of Complex numbers. |\n\nEmpty set is a set without any elements, represented by \\{\\} or \\emptyset.\n\nA set with only one element is called a singleton set. For example, \\(X = \\{a\\}\\).\nSo, Is \\( \\{\\emptyset\\} \\) an empty set?\n\n**NO**, \\( \\{\\emptyset\\} \\) is not an empty set and is a singleton set (it has element \\( \\emptyset \\) in it). Empty set is indicated by \\( \\{\\} \\) or \\( \\emptyset \\).", "id": "./materials/70.pdf" }, { "contents": "Evaluate \\( \\int_{-3}^{1} 6x^2 - 3x - 2 \\, dx \\)\n\n- I = \\([-3, 1]\\) is a closed interval.\n\n- \\( f(x) = 6x^2 - 3x - 2 \\) is continuous on I since it is a quadratic function\n\n- \\( F(x) = \\int 6x^2 - 3x - 2 \\, dx = \\int 6x^2 \\, dx - \\int 3x \\, dx - \\int 2 \\, dx \\)\n\n\\[\n= 2x^3 - \\frac{3x^2}{2} - 2x + C\n\\]\n\nThen, by using Fundamental theorem of Calculus\n\n\\[\n\\int_{-3}^{1} 6x^2 - 3x - 2 \\, dx = \\left[ 2x^3 - \\frac{3x^2}{2} - 2x \\right]_{-3}^{1}\n\\]\n\n\\[\n= 2 \\cdot x^3 \\bigg|_{-3}^{1} - 3 \\cdot \\frac{x^2}{2} \\bigg|_{-3}^{1} - 2 \\cdot x \\bigg|_{-3}^{1}\n\\]\n\n\\[\n= 2 \\cdot (1 + 27) - 3 \\cdot \\left( \\frac{1}{2} - \\frac{9}{2} \\right) - 2 \\cdot (1 + 3)\n\\]\n\n\\[\n= 60\n\\]", "id": "./materials/71.pdf" }, { "contents": "Evaluate \\( \\int_{1}^{9} \\frac{y - 1}{\\sqrt{y}} \\, dy \\)\n\n- \\( I = [1, 9] \\) is a closed interval.\n\n- \\( f(x) = \\frac{y - 1}{\\sqrt{y}} \\) is continuous on \\( I \\).\n\n- \\( F(x) = \\int \\frac{y - 1}{\\sqrt{y}} \\, dy = \\int \\frac{y}{\\sqrt{y}} \\, dy - \\int \\frac{1}{\\sqrt{y}} \\, dy \\)\n\n\\[\n= \\int \\sqrt{y} \\, dy - \\int y^{-\\frac{1}{2}} \\, dy\n\\]\n\n\\[\n= \\frac{2}{3} \\cdot y^{\\frac{3}{2}} - 2 \\cdot \\sqrt{y} + C\n\\]\n\nThen, by using Fundamental theorem of Calculus\n\n\\[\n\\int_{1}^{9} \\frac{y - 1}{\\sqrt{y}} \\, dy = \\left[ \\frac{2}{3} \\cdot y^{\\frac{3}{2}} - 2 \\cdot \\sqrt{y} \\right]_{1}^{9} \\, dx\n\\]\n\n\\[\n= \\frac{2}{3} \\cdot \\left[ y^{\\frac{3}{2}} \\right]_{1}^{9} - 2 \\cdot \\left[ \\sqrt{y} \\right]_{1}^{9} = \\frac{40}{3}\n\\]", "id": "./materials/72.pdf" }, { "contents": "Evaluate \\( \\int_{1}^{9} \\frac{y - 1}{\\sqrt{y}} \\, dy \\)\n\n- \\( I = [1, 9] \\) is a closed interval.\n\n- \\( f(x) = \\frac{y - 1}{\\sqrt{y}} \\) is continuous on \\( I \\).\n\n- \\( F(x) = \\int \\frac{y - 1}{\\sqrt{y}} \\, dy = \\int \\frac{y}{\\sqrt{y}} \\, dy - \\int \\frac{1}{\\sqrt{y}} \\, dy \\)\n\n \\[\n = \\int \\sqrt{y} \\, dy - \\int y^{-\\frac{1}{2}} \\, dy\n \\]\n\n \\[\n = \\frac{2}{3} \\cdot y^{\\frac{3}{2}} - 2 \\cdot \\sqrt{y} + C\n \\]\n\nThen, by using Fundamental theorem of Calculus\n\n\\[\n\\int_{1}^{9} \\frac{y - 1}{\\sqrt{y}} \\, dy = \\left[ \\frac{2}{3} \\cdot y^{\\frac{3}{2}} - 2 \\cdot \\sqrt{y} \\right]_{1}^{9} \\, dx\n\\]\n\n\\[\n= \\frac{2}{3} \\cdot \\left[ y^{\\frac{3}{2}} \\right]_{1}^{9} - 2 \\cdot \\left[ \\sqrt{y} \\right]_{1}^{9} = \\frac{40}{3}\n\\]", "id": "./materials/73.pdf" }, { "contents": "Evaluate \\( \\int_{0}^{1} x \\left(2x^2 + 1\\right)^2 \\, dx \\)\n\n- \\( I = [0, 1] \\) is a closed interval.\n\n- \\( f(x) = x \\left(2x^2 + 1\\right)^2 \\) is continuous on \\( I \\).\n\n- \\( F(x) = \\int x \\left(2x^2 + 1\\right)^2 \\, dx = \\frac{1}{4} \\int 4x \\left(2x^2 + 1\\right)^2 \\, dx \\)\n\n\\[\n= \\frac{(2x^2 + 1)^3}{12} + C\n\\]\n\nRemember that, \\( \\int f' f^n \\, dx = \\frac{f^{n+1}}{n+1} + C \\)\n\nThen, by using Fundamental theorem of Calculus\n\n\\[\n\\int_{0}^{1} x \\left(2x^2 + 1\\right)^2 \\, dx = \\left[ \\frac{(2x^2 + 1)^3}{12} \\right]_{0}^{1}\n\\]\n\n\\[\n= \\frac{1}{12} \\cdot \\left[ (2x^2 + 1)^3 \\right]_{0}^{1} = \\frac{13}{6}\n\\]", "id": "./materials/74.pdf" }, { "contents": "What is $A \\cup \\emptyset$ and $A \\cap \\emptyset$?\n\nRoshan Poudel\n\nInstituto Politécnico de Bragança, Bragança, Portugal\nUnion\n\n1. The union of two sets is the set containing all of the elements from both of those sets. It is represented by the symbol $\\cup$.\n\n2. $A \\cup B = \\{x \\mid x \\in A \\text{ and } x \\in B\\}$\n\nIntersection\n\n1. The intersection of two sets is the set containing just the elements that are in both of those sets. It is represented by the symbol $\\cap$.\n\n2. $A \\cap B = \\{x \\mid x \\in A \\text{ or } x \\in B\\}$\nUnion and Intersection - Venn Diagram\n\nFigure 1: $A \\cup B$\n\nFigure 2: $A \\cap B$\nIf $A = \\{1, 3, 5, 7, 9, 11, 13, 15\\}$ and $B = \\{2, 3, 5, 7, 11, 13\\}$, then:\n\n- $A \\cup B = \\{1, 2, 3, 5, 7, 9, 11, 13, 15\\}$\n- $A \\cap B = \\{3, 5, 7, 11, 13\\}$\nWhat is $A \\cup \\emptyset$ and $A \\cap \\emptyset$?\n\n**Finding $A \\cup \\emptyset$.**\n\nThe empty set is the set with no elements so, the union of any set $A$ and the $\\emptyset$ is always going to be $A$.\n\n$$A \\cup \\emptyset = A$$\n\n**Finding $A \\cap \\emptyset$.**\n\nAn empty set is a set with no elements so, the intersection of any set $A$ and $\\emptyset$ is always going to be $\\emptyset$ as there is no element simultaneously belonging to both the sets.\n\n$$A \\cap \\emptyset = \\emptyset$$\nOther Set Operations\n\n**Difference**\n\n1. The difference of any two sets $A$ and $B$ written as $A - B$ which is the set containing the elements that are in $A$ but not in $B$.\n2. $A - B = \\{x \\mid x \\in A \\text{ and } x \\notin B\\}$\n3. For two disjoint sets $A$ and $B$, $A - B = A$ and $B - A = B$.\n\n**Compliment**\n\n1. For a set $A$ in a universe $U$, the compliment of $A$ or $\\overline{A}$ is set of all the elements that are in the universe but not in $A$.\n2. $\\overline{A} = \\{x \\in U \\mid x \\notin A\\}$\nFigure 3: $A - B$\n\nFigure 4: $\\overline{A}$\nConsider $A = \\{a, e, i, o, u\\}$ and $B = \\{a, b, c, d, e\\}$, and find $A - B$ and $B - A$.\n\n- $A - B = \\{i, o, u\\}$\n- $B - A = \\{b, c, d\\}$\n\nNotice that $A - B \\neq B - A$\nIf the universe $U$ is the set of letters in the English alphabet and $A$ is the set of the consonant letters of the same alphabet, what is $\\overline{A}$?\n\n- $\\overline{A} = \\{a, e, i, o, u\\}$\n\n- Also, $\\overline{U} = \\emptyset$ and $\\overline{\\emptyset} = U$\n## Properties of Set Operations\n\n### Commutative\n\n1. \\( A \\cup B = B \\cup A \\)\n2. \\( A \\cap B = B \\cap A \\)\n\n### Associative\n\n1. \\( A \\cup (B \\cup C) = (A \\cup B) \\cup C \\)\n2. \\( A \\cap (B \\cap C) = (A \\cap B) \\cap C \\)\n\n### Idempotent\n\n1. \\( A \\cup A = A \\)\n2. \\( A \\cap A = A \\)\nDe Morgan’s Laws\n\nDe Morgan’s Law\n\n1. \\((A \\cup B) = \\overline{A} \\cap \\overline{B}\\)\n\n2. \\((A \\cap B) = \\overline{A} \\cup \\overline{B}\\)\n\nClick here to check the proof here", "id": "./materials/76.pdf" }, { "contents": "Is \\( \\{\\emptyset\\} \\) an empty set?\n\nRoshan Poudel\n\nInstituto Politécnico de Bragança, Bragança, Portugal\nA collection of objects that somehow share a common feature - the elements - is called a set. An element can be of any nature, depending on the problem under consideration, such as numbers, functions, or lines. A set can be finite or infinite.\n\n**Example**\n\n1. $A = \\{1, 3, 5, 7, 9\\}$ is an example of a finite set.\n2. $\\mathbb{Z}$, the set of the integers, is an example of a finite set.\n3. The elements of a set are not only limited to numbers, the elements of a set can be anything, $B = \\{\\text{cow, donkey, rat, horse}\\}$ is also a set.\n4. Sets can also be written in set builder notation: $A = \\{x \\in \\mathbb{N} \\mid x \\geq 4 \\text{ and } x \\leq 10\\}$ which is same as $A = \\{4, 5, 6, 7, 8, 9, 10\\}$\nProperties of Set\n\nProperties\n\n1. The order of the elements in a set doesn’t matter.\n2. If one or more elements of a set are repeated, the set remains the same.\n For example \\{1, 2, 3, 1, 2, 3, 1, 2, 3\\} is the same as just \\{1, 2, 3\\}.\n3. Two sets are considered equal if and only if each element of one set is an element of the other.\n\nSymbol \\(\\in\\) is used to denote that an element belongs to a set. For example: \\(X = \\{a, e, i, o, u\\}\\) Then, \\(a \\in X\\) but \\(b \\notin X\\) or \\(\\{a\\} \\notin X\\).\nSome important Sets\n\n| Symbol | Name |\n|--------|-------------------------------------------|\n| $\\mathbb{Z}$ | The set of integers. |\n| $\\mathbb{N}$ | The set of natural numbers. |\n| $\\mathbb{Q}$ | The set of rational numbers. |\n| $\\mathbb{R}$ | The set of real numbers. |\n| $\\mathbb{C}$ | The set of complex numbers. |\n\nThe empty set is a set without any elements, represented by $\\{\\}$ or $\\emptyset$.\n\nA set with only one element is called a singleton set. For example $X = \\{a\\}$. \nSo, Is \\( \\{\\emptyset\\} \\) an empty set?\n\n**NO**, \\( \\{\\emptyset\\} \\) is not an empty set; it is a singleton set (it has the element \\( \\emptyset \\) in it). Empty set is indicated by \\( \\{\\} \\) or \\( \\emptyset \\).", "id": "./materials/77.pdf" }, { "contents": "Determine $|\\mathcal{P}(\\mathcal{P}(\\{\\phi, \\tau\\}))|$. \n\nRoshan Poudel\n\nInstituto Politécnico de Bragança, Bragança, Portugal\nSet $A$ is a subset of set $B$ iff each element of set $A$ is also an element of set $B$. If set $A$ is a subset of set $B$ then we write as $A \\subset B$.\n\n1. If each element of set $A$ is also an element of set $B$ and $B$ may be equal to $A$, then set $A$ is an **improper subset** of set $B$.\n\n **For example:** $A = \\{1, 2, 3, 4, 5\\}$ and $B = \\{1, 2, 3, 4, 5\\}$ then $A \\subseteq B$ and $B \\subseteq A$.\n\n2. If each element of set $A$ is also element of set $B$ but set $B$ is not equal to set $A$ then Set $A$ is **proper subset** of set $B$.\n\n **For example:** $A = \\{2, 3, 4\\}$ and $B = \\{1, 2, 3, 4, 5\\}$ then $A \\subset B$ but $A \\not\\subset B$\n## Properties of Subset\n\n| | |\n|---|---|\n| 1 | A set with n elements has $2^n$ subsets. |\n| 2 | Every set is subset of itself. |\n| 3 | Empty set ($\\emptyset$) is subset of every set. |\n| 4 | $A = B$ if and only if $A \\subseteq B$ and $B \\subseteq A$. |\n| 5 | A is a subset of B if and only if their intersection is equal to A, that is, $A \\subseteq B \\iff (A \\cap B) = A$. |\n| 6 | Set A is a subset of B if and only if their union is equal to B, that is, $A \\subseteq B \\iff (A \\cup B) = B$. |\nWhat are the subsets of set $A = \\{x, y, z\\}$?\n\n- $\\emptyset$\n- $\\{x\\}$\n- $\\{y\\}$\n- $\\{z\\}$\n- $\\{x, y\\}$\n- $\\{x, z\\}$\n- $\\{y, z\\}$\n- $\\{x, y, z\\}$\n\nNotice, there are 8 subsets of set $A$ which is also the result of $2^{|A|} = 2^3 = 8$\nA set $A$ is a superset of another set $B$ if all elements of the set $B$ are elements of the set $A$. The notation for superset is $A \\supset B$.\n\n**Properties**\n\n- $A \\supset \\emptyset$.\n- Since every set is a subset of itself, then every set is also a superset of itself.\nThe set of all subsets of a set $A$ is called the power set of $A$. The power set of $A$ is denoted with the symbol $\\mathcal{P}(A)$.\n\n**Example**\n\nIf $A$ is the set $\\{1, 2, 3\\}$, then what is $\\mathcal{P}(A)$?\n\n$$\\mathcal{P}(A) = \\{\\emptyset, \\{1\\}, \\{2\\}, \\{3\\}, \\{1, 2\\}, \\{1, 3\\}, \\{2, 3\\}, \\{1, 2, 3\\}\\}$$\nAs we know, for any set $A$, $|\\mathcal{P}(A)| = 2^{|A|}$.\n\nIn this case,\n\n$$|\\{\\phi, \\tau\\}| = 2$$\n\nTherefore,\n\n$$|\\mathcal{P}(\\{\\phi, \\tau\\})| = 2^2 = 4$$\n\n$$|\\mathcal{P}(\\mathcal{P}(\\{\\phi, \\tau\\}))| = 2^4 = 16$$\n\nSo, $|\\mathcal{P}(\\mathcal{P}(\\{\\phi, \\tau\\}))| = 16$. ", "id": "./materials/78.pdf" }, { "contents": "Evaluate \\( \\int_{0}^{2} x e^{1-x^2} \\, dx \\)\n\n- I = [0, 2] is a closed interval.\n\n- \\( f(x) = x e^{1-x^2} \\) is continuous on I.\n\n- \\( F(x) = \\int x e^{1-x^2} \\, dx = -\\frac{1}{2} \\int -2x \\cdot e^{1-x^2} \\, dx \\)\n\n\\[\n= -\\frac{1}{2} \\cdot \\frac{e^{1-x^2}}{\\ln(e)} + C\n\\]\n\n\\[\n= -\\frac{e^{1-x^2}}{2} + C \\quad \\because \\ln(e) = 1\n\\]\n\nRemember that, \\( \\int f' a^f \\, dx = \\frac{a^f}{\\ln(a)} + C \\)\n\nThen, by using Fundamental theorem of Calculus\n\n\\[\n\\int_{0}^{2} x e^{1-x^2} \\, dx = \\left[ -\\frac{e^{1-x^2}}{2} \\right]_{0}^{2} = -\\frac{e^{1-4}}{2} - \\left( -\\frac{e^{1}}{2} \\right)\n\\]\n\n\\[\n= \\frac{e}{2} - \\frac{e^3}{2}\n\\]", "id": "./materials/79.pdf" }, { "contents": "Ordinanza del Presidente della Giunta Regionale N° 46 del 29 Aprile 2020\n\nOggetto:\nUlteriori misure per il contrasto ed il contenimento sul territorio regionale della diffusione del virus COVID-19 in materia di attività motoria - Revoca dell'ordinanza n. 45 del 29 aprile 2020\n\nDipartimento Proponente: DIREZIONE DIFESA DEL SUOLO E PROTEZIONE CIVILE\n\nStruttura Proponente: DIREZIONE DIFESA DEL SUOLO E PROTEZIONE CIVILE\n\nPubblicità/Pubblicazione: Atto soggetto a pubblicazione integrale (PBURT/BD)\nIL PRESIDENTE DELLA GIUNTA REGIONALE\n\nVisti gli articoli 32 e 117, comma 3, della Costituzione;\n\nVisto lo Statuto della Regione Toscana;\n\nVisto l’articolo 117, comma 1 del decreto legislativo 31 marzo 1998, n.112, in base al quale le regioni sono abilitate ad adottare provvedimenti d’urgenza in materia sanitaria;\n\nVista la legge 23 dicembre 1978, n. 833, recante “Istituzione del servizio sanitario nazionale” e, in particolare, l’articolo 32;\n\nPreso atto della delibera del Consiglio dei Ministri del 31 gennaio 2020 con la quale è stato dichiarato, per sei mesi, lo stato di emergenza sul territorio nazionale relativo al rischio sanitario connesso all’insorgenza di patologie derivanti da agenti virali trasmissibili;\n\nVista l’Ordinanza del Capo del Dipartimento della protezione civile n. 630 del 3 febbraio 2020, recante “Primi interventi urgenti di protezione civile in relazione all’emergenza relativa al rischio sanitario connesso all’insorgenza di patologie derivanti da agenti virali trasmissibili” e seguenti recanti ulteriori interventi urgenti in relazione all’emergenza in corso;\n\nRichiamato altresì il decreto del Capo del Dipartimento della Protezione civile rep. n. 630 del 27.02.2020 con cui il sottoscritto è nominato soggetto attuatore ai sensi della citata OCDPC n. 630/2020;\n\nVista l’Ordinanza del Presidente della Giunta regionale n.7 del 04 marzo 2020 avente ad oggetto “Definizione delle strutture organizzative per la gestione dell'emergenza epidemiologica da COVID-19. Revoca ordinanza n. 4/2020”;\n\nVisto il decreto legge 23 febbraio 2020, n.6 recante “Misure urgenti in materia di contenimento e gestione dell’emergenza epidemiologica da Covid 19”, convertito, con modificazioni,dalla legge 5 marzo 2020, n.13, successivamente abrogato dal decreto legge 25 marzo 2020, n.19, ad eccezione dell’articolo 3, comma 6bis, e dell’articolo 4;\n\nVisto il Decreto-legge 2 marzo 2020, n. 9, recante “Misure urgenti di sostegno per famiglie, lavoratori e imprese connesse all'emergenza epidemiologica da COVID-19”;\n\nVisto il decreto legge 25 marzo 2020, n.19 recante “Misure urgenti per fronteggiare l’emergenza epidemiologica da Covid-19”, che ai sensi dell’articolo 2, comma 3 fa salvi gli effetti prodotti e gli atti adottati sulla base dei decreti e delle ordinanze emanati ai sensi del decreto legge 23 febbraio 2020, n.6;\n\nVisto il Decreto del Presidente del Consiglio dei Ministri del 10 aprile 2020 - \"Ulteriori disposizioni attuative del decreto-legge 25 marzo 2020, n. 19, recante misure urgenti per fronteggiare l'emergenza epidemiologica da COVID-19, applicabili sull'intero territorio nazionale”;\n\nVisto, in particolare, che il Decreto del Presidente del Consiglio dei Ministri del 10 aprile 2020, sopra citato abroga il decreto del Presidente del Consiglio dei ministri 8 marzo 2020, il decreto del Presidente del Consiglio dei ministri 9 marzo 2020, il decreto del Presidente del Consiglio dei ministri 11 marzo 2020, il decreto del Presidente del Consiglio dei ministri 22 marzo 2020 e il decreto del Presidente del Consiglio dei ministri 1° Aprile 2020, imponendo misure urgenti per il contenimento del contagio sull’intero territorio nazionale fino alla data del 3 maggio 2020;\nVisto il Decreto del Presidente del Consiglio dei Ministri del 26 aprile 2020 “Ulteriori disposizioni attuative del decreto-legge 23 febbraio 2020, n. 6, recante misure urgenti in materia di contenimento e gestione dell'emergenza epidemiologica da COVID-19, applicabili sull'intero territorio nazionale”\n\nConsiderato che l’articolo 1, comma 1, lettera f) del citato DPCM 10 aprile 2020 vieta lo svolgimento dell’attività ludica o creativa all’aperto e consente lo svolgimento di attività motoria individualmente in prossimità della propria abitazione nel rispetto delle misure di distanziamento sociale;\n\nRitenuto determinante, alla luce dell’esperienza maturata, agli effetti del contenimento del contagio, la misura del distanziamento sociale almeno di 1,8 metri e l’utilizzo dei dispositivi di protezione individuale quali mascherine, con conseguente possibilità di estensione della movimentazione delle persone nel rispetto di tali condizioni;\n\nRitenuto che l’ampliamento della possibilità di spostamento nel rispetto delle sopra citate modalità, risponde ad esigenze di tutela della salute individuale e collettiva e del benessere psico-fisico dei minori;\n\nRitenuto di considerare le passeggiate all’aria aperta e l’utilizzo della bicicletta quale tipologia di attività motoria consentita nel territorio regionale;\n\nVisto anche l’articolo 1, comma 1, lettera f) del DPCM 26 aprile 2020;\n\nConsiderato, pertanto, che per non determinare conseguenze negative a danno della salute di tutti i cittadini e del benessere psico-fisico dei minori sia opportuno consentire per l'attività motoria svolgere passeggiate all’aria aperta e utilizzare la bicicletta, con partenza e rientro alla propria abitazione, nell’ambito del comune di residenza in modo individuale, da parte di genitori e figli minori, da parte di accompagnatori di persone non completamente autosufficienti, o da parte di residenti nella stessa abitazione;\n\nRitenuto che nello svolgimento delle attività motorie di cui sopra fra i genitori e figli minori o da parte di accompagnatori di persone non completamente autosufficienti, o da parte di residenti nella medesima abitazione non è necessario mantenere le misure di distanziamento sociale di almeno 1,8 metri;\n\nRitenuto che il potere di ordinanza regionale, in specie ai fini dell’adozione di misure di contenimento rigorosamente funzionali alla tutela della salute trovi tuttora fondamento negli articoli 32 e 117, comma 3, della Costituzione oltre che negli articoli 32 della l.833/1978 e 117 del d.lgs n. 112/1998;\n\nORDINA\n\nai sensi dell’articolo 32, comma 3 della legge 23 dicembre 1978, n.833 in materia di igiene e sanità pubblica le seguenti misure:\n\n1. E’ consentito per l’attività motoria svolgere passeggiate all’aria aperta e utilizzare la bicicletta, con partenza e rientro alla propria abitazione, nell’ambito del comune di residenza in modo individuale, da parte di genitori e figli minori, da parte di accompagnatori di persone non completamente autosufficienti, o da parte di residenti nella stessa abitazione;\n\n2. nello svolgimento delle attività motorie di cui al punto 1 da parte di genitori e figli minori, da parte di accompagnatori di persone non completamente autosufficienti, o da parte di residenti nella\nmedesima abitazione non è necessario mantenere le misure di distanziamento sociale di almeno 1,8 metri;\n\n**DISPOSIZIONI FINALI**\n\nL'ordinanza n. 45 del 29 aprile 2020 - Ulteriori misure per il contrasto ed il contenimento sul territorio regionale della diffusione del virus COVID-19 in materia di attività motoria è revocata e pertanto non si provvede alla pubblicazione integrale sul B.U.R.T.\n\nLa presente ordinanza ha validità dal 1 maggio e fino alla vigenza delle misure adottate dal Presidente del Consiglio dei Ministri ai sensi dell’articolo 1, comma 2, dello stesso d.l.19/2020;\n\nLa presente ordinanza, per gli adempimenti di legge, è trasmessa:\n- al Presidente del Consiglio dei Ministri e al Ministro della Salute;\n- ai Prefetti;\n- ai Sindaci;\n\nIl mancato rispetto delle misure di cui alla presente Ordinanza è sanzionato secondo quanto previsto dall’articolo 4 del d.l.19/2020;\n\nAvverso la presente ordinanza è ammesso ricorso giurisdizionale innanzi al Tribunale Amministrativo Regionale nel termine di sessanta giorni dalla pubblicazione, ovvero ricorso straordinario al Capo dello Stato entro il termine di giorni centoventi.\n\nIl presente provvedimento è pubblicato integralmente sul B.U.R.T. ai sensi degli articoli 4, 5 e 5 bis della legge regionale n. 23/2007 e nella banca dati degli atti amministrativi della Giunta regionale ai sensi dell’articolo 18 della medesima legge.\n\nIl Presidente", "id": "./materials/8.pdf" }, { "contents": "Evaluate \\( \\int_{0}^{2} 5^y \\, dy \\)\n\n**HINT:** \\( \\int f'(a^y) \\, dy = \\frac{a^y}{\\ln a} + C \\)", "id": "./materials/80.pdf" }, { "contents": "Evaluate \\( \\int_{1}^{3} \\frac{3x}{x^2 + 1} \\, dx \\)\n\n**HINT:** Use \\( \\int \\frac{f'}{f} \\, dx = \\ln |f| + C \\)\n\nWhere \\( f = x^2 + 1 \\) and \\( f' = (x^2 + 1)' = 2x \\)", "id": "./materials/81.pdf" }, { "contents": "Evaluate \\( \\int_{1}^{e} \\frac{\\ln^3(x)}{x} \\, dx \\)\n\nAt first, checking if the necessary conditions to use Fundamental theorem of calculus are fulfilled or not.\n\nConditions:\n\n- \\( I = [1, e] \\) is a closed interval.\n- \\( f(x) = \\frac{\\ln^3(x)}{x} \\) is continuous on \\( I \\).\n- \\( F(x) = \\int f(x) \\, dx = \\int \\frac{\\ln^3(x)}{x} \\, dx \\)\n\nAs we know,\n\n\\[\n\\int f' \\, f^n \\, dx = \\frac{f^{n+1}}{n+1} + C\n\\]\n\nIn this case,\n\n\\[\nf = \\ln(x) \\\\\nf' = \\frac{1}{x}\n\\]\n\nNow, using Fundamental theorem of calculus\n\n\\[\n\\int_{1}^{e} \\frac{\\ln^3(x)}{x} \\, dx = \\left[ \\frac{\\ln^4(x)}{4} \\right]_{1}^{e} \\\\\n= \\frac{\\ln^4(e)}{4} - \\frac{\\ln^4(1)}{4} \\\\\n= \\frac{1}{4}\n\\]", "id": "./materials/82.pdf" }, { "contents": "Evaluate \\( \\int_0^1 \\frac{8}{x^2 + 1} \\, dx \\)\n\n**HINT:** \\( \\int \\frac{f'}{a^2 + f^2} \\, dx = \\frac{1}{a} \\arctan\\left(\\frac{f}{a}\\right) + C \\)\n\nWhere \\( a = 1, f = x, f' = 1 \\)", "id": "./materials/83.pdf" }, { "contents": "Evaluate \\( \\int_{-1}^{0} x \\sqrt{1+x} \\, dx \\)\n\n* All the conditions required for fundamental theorem of calculus are fulfilled\n\n\\[ I(x) = \\int x \\sqrt{1+x} \\, dx \\]\n\n\\[ g(x) \\quad f(x) \\]\n\nA.C.\n\nIn this case, \\( x \\) should be \\( g(x) \\) because it reduces the degree to 0, thus avoiding the repetition of integration by parts.\n\n\\[ \\int f(x) \\cdot g(x) \\, dx = F(x) \\cdot g(x) - \\int F(x) \\cdot g'(x) \\, dx \\]\n\n\\[ g'(x) = (x)' = 1 \\]\n\n\\[ F(x) = \\int f(x) \\, dx = \\int \\sqrt{1+x} \\, dx \\]\n\n\\[ = \\frac{2}{3} (1+x)^{3/2} + C \\]\n\n\\[ I(x) = \\int x \\cdot \\sqrt{1+x} \\, dx \\]\n\n\\[ = \\frac{2x}{3} (1+x)^{3/2} - \\frac{2}{3} \\int (1+x)^{5/2} \\, dx \\]\n\n\\[ = \\frac{2x}{3} (1+x)^{3/2} - \\frac{2}{3} \\cdot \\frac{2}{5} (1+x)^{5/2} + C \\]\n\n\\[ = \\frac{2x}{3} (1+x)^{3/2} - \\frac{4}{15} (1+x)^{5/2} + C \\]\nNow,\n\n\\[ \\int_{-1}^{0} x \\sqrt{1+x} \\, dx = \\int_{-1}^{0} I(x) \\, dx \\]\n\n\\[ = \\left[ \\frac{2x}{3} (1+x)^{3/2} - \\frac{4}{15} (1+x)^{5/2} \\right]_{-1}^{0} \\]\n\n\\[ = 0 - \\frac{4}{15} - (0 - 0) \\]\n\n\\[ = -\\frac{4}{15} \\]", "id": "./materials/84.pdf" }, { "contents": "Evaluate \\( \\int_{0}^{2\\pi} (x - 4) \\sin(x) \\, dx \\)\n\n**HINT:**\n\nUse integration by parts:\n\n\\[\n\\int f(x)g(x) \\, dx = F(x)g(x) - \\int F(x)g'(x) \\, dx + C, \\text{ such that } F(x) = \\int f(x) \\, dx\n\\]\n\nTake \\( f(x) = \\sin(x) \\) and \\( g(x) = x - 4 \\)\n\nChoosing \\( g(x) = x - 4 \\) reduces the degree to 0 after \\( g'(x) \\) which prevents the repetitive use of integration by parts.\n\n**Additional note:**\n\n**LIATE**\n\nAn acronym that is very helpful to remember when using integration by parts is **LIATE**. Whichever function **comes first** in the following list should be \\( g(x) \\):\n\n| L | Logarithmic functions | \\( \\ln(x), \\log_{10}(x) \\) |\n|---|-----------------------|-----------------------------|\n| I | Inverse trig. functions | \\( \\arccos(x), \\arctan(x) \\) |\n| A | Algebraic functions | \\( x, x^4, 5x^2 \\) |\n| T | Trigonometric functions | \\( \\sin(x), \\cos(x) \\) |\n| E | Exponential Functions | \\( e^x, 2^x \\) |", "id": "./materials/85.pdf" }, { "contents": "Evaluate \\( \\int_{-3}^{1} 6x^2 - 3x - 2 \\, dx \\)\n\n- I = \\([-3, 1]\\) is a closed interval.\n\n- \\( f(x) = 6x^2 - 3x - 2 \\) is continuous on I since it is a quadratic function\n\n- \\( F(x) = \\int 6x^2 - 3x - 2 \\, dx = \\int 6x^2 \\, dx - \\int 3x \\, dx - \\int 2 \\, dx \\)\n\n\\[\n= 2x^3 - \\frac{3x^2}{2} - 2x + C\n\\]\n\nThen, by using Fundamental theorem of Calculus\n\n\\[\n\\int_{-3}^{1} 6x^2 - 3x - 2 \\, dx = \\left[ 2x^3 - \\frac{3x^2}{2} - 2x \\right]_{-3}^{1}\n\\]\n\n\\[\n= 2 \\cdot x^3 \\bigg|_{-3}^{1} - 3 \\cdot \\frac{x^2}{2} \\bigg|_{-3}^{1} - 2 \\cdot x \\bigg|_{-3}^{1}\n\\]\n\n\\[\n= 2 \\cdot (1 + 27) - 3 \\cdot \\left( \\frac{1}{2} - \\frac{9}{2} \\right) - 2 \\cdot (1 + 3)\n\\]\n\n\\[\n= 60\n\\]", "id": "./materials/86.pdf" }, { "contents": "Evaluate \\( \\int_{1}^{9} \\frac{y - 1}{\\sqrt{y}} \\, dy \\)\n\n- \\( I = [1, 9] \\) is a closed interval.\n\n- \\( f(x) = \\frac{y - 1}{\\sqrt{y}} \\) is continuous on \\( I \\).\n\n- \\( F(x) = \\int \\frac{y - 1}{\\sqrt{y}} \\, dy = \\int \\frac{y}{\\sqrt{y}} \\, dy - \\int \\frac{1}{\\sqrt{y}} \\, dy \\)\n\n\\[\n= \\int \\sqrt{y} \\, dy - \\int y^{-\\frac{1}{2}} \\, dy\n\\]\n\n\\[\n= \\frac{2}{3} \\cdot y^{\\frac{3}{2}} - 2 \\cdot \\sqrt{y} + C\n\\]\n\nThen, by using Fundamental theorem of Calculus\n\n\\[\n\\int_{1}^{9} \\frac{y - 1}{\\sqrt{y}} \\, dy = \\left[ \\frac{2}{3} \\cdot y^{\\frac{3}{2}} - 2 \\cdot \\sqrt{y} \\right]_{1}^{9}\n\\]\n\n\\[\n= \\frac{2}{3} \\cdot \\left[ y^{\\frac{3}{2}} \\right]_{1}^{9} - 2 \\cdot \\left[ \\sqrt{y} \\right]_{1}^{9} = \\frac{40}{3}\n\\]", "id": "./materials/87.pdf" }, { "contents": "Evaluate \\( \\int_0^{2\\pi} (x - 4) \\sin(x) \\, dx \\)\n\n**HINT:**\n\nUse integration by parts:\n\n\\[\n\\int f(x)g(x) \\, dx = F(x)g(x) - \\int F(x)g'(x) \\, dx + C, \\text{ such that } F(x) = \\int f(x) \\, dx\n\\]\n\nTake \\( f(x) = \\sin(x) \\) and \\( g(x) = x - 4 \\)\n\nChoosing \\( g(x) = x - 4 \\) reduces the degree to 0 after \\( g'(x) \\) which prevents the repetitive use of integration by parts.\n\n**Additional note:**\n\n**LIATE**\n\nAn acronym that is very helpful to remember when using integration by parts is **LIATE**. Whichever function **comes first** in the following list should be \\( g(x) \\):\n\n| L | Logarithmic functions | \\( \\ln(x), \\log_{10}(x) \\) |\n|---|----------------------|-----------------------------|\n| I | Inverse trig. functions | \\( \\arccos(x), \\arctan(x) \\) |\n| A | Algebraic functions | \\( x, x^4, 5x^2 \\) |\n| T | Trigonometric functions | \\( \\sin(x), \\cos(x) \\) |\n| E | Exponential Functions | \\( e^x, 2^x \\) |", "id": "./materials/88.pdf" }, { "contents": "Absolute Value\nThe absolute value of a number is its distance from zero, without considering its sign. This distance is always non-negative, signifying the magnitude of a number. Mathematically, the absolute value of a real number $x$ is written as $|x|$. If $x \\geq 0$, then $|x| = x$; if $x < 0$, then $|x| = -x$. For example, the absolute value of $-5$ is $5$, while the absolute value of $7$ is also $7$. The absolute value concept is key in mathematics for comparing magnitudes, solving equations, measuring distances, and working with inequalities.\n\nExample 1: Solve $|2x + 1| = 7$\nTo solve the equation $|2x + 1| = 7$, is necessary to consider two cases.\n\n- Case 1: The inner expression is positive, so we can remove the absolute value:\n \n $$2x + 1 = 7$$\n\n Solving for $x$, we have:\n \n $$2x = 6 \\quad \\Rightarrow \\quad x = \\frac{6}{2} = 3$$\n\n- Case 2: The inner expression is negative, so we flip the sign and remove the absolute value:\n \n $$2x + 1 = -7$$\n\n Solving for $x$, we have:\n \n $$2x = -8 \\quad \\Rightarrow \\quad x = \\frac{-8}{2} = -4$$\n\nTherefore, the solutions to the given equation are $x = 3$ and $x = -4$. \nExample 2: Solve $|x - 3| = 5$\n\nTo solve the equation $|x - 3| = 5$, it is necessary to consider two cases:\n\n- Case 1: $x - 3 = 5$\n \n $x - 3 = 5 \\Rightarrow x = 5 + 3 \\Rightarrow x = 8$\n\n- Case 2: $x - 3 = -5$\n \n $x - 3 = -5 \\Rightarrow x = -5 + 3 \\Rightarrow x = -2$\n\nTherefore, the solutions are $x = 8$ and $x = -2$.\n\nExample 3: Solve $|x - 4| < 3$\n\nTo solve the inequality $|x - 4| < 3$, consider that the absolute value inequality translates into a double inequality.\n\nRewrite the absolute value inequality into a double inequality:\n\n$-3 < x - 4 < 3$\n\nNow, solve for $x$:\n\n$-3 + 4 < x < 3 + 4$\n\n$1 < x < 7$\n\nTherefore, the solution to the inequality $|x - 4| < 3$ is:\n\n$x \\in ]1, 7[$\n\nThis means that all values of $x$ between 1 and 7 (excluding 1 and 7) satisfy the inequality.\nExample 4: Solve $|3x - 6| \\geq 9$\n\nTo solve the inequality $|3x - 6| \\geq 9$, we consider two cases:\n\n- **Case 1:** $3x - 6 \\geq 9$\n \n $3x \\geq 15 \\quad \\Rightarrow \\quad x \\geq 5$\n\n- **Case 2:** $3x - 6 \\leq -9$\n \n $3x \\leq -3 \\quad \\Rightarrow \\quad x \\leq -1$\n\nCombining these cases, the solution to $|3x - 6| \\geq 9$ is:\n\n$$x \\in (-\\infty, -1] \\cup [5, \\infty)$$\n\nThis means that $x$ must be less than or equal to $-1$ or greater than or equal to $5$. ", "id": "./materials/888.pdf" }, { "contents": "Absolute Value\nThe absolute value of a number is its distance from zero, without considering its sign. This distance is always non-negative, signifying the magnitude of a number. Mathematically, the absolute value of a real number $x$ is written as $|x|$. If $x \\geq 0$, then $|x| = x$; if $x < 0$, then $|x| = -x$. For example, the absolute value of $-5$ is $5$, while the absolute value of $7$ is also $7$. The absolute value concept is key in mathematics for comparing magnitudes, solving equations, measuring distances, and working with inequalities.\n\nExample 5: Solve $|x + 1| = |2x - 3|\nTo solve the equation $|x + 1| = |2x - 3|$, consider the following cases:\n\n- Case 1: $x + 1 = 2x - 3$\n \n $x + 1 = 2x - 3 \\Rightarrow 1 + 3 = 2x - x$\n \n $4 = x \\implies x = 4$\n\n- Case 2: $x + 1 = -(2x - 3)$\n \n $x + 1 = -2x + 3 \\Rightarrow x + 2x = 3 - 1$\n \n $3x = 2 \\implies x = \\frac{2}{3}$\n\n- Case 3: $-(x + 1) = 2x - 3$\n \n $-x - 1 = 2x - 3 \\Rightarrow -x - 2x = -3 + 1$\n \n $-3x = -2 \\implies x = \\frac{2}{3}$\n\n- Case 4: $-(x + 1) = -(2x - 3)$\n \n $-x - 1 = -2x + 3 \\Rightarrow -x + 2x = 3 + 1$\n \n $x = 4$\n\nThus, the solutions to the equation $|x + 1| = |2x - 3|$ are:\n\n$x = \\frac{2}{3}$ or $x = 4$\nExample 6: Solve $\\frac{|x+2|}{|x+1|} > 0$\n\nTo solve it, it is necessary to understand when the fraction is positive.\n\nFor the fraction\n\n$$\\frac{|x + 2|}{|x + 1|}$$\n\nto be greater than 0, both the numerator and the denominator must be non-zero and have the same sign. Since absolute values are always non-negative, the key is ensuring that neither $|x + 2|$ nor $|x + 1|$ is zero.\n\nCases to Consider:\n\n- Numerator and denominator analysis:\n $|x + 2| > 0$ and $|x + 1| > 0$ when neither $x + 2$ nor $x + 1$ is zero.\n\n- Identify Points Where the Absolute Values are Zero\n - $|x + 2| = 0$ when $x = -2$\n - $|x + 1| = 0$ when $x = -1$\n\n- The fraction $\\frac{|x+2|}{|x+1|}$ is undefined at $x = -1$ because division by zero is not allowed.\n\n- At $x = -2$, $|x + 2| = 0$, resulting in the entire fraction being zero, which does not satisfy the inequality\n\nTherefore, the solution is $x \\neq -2$ and $x \\neq -1$\n\nIn the interval notation is:\n\n$$x \\in (-\\infty, -2) \\cup (-2, -1) \\cup (-1, \\infty)$$\nExample 7: Solve \\( \\frac{x^3 + 9x^2 - 36x}{|x^3 - 1|} \\leq 0 \\).\n\nIs necessary to analyze when the rational expression is less than or equal to zero. This involves several steps:\n\n- **Step 1: Factor the Numerator**\n \n The numerator is \\( x^3 + 9x^2 - 36x \\). We can factor this polynomial as follows:\n \n \\[\n x^3 + 9x^2 - 36x = x(x^2 + 9x - 36)\n \\]\n \n Now, factor the quadratic expression \\( x^2 + 9x - 36 \\):\n \n \\[\n x^2 + 9x - 36 = (x + 12)(x - 3)\n \\]\n \n Thus,\n \n \\[\n x^3 + 9x^2 - 36x = x(x + 12)(x - 3)\n \\]\n \n So \\( \\frac{x^3 + 9x^2 - 36x}{|x^3 - 1|} \\leq 0 \\), turns into \\( \\frac{x(x + 12)(x - 3)}{|x^3 - 1|} \\leq 0 \\).\n\n- **Step 2: Analysis**\n \n - **Undefined Points:** The fraction is undefined at \\( x = 1 \\) because the denominator \\( |x^3 - 1| \\) is equal to 0 there and division by zero is not allowed.\n \n - **Zero Points:** The numerator is zero at \\( x = -12, x = 0, \\) and \\( x = 3 \\).\n\n- **Step 3: Find intervals where \\( \\frac{x(x + 12)(x - 3)}{|x^3 - 1|} \\leq 0 \\)**\n \n 1. **Intervals to Consider:**\n \n We examine the signs of \\( x(x + 12)(x - 3) \\) and \\( |x^3 - 1| \\) around the critical points \\( x = -12, x = 0, x = 1, \\) and \\( x = 3 \\).\n \n 2. **Sign Analysis:**\n \n For \\( x < -12 \\): The numerator is negative (since all factors \\( x, x + 12, \\) and \\( x - 3 \\) are negative), and the denominator is positive (since \\( x^3 - 1 \\) is negative and absolute value is positive). Thus, the fraction is negative.\n \n For \\( -12 < x < 0 \\): The numerator is positive (since \\( x \\) is negative, but \\( x + 12 \\) and \\( x - 3 \\) are positive) and the denominator is positive. Thus, the fraction is positive.\n \n For \\( 0 < x < 1 \\): The numerator is negative (since \\( x \\) and \\( x - 3 \\) are negative) and the denominator is positive. Thus, the fraction is negative.\nFor $1 < x < 3$: The numerator is negative and the denominator is positive. Thus, the fraction is negative.\n\nFor $x > 3$: The numerator is positive (since all factors are positive) and the denominator is positive. Thus, the fraction is positive.\n\nSo in the interval notation the solution is:\n\n$$x \\in (-\\infty, -12] \\cup [0, 1) \\cup (1, 3]$$", "id": "./materials/889.pdf" }, { "contents": "Evaluate \\( \\int_0^1 x(x - 1)^{14} \\, dx \\)\n\n**HINT:**\n\nUse integration by parts:\n\n\\[\n\\int f(x)g(x) \\, dx = F(x)g(x) - \\int F(x)g'(x) \\, dx + C, \\text{ such that } F(x) = \\int f(x) \\, dx\n\\]\n\nTake \\( f(x) = (x - 1)^{14} \\) and \\( g(x) = x \\)\n\nChoosing \\( g(x) = x \\) reduces the degree to 0 after \\( g'(x) \\), which prevents the repetitive use of integration by parts. Had we chosen \\( g(x) = (x - 1)^{14} \\), the degree would reduce to 13 however, the degree of \\( F(x) \\) would rise to 2 (because \\( F(x) = \\int f(x) \\, dx \\)) which forces to reuse the integration by parts repetitively.", "id": "./materials/89.pdf" }, { "contents": "Algebraic Manipulation\nAlgebraic manipulation involves applying algebraic operations, such as addition, subtraction, multiplication, and division, to reorganize equations. This process often aims to isolate a particular variable, making it easier to solve equations or understand the relationship between different variables.\n\nExamples\n1. Addition: Make $d$ the subject of the expression $b = c - d$.\n\nTo make $d$ the subject of the expression $b = c - d$, is necessary to isolate $d$ on one side of the equation. Here’s how to do it:\n\nGiven: $b = c - d$\n\nAdd $d$ to both sides to isolate $d$:\n\n$$b + d = c - d + d$$\n\n$$b + d = c - d + d$$\n\n$$b + d = c$$\n\nNow, subtract $b$ from both sides to isolate $d$:\n\n$$b + d - b = c - b$$\n\n$$b + d - b = c - b$$\n\n$$d = c - b$$\n\nSo, $d$ is the subject of the expression $b = c - d$, and it can be written as $d = c - b$. \n2. **Subtraction**: Make \\( r \\) the subject of the expression \\( p = q + r \\).\n\nTo isolate \\( r \\) in the expression \\( p = q + r \\), you can rearrange the equation by subtracting \\( q \\) from both sides. Here’s how to make \\( r \\) the subject:\n\nGiven: \\( p = q + r \\)\n\nSubtract \\( q \\) from both sides:\n\n\\[\np - q = q + r - q\n\\]\n\n\\[\np - q = q + r - q\n\\]\n\n\\[\nr = p - q\n\\]\n\nThus, the expression that makes \\( r \\) the subject is \\( r = p - q \\).\n\n3. **Multiplication**: Make \\( d \\) the subject of the expression \\( p = \\frac{\\pi d}{2} \\).\n\nTo make \\( d \\) the subject of the expression \\( p = \\frac{\\pi d}{2} \\), it is necessary to isolate \\( d \\) on one side of the equation. Here’s how to do it:\n\nGiven: \\( p = \\frac{\\pi d}{2} \\)\n\nMultiply both sides by \\( \\frac{2}{\\pi} \\) to isolate \\( d \\):\n\n\\[\n\\frac{2}{\\pi} \\cdot p = \\frac{2}{\\pi} \\cdot \\frac{\\pi d}{2}\n\\]\n\n\\[\n\\frac{2}{\\pi} \\cdot p = \\frac{2}{\\pi} \\cdot \\frac{\\pi d}{2}\n\\]\n\n\\[\n\\frac{2p}{\\pi} = d\n\\]\n\nSo, \\( d \\) is the subject of the expression \\( p = \\frac{\\pi d}{2} \\), and it can be written as \\( d = \\frac{2p}{\\pi} \\).\n\n4. **Division**: Make \\( a \\) the subject of the expression \\( mz = ad \\).\n\nTo make \\( a \\) the subject of the expression \\( mz = ad \\), it is necessary to isolate \\( a \\) on one side of the equation. Here’s how to do it:\n\nGiven: \\( mz = ad \\)\n\nDivide both sides by \\( d \\) to isolate \\( a \\):\n\n\\[\n\\frac{mz}{d} = \\frac{ad}{d} \\Rightarrow \\frac{mz}{d} = \\frac{a}{d}\n\\]\n\n\\[\n\\frac{mz}{d} = a\n\\]\n\nSo, \\( a \\) is the subject of the expression \\( mz = ad \\), and it can be written as \\( a = \\frac{mz}{d} \\).", "id": "./materials/890.pdf" }, { "contents": "Algebraic Manipulation\nAlgebraic manipulation involves applying algebraic operations, such as addition, subtraction, multiplication, and division, to reorganize equations. This process often aims to isolate a particular variable, making it easier to solve equations or understand the relationship between different variables.\n\nSpecial Cases\nI) Make \\( r \\) the subject of the expression \\( v = \\frac{2}{3} \\pi r^3 \\). To make \\( r \\) the subject of the expression \\( v = \\frac{2}{3} \\pi r^3 \\), it is necessary to isolate \\( r \\) on one side of the equation. Here’s how to do it:\n\nGiven: \\( v = \\frac{2}{3} \\pi r^3 \\)\n\nDivide both sides by \\( \\frac{2}{3} \\pi \\) to isolate \\( r^3 \\):\n\n\\[\n\\frac{v}{\\frac{2}{3} \\pi} = \\frac{\\frac{2}{3} \\pi r^3}{\\frac{2}{3} \\pi}\n\\]\n\n\\[\n\\frac{v}{\\frac{2}{3} \\pi} = \\frac{2}{3} \\pi r^3\n\\]\n\n\\[\n3v = \\frac{2}{3} \\pi r^3\n\\]\n\nNow, to solve for \\( r \\), take the cube root of both sides:\n\n\\[\n\\sqrt[3]{\\frac{3v}{2\\pi}} = \\sqrt[3]{r^3}\n\\]\n\n\\[\n\\sqrt[3]{\\frac{3v}{2\\pi}} = \\sqrt[3]{r^3}\n\\]\n\n\\[\nr = \\sqrt[3]{\\frac{3v}{2\\pi}}\n\\]\n\nSo, \\( r \\) is the subject of the expression \\( v = \\frac{2}{3} \\pi r^3 \\), and it can be written as \\( r = \\sqrt[3]{\\frac{3v}{2\\pi}} \\).\nII) Make \\( b \\) the subject of the expression \\( \\frac{p}{q} = \\sqrt{\\frac{a+2b}{a-2b}} \\). To make \\( b \\) the subject of the expression \\( \\frac{p}{q} = \\sqrt{\\frac{a+2b}{a-2b}} \\), it is necessary to isolate \\( b \\) on one side of the equation. Here’s how to do it:\n\nGiven: \\( \\frac{p}{q} = \\sqrt{\\frac{a+2b}{a-2b}} \\)\n\nSquare both sides to remove the square root:\n\n\\[\n\\left( \\frac{p}{q} \\right)^2 = \\left( \\sqrt{\\frac{a+2b}{a-2b}} \\right)^2\n\\]\n\n\\[\n\\left( \\frac{p}{q} \\right)^2 = \\frac{a+2b}{a-2b}\n\\]\n\nMultiply both sides by \\( a-2b \\) and \\( q^2 \\) to clear the fraction (or just multiply crosswise):\n\n\\[\n\\frac{p^2}{q^2} \\cdot (a-2b) \\cdot q^2 = \\frac{a+2b}{a-2b} \\cdot (a-2b) \\cdot q^2\n\\]\n\n\\[\n\\frac{p^2}{q^2} \\cdot (a-2b) \\cdot q^2 = \\frac{a+2b}{a-2b} \\cdot (a-2b) \\cdot q^2\n\\]\n\n\\[\np^2 \\cdot (a-2b) = q^2 \\cdot (a+2b)\n\\]\n\nExpand the multiplication:\n\n\\[\np^2 \\cdot a - 2 \\cdot p^2 \\cdot b = q^2 \\cdot a + 2 \\cdot q^2 \\cdot b\n\\]\n\nRearrange the terms adding \\( 2 \\cdot p^2 \\cdot b \\) and subtracting \\( q^2 \\cdot a \\) in both sides to group those with \\( b \\):\n\n\\[\np^2 \\cdot a - 2 \\cdot p^2 \\cdot b - q^2 \\cdot a + 2 \\cdot p^2 \\cdot b = q^2 \\cdot a + 2 \\cdot q^2 \\cdot b - q^2 \\cdot a + 2 \\cdot p^2 \\cdot b\n\\]\n\n\\[\np^2 \\cdot a - 2 \\cdot p^2 \\cdot b - q^2 \\cdot a + 2 \\cdot p^2 \\cdot b = q^2 \\cdot a + 2 \\cdot q^2 \\cdot b - q^2 \\cdot a + 2 \\cdot p^2 \\cdot b\n\\]\n\n\\[\np^2 \\cdot a - q^2 \\cdot a = 2 \\cdot q^2 \\cdot b + 2 \\cdot p^2 \\cdot b\n\\]\n\nNow is possible to factor out \\( b \\) and \\( a \\):\n\n\\[\n(p^2 - q^2) \\cdot a = b \\cdot 2 \\cdot (p^2 + q^2)\n\\]\nTo isolate $b$, divide both sides by $2 \\cdot (p^2 + q^2)$:\n\n$$\\frac{(p^2 - q^2) \\cdot a}{2 \\cdot (p^2 + q^2)} = b \\cdot \\frac{2(p^2 + q^2)}{2 \\cdot (p^2 + q^2)}$$\n\n$$\\frac{(p^2 - q^2) \\cdot a}{2 \\cdot (p^2 + q^2)} = b \\cdot \\frac{2(p^2 + q^2)}{2 \\cdot (p^2 + q^2)}$$\n\n$$b = \\frac{a \\cdot (p^2 - q^2)}{2 \\cdot (p^2 + q^2)}$$\n\nSo, $b$ the subject of the expression $\\frac{p}{q} = \\sqrt{\\frac{a+2b}{a-2b}}$, and it can be written as $b = \\frac{a \\cdot (p^2 - q^2)}{2 \\cdot (p^2 + q^2)}$\n\nIII) Solve for $t$: $\\frac{15}{3t^2} = \\frac{2}{5}$\n\nTo solve for $t$ in the equation $\\frac{15}{3t^2} = \\frac{2}{5}$, is necessary:\n\nGiven that there is a proportion, is possible to cross-multiply to eliminate the fractions:\n\n$$15 \\cdot 5 = 2 \\cdot (3t^2)$$\n\n$$75 = 6t^2$$\n\nNow, isolate $t^2$ by dividing both sides of the equation by 6:\n\n$$t^2 = \\frac{75}{6}$$\n\n$$t^2 = 12.5$$\n\nSince you’re interested in solving for $t$, take the square root of both sides to isolate $t$:\n\n$$t = \\pm \\sqrt{12.5}$$\n\n$$t = \\pm 3.54$$\n\nThus, the solution to the given expression is:\n\n$$t = \\pm 3.54$$", "id": "./materials/891.pdf" }, { "contents": "Inequations\nInequations, also known as inequalities, represent relationships between two quantities where they are not necessarily equal. They use symbols such as $>$, $<$, $\\geq$, and $\\leq$ to indicate whether one quantity is greater than, less than, greater than or equal to, or less than or equal to the other. Inequations can involve linear expressions, indicating a range of possible solutions, or more complex forms like absolute value inequalities, requiring consideration of multiple cases to capture all possible solutions. When solving inequalities, the focus is on isolating the variable while maintaining the relationship denoted by the inequality symbol, with special care taken when multiplying or dividing by negative numbers. Inequations are key in algebra, calculus, and optimization, providing a way to describe conditions where equality does not necessarily hold.\n\nExample 1: Solve $x^2 - 9 > 0$\nTo find the solution to the inequality $x^2 - 9 > 0$, first identify the critical points where the expression equals zero.\n\n1. $x^2 - 9 = 0 \\Rightarrow x^2 = 9 \\Rightarrow x = -3$ and $x = 3$\n\nThe points $x = -3$ and $x = 3$ divide the real number line into intervals.\n\nNext, you can test the sign of the expression $x^2 - 9$ in each interval to determine where it is positive/non-negative ($> 0$).\n\nWhen $x < -3$: In this interval, both $x^2$ and $-9$ are negative. Therefore, $x^2 - 9$ is positive.\n\nWhen $-3 < x < 3$: Here, $x^2$, the factor $(x + 3)$ is positive and the factor $(x - 3)$ is negative, so their product is negative. The interval between -3 and 3 does not satisfy the condition.\n\nWhen $x > 3$: Both $x^2$ and $-9$ are positive. Therefore, $x^2 - 9$ is positive.\n\n\\[\n\\begin{array}{cccccc}\n-\\infty & \\ldots & -3 & \\ldots & 0 & \\ldots & 3 & \\ldots & +\\infty \\\\\n>0 & =0 & <0 & =0 & >0 & \\\\\n\\uparrow & & & & & \\\\\n\\end{array}\n\\]\n\nBased on these analyses, the solution to the inequality $x^2 - 9 > 0$ is $x \\in (-\\infty, -3] \\cup [3, +\\infty)$. So, the solution is all real numbers $x$ such that $x$ is less than -3 or $x$ is greater than 3.\nExample 2: Solve $5 + \\frac{(x-2)(x+2)}{4} > 3 - x$\n\nTo solve the inequality $5 + \\frac{(x-2)(x+2)}{4} > 3 - x$, let’s first simplify the expression step by step.\n\nExpanding the expression inside the brackets:\n\n$$5 + \\frac{(x-2) \\cdot (x+2)}{4} > 3 - x$$\n\nNow, let’s simplify the expression:\n\n$$5 + \\frac{x^2 - 4}{4} > 3 - x$$\n\nWe can further simplify the expression:\n\n$$5 + \\frac{x^2 - 4}{4} + x > 3$$\n\n$$5 + \\frac{x^2 - 4}{4} + x - 3 > 0$$\n\n$$\\frac{x^2 - 4}{4} + x + 2 > 0$$\n\nNow, let’s find the common denominator:\n\n$$\\frac{x^2 - 4 + 4x + 8}{4} > 0$$\n\n$$\\frac{x^2 + 4x + 4}{4} > 0$$\n\nNext, let’s factor the numerator:\n\n$$\\frac{(x + 2)^2}{4} > 0$$\n\nSince $(x + 2)^2$ is always non-negative for all real values of $x$, the inequality holds true for all $x \\in \\mathbb{R}$.\n\nTherefore, the solution to the inequality $5 + \\frac{(x-2)(x+2)}{4} > 3 - x$ is $x \\in \\mathbb{R}$, excluding $x = -2$, because when $x = -2$, the expression becomes equal to 0, not a number greater than 0.\n\nThus, the correct solution is $x \\in \\mathbb{R} \\setminus \\{-2\\}$, indicating that all real numbers except $-2$ satisfy the inequality.", "id": "./materials/892.pdf" }, { "contents": "Inequations\nInequations, also known as inequalities, represent relationships between two quantities where they are not necessarily equal. They use symbols such as $>$, $<$, $\\geq$, and $\\leq$ to indicate whether one quantity is greater than, less than, greater than or equal to, or less than or equal to the other. Inequations can involve linear expressions, indicating a range of possible solutions, or more complex forms like absolute value inequalities, requiring consideration of multiple cases to capture all possible solutions. When solving inequalities, the focus is on isolating the variable while maintaining the relationship denoted by the inequality symbol, with special care taken when multiplying or dividing by negative numbers. Inequations are key in algebra, calculus, and optimization, providing a way to describe conditions where equality does not necessarily hold.\n\nExample 3: Solve $(x \\cdot (x - 1)) < \\frac{(x-1)(x+1)}{2}$\n\nTo solve the inequality $(x \\cdot (x - 1)) < \\frac{(x-1)(x+1)}{2}$, let’s go through the steps:\n\nExpand the inequality:\n\n$$x^2 - x < \\frac{x^2 - 1}{2}$$\n\nMultiply both sides by 2 to clear the fraction:\n\n$$2x^2 - 2x < x^2 - 1$$\n\nRearrange terms:\n\n$$2x^2 - x^2 - 2x + 1 < 0$$\n\n$$x^2 - 2x + 1 < 0$$\n\nAlthough the expression becomes zero when $x = 1$, the quadratic $x^2 - 2x + 1$ is non-negative for all real values of $x$, indicating it is never less than zero.\n\n$$\\begin{array}{c|c|c|c}\n-\\infty & \\ldots & 1 & \\ldots & +\\infty \\\\\n> 0 & = 0 & > 0 \\\\\n\\end{array}$$\n\nSince the square of any real number is non-negative and cannot be less than zero, there are no real values of $x$ that satisfy the inequality. Therefore, the solution to the inequality $(x \\cdot (x - 1)) < \\frac{(x-1)(x+1)}{2}$ is the empty set, denoted as $\\emptyset$. This means there are no real values of $x$ that make the inequality true.\nExample 4: Solve \\( \\frac{(2x+1)^2}{3} \\geq x + 2 \\)\n\nTo solve the inequality\n\n\\[\n\\frac{(2x+1)^2}{3} \\geq x + 2\n\\]\n\nExpanding the left side:\n\n\\[\n\\frac{4x^2 + 4x + 1}{3} \\geq x + 2\n\\]\n\nMultiply both sides by 3 to clear the fraction:\n\n\\[\n4x^2 + 4x + 1 \\geq 3x + 6\n\\]\n\nRearranging terms:\n\n\\[\n4x^2 + x - 5 \\geq 0\n\\]\n\nTo find the solutions to this quadratic inequality, it is possible to use the quadratic formula or by factoring the quadratic expression, now it will be done using the quadratic formula:\n\n\\[\nx = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}\n\\]\n\nFor \\( a = 4 \\), \\( b = 1 \\), and \\( c = -5 \\), we have:\n\n\\[\nx = \\frac{-1 \\pm \\sqrt{1 + 80}}{8}\n\\]\n\n\\[\nx = \\frac{-1 \\pm \\sqrt{81}}{8}\n\\]\n\n\\[\nx = \\frac{-1 \\pm 9}{8}\n\\]\n\nSo, the solutions are \\( x = -\\frac{5}{4} \\) and \\( x = 1 \\).\n\nIs necessary to test intervals in the critical points \\( x = 1 \\) and \\( x = -\\frac{5}{4} \\).\n\nWhen \\( x < -\\frac{5}{4} \\), the expression is positive.\n\nWhen \\( -\\frac{5}{4} < x < 1 \\), the expression is negative.\n\nWhen \\( x > 1 \\), the expression is positive.\n\n\\[\n\\begin{array}{cccccc}\n-\\infty & \\ldots & -\\frac{5}{4} & \\ldots & 1 & \\ldots & +\\infty \\\\\n> 0 & \\uparrow & = 0 & < 0 & = 0 & > 0 & \\uparrow\n\\end{array}\n\\]\n\nSince we want the expression to be greater than or equal to zero, the solution is:\n\n\\( x \\in (-\\infty, -\\frac{5}{4}] \\cup [1, +\\infty) \\).", "id": "./materials/893.pdf" }, { "contents": "Inequations\nInequations, also known as inequalities, represent relationships between two quantities where they are not necessarily equal. They use symbols such as $>$, $<$, $\\geq$, and $\\leq$ to indicate whether one quantity is greater than, less than, greater than or equal to, or less than or equal to the other. Inequations can involve linear expressions, indicating a range of possible solutions, or more complex forms like absolute value inequalities, requiring consideration of multiple cases to capture all possible solutions. When solving inequalities, the focus is on isolating the variable while maintaining the relationship denoted by the inequality symbol, with special care taken when multiplying or dividing by negative numbers. Inequations are key in algebra, calculus, and optimization, providing a way to describe conditions where equality does not necessarily hold.\n\nExample 5: Solve $x^2 + 4x - 12 > 0$\n\nTo solve the inequality $x^2 + 4x - 12 > 0$, is possible to use the quadratic formula or by factoring the quadratic expression:\n\nTo factorize, we find two numbers whose product is $-12$ and whose sum is $4$: the numbers $6$ and $-2$ satisfy these conditions.\n\nThus, the factorized form of the quadratic is:\n\n$$(x + 6) \\cdot (x - 2)$$\n\nSetting the factorized form to zero, we get the roots:\n\n1. $x + 6 = 0 \\Rightarrow x = -6$,\n2. $x - 2 = 0 \\Rightarrow x = 2$.\n\nGiven the roots at $x = -6$ and $x = 2$, we examine the intervals between and beyond these points to determine where the inequality holds true.\n\nWhen $x < -6$, the expression is positive.\nWhen $-6 < x < 2$, the expression is negative.\nWhen $x > 2$, the expression is positive.\n\n| $-\\infty$ | $\\ldots$ | $-6$ | $\\ldots$ | $2$ | $\\ldots$ | $+\\infty$ |\n|-----------|---------|------|---------|----|---------|---------|\n| $>$ | $\\uparrow$ | $= 0$ | $< 0$ | $= 0$ | $> 0$ | $\\uparrow$ |\n\nThus, the intervals where $x^2 + 4x - 12 > 0$ are: $x \\in (-\\infty, -6] \\cup [2, +\\infty)$\n\nThese intervals represent where the quadratic expression is positive, indicating where the inequality holds true.\nExample 6: Solve \\((x - \\frac{7}{2})^2 \\cdot (3x + 3) \\cdot (2x - 1) > 0\\)\n\nTo solve the inequality \\((x - \\frac{7}{2})^2 \\cdot (3x + 3) \\cdot (2x - 1) > 0\\), you need to find the critical points where the expression equals zero, then use these points to determine the sign of the inequality in each interval.\n\nThe critical points occur when the expression is zero. Set each factor to zero and solve for \\(x\\):\n\n1. \\(x - \\frac{7}{2} = 0 \\Rightarrow x = \\frac{7}{2}\\)\n2. \\(3x + 3 = 0 \\Rightarrow x = -1\\)\n3. \\(2x - 1 = 0 \\Rightarrow x = \\frac{1}{2}\\)\n\nThus, the critical points are: \\(x = \\frac{7}{2}, x = -1, x = \\frac{1}{2}\\).\n\nNow, consider the intervals between and beyond the critical points to determine where the expression is greater than zero.\n\nGiven the inequality has roots at \\(x = -1, x = \\frac{1}{2}, \\text{ and } x = \\frac{7}{2}\\), let’s examine the intervals:\n\n- When \\(x < -1\\), the expression is positive.\n- When \\(-1 < x < \\frac{1}{2}\\), the expression is negative.\n- When \\(\\frac{1}{2} < x < \\frac{7}{2}\\), the expression is positive.\n- When \\(x > \\frac{7}{2}\\), the expression is positive.\n\nThus, the intervals where \\((x - \\frac{7}{2})^2 \\cdot (3x + 3) \\cdot (2x - 1) > 0\\) are:\n\n\\(x \\in (-\\infty, -1] \\cup \\left[\\frac{1}{2}, \\frac{7}{2}\\right] \\cup (+\\infty)\\)\n\nThese intervals represent the regions where the given inequality holds true.", "id": "./materials/894.pdf" }, { "contents": "Inequations\nInequations, also known as inequalities, represent relationships between two quantities where they are not necessarily equal. They use symbols such as $>$, $<$, $\\geq$, and $\\leq$ to indicate whether one quantity is greater than, less than, greater than or equal to, or less than or equal to the other. Inequations can involve linear expressions, indicating a range of possible solutions, or more complex forms like absolute value inequalities, requiring consideration of multiple cases to capture all possible solutions. When solving inequalities, the focus is on isolating the variable while maintaining the relationship denoted by the inequality symbol, with special care taken when multiplying or dividing by negative numbers. Inequations are key in algebra, calculus, and optimization, providing a way to describe conditions where equality does not necessarily hold.\n\nExample 7: Solve $7x^4 > 2x^3 + 9x^2$\nTo solve the inequality $7x^4 > 2x^3 + 9x^2$, let’s first move all terms to one side to obtain a unified expression, and then analyze the resulting polynomial to find the critical points (roots) and intervals where the inequality holds true.\nMove all terms to one side to make it easier to find the roots:\n\n$$7x^4 - 2x^3 - 9x^2 > 0.$$ \n\nOne way to solve this type of inequality is to look for common factors or apply algebraic techniques to identify the critical points:\nIn this case, the common factor is $x^2$:\n\n$$x^2(7x^2 - 2x - 9).$$\n\nTo determine where the polynomial changes sign, set each factor to zero:\n\n$$x^2 = 0 \\Rightarrow x = 0.$$ \n\nFor $7x^2 - 2x - 9$ use the quadratic formula to find roots:\n\n$$x = \\frac{-(-2) \\pm \\sqrt{(-2)^2 - 4 \\times 7 \\times -9}}{2 \\times 7}.$$ \n\nSimplifying, we have:\n\n$$x = \\frac{2 \\pm \\sqrt{4 + 252}}{14} \\Rightarrow x = \\frac{2 \\pm \\sqrt{256}}{14} \\Rightarrow x = \\frac{2 \\pm 16}{14}.$$\nThis yields two roots: \\( x = \\frac{18}{14} = \\frac{9}{7} \\) and \\( x = \\frac{-14}{14} = -1 \\).\n\nThese roots indicate where the expression changes sign, allowing analysis of the intervals to determine where the inequality is greater than zero.\n\nGiven the roots \\( x = 0 \\), \\( x = \\frac{9}{7} \\), and \\( x = -1 \\), let’s consider the intervals created by these roots:\n\nWhen \\( x < -1 \\), the expression results in a positive value.\n\nWhen \\( -1 < x < 0 \\), the expression results in a negative value.\n\nWhen \\( 0 < x < \\frac{9}{7} \\), this results in a negative value.\n\nWhen \\( x > \\frac{9}{7} \\), this results in a positive value.\n\n| \\(-\\infty\\) | \\(-1\\) | \\(0\\) | \\(\\frac{9}{7}\\) | \\(+\\infty\\) |\n|-------------|-------|------|-------------|----------|\n| \\(>0\\) | \\(=0\\) | \\(<0\\) | \\(<0\\) | \\(=0\\) | \\(>0\\) |\n| ↑ | | | | | ↑ |\n\nThus, the solution for \\( 7x^4 > 2x^3 + 9x^2 \\) is \\( x \\in (-\\infty, -1] \\cup \\left[\\frac{9}{7}, +\\infty\\right) \\).\nExample 8: Solve $2x^6 - 2 > -3x^3$\n\nTo solve the inequality $2x^6 - 2 > -3x^3$, first move all terms to one side of the inequality:\n\n$$2x^6 + 3x^3 - 2 > 0.$$ \n\nTo find the critical points where the expression equals zero, set the polynomial to zero and solve for $x$:\n\n$$2x^6 + 3x^3 - 2 = 0.$$ \n\nWe can consider $x^3$ as a common term and rewrite this expression as a quadratic equation in terms of $x^3$:\n\n$$2(x^3)^2 + 3x^3 - 2 = 0.$$ \n\nTo find the roots, use the quadratic formula:\n\n$$x^3 = \\frac{-3 \\pm \\sqrt{3^2 - 4 \\times 2 \\times -2}}{2 \\times 2},$$\n\n$$x^3 = \\frac{-3 \\pm \\sqrt{25}}{4} \\Rightarrow x^3 = \\frac{-3 \\pm 5}{4}.$$ \n\nThis yields the roots:\n\n1. $x^3 = \\frac{2}{4} = \\frac{1}{2} \\Rightarrow x = \\sqrt[3]{\\frac{1}{2}} \\Rightarrow x = \\frac{1}{\\sqrt[3]{2}}$\n2. $x^3 = \\frac{-8}{4} = -2 \\Rightarrow x = -\\sqrt[3]{2}$\n\nGiven these critical points, let’s examine the intervals to determine where the original expression is greater than zero:\n\nWhen $x < -\\sqrt[3]{2}$, it results in a positive value, indicating this interval satisfies the inequality.\n\nWhen $-\\sqrt[3]{2} < x < \\frac{1}{\\sqrt[3]{2}}$ the expression results in a negative value, indicating this interval does not satisfy the inequality.\n\nWhen $x > \\frac{1}{\\sqrt[3]{2}}$ the result is positive, indicating this interval satisfies the inequality.\n\n| | $-\\infty$ | $-\\sqrt[3]{2}$ | $\\frac{1}{\\sqrt[3]{2}}$ | $+\\infty$ |\n|-------|-----------|----------------|--------------------------|-----------|\n| $>0$ | $=0$ | $<0$ | $=0$ | $>0$ |\n| ↑ | | | | ↑ |\n\nThus, the solution for $2x^6 - 2 > -3x^3$ is:\n\n$$x \\in (-\\infty, -\\sqrt[3]{2}] \\cup \\left[\\frac{1}{\\sqrt[3]{2}}, +\\infty\\right).$$", "id": "./materials/895.pdf" }, { "contents": "Simplifying algebraic expressions\nSimplifying algebraic expressions involves manipulating and rewriting algebraic terms in a simpler or more compact form without changing their original value. This can apply to linear or non-linear expressions.\n\nExample 1: Simplify the expression \\((x + y) + (3x - y)\\).\n\nTo simplify the expression \\((x + y) + (3x - y)\\), you can combine like terms.\n\ni) Remove parentheses:\nThe expression \\((x + y)\\) has no operation between its terms, so it can be rewritten as \\(x + y\\). Similarly, \\((3x - y)\\) becomes \\(3x - y\\). The combined expression is:\n\n\\[ x + y + 3x - y \\]\n\nii) Combine like terms: - Group similar terms and use cancellations as needed. Adding and subtracting like terms gives:\n\n\\[ x + y + 3x - y \\]\n\nNow, combine terms with the same variable to get the simplified expression:\n\n\\[ 4x \\]\n\nThus, the simplified expression for \\((x + y) + (3x - y)\\) is: \\(4x\\).\n\nExample 2: Simplify \\((x^2 + 3x + 5) + (x - 2)^2\\)\n\nTo simplify the expression \\((x^2 + 3x + 5) + (x - 2)^2\\), we first expand \\((x - 2)^2\\) and then combine like terms.\n\ni) Expand \\((x - 2)^2\\):\n\n\\[(x - 2) \\times (x - 2) = x^2 - 2x - 2x + 4 = x^2 - 4x + 4\\]\n\nii) Combine the expanded form with the first part of the expression:\n\n\\[(x^2 + 3x + 5) + (x^2 - 4x + 4)\\]\n\niii) Add and combine like terms:\n\n- Combine the \\(x^2\\) terms: \\(x^2 + x^2 = 2x^2\\).\n• Combine the $x$ terms: $3x - 4x = -x$.\n\n• Combine the constants: $5 + 4 = 9$.\n\niv) Rewrite the simplified expression:\n\n$$2x^2 - x + 9$$\n\nThus, the simplified expression for $(x^2 + 3x + 5) + (x - 2)^2$ is: $2x^2 - x + 9$. ", "id": "./materials/896.pdf" }, { "contents": "Simplifying algebraic expressions\n\nSimplifying algebraic expressions involves manipulating and rewriting algebraic terms in a simpler or more compact form without changing their original value. This can apply to linear or non-linear expressions.\n\nExample 3: Simplify \\( \\frac{(7x^{-3}) \\cdot (5x^2)}{5x^{-4}} \\).\n\nTo simplify \\( \\frac{(7x^{-3}) \\cdot (5x^2)}{5x^{-4}} \\), follow these steps:\n\ni) Combine the terms in the numerator, considering that in a multiplication when the bases are equal, keep the base and add the exponents \\( (a^m \\cdot a^n = a^{m+n}) \\):\n\n\\[\n7x^{-3} \\cdot 5x^2 = (7 \\cdot 5) \\cdot (x^{-3} \\cdot x^2) = 35 \\cdot x^{-1} = \\frac{35}{x}\n\\]\n\nii) Divide by the denominator: Since division with exponents is equivalent to subtracting exponents, we can rewrite \\( x^{-3} \\cdot x^2 \\) and \\( x^{-1} \\cdot x^{-4} \\):\n\n\\[\n\\frac{35 \\cdot x^{-1}}{5 \\cdot x^{-4}} = \\frac{35}{5 \\cdot x^{-4} \\cdot x^1} = \\frac{35}{5x^{-3}}\n\\]\n\niii) Simplify: After combining, we have the following:\n\n\\[\n7 \\cdot x^3 = 7x^3\n\\]\n\nThus, the simplified expression for \\( \\frac{(7x^{-3}) \\cdot (5x^2)}{5x^{-4}} \\) is: \\( 7x^3 \\).\n\nExample 4: Simplify \\( \\sqrt{y^3} \\cdot \\sqrt{y} \\).\n\nTo simplify the expression \\( \\sqrt{y^3} \\cdot \\sqrt{y} \\), you can use the rules of exponents and radicals to combine and simplify terms.\n\ni) Combine radicals:\n\nSince \\( \\sqrt{y^3} = y^{3/2} \\) and \\( \\sqrt{y} = y^{1/2} \\), you can multiply these two radicals:\n\n\\[\n\\sqrt{y^3} \\cdot \\sqrt{y} = y^{3/2} \\cdot y^{1/2}\n\\]\n\nii) Add the exponents:\n\nSince the bases are the same \\( (y) \\), you can add the exponents:\n\n\\[\ny^{3/2+1/2} = y^2\n\\]\niii) Rewrite the division:\nGiven $y^{-2}$, rewrite the division:\n\n$$\\frac{y^2}{y^{-2}}$$\n\niv) Subtract the exponents:\nSince division with the same base involves subtracting exponents, we have:\n\n$$y^{2-(-2)} = y^4$$\n\nThus, the simplified expression for $\\frac{\\sqrt{y^7} \\cdot \\sqrt{y}}{y^{-2}}$ is: $y^4$", "id": "./materials/897.pdf" }, { "contents": "Complex Numbers are an extension of the real numbers that allow us to represent numbers that include an imaginary unit, denoted by $i$, where $i^2 = -1$. This concept was introduced to solve equations where traditional real numbers weren’t sufficient, such as $x^2 = -1$.\n\nA complex number can be written in the form:\n\n$$z = a + bi,$$\n\nwhere:\n- $a$ is the real part of the complex number,\n- $b$ is the imaginary part\n- $i$ is the imaginary unit with the property $i^2 = -1$.\n\nOperations with Complex Number\n\nComplex numbers can be added, subtracted, multiplied, and divided. Here are the basic operations:\n\n- **Addition**: To add two complex numbers, add the real parts and the imaginary parts separately:\n $$(a + bi) + (c + di) = (a + c) + (b + d)i.$$\n\n- **Subtraction**: Similarly, subtract the real parts and the imaginary parts separately:\n $$(a + bi) - (c + di) = (a - c) + (b - d)i.$$\n\n- **Multiplication**: Multiplying complex numbers requires using the distributive property and the fact that $i^2 = -1$:\n $$(a + bi) \\times (c + di) = (ac - bd) + (ad + bc)i.$$\n\n- **Division**: To divide complex numbers, multiply the numerator and denominator by the conjugate of the denominator, then simplify:\n $$\\frac{a + bi}{c + di} = \\frac{(a + bi) \\times (c - di)}{(c + di) \\times (c - di)} = \\frac{ac + bd}{c^2 + d^2} + \\frac{bc - ad}{c^2 + d^2}i.$$", "id": "./materials/898.pdf" }, { "contents": "Complex Plane is a two-dimensional representation of complex numbers, where each complex number is plotted as a point with a real part and an imaginary part.\n\nIn the complex plane:\n\n- The horizontal axis (x-axis) represents the real part, also known as Re(z).\n- The vertical axis (y-axis) represents the imaginary part, referred to as Im(z).\n\nA complex number $z = a + bi$ can be plotted as the point $(a, b)$ in the complex plane, where:\n\n- $a$ is the real part, determining the x-coordinate.\n- $b$ is the imaginary part, determining the y-coordinate.\n\nExamples\n\n1. Consider the complex number $z = 3 + 4i$.\n\n In the complex plane, this number is plotted as the point $(3, 4)$, where:\n\n - The real part is 3.\n - The imaginary part is 4.\n\n To visualize this, you would move 3 units along the x-axis (representing the real part), then 4 units along the y-axis (representing the imaginary part). This places the point at $(3, 4)$. \n2. Consider the complex number \\( z = -5i \\).\n\nIn the complex plane, this number is plotted as the point \\((0, -5)\\), where:\n\n- The real part is 0.\n- The imaginary part is \\(-5\\).\n\nTo visualize this, you would start at the origin, move 0 units along the x-axis (representing the real part), then move \\(-5\\) units along the y-axis (representing the imaginary part), indicating a downward shift. This places the point at \\((0, -5)\\).\n\n3. Consider the complex number \\( z = 7i - 6 \\).\n\nIn the complex plane, this number is plotted as the point \\((-6, 7)\\), where:\n\n- The real part is \\(-6\\).\n- The imaginary part is 7.\n\nTo visualize this, you would start at the origin, move \\(-6\\) units along the x-axis (indicating a leftward shift), then move 7 units along the y-axis (representing the imaginary part), indicating an upward shift. This places the point at \\((-6, 7)\\).", "id": "./materials/899.pdf" }, { "contents": "Turn a Word Doc into a PDF\n\nPDF stands for portable document format. It is a file type (.pdf) just as a Microsoft Word document is a text document (.doc).\n\nPDF is the preferred file type for online publishing because unlike a Word doc, which can be modified, PDFs preserve text and formatting and are easily downloaded to look exactly as it does online.\n\nHere’s how to quickly make a PDF from a Word file:\n\n1. Use a file name that’s all lowercase, inserting hyphens for spaces: vista-community-college.doc\n\n2. Open each Word doc and then for each doc select Print under File (in the main menu bar) just like you were going to print the page.\n\n3. Notice the PDF button on the bottom far left side of the window (see diagram). Select PDF and a drop-down menu appears with Save as PDF as the first option. Select it.\n\n4. Word will now create a PDF file where you want it on your computer (either on your desktop, in a selected folder, or on an external device). Notice now that the file name has changed its extension (vista-community-college.pdf). You now have a PDF, as well as your original Word doc.\n\nFor multiple Word docs, repeat steps 1 through 4 for each doc. Attach the PDF to an email just as you would a Word doc or other file attachment.\n\nViewers can download the free Adobe Reader software to view PDFs or use another image viewer, like Apple’s Preview or Microsoft Reader.", "id": "./materials/9.pdf" }, { "contents": "Evaluate \\[ \\int_{0}^{\\frac{3\\pi}{2}} y \\csc^2(y) \\, dy \\]\n\n**HINT:** When solving problems related to definite integrals, the first step is to check if all the conditions to use Fundamental Theorem of Calculus are met or not. This question is an example where one of the condition (Continuity condition) is not met.", "id": "./materials/90.pdf" }, { "contents": "Conjugate number\n\nThe conjugate of a complex number \\( z = a + bi \\) is given by \\( z^* = a - bi \\). This operation changes the sign of the imaginary part while leaving the real part unchanged.\n\nThe complex conjugate has important properties. For example, when a complex number is multiplied by its conjugate, the result is always a real number. This can be shown mathematically as:\n\n\\[\nz \\times z^* = (a + bi) \\times (a - bi) = a^2 + b^2.\n\\]\n\nIn ordered pair notation, where a complex number is represented as \\( z = (a, b) \\), the complex conjugate is represented as:\n\n\\[\nz^* = (a, -b).\n\\]\n\nThese properties make the complex conjugate a valuable tool in various mathematical and engineering contexts, allowing the simplification of expressions and the calculation of magnitudes in complex arithmetic.\n\n- **Examples:**\n 1. If \\( z = 1 + i \\), then \\( z^* = 1 - i \\).\n 2. If \\( z = i \\), then \\( z^* = -i \\).\n 3. If \\( z = -1 - 5i \\), then \\( z^* = -1 + 5i \\).\n\n- **Properties**\n\n Given two complex numbers in algebraic form, \\( z = a + bi \\) and \\( w = c + di \\), several properties related to complex conjugates can be derived.\n\n 1. A complex number equals its conjugate:\n\n The complex conjugate of \\( z = a + bi \\) is \\( z^* = a - bi \\). If a complex number is equal to its complex conjugate, it must be a real number, as the imaginary parts cancel out:\n\n If \\( z = z^* \\), then \\( a + bi = a - bi \\), which is true only when \\( b = 0 \\). Thus, the complex number is real, or \\( z \\in \\mathbb{R} \\).\n\n 2. Conjugate of the sum of two Complex Numbers:\n\n Given \\( z = a + bi \\) and \\( w = c + di \\), the complex conjugate of their sum is equivalent to the sum of their complex conjugates:\n\\((z + w)^*\\) is the complex conjugate of \\(z + w\\):\n\n\\[(a + bi + c + di)^* = (a + c) - (b + d)i.\\]\n\nThe sum of the complex conjugates of \\(z\\) and \\(w\\) is:\n\n\\[(z^* + w^*) = (a - bi) + (c - di) = (a + c) - (b + d)i.\\]\n\nThus, \\((z + w)^* = z^* + w^*\\).\n\n3. Conjugate of the product of two complex numbers:\n\nGiven \\(z = a + bi\\) and \\(w = c + di\\), the complex conjugate of their product is equivalent to the product of their complex conjugates:\n\n\\((z \\cdot w)^*:\\)\n\n\\[((a + bi) \\cdot (c + di))^* = ((ac - bd) + (ad + bc)i)^* = (ac - bd) - (ad + bc)i.\\]\n\nThe product of \\(z^*\\) and \\(w^*\\) is:\n\n\\[(z^* \\cdot w^*) = (a - bi) \\cdot (c - di) = (ac - bd) - (ad + bc)i.\\]\n\nThus, \\((z \\cdot w)^* = z^* \\cdot w^*\\).\n\n4. Multiplication of a Complex Number by Its Conjugate:\n\nGiven \\(z = a + bi\\), the product of a complex number with its conjugate is always a real number:\n\n\\(z \\cdot z^*:\\)\n\n\\[(a + bi) \\cdot (a - bi) = a^2 + b^2,\\]\n\nwhich is always real.\n\n5. Conjugate of the quotient of two complex numbers:\n\nThe complex conjugate of the quotient of two complex numbers is equivalent to the quotient of their complex conjugates:\n\n\\[\\left(\\frac{z}{w}\\right)^* = \\frac{z^*}{w^*}.\\]", "id": "./materials/900.pdf" }, { "contents": "**Exponential Form:** Using Euler’s formula, which states $e^{i\\theta} = \\cos(\\theta) + i\\sin(\\theta)$, the polar form can be written as:\n\n$$z = r \\cdot e^{i\\theta}.$$ \n\n**Example:** Find the exponential form of the complex number $z = \\frac{-i}{2i+2}$\n\nFirst it is necessary to simplify the expression. Here are the steps:\n\n1. Simplify the complex fraction factorizing it:\n\n $$z = \\frac{-i}{2i+2} \\Rightarrow z = \\frac{-i}{2(i+1)}$$\n\n Simplify by dividing the numerator and denominator by 2:\n\n $$z = \\frac{-i}{2} \\cdot \\frac{1}{i+1}$$\n\n2. Multiply the numerator and the denominator by the conjugate of the denominator:\n\n The conjugate of $i+1$ is $i-1$. So we have:\n\n $$z = \\frac{-i}{2} \\cdot \\frac{1}{i+1} \\cdot \\frac{i-1}{i-1}$$\n\n Simplify the denominator:\n\n $$(i+1)(i-1) = i^2 - 1 = -1 - 1 = -2$$\n\n Now we have:\n\n $$z = \\frac{-i}{2} \\cdot \\frac{i-1}{-2} \\Rightarrow \\frac{-i(i-1)}{-4}$$\n\n3. Simplify the Numerator\n\n Expand and simplify:\n\n $$-i(i-1) = -i^2 + i = 1 + i$$\n\n Thus:\n\n $$z = \\frac{1 + i}{-4}$$\nSo:\n\n\\[ z = -\\frac{1}{4} - \\frac{i}{4} \\]\n\n4. Identify the Quadrant\n\nThe complex number \\(-\\frac{1}{4} - \\frac{i}{4}\\) is in the third quadrant because both the real part and the imaginary part are negative.\n\n5. Calculate the Argument\n\nThe general argument for a complex number \\(x + yi\\) is:\n\n\\[ \\theta = \\tan^{-1}\\left(\\frac{y}{x}\\right) \\]\n\nHere, \\(x = -\\frac{1}{4}\\) and \\(y = -\\frac{1}{4}\\), so:\n\n\\[ \\theta = \\tan^{-1}\\left(\\frac{-\\frac{1}{4}}{-\\frac{1}{4}}\\right) = \\tan^{-1}(1) = \\frac{\\pi}{4} \\]\n\nSince the number is in the third quadrant, we need to adjust the angle to reflect this. The correct angle for the third quadrant is:\n\n\\[ \\theta = \\pi + \\frac{\\pi}{4} = \\frac{5\\pi}{4} \\]\n\nAlternatively, this angle can also be expressed as a negative angle:\n\n\\[ \\theta = -\\frac{3\\pi}{4} \\]\n\nSo the argument for the complex number \\(-\\frac{1}{4} - \\frac{i}{4}\\) is indeed \\(-\\frac{3\\pi}{4}\\).\n\n6. Exponential Form\n\nUsing Euler’s formula, the polar form \\(z = re^{i\\theta}\\), the exponential form of \\(z = -\\frac{1}{4} - \\frac{i}{4}\\) is:\n\n\\[ z = \\frac{\\sqrt{2}}{4} e^{-\\frac{3\\pi}{4}} \\]", "id": "./materials/901.pdf" }, { "contents": "Distance in the Complex Plane\nGiven two complex numbers \\( z_0 = x_0 + y_0i \\) and \\( z = x + yi \\), the distance between them in the complex plane is given by:\n\n\\[\n|z - z_0| = \\sqrt{(x - x_0)^2 + (y - y_0)^2}.\n\\]\n\n- Circles in the Complex Plane:\n A circle in the complex plane is defined by an equation of the form \\( |z - z_0| = \\delta \\), where \\( \\delta > 0 \\) is the radius and \\( z_0 = x_0 + y_0i \\) is the center.\n This equation describes all points in the complex plane that are at a distance \\( \\delta \\) from the center \\( z_0 \\).\n\n- Ring in the Complex Plane:\n A ring is a region defined by the inequality \\( \\beta \\leq |z - z_0| \\leq \\delta \\).\n This set includes all points between two concentric circles with radii \\( \\beta \\) and \\( \\delta \\), centered at \\( z_0 \\).\n\n- Open Sets in the Complex Plane:\n An open set is a region where the boundary is not included. For example, the set defined by \\( |z - z_0| > \\delta \\) is an open set, describing all points outside a circle with radius \\( \\delta \\).\n\nExamples of Regions in the Complex Plane\n- Circle with Radius 2:\n Equation \\( |z| = 2 \\) represents a circle with radius 2 centered at the origin.\n\n- Circle with Center 1 + i:\n The equation \\( |z - 1 - i| = 1 \\) describes a circle with radius 1 centered at \\( 1 + i \\).\n• Ring with Radius 2 and 5:\nThe inequality $2 < |z| < 5$ describes a ring where the smallest circle has a radius of 2, and the largest circle has a radius of 5. This ring does not include the boundary circles due to the strict inequalities.\n\n• Open Set Outside a Circle with Radius 3:\nThe inequality $|z| > 3$ represents the set of points outside a circle with radius 3, centered at the origin.\n\nRegions with Real and Imaginary Parts\nWhen considering only the real or imaginary parts, regions may no longer be circular:\n\n• Plane with $-1 < x < 1$:\nThe inequality $-1 < \\text{Re}(z) < 1$ represents all points between $-1 < x < 1$ and any $y$.\n\n• Half-Plane with $x > 0$:\nInequality $\\text{Re}(z) \\geq 0$ represents all points with $x > 0$ and any $y$.\n\n• Plane with $-1 < y < 1$:\nThe inequality $-1 < \\text{Im}(z) < 1$ represents all points between $-1 < y < 1$ and any $x$. ", "id": "./materials/902.pdf" }, { "contents": "**Polar Form**: A complex number can be expressed using its magnitude \\((r)\\) and argument \\((\\theta)\\), where \\(r = \\sqrt{a^2 + b^2}\\) and \\(\\theta = \\arctan\\left(\\frac{b}{a}\\right)\\):\n\n\\[\nz = r(\\cos(\\theta) + i \\sin(\\theta)).\n\\]\n\n**Example** Find the polar form of the complex number \\(z = 1 + i\\)\n\nFollowing the steps:\n\n1. Calculate the modulus \\(r\\):\n \n The modulus \\(r\\) of a complex number \\(z = a + bi\\) is given by:\n \n \\[\nr = \\sqrt{a^2 + b^2}\n \\]\n \n For \\(z = 1 + i\\):\n \n \\[\nr = \\sqrt{1^2 + 1^2} = \\sqrt{1 + 1} = \\sqrt{2}\n \\]\n\n2. Calculate the argument \\(\\theta\\):\n \n The argument \\(\\theta\\) (in radians) of a complex number \\(z = a + bi\\) is given by:\n \n \\[\n\\theta = \\tan^{-1}\\left(\\frac{b}{a}\\right)\n\\]\n \n For \\(z = 1 + i\\):\n \n \\[\n\\theta = \\tan^{-1}\\left(\\frac{1}{1}\\right) = \\tan^{-1}(1)\n\\]\n \n The arctangent of 1 is \\(\\frac{\\pi}{4}\\) radians.\n\n3. Write the polar form:\n \n Knowing that \\(r = \\sqrt{2}\\) and \\(\\theta = \\frac{\\pi}{4}\\) and that the polar form of a complex number is \\(z = r(\\cos \\theta + i \\sin \\theta)\\).\n \n Therefore, the polar form of \\(z = 1 + i\\) is:\n \n \\[\nz = \\sqrt{2} \\left(\\cos \\frac{\\pi}{4} + i \\sin \\frac{\\pi}{4}\\right)\n\\]\n\n Alternatively, this can also be expressed using Euler’s formula as:\n \n \\[\nz = \\sqrt{2} e^{i \\frac{\\pi}{4}}\n\\]", "id": "./materials/903.pdf" }, { "contents": "Powers of $i$\n\nIn the complex number system, the imaginary unit $i$ is defined such that $i^2 = -1$. The concept of powers of $i$ is central to complex arithmetic and is used extensively in mathematics.\n\n1. Basic Powers of $i$\n\nGiven that $i^2 = -1$, you can derive the following patterns for the powers of $i$:\n\n- $i^0 = 1$: By definition, any number raised to the power of zero is one.\n- $i^1 = i$: The imaginary unit itself.\n- $i^2 = -1$: By definition.\n- $i^3 = i^2 \\times i = -1 \\times i = -i$: Multiplying $i^2$ by $i$.\n- $i^4 = i^2 \\times i^2 = (-1) \\times (-1) = 1$: Returning to the starting point.\n\n2. Cycle of Powers\n\nGiven the above, it’s clear that the powers of $i$ follow a repeating pattern every four iterations:\n\n- $i^0 = 1$.\n- $i^1 = i$.\n- $i^2 = -1$.\n- $i^3 = -i$.\n- $i^4 = 1$.\n\nThus, for any power of $i^n$, the result will be one of these four values, depending on the remainder when $n$ is divided by 4.\n\n3. General Formula for Powers of $i$\n\nThe periodicity in the powers of $i$ means that the general formula to determine $i^n$ involves the modulo operation. Given $n$, the remainder when $n$ is divided by 4 determines the power of $i^n$:\n\n- If $n \\mod 4 = 0$, then $i^n = 1$.\n- If $n \\mod 4 = 1$, then $i^n = i$.\n- If $n \\mod 4 = 2$, then $i^n = -1$.\n- If $n \\mod 4 = 3$, then $i^n = -i$. \n4. Examples to illustrate this concept:\n\n- \\( i^7 \\): When 7 is divided by 4, the quotient is 1 and the remainder is 3.\n Then \\( 7 \\mod 4 = 3 \\), the result is \\( i^7 = i^3 = -i \\).\n- \\( i^{10} \\): When 10 is divided by 4, the quotient is 2 and the remainder is 2.\n So, \\( 10 \\mod 4 = 2 \\) results in \\( i^{10} = i^2 = -1 \\).\n- \\( i^{15} \\): When 15 is divided by 4, the quotient is 3 and the remainder is 3.\n So, \\( 15 \\mod 4 = 3 \\) results in \\( i^{15} = i^3 = -i \\).\n- \\( i^5 \\): When 5 is divided by 4, the quotient is 1 and the remainder is 1.\n Then \\( 5 \\mod 4 = 1 \\), the result is \\( i^5 = i^1 = i \\).\n\n**Exercise 1:** Solve \\( i^{148} \\)\n\nWhen 148 is divided by 4, the quotient is 37 and the remainder is 0.\nThen \\( 148 \\mod 4 = 0 \\), the result is \\( i^{148} = i^0 = 1 \\).\n\n**Exercise 2:** Simplify \\( i^3 + i^6 + i^9 \\)\n\nTo simplify \\( i^3 + i^6 + i^9 \\), let’s determine the values of each power of \\( i \\) using the modulo operation with 4:\n\n- Find \\( i^3 \\) Since \\( 3 \\mod 4 = 3 \\), we know that \\( i^3 = -i \\).\n- Find \\( i^6 \\) Since \\( 6 \\mod 4 = 2 \\), the value of \\( i^6 = i^2 = -1 \\).\n- Find \\( i^9 \\) Since \\( 9 \\mod 4 = 1 \\), the value of \\( i^9 = i^1 = i \\).\n\nNow that we know the individual values, we can simplify the expression \\( i^3 + i^6 + i^9 \\):\n\n\\[\ni^3 + i^6 + i^9 = (-i) + (-1) + (i) = -i - 1 + i.\n\\]\n\nThe \\( i \\) terms cancel out, leaving:\n\n\\[\n-i + i - 1 = -1.\n\\]\n\nTherefore, the simplified result for \\( i^3 + i^6 + i^9 \\) is: \\(-1\\).", "id": "./materials/904.pdf" }, { "contents": "Face aos respetivos registos, certifico que o estudante abaixo identificado, obteve aprovação nas unidades curriculares referidas, nas datas e com as classificações indicadas.\n\nEstudante: 21485 - Tânia Cristina Ferreira da Silva\n\nCartão de cidadão nº: 15696674\n\nPlano: 1 Curso: MA81 - Mestrado - 2.º ciclo - Cibersegurança\n\nEscola Superior de Tecnologia e Gestão do Instituto Politécnico de Viana do Castelo\n\n| A/S Cur. | Unidade Curricular | Nota | Nota (extenso) | ECTS | Conclusão da UC | Desc. época avaliação |\n|---------|---------------------------------------------------------|------|----------------|------|----------------|----------------------|\n| 1 | Criptografia Aplicada | 16 | dezessete valores | 5 | 12-01-2022 | Avaliação por Exame Normal |\n| 1 | Segurança de Redes e Sistemas | 14 | catorze valores | 5 | 06-02-2022 | Avaliação por Exame Normal |\n| 1 | Auditoria e Conformidade em Cibersegurança | 16 | dezessete valores | 4 | 20-06-2022 | Avaliação por Exame Normal |\n| 1 | Gestão da Segurança da Informação | 17 | dezessete valores | 5 | 28-01-2022 | Avaliação por Exame Normal |\n| 1 | Segurança no Software | 16 | dezessete valores | 5 | 31-01-2022 | Avaliação por Exame Normal |\n| 1 | Segurança de Sistemas Cibernéticos | 12 | doze valores | 4 | 10-02-2022 | Avaliação por Exame Normal |\n| 1 | Estratégias de Defesa na Administração de Sistemas | 15 | quinze valores | 6 | 12-02-2022 | Avaliação por Exame Normal |\n| 1 | Cibercrime e Análise Forense Digital | 13 | treze valores | 6 | 28-06-2022 | Avaliação por Exame Normal |\n| 1 | Hacking Ótico | 17 | dezessete valores | 6 | 28-06-2022 | Avaliação por Exame Normal |\n| 1 | Privacidade e Proteção de Dados | 15 | quinze valores | 3 | 20-06-2022 | Avaliação por Exame Normal |\n| 1 | Engenharia Social | 17 | dezessete valores | 3 | 14-07-2022 | Avaliação por Exame Normal |\n| 1 | Gestão de Identidade Digital | 18 | dezoito valores | 3 | 15-07-2022 | Avaliação por Exame Normal |\n| 1 | Análise de Dados e CiberInteligência | 19 | dezanove valores | 5 | 16-07-2022 | Avaliação por Exame Normal |\n| 2 | Metodologia de Investigação e Gestão de Projetos | 17 | dezessete valores | 3 | 24-01-2023 | Avaliação por Exame Normal |\n| 2 | Dissertação/Projeto/Fotógrafo | 19 | dezanove valores | 57 | 15-01-2024 | Avaliação por Exame Normal - 1º Admitimento - Mestrado |\n\nMais certifico que, nos termos do Decreto-Lei n.º 74/2006 de 24 de março, republicado pelo Decreto-Lei n.º 65/2018 de 16 de agosto, com a aprovação nas unidades curriculares referidas que integra o ciclo de estudos, concluíu a 15 de janeiro de 2024, com a média de 17 (dezessete valores) – resultante do arredondamento de 17,3 (dezessete valores e três décimas) o curso: MA81 - Mestrado - 2.º ciclo - Cibersegurança.\n\nA avaliação e aprovação nas unidades referidas, são objeto de certificação e eventual creditação nos limites fixados na alínea c) do n.º 1 do artigo 45.º do referido diploma legal.\n\nO presente vai autenticado com o selo branco desta escola, e contém 01 página(s).\n\nEscola Superior de Tecnologia e Gestão do Instituto Politécnico de Viana do Castelo, em 16 de abril de 2024.\n\nA Diretora\n\n[Assinatura]\n\nProf. Doutora Mafalda Lopes Laranjo\n\nEmitted por, Arminda Moreira\nConferido por, [Assinatura]", "id": "./materials/905.pdf" }, { "contents": "Bisection Method\nThe bisection method is a simple and robust numerical technique used to find the roots of a continuous function. It works by iteratively narrowing down an interval where a root is known to exist, based on the Intermediate Value Theorem. This theorem states that if a continuous function changes sign over an interval, then there is at least one root within that interval. The method is particularly useful when an analytical solution is difficult or impossible to obtain.\n\nSteps of the Bisection Method\n\n1. **Initial Interval Selection**: Choose initial points $a$ and $b$ such that $f(a) \\cdot f(b) < 0$.\n\n2. **Midpoint Calculation**: Compute the midpoint $c = \\frac{a+b}{2}$.\n\n3. **Interval Reduction**: Determine whether the root lies in $[a, c]$ or $[c, b]$ by checking the sign of $f(c)$:\n - If $f(a) \\cdot f(c) < 0$, then the root is in $[a, c]$.\n - If $f(b) \\cdot f(c) < 0$, then the root is in $[c, b]$.\n\n4. **Iteration**: Repeat the process until the interval is sufficiently small.\nProblem: Solving $x^3 + 4x^2 = 10$ in the Interval $[1, 2]$\n\na) What is the approximation after 5 iterations?\n\n1. Initial interval: $[1, 2]$\n - Compute $f(1) = 1^3 + 4(1)^2 - 10 = -5$\n - Compute $f(2) = 2^3 + 4(2)^2 - 10 = 14$\n - Since $f(1) \\cdot f(2) < 0$, a root exists in $[1, 2]$.\n\n2. Iteration 1:\n - $c = \\frac{1+2}{2} = 1.5$\n - Compute $f(1.5) = 1.5^3 + 4(1.5)^2 - 10 = 2.375$\n - Since $f(1) \\cdot f(1.5) < 0$, new interval is $[1, 1.5]$.\n\n3. Iteration 2:\n - $c = \\frac{1+1.5}{2} = 1.25$\n - Compute $f(1.25) = 1.25^3 + 4(1.25)^2 - 10 = -1.796875$\n - Since $f(1.25) \\cdot f(1.5) < 0$, new interval is $[1.25, 1.5]$.\n\n4. Iteration 3:\n - $c = \\frac{1.25+1.5}{2} = 1.375$\n - Compute $f(1.375) = 1.375^3 + 4(1.375)^2 - 10 = 0.162109$\n - Since $f(1.25) \\cdot f(1.375) < 0$, new interval is $[1.25, 1.375]$.\n\n5. Iteration 4:\n - $c = \\frac{1.25+1.375}{2} = 1.3125$\n - Compute $f(1.3125) = 1.3125^3 + 4(1.3125)^2 - 10 = -0.848388$\n - Since $f(1.3125) \\cdot f(1.375) < 0$, new interval is $[1.3125, 1.375]$.\n\n6. Iteration 5:\n - $c = \\frac{1.3125+1.375}{2} = 1.34375$\n - Compute $f(1.34375) = 1.34375^3 + 4(1.34375)^2 - 10 = -0.350982$\n - Since $f(1.34375) \\cdot f(1.375) < 0$, new interval is $[1.34375, 1.375]$.\n\nApproximation after 5 iterations: 1.34375\nb) What are the conditions for the bisection method to converge?\n\nConditions for Convergence:\n\n- The function \\( f(x) \\) must be continuous on the interval \\([a, b]\\).\n- The initial interval \\([a, b]\\) must be chosen such that \\( f(a) \\cdot f(b) < 0 \\), indicating that the function changes sign over the interval. This ensures there is at least one root within the interval.\n\nc) What is the minimum number of iterations to find an approximation to the solution, using the bisection method, with an absolute error smaller than \\(10^{-4}\\)?\n\nTo determine the minimum number of iterations required to achieve an absolute error smaller than \\(10^{-4}\\), we use the formula for the interval width after \\(n\\) iterations:\n\n\\[\n\\text{Width after } n \\text{ iterations} = \\frac{b - a}{2^n}\n\\]\n\nGiven that \\(b - a = 1\\) (since the interval is \\([1, 2]\\)), we want the interval width to be less than \\(10^{-4}\\):\n\n\\[\n\\frac{1}{2^n} < 10^{-4}\n\\]\n\nTaking the logarithm (base 10) of both sides:\n\n\\[\n\\log_{10} \\left( \\frac{1}{2^n} \\right) < \\log_{10}(10^{-4})\n\\]\n\n\\[-n \\log_{10}(2) < -4\n\\]\n\nSolving for \\(n\\):\n\n\\[\nn > \\frac{4}{\\log_{10}(2)}\n\\]\n\nSince \\(\\log_{10}(2) \\approx 0.3010\\):\n\n\\[\nn > \\frac{4}{0.3010} \\approx 13.2877\n\\]\n\nTherefore, the minimum number of iterations \\(n\\) must be: \\(n \\geq 14\\)\n\nMinimum number of iterations: 14", "id": "./materials/906.pdf" }, { "contents": "Gauss-Seidel\nThe Gauss-Seidel method is an iterative technique for solving a system of linear equations. The method improves upon the Jacobi method by using the most recent updates of the variables as soon as they are available.\n\nSteps of the Gauss-Seidel Method:\n\n- Formulate the System of Equations: Start with a system of linear equations in the form $Ax = b$:\n\n\\[\n\\begin{align*}\n a_{11}x_1 + a_{12}x_2 + \\cdots + a_{1n}x_n &= b_1 \\\\\n a_{21}x_1 + a_{22}x_2 + \\cdots + a_{2n}x_n &= b_2 \\\\\n \\vdots \\\\\n a_{n1}x_1 + a_{n2}x_2 + \\cdots + a_{nn}x_n &= b_n\n\\end{align*}\n\\]\n\n- Rewrite Each Equation: Rewrite each equation to solve for one of the variables in terms of the others. For example, solve the $i$-th equation for $x_i$:\n\n\\[\nx_i = \\frac{1}{a_{ii}} \\left( b_i - \\sum_{j \\neq i} a_{ij}x_j \\right)\n\\]\n\n- Initial Guess: Choose an initial guess for the values of the unknowns $(x_1^{(0)}, x_2^{(0)}, \\ldots, x_n^{(0)})$.\n\n- Iterative Process: Update each variable sequentially using the most recent values. For the $k$-th iteration:\n\n\\[\nx_i^{(k+1)} = \\frac{1}{a_{ii}} \\left( b_i - \\sum_{j=1}^{i-1} a_{ij}x_j^{(k+1)} - \\sum_{j=i+1}^{n} a_{ij}x_j^{(k)} \\right)\n\\]\n\nHere, $x_j^{(k+1)}$ is the updated value and $x_j^{(k)}$ is the previous iteration’s value.\n\n- Convergence Check: Continue the iterations until the solution converges, i.e., the changes in the variables between successive iterations are smaller than a chosen tolerance level:\n\n\\[\n|x_i^{(k+1)} - x_i^{(k)}| < \\epsilon \\quad \\text{for all } i\n\\]\nProblem\n\nGiven the system of equations, find the solution approximation after 1 iteration of the Gauss-Seidel method:\n\n\\[\n\\begin{align*}\n60x_1 - 30x_2 - 20x_3 &= -400 \\\\\n-30x_1 + 180x_2 - 60x_3 &= 400 \\\\\n-20x_1 - 60x_2 + 150x_3 &= 300\n\\end{align*}\n\\]\n\nwe can rewrite the equations in the form suitable for the method. Starting with an initial approximation \\((x_1^{(0)}, x_2^{(0)}, x_3^{(0)}) = (0, 0, 0)\\):\n\n1. **Initial Values:**\n \\((x_1^{(0)}, x_2^{(0)}, x_3^{(0)}) = (0, 0, 0)\\)\n\n2. **Step 1:** Write the equations in a form suitable for the Gauss-Seidel method\n\n The given system of equations, we need to rewrite each equation to solve for one of the variables in terms of the others:\n\n \\[\n \\begin{align*}\n x_1 &= \\frac{1}{60} (-400 + 30x_2 + 20x_3) \\\\\n x_2 &= \\frac{1}{180} (400 + 30x_1 + 60x_3) \\\\\n x_3 &= \\frac{1}{150} (300 + 20x_1 + 60x_2)\n \\end{align*}\n \\]\n\n3. **Step 2:** Iterative Process\n\n We use the Gauss-Seidel update rules to compute the next approximations. We will perform one iteration to demonstrate the process.\n\n - **Iteration 1:**\n 1. Update \\(x_1\\):\n\n \\[\n x_1^{(1)} = \\frac{-400 + 30x_2^{(0)} + 20x_3^{(0)}}{60} = \\frac{-400 + 30 \\cdot 0 + 20 \\cdot 0}{60} = \\frac{-400}{60} = -6.6667\n \\]\n\n 2. Update \\(x_2\\):\n\n \\[\n x_2^{(1)} = \\frac{400 + 30x_1^{(1)} + 60x_3^{(0)}}{180} = \\frac{400 + 30 \\cdot (-6.6667) + 60 \\cdot 0}{180} = \\frac{400 - 200}{180} = 1.1111\n \\]\n3. Update $x_3$:\n\n$$x_3^{(1)} = \\frac{300 + 20x_1^{(1)} + 60x_2^{(1)}}{150} = \\frac{300 + 20 \\cdot -6.6667 + 60 \\cdot 1.1111}{150}$$\n\n$$x_3^{(1)} = \\frac{300 - 133.334 + 66.666}{150} = 1.1111$$\n\n- **Iteration 2**\n\n1. Update $x_1$:\n\n$$x_1^{(2)} = \\frac{-400 + 30x_2^{(1)} + 20x_3^{(1)}}{60} = \\frac{-400 + 30 \\cdot 1.1111 + 20 \\cdot 1.1111}{60}$$\n\n$$x_1^{(2)} = \\frac{-400 + 33.333 + 22.222}{60} = -5.7245$$\n\n2. Update $x_2$:\n\n$$x_2^{(2)} = \\frac{400 + 30x_1^{(2)} + 60x_3^{(1)}}{180} = \\frac{400 + 30 \\cdot -5.7245 + 60 \\cdot 1.1111}{180}$$\n\n$$x_2^{(2)} = \\frac{400 - 171.735 + 66.666}{180} = 1.6213$$\n\n3. Update $x_3$:\n\n$$x_3^{(2)} = \\frac{300 + 20x_1^{(2)} + 60x_2^{(2)}}{150} = \\frac{300 + 20 \\cdot -5.7245 + 60 \\cdot 1.6213}{150}$$\n\n$$x_3^{(2)} = \\frac{300 - 114.49 + 97.278}{150} = 1.606$$\n\n- **Iteration 3**\n\n1. Update $x_1$:\n\n$$x_1^{(3)} = \\frac{-400 + 30x_2^{(2)} + 20x_3^{(2)}}{60} = \\frac{-400 + 30 \\cdot 1.6213 + 20 \\cdot 1.606}{60}$$\n\n$$x_1^{(3)} = \\frac{-400 + 48.639 + 32.12}{60} = -5.3184$$\n\n2. Update $x_2$:\n\n$$x_2^{(3)} = \\frac{400 + 30x_1^{(3)} + 60x_3^{(2)}}{180} = \\frac{400 + 30 \\cdot -5.3184 + 60 \\cdot 1.606}{180}$$\n\\[ x_2^{(3)} = \\frac{400 - 159.552 + 96.36}{180} = 1.8755 \\]\n\n3. Update \\( x_3 \\):\n\n\\[ x_3^{(3)} = \\frac{300 + 20x_1^{(3)} + 60x_2^{(3)}}{150} = \\frac{300 + 20 \\cdot -5.3184 + 60 \\cdot 1.8755}{150} \\]\n\\[ x_3^{(3)} = \\frac{300 - 106.368 + 112.53}{150} = 2.0558 \\]\n\n- Results after 3 Iterations:\n\n\\[ (x_1^{(3)}, x_2^{(3)}, x_3^{(3)}) = (-5.3184, 1.8755, 2.0558) \\]\n\nThese values are the approximations of \\( x_1 \\), \\( x_2 \\), and \\( x_3 \\) after 3 iterations of the Gauss-Seidel method starting from the initial guess \\((0, 0, 0)\\). This method converges towards the solution, and the accuracy improves with more iterations.", "id": "./materials/907.pdf" }, { "contents": "Gauss-Seidel\nThe Gauss-Seidel method is an iterative technique for solving a system of linear equations. The method improves upon the Jacobi method by using the most recent updates of the variables as soon as they are available.\n\nSteps of the Gauss-Seidel Method:\n\n- Formulate the System of Equations: Start with a system of linear equations in the form $Ax = b$:\n\n$$\n\\begin{align*}\n a_{11}x_1 + a_{12}x_2 + \\cdots + a_{1n}x_n &= b_1 \\\\\n a_{21}x_1 + a_{22}x_2 + \\cdots + a_{2n}x_n &= b_2 \\\\\n \\vdots \\\\\n a_{n1}x_1 + a_{n2}x_2 + \\cdots + a_{nn}x_n &= b_n\n\\end{align*}\n$$\n\n- Rewrite Each Equation: Rewrite each equation to solve for one of the variables in terms of the others. For example, solve the $i$-th equation for $x_i$:\n\n$$\nx_i = \\frac{1}{a_{ii}} \\left( b_i - \\sum_{j \\neq i} a_{ij}x_j \\right)\n$$\n\n- Initial Guess: Choose an initial guess for the values of the unknowns $(x_1^{(0)}, x_2^{(0)}, \\ldots, x_n^{(0)})$.\n\n- Iterative Process: Update each variable sequentially using the most recent values. For the $k$-th iteration:\n\n$$\nx_i^{(k+1)} = \\frac{1}{a_{ii}} \\left( b_i - \\sum_{j=1}^{i-1} a_{ij}x_j^{(k+1)} - \\sum_{j=i+1}^{n} a_{ij}x_j^{(k)} \\right)\n$$\n\nHere, $x_j^{(k+1)}$ is the updated value and $x_j^{(k)}$ is the previous iteration’s value.\n\n- Convergence Check: Continue the iterations until the solution converges, i.e., the changes in the variables between successive iterations are smaller than a chosen tolerance level:\n\n$$\n|x_i^{(k+1)} - x_i^{(k)}| < \\epsilon \\quad \\text{for all } i\n$$\nProblem\n\nGiven the system of equations, find the solution approximation after 1 iteration of the Gauss-Seidel method:\n\n\\[\n\\begin{align*}\n60x_1 - 30x_2 - 20x_3 &= -400 \\\\\n-30x_1 + 180x_2 - 60x_3 &= 400 \\\\\n-20x_1 - 60x_2 + 150x_3 &= 300\n\\end{align*}\n\\]\n\nwe can rewrite the equations in the form suitable for the method. Starting with an initial approximation \\((x_1^{(0)}, x_2^{(0)}, x_3^{(0)}) = (0, 0, 0)\\):\n\n1. **Initial Values:**\n \\((x_1^{(0)}, x_2^{(0)}, x_3^{(0)}) = (0, 0, 0)\\)\n\n2. **Step 1:** Write the equations in a form suitable for the Gauss-Seidel method\n\n The given system of equations, we need to rewrite each equation to solve for one of the variables in terms of the others:\n\n \\[\n \\begin{align*}\n x_1 &= \\frac{1}{60} (-400 + 30x_2 + 20x_3) \\\\\n x_2 &= \\frac{1}{180} (400 + 30x_1 + 60x_3) \\\\\n x_3 &= \\frac{1}{150} (300 + 20x_1 + 60x_2)\n \\end{align*}\n \\]\n\n3. **Step 2:** Iterative Process\n\n We use the Gauss-Seidel update rules to compute the next approximations. We will perform one iteration to demonstrate the process.\n\n - **Iteration 1:**\n 1. Update \\(x_1\\):\n\n \\[\n x_1^{(1)} = \\frac{-400 + 30x_2^{(0)} + 20x_3^{(0)}}{60} = \\frac{-400 + 30 \\cdot 0 + 20 \\cdot 0}{60} = \\frac{-400}{60} = -6.6667\n \\]\n\n 2. Update \\(x_2\\):\n\n \\[\n x_2^{(1)} = \\frac{400 + 30x_1^{(1)} + 60x_3^{(0)}}{180} = \\frac{400 + 30 \\cdot (-6.6667) + 60 \\cdot 0}{180} = \\frac{400 - 200}{180} = 1.1111\n \\]\n3. Update $x_3$:\n\n$$x_3^{(1)} = \\frac{300 + 20x_1^{(1)} + 60x_2^{(1)}}{150} = \\frac{300 + 20 \\cdot -6.6667 + 60 \\cdot 1.1111}{150}$$\n\n$$x_3^{(1)} = \\frac{300 - 133.334 + 66.666}{150} = 1.1111$$\n\n- **Iteration 2**\n\n1. Update $x_1$:\n\n$$x_1^{(2)} = \\frac{-400 + 30x_2^{(1)} + 20x_3^{(1)}}{60} = \\frac{-400 + 30 \\cdot 1.1111 + 20 \\cdot 1.1111}{60}$$\n\n$$x_1^{(2)} = \\frac{-400 + 33.333 + 22.222}{60} = -5.7245$$\n\n2. Update $x_2$:\n\n$$x_2^{(2)} = \\frac{400 + 30x_1^{(2)} + 60x_3^{(1)}}{180} = \\frac{400 + 30 \\cdot -5.7245 + 60 \\cdot 1.1111}{180}$$\n\n$$x_2^{(2)} = \\frac{400 - 171.735 + 66.666}{180} = 1.6213$$\n\n3. Update $x_3$:\n\n$$x_3^{(2)} = \\frac{300 + 20x_1^{(2)} + 60x_2^{(2)}}{150} = \\frac{300 + 20 \\cdot -5.7245 + 60 \\cdot 1.6213}{150}$$\n\n$$x_3^{(2)} = \\frac{300 - 114.49 + 97.278}{150} = 1.606$$\n\n- **Iteration 3**\n\n1. Update $x_1$:\n\n$$x_1^{(3)} = \\frac{-400 + 30x_2^{(2)} + 20x_3^{(2)}}{60} = \\frac{-400 + 30 \\cdot 1.6213 + 20 \\cdot 1.606}{60}$$\n\n$$x_1^{(3)} = \\frac{-400 + 48.639 + 32.12}{60} = -5.3184$$\n\n2. Update $x_2$:\n\n$$x_2^{(3)} = \\frac{400 + 30x_1^{(3)} + 60x_3^{(2)}}{180} = \\frac{400 + 30 \\cdot -5.3184 + 60 \\cdot 1.606}{180}$$\n\\[ x_2^{(3)} = \\frac{400 - 159.552 + 96.36}{180} = 1.8755 \\]\n\n3. Update \\( x_3 \\):\n\n\\[ x_3^{(3)} = \\frac{300 + 20x_1^{(3)} + 60x_2^{(3)}}{150} = \\frac{300 + 20 \\cdot -5.3184 + 60 \\cdot 1.8755}{150} \\]\n\\[ x_3^{(3)} = \\frac{300 - 106.368 + 112.53}{150} = 2.0558 \\]\n\n- Results after 3 Iterations:\n\n\\[ (x_1^{(3)}, x_2^{(3)}, x_3^{(3)}) = (-5.3184, 1.8755, 2.0558) \\]\n\nThese values are the approximations of \\( x_1, x_2, \\) and \\( x_3 \\) after 3 iterations of the Gauss-Seidel method starting from the initial guess \\((0, 0, 0)\\). This method converges towards the solution, and the accuracy improves with more iterations.", "id": "./materials/908.pdf" }, { "contents": "Jacobi Method\nThe Jacobi method is an iterative algorithm used to solve a system of linear equations. The general form of a system of linear equations is:\n\n\\[ Ax = b \\]\n\nwhere \\( A \\) is the coefficient matrix, \\( x \\) is the vector of unknowns, and \\( b \\) is the right-hand side vector.\n\nFor a given system of equations:\n\n\\[\n\\begin{align*}\n a_{11}x_1 + a_{12}x_2 + \\cdots + a_{1n}x_n &= b_1 \\\\\n a_{21}x_1 + a_{22}x_2 + \\cdots + a_{2n}x_n &= b_2 \\\\\n \\vdots \\\\\n a_{n1}x_1 + a_{n2}x_2 + \\cdots + a_{nn}x_n &= b_n\n\\end{align*}\n\\]\n\nthe Jacobi method updates the value of each unknown \\( x_i \\) based on the previous values of all the other unknowns. The formula for updating \\( x_i \\) is:\n\n\\[\nx_i^{(k+1)} = \\frac{1}{a_{ii}} \\left( b_i - \\sum_{j \\neq i} a_{ij}x_j^{(k)} \\right)\n\\]\n\nwhere \\( k \\) denotes the iteration step.\nProblem\nGiven the system of equations, find the solution approximation after 2 iterations of the Jacobi method:\n\n\\[\n\\begin{align*}\n60x_1 - 30x_2 - 20x_3 &= -400 \\\\\n-30x_1 + 180x_2 - 60x_3 &= 400 \\\\\n-20x_1 - 60x_2 + 150x_3 &= 300\n\\end{align*}\n\\]\n\nwe can rewrite the equations in the form suitable for the Jacobi method. Starting with an initial approximation \\((x_1^{(0)}, x_2^{(0)}, x_3^{(0)}) = (0, 0, 0)\\):\n\n1. Initial Values:\n \\[x_1^{(0)} = 0, \\quad x_2^{(0)} = 0, \\quad x_3^{(0)} = 0\\]\n\n2. First Equation:\n \\[x_1^{(k+1)} = \\frac{1}{60} \\left( -400 + 30x_2^{(k)} + 20x_3^{(k)} \\right)\\]\n\n3. Second Equation:\n \\[x_2^{(k+1)} = \\frac{1}{180} \\left( 400 + 30x_1^{(k)} + 60x_3^{(k)} \\right)\\]\n\n4. Third Equation:\n \\[x_3^{(k+1)} = \\frac{1}{150} \\left( 300 + 20x_1^{(k)} + 60x_2^{(k)} \\right)\\]\n\n5. First Iteration \\((k = 0)\\):\n \\[x_1^{(1)} = \\frac{1}{60} \\left( -400 + 30 \\cdot 0 + 20 \\cdot 0 \\right) = \\frac{-400}{60} = -\\frac{20}{3} \\approx -6.6667\\]\n \\[x_2^{(1)} = \\frac{1}{180} \\left( 400 + 30 \\cdot 0 + 60 \\cdot 0 \\right) = \\frac{400}{180} = \\frac{20}{9} \\approx 2.2222\\]\n \\[x_3^{(1)} = \\frac{1}{150} \\left( 300 + 20 \\cdot 0 + 60 \\cdot 0 \\right) = \\frac{300}{150} = 2\\]\n\nSo after the first iteration, we have:\n\\[x_1^{(1)} \\approx -6.6667, \\quad x_2^{(1)} \\approx 2.2222, \\quad x_3^{(1)} = 2\\]\n6. **Second Iteration** \\((k = 1)\\):\n\n\\[\nx_1^{(2)} = \\frac{1}{60} (-400 + 30 \\cdot 2.2222 + 20 \\cdot 2) = \\frac{1}{60} (-400 + 66.666 + 40) = \\frac{-293.334}{60} \\approx -4.889\n\\]\n\n\\[\nx_2^{(2)} = \\frac{1}{180} (400 + 30 \\cdot -6.6667 + 60 \\cdot 2) = \\frac{1}{180} (400 - 200 + 120) = \\frac{320}{180} \\approx 1.778\n\\]\n\n\\[\nx_3^{(2)} = \\frac{1}{150} (300 + 20 \\cdot -6.6667 + 60 \\cdot 2.2222) = \\frac{1}{150} (300 - 133.334 + 133.332) = \\frac{299.998}{150} \\approx 2\n\\]\n\nSo after the second iteration, we have:\n\n\\[\nx_1^{(2)} \\approx -4.889, \\quad x_2^{(2)} \\approx 1.778, \\quad x_3^{(2)} \\approx 2\n\\]\n\nThus, the solution approximation after 2 iterations of the Jacobi method is:\n\n\\[\n(x_1, x_2, x_3) \\approx (-4.889, 1.778, 2)\n\\]", "id": "./materials/909.pdf" }, { "contents": "Evaluate \\( \\int_{1}^{4} \\ln \\left( \\frac{x}{2} \\right) \\, dx \\)\n\n**HINT:**\n\nUse integration by parts:\n\n\\[\n\\int f(x)g(x) \\, dx = F(x)g(x) - \\int F(x)g'(x) \\, dx + C, \\text{ such that } F(x) = \\int f(x) \\, dx\n\\]\n\nThis integral is not a common integral so we cannot find it in the integral table. Therefore, it is necessary to introduce something in the integrand in order to solve it.\n\n\\[\n\\int_{1}^{4} \\ln \\left( \\frac{x}{2} \\right) \\, dx = \\int_{1}^{4} 1 \\cdot \\ln \\left( \\frac{x}{2} \\right) \\, dx\n\\]\n\nNow, take \\( f(x) = 1 \\) and \\( g(x) = \\ln \\frac{x}{2} \\).\n\nWe chose \\( g(x) = \\ln \\frac{x}{2} \\) because we know it’s derivative.\n\nFollow the LIATE rule.", "id": "./materials/91.pdf" }, { "contents": "Lagrange’s polynomial\nThe Lagrange interpolation polynomial is a method used to find the polynomial of the lowest degree that passes through a given set of points. Given a set of \\( n + 1 \\) data points \\((x_0, y_0), (x_1, y_1), \\ldots, (x_n, y_n)\\), the Lagrange interpolation polynomial \\( P(x) \\) is constructed as a linear combination of Lagrange basis polynomials \\( L_i(x) \\).\n\nSteps to Construct the Polynomial:\n\n1. Identify the Data Points: Collect the data points\n\n\\[\n(x_0, y_0), (x_1, y_1), \\ldots, (x_n, y_n)\n\\]\n\n2. Construct Lagrange Basis Polynomials: For each \\( i \\) from 0 to \\( n \\), construct the basis polynomial \\( L_i(x) \\):\n\n\\[\nL_i(x) = \\prod_{0 \\leq j \\leq n, j \\neq i} \\frac{x - x_j}{x_i - x_j}\n\\]\n\n3. Form the Interpolation Polynomial: Combine the basis polynomials weighted by the corresponding \\( y_i \\) values:\n\n\\[\nP(x) = \\sum_{i=0}^{n} y_i L_i(x)\n\\]\nProblem\nConsider the table\n\n| x | 1 | 2 | 3 |\n|---|---|---|---|\n| y | 42 | 42.4 | 42.7 |\n\nUsing the Lagrange polynomial, find the value of \\( P(1.5) \\).\n\nThe following steps are:\n\n1. **Data Points**: \\((x_0, y_0) = (1, 42), (x_1, y_1) = (2, 42.4), (x_2, y_2) = (3, 42.7)\\)\n\n2. **Lagrange Polynomial**: The Lagrange interpolation polynomial \\( P(x) \\) and the Lagrange basis polynomials \\( L_i(x) \\) are given by:\n\n\\[\nP(x) = \\sum_{i=0}^{n} y_i L_i(x) \\quad L_i(x) = \\prod_{0 \\leq j \\leq n \\atop j \\neq i} \\frac{x - x_j}{x_i - x_j}\n\\]\n\n3. **Constructing the Basis Polynomials**: For \\( n = 2 \\), we have three basis polynomials:\n\n\\[\nL_0(x) = \\frac{(x - x_1)(x - x_2)}{(x_0 - x_1)(x_0 - x_2)} = \\frac{(x - 2)(x - 3)}{(1 - 2)(1 - 3)} = \\frac{(x - 2)(x - 3)}{2}\n\\]\n\n\\[\nL_1(x) = \\frac{(x - x_0)(x - x_2)}{(x_1 - x_0)(x_1 - x_2)} = \\frac{(x - 1)(x - 3)}{(2 - 1)(2 - 3)} = \\frac{(x - 1)(x - 3)}{-1} = -(x - 1)(x - 3)\n\\]\n\n\\[\nL_2(x) = \\frac{(x - x_0)(x - x_1)}{(x_2 - x_0)(x_2 - x_1)} = \\frac{(x - 1)(x - 2)}{(3 - 1)(3 - 2)} = \\frac{(x - 1)(x - 2)}{2}\n\\]\n\n4. **Forming the Interpolation Polynomial**:\n\n\\[\nP(x) = y_0 L_0(x) + y_1 L_1(x) + y_2 L_2(x)\n\\]\n\n\\[\nP(x) = 42 \\left( \\frac{(x - 2)(x - 3)}{2} \\right) - 42.4(x - 1)(x - 3) + 42.7 \\left( \\frac{(x - 1)(x - 2)}{2} \\right)\n\\]\n\nIf this polynomial is simplified, the Lagrange interpolation polynomial \\( P(x) \\) is:\n\n\\[\nP(x) = -0.05x^2 + 0.55x + 41.5\n\\]\n5. **Evaluate** $P(1.5)$: Substitute $x = 1.5$ into the polynomial:\n\n\\[\nL_0(1.5) = \\frac{(1.5 - 2)(1.5 - 3)}{2} = \\frac{(-0.5)(-1.5)}{2} = \\frac{0.75}{2} = 0.375\n\\]\n\n\\[\nL_1(1.5) = -(1.5 - 1)(1.5 - 3) = -(0.5)(-1.5) = -(-0.75) = 0.75\n\\]\n\n\\[\nL_2(1.5) = \\frac{(1.5 - 1)(1.5 - 2)}{2} = \\frac{(0.5)(-0.5)}{2} = \\frac{-0.25}{2} = -0.125\n\\]\n\nTherefore:\n\n\\[\nP(x) = y_0L_0(x) + y_1L_1(x) + y_2L_2(x)\n\\]\n\n\\[\nP(1.5) = 42 \\cdot 0.375 + 42.4 \\cdot 0.75 + 42.7 \\cdot (-0.125)\n\\]\n\n\\[\nP(1.5) = 15.75 + 31.8 - 5.3375 = 42.2125\n\\]\n\nSo, the value of $P(1.5)$ using the Lagrange interpolation polynomial is 42.2125.", "id": "./materials/910.pdf" }, { "contents": "Least Square Method\nThe least squares method is a statistical technique used to determine the best-fitting line or curve to a set of data points by minimizing the sum of the squares of the vertical deviations (residuals) from each data point to the line or curve. For a polynomial of degree 1, this method finds the best-fitting linear equation of the form $y = ax + b$.\n\nSteps to solve a problem given the data points:\n\n1. Set up the normal equations:\n - The linear equation is $y = ax + b$.\n - To find the coefficients $a$ and $b$ that minimize the sum of the squared residuals is necessary to use the following normal equations:\n \n $$a \\cdot S_x + b \\cdot n = S_y$$\n $$a \\cdot S_{xx} + b \\cdot S_x = S_{xy}$$\n\n2. Calculate the sums needed:\n \n $$S_x = \\sum x_i, \\quad S_y = \\sum y_i, \\quad S_{xx} = \\sum x_i^2, \\quad S_{xy} = \\sum x_i y_i$$\n\n Where:\n - $S_x$ and $S_y$ are the sums of the $x$-values and $y$-values, respectively.\n - $S_{xx}$ is the sum of the squares of the $x$-values.\n - $S_{xy}$ is the sum of the product of the $x$ and $y$ values.\n\n3. Insert the values and solve for $a$ and $b$. \n**Problem** Consider the table:\n\n| $x$ | $f(x)$ |\n|-----|--------|\n| 2 | 0.5 |\n| 2.5 | 0.4 |\n| 4 | 0.25 |\n\nUsing the least squares method, find the polynomial of degree 1.\n\n1. Given the data points:\n \n $(x_1, y_1) = (2, 0.5)$\n \n $(x_2, y_2) = (2.5, 0.4)$\n \n $(x_3, y_3) = (4, 0.25)$\n\n2. Calculate the required sums:\n \n $S_x = 2 + 2.5 + 4 = 8.5$\n \n $S_y = 0.5 + 0.4 + 0.25 = 1.15$\n \n $S_{xx} = 2^2 + 2.5^2 + 4^2 = 4 + 6.25 + 16 = 26.25$\n \n $S_{xy} = (2 \\cdot 0.5) + (2.5 \\cdot 0.4) + (4 \\cdot 0.25) = 1 + 1 + 1 = 3$\n\n3. Set up the normal equations:\n \n $n = 3$ (number of data points)\n \n $a \\cdot S_x + b \\cdot n = S_y$\n \n $a \\cdot S_{xx} + b \\cdot S_x = S_{xy}$\n\n Substitute the values:\n \n $8.5a + 3b = 1.15$ (1)\n \n $26.25a + 8.5b = 3$ (2)\n\n4. Solve the system of equations:\n\n From (1):\n \n $b = \\frac{1.15 - 8.5a}{3}$\n\n Substitute $b$ in (2):\n \n $26.25a + 8.5 \\left( \\frac{1.15 - 8.5a}{3} \\right) = 3$\n \n $26.25a + \\frac{8.5 \\cdot 1.15 - 8.5 \\cdot 8.5a}{3} = 3$\n\\[26.25a + \\frac{9.775 - 72.25a}{3} = 3\\]\n\\[26.25a + 3.2583 - 24.0833a = 3\\]\n\\[2.1667a + 3.2583 = 3\\]\n\\[2.1667a = -0.2583\\]\n\\[a \\approx -0.1192\\]\n\nSubstitute \\(a\\) back into the equation for \\(b\\):\n\\[b = \\frac{1.15 - 8.5(-0.1192)}{3}\\]\n\\[b = \\frac{1.15 + 1.0132}{3}\\]\n\\[b \\approx 0.7211\\]\n\nThe polynomial of degree 1 \\(P_1(x)\\) that best fits the given data using the least squares method is:\n\\[y = ax + b\\]\n\\[P_1(x) = -0.1192x + 0.7211\\]", "id": "./materials/911.pdf" }, { "contents": "Newton’s polynomial\nNewton’s interpolation polynomial is a method for constructing a polynomial that passes through a given set of data points. It is built incrementally using divided differences.\n\nSteps to Construct Newton’s Interpolation Polynomial\n\n- **Given Data Points:**\n Suppose that we have data points \\((x_0, y_0), (x_1, y_1), \\ldots, (x_n, y_n)\\).\n\n- **Compute Divided Differences:**\n - Zero-order divided difference:\n \\[ f[x_i] = y_i \\]\n - First-order divided difference\n \\[ f[x_i, x_{i+1}] = \\frac{f[x_{i+1}] - f[x_i]}{x_{i+1} - x_i} \\]\n - Second-order divided difference:\n \\[ f[x_i, x_{i+1}, x_{i+2}] = \\frac{f[x_{i+1}, x_{i+2}] - f[x_i, x_{i+1}]}{x_{i+2} - x_i} \\]\n - Continue this process up to \\(n\\)-th order.\n\n- **Construct the Polynomial:**\n The Newton’s interpolation polynomial is:\n \\[\nP(x) = f[x_0] + f[x_0, x_1](x - x_0) + f[x_0, x_1, x_2](x - x_0)(x - x_1) + \\ldots\n \\]\n \\[\n \\ldots + f[x_0, x_1, \\ldots, x_n](x - x_0)(x - x_1) \\cdots (x - x_{n-1})\n \\]\nProblem\nConsider the table:\n\n| $x$ | 2 | 2.5 | 4 |\n|-----|-----|-----|-----|\n| $f(x)$ | 0.5 | 0.4 | 0.25 |\n\nTo find the Newton’s interpolation polynomial $P(x)$:\n\n1. Compute the Divided Differences:\n\n $f[2] = 0.5$\n $f[2.5] = 0.4$\n $f[4] = 0.25$\n\n First-order divided differences:\n\n $f[x_i, x_{i+1}] = \\frac{f[x_{i+1}] - f[x_i]}{x_{i+1} - x_i}$\n\n $f[2, 2.5] = \\frac{0.4 - 0.5}{2.5 - 2} = -0.2$\n $f[2.5, 4] = \\frac{0.25 - 0.4}{4 - 2.5} = -0.1$\n\n Second-order divided difference:\n\n $f[x_i, x_{i+1}, x_{i+2}] = \\frac{f[x_{i+1}, x_{i+2}] - f[x_i, x_{i+1}]}{x_{i+2} - x_i}$\n\n $f[2, 2.5, 4] = \\frac{-0.1 - (-0.2)}{4 - 2} = 0.05$\n\n In short:\n\n | $x$ | $f[x_i]$ | $f[x_i, x_{i+1}]$ | $f[x_i, x_{i+1}, x_{i+2}]$ |\n |-----|----------|-------------------|-----------------------------|\n | 2 | 0.5 | | |\n | 2.5 | 0.4 | $\\frac{0.4 - 0.5}{2.5 - 2} = -0.2$ | |\n | 4 | 0.25 | $\\frac{0.25 - 0.4}{4 - 2.5} = -0.1$ | $\\frac{-0.1 - (-0.2)}{4 - 2} = 0.05$ |\n\n2. Construct the Polynomial:\n\n $P(x) = f[2] + f[2, 2.5](x - 2) + f[2, 2.5, 4](x - 2)(x - 2.5)$\n $= 0.5 - 0.2(x - 2) + 0.05(x - 2)(x - 2.5)$\n3. Simplify the Polynomial:\n\n\\[ P(x) = 0.5 - 0.2(x - 2) + 0.05(x - 2)(x - 2.5) \\]\n\\[ = 0.5 - 0.2x + 0.4 + 0.05(x^2 - 4.5x + 5) \\]\n\\[ = 0.5 - 0.2x + 0.4 + 0.05x^2 - 0.225x + 0.25 \\]\n\\[ = 0.05x^2 - 0.425x + 1.15 \\]\n\nTherefore, the Newton’s interpolation polynomial is:\n\n\\[ P(x) = 0.05x^2 - 0.425x + 1.15 \\]", "id": "./materials/912.pdf" }, { "contents": "Newton’s Method\nNewton’s method is an iterative numerical technique to find approximations of the roots (or zeros) of a real-valued function. The basic idea is to start with an initial guess $x_0$ and then use the tangent line at that point to find a better approximation. The formula for Newton’s method is:\n\n$$x_{n+1} = x_n - \\frac{f(x_n)}{f'(x_n)}$$\n\nWhere:\n\n- $x_n$ is the current approximation,\n- $x_{n+1}$ is the next approximation,\n- $f(x)$ is the function,\n- $f'(x)$ is the derivative of the function.\nProblem\na) Find $x_3$ for $x^3 + 5 = 9x$ using Newton’s method starting with $x_0 = 2.75$\n\n1. Initial Values:\n\n$$x_0 = 2.75, \\quad f(x_0) = 1.046875, \\quad f'(x_0) = 13.6875$$\n\n2. Iteration 1: Compute $x_1$:\n\n$$x_1 = x_0 - \\frac{f(x_0)}{f'(x_0)} = 2.75 - \\frac{1.046875}{13.6875} \\approx 2.673515$$\n\n$$f(x_1) = (2.673515)^3 + 5 - 9 \\cdot 2.673515 \\approx 0.047813$$\n\n$$f'(x_1) = 3 \\cdot (2.673515)^2 - 9 \\approx 12.44306$$\n\n3. Iteration 2: Compute $x_2$:\n\n$$x_2 = x_1 - \\frac{f(x_1)}{f'(x_1)} = 2.673515 - \\frac{0.047813}{12.44306} \\approx 2.669673$$\n\n$$f(x_2) = (2.669673)^3 + 5 - 9 \\cdot 2.669673 \\approx 0.00011837$$\n\n$$f'(x_2) = 3 \\cdot (2.669673)^2 - 9 \\approx 12.3814$$\n\n4. Iteration 3:\n\nCompute $x_3$:\n\n$$x_3 = x_2 - \\frac{f(x_2)}{f'(x_2)} = 2.669673 - \\frac{0.00011837}{12.3814} \\approx 2.669663$$\n\nb) Considering the equation $f(x) = x^3 + 4x^2 - 10$, solve the problem using Newton’s method considering $x_0 = 1.0$ and find $x_4$.\n\n1. Initial Values:\n\n$$x_0 = 1.0$$\n\n$$f(x_0) = 1^3 + 4(1)^2 - 10 = 1 + 4 - 10 = -5$$\n\n$$f'(x_0) = 3(1)^2 + 8(1) = 3 + 8 = 11$$\n\n$$x_1 = x_0 - \\frac{f(x_0)}{f'(x_0)} = 1.0 - \\frac{-5}{11} = 1.0 + 0.4545455 = 1.4545455$$\n2. Iteration 1:\n\n\\[ x_1 = 1.4545455 \\]\n\\[ f(x_1) = (1.4545455)^3 + 4(1.4545455)^2 - 10 \\approx 1.5401953 \\]\n\\[ f'(x_1) = 3(1.4545455)^2 + 8(1.4545455) \\approx 17.9834711 \\]\n\\[ x_2 = x_1 - \\frac{f(x_1)}{f'(x_1)} = 1.4545455 - \\frac{1.5401953}{17.9834711} \\approx 1.3689004 \\]\n\n3. Iteration 2:\n\n\\[ x_2 = 1.3689004 \\]\n\\[ f(x_2) = (1.3689004)^3 + 4(1.3689004)^2 - 10 \\approx 0.0607197 \\]\n\\[ f'(x_2) = 3(1.3689004)^2 + 8(1.3689004) \\approx 16.5728681 \\]\n\\[ x_3 = x_2 - \\frac{f(x_2)}{f'(x_2)} = 1.3689004 - \\frac{0.0607197}{16.5728681} \\approx 1.3652366 \\]\n\n4. Iteration 3:\n\n\\[ x_3 = 1.3652366 \\]\n\\[ f(x_3) = (1.3652366)^3 + 4(1.3652366)^2 - 10 \\approx 0.0001088 \\]\n\\[ f'(x_3) = 3(1.3652366)^2 + 8(1.3652366) \\approx 16.5135057 \\]\n\\[ x_4 = x_3 - \\frac{f(x_3)}{f'(x_3)} = 1.3652366 - \\frac{0.0001088}{16.5135057} \\approx 1.3652300 \\]", "id": "./materials/913.pdf" }, { "contents": "Three-Points Rule\nThe Three-Points Rule is a specific variant of numerical integration that uses three points within each subinterval to estimate the integral. It is less commonly used than Trapezoidal and Simpson’s rules but can provide a more accurate approximation under certain conditions.\n\nSteps:\n\n- Similar to Trapezoidal and Simpson’s rules, divide the interval \\([a, b]\\) into \\(n\\) equal subintervals, each of width \\(h = \\frac{b-a}{n}\\).\n- Calculate the function values at three points within each subinterval: typically at \\(x_i, x_{i+0.5},\\) and \\(x_{i+1}\\), where \\(x_i = a + i \\cdot h\\).\n- Apply the Three-Point Rule formula to each subinterval:\n\n\\[\nf'(x_0) \\approx \\frac{1}{2h} \\cdot [-3f(x_0) + 4f(x_0 + h) - f(x_0 + 2h)]\n\\]\n\nwhere \\(h\\) is the step size between the points.\n- Sum up the contributions from all subintervals to obtain the numerical approximation of the integral.\nProblem Approximate $f'(2)$ using the three-point rule, given the table of values:\n\n| $x$ | 2.0 | 2.5 | 3.0 | 3.5 | 4.0 | 4.5 |\n|-----|-----|-----|-----|-----|-----|-----|\n| $f(x)$ | 0.5 | 0.4 | 0.3333 | 0.2857 | 0.25 | 0.2222 |\n\nIs possible to use the forward difference formula for the derivative at $x = 2.0$, which is:\n\n$$f'(x_0) \\approx \\frac{1}{2h} \\cdot [-3f(x_0) + 4f(x_0 + h) - f(x_0 + 2h)]$$\n\nHere, $h = 0.5$. To approximate $f'(2.0)$, is possible to use the points $x = 2.0$, $x = 2.5$, and $x = 3.0$. Plugging these into the formula:\n\n$$f'(2.0) \\approx \\frac{-3f(2.0) + 4f(2.5) - f(3.0)}{2 \\cdot 0.5}$$\n\nSubstituting the given function values:\n\n$$f'(2.0) \\approx \\frac{-3(0.5) + 4(0.4) - 0.3333}{1}$$\n\nSimplifying the numerator:\n\n$$f'(2.0) \\approx \\frac{-1.5 + 1.6 - 0.3333}{1} = \\frac{0.1 - 0.3333}{1} = -0.2333$$\n\nThus, the approximate value of $f'(2.0)$ using the three-point rule is:\n\n$$f'(2.0) \\approx -0.2333$$", "id": "./materials/914.pdf" }, { "contents": "Trapezoidal Rule:\nThe Trapezoidal Rule is a simple and effective method for approximating a definite integral of a function $f(x)$ over an interval $[a, b]$. It approximates the area under the curve by dividing the interval into small segments and approximating the area under each segment using trapezoids.\n\nSteps:\n\n- Divide the interval $[a, b]$ into $n$ equal subintervals, each of width $h = \\frac{b-a}{n}$.\n- Calculate the function values at the boundaries of these subintervals: $f(x_0), f(x_1), ..., f(x_n)$ where $x_i = a + i \\cdot h$.\n- Apply the trapezoidal rule formula to each subinterval:\n \\[\n \\int_a^b f(x) \\, dx \\approx \\frac{h}{2} \\left[ f(x_0) + 2f(x_1) + 2f(x_2) + ... + 2f(x_{n-1}) + f(x_n) \\right]\n \\]\n- Simplify the formula to get the numerical approximation of the integral.\nProblem\nConsider the table:\n\n| $x$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 |\n|-----|-----|-----|-----|-----|-----|-----|-----|\n| $f(x)$ | 961.5 | 466.7 | 662.8 | 620.1 | 757.2 | 807.6 | 761.1 |\n\nUsing the Trapezoidal Rule, find $I = \\int_{1}^{7} f(x) \\, dx$.\n\n1. Divide the interval $[1, 7]$ into equal subintervals:\n\n $h = \\frac{7 - 1}{6} = 1$\n\n2. Calculate the sum of function values excluding the endpoints:\n\n \\[\n \\sum_{i=2}^{6} f(x_i) = f(2) + f(3) + f(4) + f(5) + f(6)\n \\]\n\n \\[\n = 466.7 + 662.8 + 620.1 + 757.2 + 807.6\n \\]\n\n \\[\n = 3314.4\n \\]\n\n3. Apply the Trapezoidal Rule formula:\n\n \\[\n I \\approx \\frac{h}{2} [f(a) + 2 \\sum_{i=2}^{6} f(x_i) + f(b)]\n \\]\n\n Where $a = 1$ and $b = 7$.\n\n \\[\n I \\approx \\frac{1}{2} [f(1) + 2 \\cdot 3314.4 + f(7)]\n \\]\n\n \\[\n I \\approx \\frac{1}{2} [961.5 + 2 \\cdot 3314.4 + 761.1]\n \\]\n\n \\[\n I \\approx \\frac{1}{2} [961.5 + 6628.8 + 761.1]\n \\]\n\n \\[\n I \\approx \\frac{1}{2} \\cdot 8351.4\n \\]\n\n \\[\n I \\approx 4175.7\n \\]\n\nTherefore, using the Trapezoidal Rule, the approximate value of $I = \\int_{1}^{7} f(x) \\, dx$ is 4175.7", "id": "./materials/915.pdf" }, { "contents": "HINT:\n\nFind \\( \\int e^x \\sin x \\, dx \\)\n\n**Solution** Take \\( e^x \\) as the first function and \\( \\sin x \\) as second function. Then, integrating by parts, we have\n\n\\[\nI = \\int e^x \\sin x \\, dx = e^x (-\\cos x) + \\int e^x \\cos x \\, dx\n\\]\n\n\\[\n= -e^x \\cos x + I_1 \\quad \\text{(say)}\n\\]\n\nTaking \\( e^x \\) and \\( \\cos x \\) as the first and second functions, respectively, in \\( I_1 \\), we get\n\n\\[\nI_1 = e^x \\sin x - \\int e^x \\sin x \\, dx\n\\]\n\nSubstituting the value of \\( I_1 \\) in (1), we get\n\n\\[\nI = -e^x \\cos x + e^x \\sin x - I \\quad \\text{or} \\quad 2I = e^x (\\sin x - \\cos x)\n\\]\n\nHence,\n\n\\[\nI = \\int e^x \\sin x \\, dx = \\frac{e^x}{2} (\\sin x - \\cos x) + C\n\\]\n\n**Alternatively**, above integral can also be determined by taking \\( \\sin x \\) as the first function and \\( e^x \\) the second function.", "id": "./materials/92.pdf" }, { "contents": "Evaluate $\\int_0^1 e^{-3x} x^2 \\, dx$\n\n**HINT:**\n\nUse integration by parts:\n\n$$\\int f(x)g(x) \\, dx = F(x)g(x) - \\int F(x)g'(x) \\, dx + C,$$\n\nsuch that $F(x) = \\int f(x) \\, dx$\n\n**Follow LIATE rule**\n\nAccording to LIATE rule, $f(x) = e^{-3x}$ and $g(x) = x^2$", "id": "./materials/93.pdf" }, { "contents": "Similar problem,\n\n\\[ \\int \\cos^2(5x) \\, dx, \\]\n\nA.C.1\n\nFrom the integration table,\n\n\\[ \\int \\cos^2(mx) \\, dx = \\frac{1}{2m} \\left( mx + \\sin(mx) \\cos(mx) \\right) + C \\]\n\nA.C.1, For \\( m = 5 \\),\n\n\\[ \\int \\cos^2(5x) \\, dx = \\frac{1}{10} \\left( 5x + \\sin(5x) \\cos(5x) \\right) + C \\]\n\nFor \\( \\pi \\)\n\n\\[ \\int_0^\\pi \\cos^2(5x) \\, dx = \\left[ \\frac{1}{10} \\left( 5x + \\sin(5x) \\cos(5x) \\right) \\right]_0^\\pi \\]\n\n\\[ = \\left[ \\frac{1}{10} \\left( 5\\pi + \\sin(5\\pi) \\cos(5\\pi) \\right) \\right] - \\left[ \\frac{1}{10} \\left( 0 + \\sin(0) \\cos(0) \\right) \\right] \\]\n\n\\[ = \\frac{1}{10} \\left( 5\\pi + (0) \\cdot (-1) \\right) \\]\n\n\\[ = \\frac{5\\pi}{10} = \\frac{\\pi}{2} \\]", "id": "./materials/94.pdf" }, { "contents": "Arithmetic Sequence\n\nAn arithmetic sequence, also known as an arithmetic progression, is a sequence of numbers in which the difference between any two successive terms is constant, with a pattern that repeats itself with each term. This difference is called the \"common difference\" and is denoted by $d$. If the first term of the sequence was denoted by $a_1$, the sequence can be written as:\n\n$$a_1, a_1 + d, a_1 + 2d, a_1 + 3d, \\ldots$$\n\nThe $n$-th term of an arithmetic sequence can be determined using the general formula:\n\n$$a_n = a_1 + (n - 1) \\times d$$\n\nWhere:\n\n- $a_n$ is the $n$-th term.\n- $a_1$ is the first term.\n- $d$ is the common difference.\n- $n$ is the position of the term in the sequence.\n\nExample: In the arithmetic sequence $3, 8, 13, 18, 23, \\ldots$ the first term $a_1$ is 3 and 5 is the common difference and the pattern since it repeats itself every term as: $3 + 5 = 8$, $8 + 5 = 13$, and so on.\n\nClosed Formula\n\nA closed formula is an explicit expression that allows the computation of the $n$-th term of a sequence directly, without needing to refer to previous terms. For arithmetic sequences, the closed formula is derived from the general form mentioned above $a_n = a_1 + (n - 1) \\times d$\n\nExample: Using the arithmetic sequence example above with $a_1 = 3$ and $d = 5$, the closed formula is:\n\n$$a_n = 3 + (n - 1) \\times 5 \\Rightarrow a_n = 3 + 5n - 5 \\Rightarrow a_n = 5n - 2$$\n\nThis formula allows to compute, for instance, the 10th term directly:\n\n$$a_{10} = 5 \\times 10 - 2 = 50 - 2 = 48$$\nProblems\n\na) If \\( u_1 = -1 \\) and \\( u_8 = 20 \\), find the common ratio and the general term of the arithmetic sequence \\((u_n)_{n \\in \\mathbb{N}}\\).\n\n1. Use the general form of an arithmetic sequence:\n \\[ u_n = u_1 + (n - 1) \\times d \\]\n\n2. Plug in the given terms to form equations:\n - For \\( u_1 \\):\n \\[ u_1 = -1 \\]\n - For \\( u_8 \\):\n \\[ u_8 = -1 + (8 - 1) \\times d = 20 \\]\n \\[ -1 + 7d = 20 \\]\n\n3. Solve for the common difference \\( d \\):\n \\[ 7d = 20 + 1 \\Rightarrow 7d = 21 \\]\n \\[ d = 3 \\]\n\n4. Find the general term \\( u_n \\):\n \\[ u_n = -1 + (n - 1) \\times 3 \\Rightarrow u_n = -1 + 3n - 3 \\]\n \\[ u_n = 3n - 4 \\]\n\nb) For an arithmetic sequence \\((u_n)_{n \\in \\mathbb{N}}\\) with common ratio 4, it is known that \\( u_{12} = 15 \\). Verify if 139 and 150 are terms of the sequence.\n\n1. Find the first term \\( u_1 \\):\n - For \\( u_{12} \\):\n \\[ u_n = u_1 + (n - 1) \\times d \\]\n \\[ u_{12} = u_1 + (12 - 1) \\times 4 = 15 \\]\n \\[ u_1 + 44 = 15 \\Rightarrow u_1 = 15 - 44 \\]\n \\[ u_1 = -29 \\]\n\n2. Write the general term \\( u_n \\):\n \\[ u_n = -29 + (n - 1) \\times 4 \\Rightarrow u_n = -29 + 4n - 4 \\]\n \\[ u_n = 4n - 33 \\]\n3. Verify if 139 is a term in the sequence:\n - Set \\( u_n = 139 \\):\n \\[\n 4n - 33 = 139 \\\\\n 4n = 139 + 33 \\\\\n 4n = 172 \\\\\n n = 43\n \\]\n Since \\( n = 43 \\) is an integer, 139 is the \\( u_{43} \\) term of the sequence.\n\n4. Verify if 150 is a term in the sequence:\n - Set \\( u_n = 150 \\):\n \\[\n 4n - 33 = 150 \\\\\n 4n = 150 + 33 \\\\\n 4n = 183 \\\\\n n = 45.75\n \\]\n Since \\( n = 45.75 \\) is not an integer, 150 is not a term of the sequence.\n\nc) Solve the equation \\( 9 + 14 + 19 + 24 + \\cdots + x = 1239 \\) where \\( 9, 14, 19, 24, \\ldots, x \\) are consecutive terms of an arithmetic progression.\n\n1. Identify the Common Difference:\n From the sequence, we can see that the common difference \\( (d) \\) is:\n \\[\n d = 14 - 9 = 5\n \\]\n\n2. Define the General Term given by:\n \\[\n a_n = a + (n - 1) \\times d\n \\]\n Substituting the known values:\n \\[\n a_n = 9 + (n - 1) \\times 5 \\Rightarrow a_n = 9 + 5n - 5 \\\\\n a_n = 5n + 4\n \\]\n\n3. Find the Number of Terms \\( (n) \\):\n The sum of the first \\( n \\) terms of an arithmetic sequence is given by:\n \\[\n S_n = \\frac{n}{2} (2a_1 + (n - 1) \\times d)\n \\]\nKnowing that the sum $S_n$ is 1239:\n\n$$1239 = \\frac{n}{2} (2 \\times 9 + (n - 1) \\times 5) \\Rightarrow 1239 = \\frac{n}{2} (18 + 5n - 5)$$\n\n$$1239 = \\frac{n}{2} (5n + 13) \\Rightarrow 1239 = \\frac{n}{2} \\times 5n + \\frac{n}{2} \\times 13$$\n\n$$1239 = \\frac{5n^2}{2} + \\frac{13n}{2}$$\n\nMultiplying through by 2 to clear the fraction and rearranging:\n\n$$2478 = 5n^2 + 13n \\Rightarrow 5n^2 + 13n - 2478 = 0$$\n\n4. Solve the Quadratic Equation:\nTo solve $5n^2 + 13n - 2478 = 0$, we use the quadratic formula $n = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}$, where $a = 5$, $b = 13$, and $c = -2478$:\n\n$$n = \\frac{-13 \\pm \\sqrt{13^2 - 4 \\times 5 \\times (-2478)}}{2 \\times 5} \\Rightarrow n = \\frac{-13 \\pm \\sqrt{169 + 4 \\times 5 \\times 2478}}{10}$$\n\n$$n = \\frac{-13 \\pm \\sqrt{169 + 49560}}{10} \\Rightarrow n = \\frac{-13 \\pm \\sqrt{49729}}{10}$$\n\n$$n = \\frac{-13 \\pm 223}{10}$$\n\nTaking the positive root since $n$ must be positive:\n\n$$n = \\frac{-13 + 223}{10} \\Rightarrow n = \\frac{210}{10} \\Rightarrow n = 21$$\n\n5. Find the Last Term ($x$):\nThe last term $x$ is the 21st term of the sequence:\n\n$$a_{21} = 9 + (21 - 1) \\times 5$$\n\n$$a_{21} = 9 + 20 \\times 5 \\Rightarrow a_{21} = 9 + 100$$\n\n$$a_{21} = 109$$\n\nThe last term $x$ of the arithmetic progression that sums to 1239 is $x = 109$. ", "id": "./materials/943.pdf" }, { "contents": "Generating Functions\n\nThe method of generating functions is a powerful tool for solving recurrence relations. It involves transforming a sequence into a function (the generating function) and then manipulating this function to find a closed form for the sequence.\n\n- Define the Generating Function\n For a given sequence \\( \\{a_n\\} \\), the generating function \\( A(x) \\) is defined as:\n \\[\n A(x) = \\sum_{n=0}^{\\infty} a_n x^n\n \\]\n This is a formal power series where the coefficients of \\( x^n \\) are the terms of the sequence \\( a_n \\).\n\n- Express the Recurrence Relation\n Write the given recurrence relation in terms of the generating function. For a recurrence relation of the form:\n \\[\n a_n = c_1 a_{n-1} + c_2 a_{n-2} + \\ldots + c_k a_{n-k}\n \\]\n multiply both sides by \\( x^n \\) and sum over all \\( n \\geq k \\):\n \\[\n \\sum_{n=k}^{\\infty} a_n x^n = \\sum_{n=k}^{\\infty} \\left( c_1 a_{n-1} x^n + c_2 a_{n-2} x^n + \\ldots + c_k a_{n-k} x^n \\right)\n \\]\n\n- Manipulate the Series\n Shift the indices on the right-hand side so that the series starts from \\( n = 0 \\):\n \\[\n \\sum_{n=k}^{\\infty} a_n x^n = c_1 x \\sum_{n=k-1}^{\\infty} a_n x^n + c_2 x^2 \\sum_{n=k-2}^{\\infty} a_n x^n + \\ldots + c_k x^k \\sum_{n=0}^{\\infty} a_n x^n\n \\]\n\n- Incorporate Initial Conditions\n Include the initial terms of the sequence separately, as they are not covered in the sums starting from \\( n = k \\). For example, if the initial conditions are \\( a_0, a_1, \\ldots, a_{k-1} \\), adjust the generating function equation accordingly.\n• Solve for the Generating Function\n\nCombine and solve the resulting equation for $A(x)$. This typically involves algebraic manipulation and sometimes partial fractions decomposition.\n\nOnce you have $A(x)$, expand it as a power series to find the general term $a_n$. This may involve recognizing standard power series expansions or using techniques like partial fraction decomposition.\n\n**Problem**\n\nConsider the recurrence relation:\n\n$$a_n = 4a_{n-1} - 4a_{n-2}, \\quad \\forall n \\geq 2$$\n\nWith initial conditions:\n\n$$a_0 = 1, \\quad a_1 = -4$$\n\n1. Define the Generating Function:\n\n$$A(x) = \\sum_{n=0}^{\\infty} a_n x^n$$\n\n2. Express the Recurrence Relation:\n\n$$\\sum_{n=2}^{\\infty} a_n x^n = \\sum_{n=2}^{\\infty} (4a_{n-1} - 4a_{n-2}) x^n$$\n\n3. Manipulate the Series:\n\n$$\\sum_{n=2}^{\\infty} a_n x^n = 4x \\sum_{n=1}^{\\infty} a_n x^n - 4x^2 \\sum_{n=0}^{\\infty} a_n x^n$$\n\n$$A(x) - a_0 - a_1 x = 4x(A(x) - a_0) - 4x^2 A(x)$$\n\n4. Include Initial Conditions:\n\n$$A(x) - 1 + 4x = 4x A(x) - 4x - 4x^2 A(x)$$\n\n5. Solve for $A(x)$:\n\n$$A(x) - 4x A(x) + 4x^2 A(x) = 1 - 8x$$\n\n$$A(x)(1 - 4x + 4x^2) = 1 - 8x$$\n\n$$A(x) = \\frac{1 - 8x}{(1 - 2x)^2}$$\n6. Expand and Find the Closed Form: Using partial fraction decomposition:\n\n\\[ A(x) = \\frac{4}{1 - 2x} - \\frac{3}{(1 - 2x)^2} \\]\n\n\\[ A(x) = 4 \\sum_{n=0}^{\\infty} (2x)^n - 3 \\sum_{n=0}^{\\infty} (n + 1)(2x)^n \\]\n\n\\[ A(x) = \\sum_{n=0}^{\\infty} 2^n (4 - 3(n + 1)) x^n \\]\n\n\\[ a_n = 2^n(1 - 3n) \\]\n\nThe closed form for the sequence is:\n\n\\[ a_n = 2^n(1 - 3n) \\]", "id": "./materials/944.pdf" }, { "contents": "Geometric Sequence\n\nA geometric sequence, or geometric progression, is a sequence of numbers in which each term after the first is found by multiplying the previous term by a constant, called the common ratio.\n\nThe general form of a geometric sequence can be written as:\n\n\\[ a, ar, ar^2, ar^3, \\ldots \\]\n\nThe \\( n \\)-th term of a geometric sequence can be determined using the closed formula:\n\n\\[ u_n = a \\times r^{n-1} \\]\n\nwhere:\n\n- \\( u_n \\) is the \\( n \\)-th term of the sequence.\n- \\( a \\) is the first term.\n- \\( r \\) is the common ratio.\n- \\( n \\) is the position of the term in the sequence.\n\nExample: For a geometric sequence where the first term \\( a \\) is 2 and the common ratio \\( r \\) is 3, the sequence would be:\n\n\\[ 2, 2 \\times 3, 2 \\times 3^2, 2 \\times 3^3, \\ldots \\]\n\n\\[ 2, 6, 18, 54, \\ldots \\]\nProblems:\n\na) What is the fifth term of the sequence \\((u_n)_{n \\in \\mathbb{N}}\\), knowing that \\(u_4 = 32\\) and \\(u_8 = 8192\\).\n\n1. Using the closed formula:\n \\[ u_n = a \\times r^{n-1} \\]\n\n2. Set up equations using given values:\n For \\(u_4\\):\n \\[ u_4 = a \\times r^{4-1} \\]\n \\[ 32 = a \\times r^3 \\quad (1) \\]\n For \\(u_8\\):\n \\[ u_8 = a \\times r^{8-1} \\]\n \\[ 8192 = a \\times r^7 \\quad (2) \\]\n\n3. Solve for \\(r\\):\n Divide equation (2) by equation (1) to eliminate \\(a\\):\n \\[ \\frac{u_8}{u_4} = \\frac{a \\times r^7}{a \\times r^3} \\]\n \\[ \\frac{8192}{32} = r^4 \\Rightarrow r^4 = 256 \\]\n \\[ r = \\sqrt[4]{256} \\]\n \\[ r = 4 \\]\n\n4. Find \\(a\\):\n Substitute \\(r = 4\\) into equation (1):\n \\[ 32 = a \\times 4^3 \\Rightarrow 32 = a \\times 64 \\]\n \\[ a = \\frac{32}{64} \\Rightarrow a = \\frac{1}{2} \\]\n\n5. Find \\(u_5\\):\n Use the general formula with \\(n = 5\\):\n \\[ u_5 = a \\times r^{5-1} \\Rightarrow u_5 = \\frac{1}{2} \\times 4^4 \\]\n \\[ u_5 = \\frac{1}{2} \\times 256 \\]\n \\[ u_5 = 128 \\]\n\nThe fifth term of the geometric sequence is \\(u_5 = 128\\).\nb) Consider that after purchasing a car, its value depreciates by 15% annually. Mr. Smith just bought a new car for €30,000. What is the approximate value of his car if he sells it in five years?\n\n1. Understand the geometric sequence:\nThe value of the car depreciates geometrically, meaning that each year the car’s value is multiplied by a constant ratio. In this case, the ratio is $1 - r$, where $r$ is the depreciation rate.\n\n2. Set up the geometric sequence:\nThe value of the car after $n$ years can be represented as the $n$-th term of a geometric sequence. The general formula for the $n$-th term of a geometric sequence is:\n\n$$V_n = V_0(1 - r)^n$$\n\nHere, $n$ is the number of years.\n\n3. Substitute the values into the formula:\n\n$$V_5 = 30,000 \\times (1 - 0.15)^5$$\n\n$$V_5 = 30,000 \\times 0.85^5$$\n\n$$V_5 = 30,000 \\times 0.4437 \\approx 13,311$$\n\nThe approximate value of the car after five years is €13,311.", "id": "./materials/946.pdf" }, { "contents": "Linear Recurrence\n\nA linear recurrence relation is an equation that expresses the $n$-th term of a sequence as a linear combination of previous terms.\n\nHomogeneous Linear Recurrence Relation\n\nA homogeneous linear recurrence relation is one in which every term on the right-hand side of the equation is a linear combination of previous terms, with no additional non-zero constants or functions.\n\nGeneral Form:\n\n$$a_n = c_1a_{n-1} + c_2a_{n-2} + \\cdots + c_ka_{n-k}$$\n\nwhere $c_1, c_2, \\ldots, c_k$ are constants.\n\nNon-Homogeneous Linear Recurrence Relation\n\nA non-homogeneous linear recurrence relation includes additional terms that are not linear combinations of the previous terms, such as constant terms, polynomials, or exponential functions.\n\nGeneral Form:\n\n$$a_n = c_1a_{n-1} + c_2a_{n-2} + \\cdots + c_ka_{n-k} + f(n)$$\n\nwhere $f(n)$ is a function of $n$ that is not a combination of previous $a$ terms.\n\nKey Differences\n\n- **Homogeneous vs. Non-Homogeneous:**\n - Homogeneous: Only involves linear combinations of previous terms. No additional terms.\n - Non-Homogeneous: Includes additional terms that are functions of $n$ (e.g., constants, polynomials, exponentials).\n\n- **Solution Approach:**\n - For homogeneous recurrences, solve the characteristic equation and apply initial conditions.\n - For non-homogeneous recurrences, solve the associated homogeneous recurrence first, then find a particular solution to the non-homogeneous part, and combine them.\nProblems\na) Solve the homogeneous linear recurrence relation:\n\n\\[ a_n = 2a_{n-1} - a_{n-2} \\]\n\nwith initial conditions \\( a_0 = 0 \\) and \\( a_1 = 3 \\).\n\n1. Find the Characteristic Equation\n Assume the solution has the form \\( a_n = r^n \\). Substituting this into the recurrence relation gives:\n \\[ r^n = 2r^{n-1} - r^{n-2}. \\]\n Dividing through by \\( r^{n-2} \\) (assuming \\( r \\neq 0 \\)) gives:\n \\[ r^2 = 2r - 1. \\]\n Rearranging this equation results in the characteristic equation:\n \\[ r^2 - 2r + 1 = 0. \\]\n\n2. Solve the Characteristic Equation\n Solve the quadratic equation \\( r^2 - 2r + 1 = 0 \\). This can be factored as:\n \\[ (r - 1)^2 = 0. \\]\n The root of this equation is \\( r = 1 \\) with multiplicity 2.\n\n3. Write the General Solution\n Since the characteristic root \\( r = 1 \\) has multiplicity 2, the general solution to the recurrence relation is:\n \\[ a_n = (C_1 + C_2n) \\times 1^n. \\]\n This simplifies to:\n \\[ a_n = C_1 + C_2n. \\]\n\n4. Determine the Constants Using Initial Conditions\n Now use the initial conditions to find the constants \\( C_1 \\) and \\( C_2 \\).\n 1. For \\( n = 0 \\):\n \\[ a_0 = C_1 + C_2 \\times 0 = C_1. \\]\n Since \\( a_0 = 0 \\), it follows that \\( C_1 = 0 \\).\n 2. For \\( n = 1 \\):\n \\[ a_1 = C_1 + C_2 \\times 1 = C_1 + C_2. \\]\n Since \\( a_1 = 3 \\) and \\( C_1 = 0 \\), it follows that \\( C_2 = 3 \\).\n5. Write the Final Solution\nSubstitute the values of $C_1$ and $C_2$ into the general solution:\n\n$$a_n = 0 + 3n \\Rightarrow a_n = 3n$$\n\nb) Solve the non-homogeneous linear recurrence relation:\n\n$$a_n = 2a_{n-1} - a_{n-2} + 2^n$$\n\nwith initial conditions $a_1 = 0$ and $a_2 = 1$.\n\n1. Solve the Homogeneous Recurrence Relation:\nThe homogeneous part is: $a_n = 2a_{n-1} - a_{n-2}$\nThe characteristic equation is: $r^2 - 2r + 1 = 0 \\Rightarrow (r - 1)^2 = 0$\nThe characteristic root is $r = 1$ with multiplicity 2. Thus, the general solution to the homogeneous recurrence relation is:\n\n$$a_n^{(h)} = C_1 + C_2n$$\n\n2. Find a Particular Solution for the Non-Homogeneous term:\nFor the particular solution, since the non-homogeneous term is $2^n$, we guess a solution of the form:\n\n$$a_n^{(p)} = A \\times 2^n$$\n\nSubstitute $a_n^{(p)} = A \\times 2^n$ into the non-homogeneous recurrence relation:\n\n$$a_n = 2a_{n-1} - a_{n-2} + 2^n$$\n\n$$A \\times 2^n = 2 \\times (A \\times 2^{n-1}) - (A \\times 2^{n-2}) + 2^n$$\n\nSimplifying:\n\n$$A \\times 2^n = 2 \\times \\frac{A \\times 2^n}{2} - \\frac{A \\times 2^n}{4} + 2^n \\Rightarrow A \\times 2^n = A \\times 2^n - \\frac{A \\times 2^n}{4} + 2^n$$\n\n$$A \\times 2^n = \\left(A - \\frac{A}{4} + 1\\right) \\times 2^n \\Rightarrow A = A - \\frac{A}{4} + 1$$\n\nSolving for $A$:\n\n$$0 = -\\frac{A}{4} + 1 \\Rightarrow \\frac{A}{4} = 1 \\Rightarrow A = 4$$\n\nThus, a particular solution is: $a_n^{(p)} = 4 \\times 2^n$\n3. Combine Solutions:\n\nThe general solution to the recurrence relation is:\n\n\\[ a_n = a_n^{(h)} + a_n^{(p)} \\Rightarrow a_n = C_1 + C_2n + 4 \\times 2^n \\]\n\n4. Apply Initial Conditions:\n\nUse initial conditions \\( a_1 = 0 \\) and \\( a_2 = 1 \\) to find \\( C_1 \\) and \\( C_2 \\).\n\nFor \\( n = 1 \\):\n\n\\[ a_1 = C_1 + C_2 \\times 1 + 4 \\times 2^1 \\Rightarrow 0 = C_1 + C_2 + 8 \\]\n\\[ C_1 + C_2 = -8 \\quad (1) \\]\n\nFor \\( n = 2 \\):\n\n\\[ a_2 = C_1 + C_2 \\times 2 + 4 \\times 2^2 \\Rightarrow 1 = C_1 + 2C_2 + 16 \\]\n\\[ C_1 + 2C_2 = -15 \\quad (2) \\]\n\nSubtract equation (1) from equation (2):\n\n\\[ (C_1 + 2C_2) - (C_1 + C_2) = -15 - (-8) \\]\n\\[ C_2 = -7 \\]\n\nReplacing \\( C_2 = -7 \\) into equation (1) is found \\( C_1 = -1 \\).\n\nReplacing \\( C_1 \\) and \\( C_2 \\) in the recurrence relation:\n\n\\[ a_n = C_1 + C_2n + 4 \\times 2^n \\]\n\\[ a_n = -1 - 7n + 4 \\times 2^n \\]\n\nIn terms of \\( 2^{n+2} \\) for simplicity:\n\n\\[ a_n = -1 - 7n + 4 \\times 2^n = -1 - 7n + 2^{n+2} \\]", "id": "./materials/947.pdf" }, { "contents": "Sum of an Arithmetic Progression\n\nAn arithmetic progression (or arithmetic sequence) is a sequence of numbers in which the difference between any two consecutive terms is constant. This difference is called the \"common difference\" and is denoted by $d$. If we denote the first term of the sequence by $a_1$, the $n$-th term of the arithmetic sequence can be written as:\n\n$$a_n = a_1 + (n - 1)d$$\n\nThe sum of the first $n$ terms of an arithmetic progression is given by the formula:\n\n$$S_n = \\frac{n}{2} (2a_1 + (n - 1)d)$$\n\nAlternatively, this can be written as:\n\n$$S_n = \\frac{n}{2} (a_1 + a_n)$$\n\nwhere $a_n$ is the $n$-th term of the sequence.\nProblems\n\na) Solve the equation $9 + 14 + 19 + 24 + + x = 1239$ where $9, 14, 19, 24, \\ldots, x$ are consecutive terms of an arithmetic progression.\n\n1. Identify the common difference:\n From the sequence, we can see that the common difference ($d$) is:\n \n $$d = 14 - 9 = 5$$\n\n2. Define the general term given by:\n \n $$a_n = a + (n - 1)d$$\n\n Substituting the known values:\n \n $$a_n = 9 + (n - 1) \\times 5 \\Rightarrow a_n = 9 + 5n - 5$$\n \n $$a_n = 5n + 4$$\n\n3. Find the number of terms ($n$):\n The sum of the first $n$ terms of an arithmetic sequence is given by:\n \n $$S_n = \\frac{n}{2} (2a_1 + (n - 1)d)$$\n\n Knowing that the sum $S_n$ is 1239:\n \n $$1239 = \\frac{n}{2} (2 \\times 9 + (n - 1) \\times 5) \\Rightarrow 1239 = \\frac{n}{2} (18 + 5n - 5)$$\n \n $$1239 = \\frac{n}{2} (5n + 13) \\Rightarrow 1239 = \\frac{n}{2} \\times 5n + \\frac{n}{2} \\times 13$$\n \n $$1239 = \\frac{5n^2}{2} + \\frac{13n}{2}$$\n\n Multiplying through by 2 to clear the fraction and rearranging:\n \n $$2478 = 5n^2 + 13n \\Rightarrow 5n^2 + 13n - 2478 = 0$$\n\n4. Solve the quadratic equation:\n To solve $5n^2 + 13n - 2478 = 0$, we use the quadratic formula $n = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}$, where $a = 5$, $b = 13$, and $c = -2478$:\n \n $$n = \\frac{-13 \\pm \\sqrt{13^2 - 4 \\times 5 \\times (-2478)}}{2 \\times 5} \\Rightarrow n = \\frac{-13 \\pm \\sqrt{169 + 4 \\times 5 \\times 2478}}{10}$$\n\\[ n = \\frac{-13 \\pm \\sqrt{169 + 49560}}{10} \\Rightarrow n = \\frac{-13 \\pm \\sqrt{49729}}{10} \\]\n\n\\[ n = \\frac{-13 \\pm 223}{10} \\Rightarrow n = \\frac{210}{10} \\Rightarrow n = 21 \\]\n\nTaking the positive root since \\( n \\) must be positive:\n\n\\[ n = \\frac{-13 + 223}{10} \\Rightarrow n = \\frac{210}{10} \\Rightarrow n = 21 \\]\n\n5. Find the last term \\( x \\):\nThe last term \\( x \\) is the 21st term of the sequence:\n\n\\[ a_{21} = 9 + (21 - 1) \\times 5 \\]\n\n\\[ a_{21} = 9 + 20 \\times 5 \\Rightarrow a_{21} = 9 + 100 \\]\n\n\\[ a_{21} = 109 \\]\n\nThe last term \\( x \\) of the arithmetic progression that sums to 1239 is \\( x = 109 \\).\n\nb) Solve for the sum of the first 20 terms of the sequence given by \\( a_n = 5n - 2 \\).\n\n1. First, identify the first term \\( (a_1) \\) and the common difference \\( (d) \\):\n\n - The first term \\( a_1 \\) is found by substituting \\( n = 1 \\) into the general term formula:\n \\[ a_1 = 5(1) - 2 = 3 \\]\n\n - The common difference \\( d \\) is the difference between the second term and the first term. We find the second term \\( a_2 \\) by substituting \\( n = 2 \\):\n \\[ a_2 = 5(2) - 2 = 8 \\]\n\n So, the common difference \\( d \\) is:\n \\[ d = a_2 - a_1 = 8 - 3 = 5 \\]\n\n2. The sum \\( S_n \\) of the first \\( n \\) terms of an arithmetic sequence can be calculated using the formula:\n\n\\[ S_n = \\frac{n}{2} (2a_1 + (n - 1)d) \\]\n\nSubstitute \\( n = 20 \\), \\( a_1 = 3 \\), and \\( d = 5 \\) into the formula:\n\n\\[ S_{20} = \\frac{20}{2} (2(3) + (20 - 1) \\times 5) \\]\nSimplify the expression:\n\n\\[ S_{20} = 10 (6 + 19 \\times 5) \\]\n\\[ S_{20} = 10 \\times 101 = 1010 \\]\n\nTherefore, the sum of the first 20 terms of the sequence \\( a_n = 5n - 2 \\) is 1010.", "id": "./materials/948.pdf" }, { "contents": "• **Set**\n A set is a collection of distinct objects, called elements or members.\n For example, the set $A = \\{a, b, c\\}$ contains the elements $a$, $b$, and $c$.\n\n• **Element of a Set**\n An element of a set is an individual object within that set. If an object $x$ belongs to set $A$, we write $x \\in A$.\n For example, in the set $A = \\{a, b, c\\}$, $a$ is an element of $A$, so we write $a \\in A$.\n\n• **Subset**\n A subset is a set where all of its elements are also elements of another set. If set $A$ is a subset of set $B$, we write $A \\subseteq B$.\n For example, if $A = \\{a, b\\}$ and $B = \\{a, b, c\\}$, then $A$ is a subset of $B$, because every element of $A$ is also in $B$.\n\n• **Proper Subset**\n A proper subset is a subset that is strictly smaller than another set, meaning it contains some, but not all, elements of that set. If set $A$ is a proper subset of set $B$, we write $A \\subset B$.\n For example, $A = \\{a, b\\}$ is a proper subset of $B = \\{a, b, c\\}$, because $A$ does not contain all the elements of $B$.\n\n• **Superset**\n A superset is a set that contains all the elements of another set. If set $B$ is a superset of set $A$, we write $B \\supseteq A$.\n For example, $B = \\{a, b, c\\}$ is a superset of $A = \\{a, b\\}$, because $B$ contains all the elements of $A$.\n\n• **Power Set**\n The power set of a set $A$ is the set of all possible subsets of $A$, including the empty set and $A$ itself. The power set is denoted as $P(A)$.\n For example, if $A = \\{a, b, c\\}$, then the power set include the subsets with one element $\\{a\\}$, $\\{b\\}$, $\\{c\\}$; subsets with two elements $\\{a, b\\}$, $\\{a, c\\}$, $\\{b, c\\}$; the empty set $\\emptyset$ and the set $A$ itself.\n So $P(A) = \\{\\emptyset, \\{a\\}, \\{b\\}, \\{c\\}, \\{a, b\\}, \\{a, c\\}, \\{b, c\\}, \\{a, b, c\\}\\}$. \n\nBy: Ana Julia B. Bezerra\n• **Cardinality**\n\nThe cardinality of a set is the number of elements in the set.\n\nFor example, if \\( A = \\{a, b, c\\} \\), the cardinality of \\( A \\), that has 3 elements, denoted \\(|A|\\), is 3.\n\n• **Principle of Inclusion-Exclusion (PIE)**\n\nThe Principle of Inclusion-Exclusion (PIE) is used to calculate the cardinality (size) of the union of multiple sets. It provides a way to avoid overcounting when adding the sizes of overlapping sets.\n\nFor two sets \\( A \\) and \\( B \\), the formula for the union using PIE is:\n\n\\[\n|A \\cup B| = |A| + |B| - |A \\cap B|\n\\]\n\nThis formula counts the elements in \\( A \\) and \\( B \\), but since elements in the intersection \\( A \\cap B \\) are counted twice (once in \\( A \\) and once in \\( B \\)), it is necessary to subtract the size of the intersection.\n\nFor three sets \\( A \\), \\( B \\), and \\( C \\), the formula extends to:\n\n\\[\n|A \\cup B \\cup C| = |A| + |B| + |C| - |A \\cap B| - |B \\cap C| - |A \\cap C| + |A \\cap B \\cap C|\n\\]\n\nThis formula ensures that the overlap between each pair of sets is subtracted once, and the triple overlap \\( A \\cap B \\cap C \\) is added back to correct for the fact that it was subtracted three times.\nProblems\n\na) Is \\( \\{a, b, c\\} \\) a set of \\( A = \\{a, b, c\\} \\) and of \\( B = \\{a, b, c, d, e, f\\} \\)?\n\n- \\( \\{a, b, c\\} \\) is a subset of \\( B = \\{a, b, c, d, e, f\\} \\) because all elements of this set are in \\( B \\).\n- \\( \\{a, b, c\\} \\) is also a superset of \\( A = \\{a, b, c\\} \\) because it contains all elements of \\( A \\).\n\nb) If \\( A \\) and \\( B \\) are two sets such that \\( A \\subseteq B \\), when does \\( A \\in P(B) \\)?\n\n- By definition, the power set \\( P(B) \\) is the set of all subsets of \\( B \\).\n- If \\( A \\subseteq B \\), then \\( A \\) is one of the subsets of \\( B \\).\n- Therefore, \\( A \\) belongs to the power set of \\( B \\).\n\nSo \\( A \\in P(B) \\) always \\( A \\subseteq B \\). This is true for all subsets of \\( B \\).\n\nc) If \\( A \\) is a set such that \\( |A| = 6 \\), what is \\( |P(A)| \\)?\n\nApply the formula: The number of subsets of a set with \\( n \\) elements is \\( 2^n \\).\n\nSubstitute \\( n = 6 \\):\n\n\\[\n|P(A)| = 2^6\n\\]\n\nCalculate \\( 2^6 \\):\n\n\\[\n2^6 = 64\n\\]\n\nd) If \\( |X \\cap Y| = 7 \\) (elements that are in \\( Y \\) but not in \\( X \\)), \\( |X \\cap \\overline{Y}| = 8 \\) (number of elements that are in \\( X \\) but not in \\( Y \\)) and \\( |X \\cap Y| = 5 \\) (elements that are in both \\( X \\) and \\( Y \\)), what is \\( |X \\cup Y| \\)?\n\nUsing the principle of inclusion-exclusion: \\( |X \\cup Y| = |X| + |Y| - |X \\cap Y| \\)\n\nFirst, calculate \\( |X| \\) and \\( |Y| \\):\n\n\\[\n|X| = |X \\cap Y| + |X \\cap \\overline{Y}| = 5 + 8 = 13\n\\]\n\\[\n|Y| = |X \\cap Y| + |X \\cap \\overline{Y}| = 5 + 7 = 12\n\\]\n\nNow, apply the principle of inclusion-exclusion to find \\( |X \\cup Y| \\):\n\n\\[\n|X \\cup Y| = |X| + |Y| - |X \\cap Y| = 13 + 12 - 5 = 20\n\\]", "id": "./materials/949.pdf" }, { "contents": "Similar problem,\n\n\\[ \\int \\cos^2(5x) \\, dx, \\]\n\nA.C.1\n\nFrom the integration table,\n\n\\[ \\int \\cos^2(mx) \\, dx = \\frac{1}{2m} (mx + \\sin(mx) \\cos(mx)) + C \\]\n\nA.C.1, For \\( m = 5 \\),\n\n\\[ \\int \\cos^2(5x) \\, dx = \\frac{1}{10} (5x + \\sin(5x) \\cos(5x)) + C \\]\n\nFor \\( \\pi \\)\n\n\\[ \\int_0^\\pi \\cos^2(5x) \\, dx = \\left[ \\frac{1}{10} (5x + \\sin(5x) \\cos(5x)) \\right]_0^\\pi \\]\n\n\\[ = \\left[ \\frac{1}{10} (5\\pi + \\sin(5\\pi) \\cos(5\\pi)) \\right] - \\left[ \\frac{1}{10} (0 + \\sin(0) \\cos(0)) \\right] \\]\n\n\\[ = \\frac{1}{10} (5\\pi + (0) \\cdot (-1)) \\]\n\n\\[ = \\frac{5\\pi}{10} = \\frac{\\pi}{2} \\]", "id": "./materials/95.pdf" }, { "contents": "Set Operations\n\n- **Set**\n A set is a collection of distinct objects, called elements or members.\n For example, the set $A = \\{a, b, c\\}$ contains the elements $a$, $b$, and $c$.\n\n- **Element of a Set**\n An element (or member) of a set is any individual object contained within a set.\n If $x$ is an element of set $A$, we write $x \\in A$.\n For example, if $A = \\{1, 2, 3\\}$, then 1 is an element of $A$, i.e., $1 \\in A$.\n Also, if $A = \\{\\{a, b, c\\}, \\{a\\}, \\{b, c\\}, d, e, f\\}$, this set contains the following elements:\n a set: $\\{a, b, c\\}$; a set: $\\{a\\}$; a set: $\\{b, c\\}$ and individual elements: $d, e, f$.\n\n- **Complement of a Set**\n The complement of a set $A$ (denoted as $A'$ or $A^c$) is the set of all elements in the universal set $U$ that are not in $A$.\n For example, if $U = \\{1, 2, 3, 4, 5\\}$ and $A = \\{1, 3, 5\\}$, then the complement of $A$, $A^c$, would be $\\{2, 4\\}$.\n\n- **Union of Sets**\n The union of two sets $A$ and $B$ (denoted as $A \\cup B$) is the set containing all elements that are in either $A$ or $B$, or in both.\n For example, if $A = \\{1, 2, 3\\}$ and $B = \\{3, 4, 5\\}$, then $A \\cup B = \\{1, 2, 3, 4, 5\\}$.\n\n- **Intersection of Sets**\n The intersection of two sets $A$ and $B$ (denoted as $A \\cap B$) is the set containing all elements that are both in $A$ and in $B$.\n For example, if $A = \\{1, 2, 3\\}$ and $B = \\{3, 4, 5\\}$, then $A \\cap B = \\{3\\}$.\n\n- **Set Difference**\n The difference between two sets $A$ and $B$ (denoted as $A - B$) is the set of elements that are in $A$ but not in $B$.\n For example, if $A = \\{1, 2, 3\\}$ and $B = \\{3, 4, 5\\}$, then $A - B = \\{1, 2\\}$.\nProblem\n\na) There were 150 students in a class. Among them, 75 liked Calculus, 80 liked Algebra, and 50 liked Programming. Moreover, it was found that 20 liked only Programming and Algebra, 25 liked only Calculus and Algebra, 15 liked only Calculus and Programming, and 25 liked none of the subjects. Calculate what percentage of students liked all 3 subjects.\n\n1. Let $C$ be the set of students who liked Calculus, $A$ be the set who liked Algebra, and $P$ be the set who liked Programming.\n\n2. From the problem:\n\n $|C| = 75$, $|A| = 80$, $|P| = 50$\n $|P \\cap A - (P \\cap A \\cap C)| = 20,$\n $|C \\cap A - (P \\cap A \\cap C)| = 25,$\n $|C \\cap P - (P \\cap A \\cap C)| = 15,$\n $|C' \\cap A' \\cap P'| = 25$ (students who liked none)\n $|P \\cap A \\cap C| = x$ (students who liked all 3 subjects, which is necessary to find),\n\n3. Using the inclusion-exclusion principle:\n\n $|C \\cup A \\cup P| = |C| + |A| + |P| - |C \\cap A| - |C \\cap P| - |A \\cap P| + |C \\cap A \\cap P|$\n\nSince $|C \\cup A \\cup P| = 150 - 25 = 125$:\n\n $125 = 75 + 80 + 50 - (25 + x) - (15 + x) - (20 + x) + x$\n\nSimplifying:\n\n $125 = 205 - 60 - 3x + x$\n\nFurther simplification:\n\n $x = 10$\n\nSo, 10 students out of 150 in the room liked all 3 subjects. The percentage is:\n\n $\\text{Percentage} = \\frac{10}{150} \\times 100 = 6.666\\%$\nb) Let $U = \\{1, 2, 3, 4, \\ldots, 10\\}$, $A = \\{1, 3, 5, 7\\}$, $B = \\{2, 4, 6, 8\\}$, and $C = \\{9, 10\\}$. Find $(A \\cup B) - (B \\cap C)$.\n\n1. Union $A \\cup B$:\n \n $$A \\cup B = \\{1, 3, 5, 7\\} \\cup \\{2, 4, 6, 8\\} = \\{1, 2, 3, 4, 5, 6, 7, 8\\}$$\n\n2. Intersection $B \\cap C$:\n \n $$B \\cap C = \\{2, 4, 6, 8\\} \\cap \\{9, 10\\} = \\emptyset$$ (no common elements)\n\n3. Set Difference $(A \\cup B) - (B \\cap C)$: Since $B \\cap C = \\emptyset$\n \n $$(A \\cup B) - (B \\cap C) = (A \\cup B) - \\emptyset = A \\cup B = \\{1, 2, 3, 4, 5, 6, 7, 8\\}$$\n\nThus, the result of $(A \\cup B) - (B \\cap C)$ is $\\{1, 2, 3, 4, 5, 6, 7, 8\\}$.\n\nc) Given $U = \\{1, 2, 3, 4, \\ldots, 20\\}$, $A = \\{x \\mid 5 < x \\leq 10\\}$, $B = \\{x \\mid 8 \\leq x \\leq 15\\}$, and $C = \\{x \\mid 1 \\leq x \\leq 5\\}$, find $(A \\cap B \\cap C) \\cup C$.\n\n1. Defining the sets:\n \n $U = \\{1, 2, 3, 4, \\ldots, 20\\}$ this set contains all integers from 1 to 20.\n \n $A = \\{x \\mid 5 < x \\leq 10\\}$ this means $A$ contains all elements greater than 5 and less than or equal to 10: $A = \\{6, 7, 8, 9, 10\\}$\n \n $B = \\{x \\mid 8 \\leq x \\leq 15\\}$ this means $B$ contains all elements between 8 and 15, inclusive: $B = \\{8, 9, 10, 11, 12, 13, 14, 15\\}$\n \n $C = \\{x \\mid 1 \\leq x \\leq 5\\}$ this means $C$ contains all elements from 1 to 5, inclusive: $C = \\{1, 2, 3, 4, 5\\}$\n\n2. Finding the Intersection $A \\cap B \\cap C$:\n \n $A = \\{6, 7, 8, 9, 10\\}$\n \n $B = \\{8, 9, 10, 11, 12, 13, 14, 15\\}$\n \n The common elements between $A$ and $B$ are $\\{8, 9, 10\\}$. So, $A \\cap B = \\{8, 9, 10\\}$.\n \n $A \\cap B = \\{8, 9, 10\\}$\n \n $C = \\{1, 2, 3, 4, 5\\}$\n \n There are no common elements between $A \\cap B$ and $C$, so the intersection is the empty set: $A \\cap B \\cap C = \\emptyset$. \n\n3\n3. Find \\((A \\cap B \\cap C) \\cup C:\\)\n\nSince \\(A \\cap B \\cap C = \\emptyset\\), the union will just be the set \\(C:\\)\n\n\\((A \\cap B \\cap C) \\cup C = \\emptyset \\cup C = C.\\)\n\nSo, \\((A \\cap B \\cap C) \\cup C = \\{1, 2, 3, 4, 5\\}\\)", "id": "./materials/950.pdf" }, { "contents": "Binary Relations\n\n- **Cartesian Product**\n The Cartesian product of two sets $A$ and $B$, denoted $A \\times B$, is the set of all ordered pairs $(a, b)$ where $a \\in A$ and $b \\in B$.\n For example, if $A = \\{1, 2\\}$ and $B = \\{a, b\\}$, $A \\times B$ is $\\{(1, a), (1, b), (2, a), (2, b)\\}$.\n\n- **Binary Relation**\n A binary relation on a set $A$ is a collection of ordered pairs $(a, b)$ where $a$ and $b$ are elements of $A$. In formal terms, a binary relation $R$ from $A$ to $B$ is a subset of the Cartesian product $A \\times B$.\n For instance, if $A = \\{1, 2\\}$ and $B = \\{a, b\\}$, then $A \\times B$ consists of the pairs $(1, a)$, $(1, b)$, $(2, a)$, and $(2, b)$, and any subset of these pairs represents a binary relation from $A$ to $B$.\n\n- **Reflexive Binary Relation**\n A binary relation $R$ on a set $A$ is reflexive if every element in $A$ is related to itself. Formally, $R$ is reflexive if for all $a \\in A$, the pair $(a, a) \\in R$.\n For example, if $A = \\{1, 2\\}$ and $R = \\{(1, 1), (2, 2)\\}$, then $R$ is reflexive because every element in $A$ is related to itself.\n\n- **Symmetric Binary Relation**\n A binary relation $R$ on a set $A$ is symmetric if whenever $(a, b) \\in R$, then $(b, a) \\in R$ as well. In other words, if one element is related to another, then the second element must be related back to the first.\n For example, if $R = \\{(1, 2), (2, 1)\\}$ on $A = \\{1, 2\\}$, then $R$ is symmetric because both $(1, 2)$ and $(2, 1)$ are present.\n\n- **Antisymmetric Binary Relation**\n A binary relation $R$ on a set $A$ is antisymmetric if whenever $(a, b) \\in R$ and $(b, a) \\in R$, then $a$ must equal $b$. This means that if two elements are mutually related, they must be the same element.\n For example, if $R = \\{(1, 2), (2, 1), (1, 1)\\}$ on $A = \\{1, 2\\}$, then $R$ is not antisymmetric because $(1, 2)$ and $(2, 1)$ are in $R$, but 1 and 2 are not equal.\n• **Transitive Binary Relation**\n\nA binary relation $R$ on a set $A$ is transitive if whenever $(a, b) \\in R$ and $(b, c) \\in R$, then $(a, c) \\in R$ as well. In other words, if an element $a$ is related to $b$, and $b$ is related to $c$, then $a$ must be related to $c$.\n\nFor example, if $R = \\{(1, 2), (2, 3), (1, 3)\\}$ on $A = \\{1, 2, 3\\}$, then $R$ is transitive because $(1, 2)$ and $(2, 3)$ imply $(1, 3)$.\n\nOn the other hand, a relation $R$ in a set $A$ is considered non-transitive if there are elements $a$, $b$, and $c$ in $A$ such that $(a, b) \\in R$ and $(b, c) \\in R$, but $(a, c) \\notin R$. In other words, the relation does not satisfy the transitivity condition when there is at least one example in which two consecutive relations do not imply a direct relation between the first and last elements.\n\nFor example, if $R$ includes $(1, 2)$ and $(2, 3)$, but does not include $(1, 3)$, then $R$ is not transitive.\n\n• **Equivalence Relation**\n\nAn equivalence relation on a set $A$ is a binary relation that is reflexive, symmetric, and transitive. In other words, for a relation $R$ to be an equivalence relation, it must satisfy the following conditions:\n\n- **Reflexive**: For every $a \\in A$, $(a, a) \\in R$.\n- **Symmetric**: If $(a, b) \\in R$, then $(b, a) \\in R$.\n- **Transitive**: If $(a, b) \\in R$ and $(b, c) \\in R$, then $(a, c) \\in R$.\n\nFor example, the relation \"is equal to\" on any set $A$ is an equivalence relation because it is reflexive (every element is equal to itself), symmetric (if $a = b$, then $b = a$), and transitive (if $a = b$ and $b = c$, then $a = c$).\nProblems\n\na) Determine $P(\\{a, b\\}) \\times \\{c, d\\}$.\n\n1. Find the Power Set $P(\\{a, b\\})$:\n The power set $P(\\{a, b\\})$ is the set of all subsets of $\\{a, b\\}$. The subsets are:\n - The empty set $\\emptyset$\n - The single-element subsets $\\{a\\}$ and $\\{b\\}$\n - The subset containing both elements $\\{a, b\\}$\n So: $P(\\{a, b\\}) = \\{\\emptyset, \\{a\\}, \\{b\\}, \\{a, b\\}\\}$\n\n2. Calculate the Cartesian Product $P(\\{a, b\\}) \\times \\{c, d\\}$:\n This consists of all ordered pairs $(X, y)$ where $X \\in P(\\{a, b\\})$ and $y \\in \\{c, d\\}$.\n List all the pairs:\n - For $X = \\emptyset$: $(\\emptyset, c), (\\emptyset, d)$\n - For $X = \\{a\\}$: $(\\{a\\}, c), (\\{a\\}, d)$\n - For $X = \\{b\\}$: $(\\{b\\}, c), (\\{b\\}, d)$\n - For $X = \\{a, b\\}$: $(\\{a, b\\}, c), (\\{a, b\\}, d)$\n Combining all these pairs:\n $P(\\{a, b\\}) \\times \\{c, d\\} = \\{(\\emptyset, c), (\\emptyset, d), (\\{a\\}, c), (\\{a\\}, d), (\\{b\\}, c), (\\{b\\}, d), (\\{a, b\\}, c), (\\{a, b\\}, d)\\}$\n\nb) If $A = \\{1, 2, 3\\}$, what makes $A$ reflexive but not transitive?\n\n1. Reflexive Property:\n A relation $R$ is reflexive if for every element $a \\in A$, the pair $(a, a)$ is in $R$.\n - $(1, 1)$ is in $R$.\n - $(2, 2)$ is in $R$.\n - $(3, 3)$ is in $R$.\n Since all elements of $A$ (that is, 1, 2, 3) have their corresponding pairs $(a, a)$ in $R$, the relation is reflexive.\n\n2. Transitive Property:\n A relation $R$ is transitive if whenever $(a, b) \\in R$ and $(b, c) \\in R$, then $(a, c)$ must also be in $R$. So, if one of them is missing, it is not transitive.\n• If \\( R = \\{(1, 1), (2, 2), (3, 3), (1, 3), (3, 2), (1, 2)\\} \\), with all the elements of \\( A \\) having their corresponding pairs in \\( R \\), makes it reflexive, and with \\((a, b) \\in R \\) and \\((b, c) \\in R \\) and \\((a, c) \\in R \\), makes it transitive.\n\n• On the other side, \\( R = \\{(1, 1), (2, 2), (3, 3), (1, 3), (3, 2)\\} \\), has all the corresponding pairs in \\( R \\), but only \\((1, 3) \\in R \\) and \\((3, 2) \\in R \\). By transitivity, it would be expected that \\((1, 2) \\) would be in \\( R \\). However, \\((1, 2) \\notin R \\).\n\nTherefore, this relation is not transitive.\n\nIn summary, a relation on \\( A = \\{1, 2, 3\\} \\) is reflexive but not transitive if it includes all reflexive pairs \\((a, a)\\) but fails to include the necessary pairs to satisfy the transitivity condition for some combinations of \\((a, b)\\) and \\((b, c)\\).\n\nc) Let \\( A = \\{\\emptyset, \\tau\\} \\) and \\( B = \\{a, b, c, d\\} \\). How many elements can \\((A \\times B)^3\\) have?\n\n1. Find the number of elements in \\( A \\times B \\):\n\n The Cartesian product \\( A \\times B \\) consists of all ordered pairs \\((a_1, b_1)\\) such that \\(a_1 \\in A\\) and \\(b_1 \\in B\\). The number of elements in \\( A \\times B \\) is calculated by multiplying the number of elements in \\( A \\) and \\( B \\).\n\n • Number of elements in \\( A = 2 \\) (since \\( A = \\{\\emptyset, \\tau\\} \\))\n • Number of elements in \\( B = 4 \\) (since \\( B = \\{a, b, c, d\\} \\))\n\n Thus, the number of elements in \\( A \\times B \\) is:\n\n \\[ |A \\times B| = |A| \\times |B| = 2 \\times 4 = 8 \\]\n\n2. Find the number of elements in \\((A \\times B)^3\\):\n\n The expression \\((A \\times B)^3\\) represents the Cartesian product of \\( A \\times B \\) by itself three times. This means that each element in \\((A \\times B)^3\\) is an ordered triple \\(((a_1, b_1), (a_2, b_2), (a_3, b_3))\\), where each \\((a_i, b_i)\\) is an element of \\( A \\times B \\).\n\n Since \\( A \\times B \\) has 8 elements, the number of elements in \\((A \\times B)^3\\) is:\n\n \\[ |(A \\times B)^3| = |A \\times B|^3 = 8^3 = 512 \\]\n\n The number of elements in \\((A \\times B)^3\\) is 512.", "id": "./materials/951.pdf" }, { "contents": "Venn Diagram\n\nA Venn diagram is a graphical way of representing sets and their relationships. Each set is typically represented by a circle, and overlaps between the circles represent intersections of sets. Venn diagrams are useful for visualizing unions, intersections, and differences of sets.\nProblems\n\na) Illustrate the intersection and union of $A = \\{1, 2, 3, 4\\}$ with $B = \\{3, 4, 5, 6, 7\\}$.\n\nSo $A \\cup B = \\{1, 2, 3, 4, 5, 6, 7\\}$ and $A \\cap B = \\{3, 4\\}$.\n\nb) From the given Venn diagram, determine $(P - Q) \\cup (R - Q)$.\n\nUnderstanding what is being represented in this diagram:\n\nSo $(P - Q) \\cup (R - Q)$ is $\\{a, i, j, b, l, k, g\\}$", "id": "./materials/952.pdf" }, { "contents": "Magnitude of Vectors: the magnitude (or length) of a vector is a measure of its size, and it’s calculated using the components of the vector in a given coordinate system. Let a vector \\( \\mathbf{v} \\) be represented in 3D space by its components:\n\n\\[\n\\mathbf{v} = \\langle a, b, c \\rangle\n\\]\n\nThe magnitude (denoted as \\( \\| \\mathbf{v} \\| \\)) is given by the formula:\n\n\\[\n\\| \\mathbf{v} \\| = \\sqrt{a^2 + b^2 + c^2}\n\\]\n\nThis formula is derived from the Pythagorean theorem. In 2D space, the magnitude would simply be:\n\n\\[\n\\| \\mathbf{v} \\| = \\sqrt{a^2 + b^2}\n\\]\n\nThe magnitude represents the distance from the origin to the point \\((v_1, v_2, v_3)\\) in space.\nProblems\n\na) Calculate the magnitude of the vector \\( \\mathbf{u} = (-1, 2, 1) \\)\nUsing the formula for the magnitude of a vector in 3D space:\n\n\\[\n\\| \\mathbf{u} \\| = \\sqrt{u_1^2 + u_2^2 + u_3^2}\n\\]\n\nSubstituting the components of \\( \\mathbf{u} = (-1, 2, 1) \\):\n\n\\[\n\\| \\mathbf{u} \\| = \\sqrt{(-1)^2 + 2^2 + 1^2} = \\sqrt{1 + 4 + 1} = \\sqrt{6}\n\\]\n\nThus, the magnitude of \\( \\mathbf{u} \\) is:\n\n\\[\n\\| \\mathbf{u} \\| = \\sqrt{6}\n\\]\n\nb) Calculate the magnitude of the vector \\( \\mathbf{v} = (1, 5) \\)\nIn 2D space, using the formula for the magnitude of a vector:\n\n\\[\n\\| \\mathbf{v} \\| = \\sqrt{v_1^2 + v_2^2}\n\\]\n\nSubstituting the components of \\( \\mathbf{v} = (1, 5) \\):\n\n\\[\n\\| \\mathbf{v} \\| = \\sqrt{1^2 + 5^2} = \\sqrt{1 + 25} = \\sqrt{26}\n\\]\n\nThus, the magnitude of \\( \\mathbf{v} \\) is:\n\n\\[\n\\| \\mathbf{v} \\| = \\sqrt{26}\n\\]\n\nReferences: Silva, C. e Medeiros, E. (2018). Geometria analítica. ISBN 9788595028739.", "id": "./materials/953.pdf" }, { "contents": "Vectors and Operations\n\n1. Cross Product (Vector Product)\n\nThe cross product of two vectors results in a new vector that is orthogonal to both of the original vectors. This operation is defined only in three dimensions.\n\nGiven two vectors \\( \\mathbf{a} = \\langle a_1, a_2, a_3 \\rangle \\) and \\( \\mathbf{b} = \\langle b_1, b_2, b_3 \\rangle \\), the product of two vectors can be determined using the matrix method by calculating its determinant:\n\n\\[\n\\mathbf{b} \\times \\mathbf{a} = \\begin{vmatrix}\n\\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\\nb_1 & b_2 & b_3 \\\\\na_1 & a_2 & a_3\n\\end{vmatrix} = \\mathbf{i} \\begin{vmatrix} b_2 & b_3 \\\\ a_2 & a_3 \\end{vmatrix} - \\mathbf{j} \\begin{vmatrix} b_1 & b_3 \\\\ a_1 & a_3 \\end{vmatrix} + \\mathbf{k} \\begin{vmatrix} b_1 & b_2 \\\\ a_1 & a_2 \\end{vmatrix}\n\\]\n\nThe cross product can also be written using the following component-wise formula:\n\n\\[\n\\mathbf{b} \\times \\mathbf{a} = \\langle b_2a_3 - b_3a_2, b_3a_1 - b_1a_3, b_1a_2 - b_2a_1 \\rangle\n\\]\n\nIf the order of the cross-product changes:\n\n\\[\n\\mathbf{a} \\times \\mathbf{b} = \\langle a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1 \\rangle\n\\]\n\nBoth the matrix method and the component-by-component formula represent the cross product of the vectors \\( \\mathbf{a} \\) and \\( \\mathbf{b} \\), giving the same result in two different forms.\n\n2. Scalar Product (Dot Product)\n\nThe scalar product, also called the dot product, gives a scalar (number) rather than a vector. Given two vectors \\( \\mathbf{a} = \\langle a_1, a_2, a_3 \\rangle \\) and \\( \\mathbf{b} = \\langle b_1, b_2, b_3 \\rangle \\), the scalar product is:\n\n\\[\n\\mathbf{a} \\cdot \\mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3\n\\]\n\nKey properties of the scalar product:\n\na) Magnitude: The magnitude of the dot product is given by:\n\n\\[\n\\mathbf{a} \\cdot \\mathbf{b} = \\|\\mathbf{a}\\|\\|\\mathbf{b}\\| \\cos \\theta\n\\]\n\nwhere \\( \\theta \\) is the angle between \\( \\mathbf{a} \\) and \\( \\mathbf{b} \\).\n\nb) Orthogonality: If \\( \\mathbf{a} \\cdot \\mathbf{b} = 0 \\), then the vectors \\( \\mathbf{a} \\) and \\( \\mathbf{b} \\) are orthogonal (perpendicular).\n\nc) Applications: The dot product is useful in calculating angles between vectors and projections of one vector onto another.\nProblems\n\na) Consider the vectors \\( \\mathbf{u} = (1, -1, 1) \\) and \\( \\mathbf{v} = (2, -2, 0) \\). Find the cross product \\( \\mathbf{v} \\times \\mathbf{u} \\) and its magnitude:\n\n1. Matrix Method\n\nTo find the cross product using the matrix method, we compute the determinant of the following matrix:\n\n\\[\n\\mathbf{v} \\times \\mathbf{u} = \\begin{vmatrix}\n\\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\\n2 & -2 & 0 \\\\\n1 & -1 & 1\n\\end{vmatrix}\n\\]\n\nNow, calculate the 2x2 determinants:\n\n1. For the \\( \\mathbf{i} \\) component:\n\n\\[\n\\mathbf{i} \\begin{vmatrix}\n-2 & 0 \\\\\n-1 & 1\n\\end{vmatrix} = \\mathbf{i}((-2)(1) - (0)(-1)) = \\mathbf{i}(-2)\n\\]\n\n2. For the \\( \\mathbf{j} \\) component:\n\n\\[\n-\\mathbf{j} \\begin{vmatrix}\n2 & 0 \\\\\n1 & 1\n\\end{vmatrix} = -\\mathbf{j}((2)(1) - (0)(1)) = -\\mathbf{j}(2)\n\\]\n\n3. For the \\( \\mathbf{k} \\) component:\n\n\\[\n\\mathbf{k} \\begin{vmatrix}\n2 & -2 \\\\\n1 & -1\n\\end{vmatrix} = \\mathbf{k}((2)(-1) - (-2)(1)) = \\mathbf{k}(-2 + 2) = \\mathbf{k}(0)\n\\]\n\nThus, the cross product is \\( \\mathbf{v} \\times \\mathbf{u} = -2\\mathbf{i} - 2\\mathbf{j} + 0\\mathbf{k} \\)\n\nIn vector notation, this is \\( \\mathbf{v} \\times \\mathbf{u} = (-2, -2, 0) \\)\n\n2. Component-Wise Formula\n\nThe cross product of two vectors \\( \\mathbf{a} = (a_1, a_2, a_3) \\) and \\( \\mathbf{b} = (b_1, b_2, b_3) \\) is given by:\n\n\\[\n\\mathbf{a} \\times \\mathbf{b} = \\langle a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1 \\rangle\n\\]\n\nFor \\( \\mathbf{u} = (1, -1, 1) \\) as \\( \\mathbf{b} \\) and \\( \\mathbf{v} = (2, -2, 0) \\) as \\( \\mathbf{a} \\), we calculate each component:\n\n\\[\n\\mathbf{a} \\times \\mathbf{b} = \\langle (-2 \\times 1) - 0, 0 - (2 \\times 1), (2 \\times (-1)) - ((-2) \\times 1) \\rangle\n\\]\n\n\\[\n\\mathbf{a} \\times \\mathbf{b} = \\langle -2, -2, 0 \\rangle\n\\]\n\nThus, the cross product is \\( \\mathbf{v} \\times \\mathbf{u} = (-2, -2, 0) \\)\n3. Magnitude of the Cross Product\n\nThe magnitude of a vector \\((x, y, z)\\) is given by:\n\n\\[\n\\|v \\times u\\| = \\sqrt{x^2 + y^2 + z^2}\n\\]\n\nFor \\((-2, -2, 0)\\), the magnitude is:\n\n\\[\n\\|v \\times u\\| = \\sqrt{(-2)^2 + (-2)^2 + 0^2} = \\sqrt{4 + 4 + 0} = \\sqrt{8} = 2\\sqrt{2}\n\\]\n\nSo, the cross product \\(v \\times u\\) is \\((-2, -2, 0)\\), and its magnitude is \\(2\\sqrt{2}\\).\n\nb) Compute the scalar product \\(u \\cdot v\\) and the magnitude \\(|u|\\). Consider the unusual scalar product of \\(\\mathbb{R}^3\\) defined by \\((x_1, y_1) \\cdot (x_2, y_2) = 4x_1x_2 + 3y_1y_2\\) and consider the vectors \\(u = (1, -4)\\), \\(v = (3, 2)\\).\n\n1. Find the scalar product \\(u \\cdot v\\):\n\nUsing the given formula:\n\n\\[\nu \\cdot v = 4x_1x_2 + 3y_1y_2\n\\]\n\n\\[\nu \\cdot v = 4(1)(3) + 3(-4)(2)\n\\]\n\n\\[\nu \\cdot v = 12 + (-24) = -12\n\\]\n\nSo, the scalar product is \\(u \\cdot v = -12\\)\n\n2. Compute the magnitude \\(|u|\\):\n\nThe magnitude \\(|u|\\) is defined by the scalar product of \\(u\\) with itself, using the unusual scalar product definition \\((x_1, y_1) \\cdot (x_2, y_2) = 4x_1x_2 + 3y_1y_2\\):\n\n\\[\n|u| = \\sqrt{u \\cdot u}\n\\]\n\nFirst, compute \\(u \\cdot u\\):\n\n\\[\nu \\cdot u = 4(1)(1) + 3(-4)(-4) = 4(1) + 3(16) = 4 + 48 = 52\n\\]\n\nNow, the magnitude is:\n\n\\[\n|u| = \\sqrt{52} = 2\\sqrt{13}\n\\]\n\nReferences: Silva, C. e Medeiros, E. (2018). Geometria analítica. ISBN 9788595028739.", "id": "./materials/954.pdf" }, { "contents": "Scalar Triple Product\n\nGiven three vectors \\( \\mathbf{a}, \\mathbf{b}, \\) and \\( \\mathbf{c} \\) in \\( \\mathbb{R}^3 \\), the scalar triple product is defined as:\n\n\\[\n\\mathbf{a} \\cdot (\\mathbf{b} \\times \\mathbf{c})\n\\]\n\nWhere:\n- \\( \\mathbf{a} \\cdot \\) represents the dot product (scalar product).\n- \\( \\mathbf{b} \\times \\mathbf{c} \\) represents the cross product (vector product) between vectors \\( \\mathbf{b} \\) and \\( \\mathbf{c} \\).\n\nThe scalar triple product can be expressed as the determinant of a \\( 3 \\times 3 \\) matrix too, with each row representing the components of one of the vectors:\n\n\\[\n\\mathbf{a} \\cdot (\\mathbf{b} \\times \\mathbf{c}) = \\det \\begin{vmatrix} a_1 & a_2 & a_3 \\\\ b_1 & b_2 & b_3 \\\\ c_1 & c_2 & c_3 \\end{vmatrix}\n\\]\n\nWhere \\( \\mathbf{a} = (a_1, a_2, a_3); \\mathbf{b} = (b_1, b_2, b_3) \\) and \\( \\mathbf{c} = (c_1, c_2, c_3) \\).\n\nThis determinant form provides an efficient way to compute the scalar triple product algebraically.\n\nThe sign of the scalar triple product indicates the orientation of the vectors:\n- If the result is positive, the vectors form a right-handed system.\n- If the result is negative, the vectors form a left-handed system.\n- A result of zero means the vectors are coplanar, and thus no distinct handedness exists.\n\nThe scalar triple product also calculates the volume of the parallelepiped formed by the three vectors. If the result is zero, it means that the three vectors are coplanar and do not enclose any volume. The formula can be written as:\n\n\\[\nV = |\\mathbf{a} \\cdot (\\mathbf{b} \\times \\mathbf{c})|\n\\]\n\n- \\( \\mathbf{b} \\times \\mathbf{c} \\) gives a vector perpendicular to the plane formed by \\( \\mathbf{b} \\) and \\( \\mathbf{c} \\), and its magnitude is equal to the area of the parallelogram spanned by \\( \\mathbf{b} \\) and \\( \\mathbf{c} \\).\n- \\( \\mathbf{a} \\cdot (\\mathbf{b} \\times \\mathbf{c}) \\) projects the vector \\( \\mathbf{a} \\) onto this normal vector, giving the height of the parallelepiped.\nProblem\n\na) Compute the scalar triple product $\\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w})$, given\n$\\mathbf{u} = (1, 3, 2), \\quad \\mathbf{v} = (-1, 1, 2), \\quad \\mathbf{w} = (0, 1, 1)$\n\n1. Cross product $\\mathbf{v} \\times \\mathbf{w}$ using the determinant method:\n\n$$\\mathbf{v} \\times \\mathbf{w} = \\begin{vmatrix} i & j & k \\\\ -1 & 1 & 2 \\\\ 0 & 1 & 1 \\end{vmatrix}$$\n\nExpanding the determinant:\n\n$$\\mathbf{v} \\times \\mathbf{w} = i \\begin{vmatrix} 1 & 2 \\\\ 0 & 1 \\end{vmatrix} - j \\begin{vmatrix} -1 & 2 \\\\ 0 & 1 \\end{vmatrix} + k \\begin{vmatrix} -1 & 1 \\\\ 0 & 1 \\end{vmatrix}$$\n\nNow, compute the 2x2 determinants: - For the $i$-component:\n\n$$\\begin{vmatrix} 1 & 2 \\\\ 0 & 1 \\end{vmatrix} = (1)(1) - (2)(0) = 1 - 2 = -1$$\n\n- For the $j$-component:\n\n$$\\begin{vmatrix} -1 & 2 \\\\ 0 & 1 \\end{vmatrix} = (-1)(1) - (2)(0) = -1$$\n\n- For the $k$-component:\n\n$$\\begin{vmatrix} -1 & 1 \\\\ 0 & 1 \\end{vmatrix} = (-1)(1) - (1)(0) = -1$$\n\nSo, the cross product is $\\mathbf{v} \\times \\mathbf{w} = -\\mathbf{i} + \\mathbf{j} - \\mathbf{k}$\n\nIn vector notation, this is $\\mathbf{v} \\times \\mathbf{w} = (-1, 1, -1)$\n\n2. Scalar product $\\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w})$\n\n$$\\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w}) = (1, 3, 2) \\cdot (-1, 1, -1)$$\n\n$$\\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w}) = (1)(-1) + (3)(1) + (2)(-1)$$\n\n$$\\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w}) = -1 + 3 - 2 = 0$$\n\nThe scalar triple product is $\\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w}) = 0$\n\nAlso, since the scalar product $\\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w}) = 0$, then the vectors are perpendicular and $\\mathbf{v} \\times \\mathbf{w}$ is normal to $\\mathbf{u}$.\n\nReferences: Silva, C. e Medeiros, E. (2018). Geometria analítica. ISBN 9788595028739.", "id": "./materials/955.pdf" }, { "contents": "Relative Position of Vectors\nThe relative position of vectors involves understanding the spatial relationships between two or more vectors. This concept is useful for determining whether vectors are parallel, perpendicular, or form specific geometric configurations.\n\n1. Parallel (Collinear) Vectors\nTwo vectors are parallel if they point in the same or exactly opposite directions, but with potentially different magnitudes.\n\nTwo vectors \\( \\mathbf{a} \\) and \\( \\mathbf{b} \\) are parallel if there exists a scalar \\( k \\) such that:\n\n\\[\n\\mathbf{a} = k \\mathbf{b}\n\\]\n\nwhere \\( k \\) is a non-zero scalar.\n\nIf \\( k \\) is positive, the vectors point in the same direction; if \\( k \\) is negative, they point in opposite directions.\n\nAlso, it is possible to check if two vectors are parallel calculating the cross product:\n\n\\[\n\\mathbf{a} \\times \\mathbf{b} = 0\n\\]\n\nIf the cross product is zero, the vectors are parallel.\n\nTwo or more vectors are collinear if they lie along the same line. For vectors to be collinear, they must not only be parallel but also either share a common point or exist on the same line in space. All collinear vectors are parallel, but not all parallel vectors are collinear.\n\nIf the ratio of their components is constant, it is parallel and collinear:\n\n\\[\n\\frac{a_1}{b_1} = \\frac{a_2}{b_2} = \\frac{a_3}{b_3}\n\\]\n2. **Perpendicular (Orthogonal/Normal) Vectors**\n\nTwo vectors are perpendicular (or orthogonal or normal) if they meet at a right angle, and the angle between them is 90 degrees.\n\nTwo vectors \\( \\mathbf{a} \\) and \\( \\mathbf{b} \\) are perpendicular if their dot product is zero:\n\n\\[\n\\mathbf{a} \\cdot \\mathbf{b} = 0\n\\]\n\nor\n\n\\[\na_1 b_1 + a_2 b_2 + a_3 b_3 = 0\n\\]\n\nIn geometric terms, the dot product of two perpendicular vectors is zero because the cosine of 90 degrees is zero. This relationship implies that the vectors are at right angles to each other.\n\n3. **Coplanar Vectors**\n\nVectors are coplanar if they lie within the same plane. In 3D space, three vectors \\( \\mathbf{a}, \\mathbf{b}, \\) and \\( \\mathbf{c} \\) are coplanar if their scalar triple product is zero:\n\n\\[\n\\mathbf{a} \\cdot (\\mathbf{b} \\times \\mathbf{c}) = 0\n\\]\n\nHere, \\( \\mathbf{b} \\times \\mathbf{c} \\) is the cross product of \\( \\mathbf{b} \\) and \\( \\mathbf{c} \\), and \\( \\mathbf{a} \\cdot (\\mathbf{b} \\times \\mathbf{c}) \\) represents the volume of the parallelepiped formed by these vectors. A zero volume implies that the vectors are coplanar.\n\nIf three vectors are coplanar, you can place them on a single flat surface (plane) without lifting any of them out of that plane.\nProblems\n\na) Determine whether the vectors \\( u = (-1, 2, 1) \\) and \\( v = (3, -6, -3) \\) are collinear or orthogonal.\n\n1. Check if the vectors are orthogonal: Vectors are orthogonal (perpendicular) if their scalar product (dot product) is zero. The scalar product of two vectors is given by:\n\n\\[\n\\mathbf{u} \\cdot \\mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3\n\\]\n\nFor \\( u = (-1, 2, 1) \\) and \\( v = (3, -6, -3) \\), we compute the scalar product:\n\n\\[\n\\mathbf{u} \\cdot \\mathbf{v} = (-1)(3) + (2)(-6) + (1)(-3)\n\\]\n\n\\[\n\\mathbf{u} \\cdot \\mathbf{v} = -3 + (-12) + (-3) = -18\n\\]\n\nSince \\( \\mathbf{u} \\cdot \\mathbf{v} = -18 \\), the vectors are not orthogonal because the dot product is not zero.\n\n2. Check if the vectors are collinear: Vectors are collinear if one is a scalar multiple of the other. To check if two vectors are collinear, we see if the components of \\( \\mathbf{u} \\) and \\( \\mathbf{v} \\) are proportional. Specifically, if:\n\n\\[\n\\frac{u_1}{v_1} = \\frac{u_2}{v_2} = \\frac{u_3}{v_3}\n\\]\n\nFor \\( u = (-1, 2, 1) \\) and \\( v = (3, -6, -3) \\):\n\n\\[\n\\frac{-1}{3} = \\frac{2}{-6} = \\frac{1}{-3}\n\\]\n\nSimplifying:\n\n\\[\n\\frac{-1}{3} = \\frac{-1}{3} = \\frac{-1}{3}\n\\]\n\nSince all the ratios are equal, the vectors are collinear.\n\nb) Given the vectors \\( u = (2, 1, 1) \\), \\( v = (1, 2, -2) \\), find the cross product \\( \\mathbf{u} \\times \\mathbf{v} \\) and determine whether it is parallel or perpendicular to \\( \\mathbf{u} \\) and \\( \\mathbf{v} \\)\n\n1. Compute the Cross Product \\( \\mathbf{u} \\times \\mathbf{v} \\) The wise formula for the cross product of two vectors is:\n\n\\[\n\\mathbf{u} \\times \\mathbf{v} = (u_2v_3 - u_3v_2, u_3v_1 - u_1v_3, u_1v_2 - u_2v_1)\n\\]\n\n1. \\( x \\)-component:\n\n\\[\n(1)(-2) - (1)(2) = -2 - 2 = -4\n\\]\n2. y-component:\n\n\\[(1)(1) - (2)(-2) = 1 + 4 = 5\\]\n\n3. z-component:\n\n\\[(2)(2) - (1)(1) = 4 - 1 = 3\\]\n\nThus, the cross product is:\n\n\\[\\mathbf{u} \\times \\mathbf{v} = (-4, 5, 3)\\]\n\n2. Determine the Relationship\n\n- Perpendicular (Normal): The cross product of two vectors is always perpendicular to both vectors. Since the cross product gives a vector that is perpendicular to both \\(\\mathbf{u}\\) and \\(\\mathbf{v}\\), the result \\((-4, 5, 3)\\) is perpendicular to both \\(\\mathbf{u}\\) and \\(\\mathbf{v}\\).\n\n- Parallel: The cross product of two vectors is zero if and only if the vectors are parallel. Since \\(\\mathbf{u} \\times \\mathbf{v} \\neq 0\\), the vectors are not parallel.\n\nSo, the cross product \\(\\mathbf{u} \\times \\mathbf{v} = (-4, 5, 3)\\) is **perpendicular (or normal)** to both \\(\\mathbf{u}\\) and \\(\\mathbf{v}\\).\n\nReferences: Silva, C. e Medeiros, E. (2018). Geometria analítica. ISBN 9788595028739.", "id": "./materials/956.pdf" }, { "contents": "Constant Coefficient Ordinary Differential Equation\n\nA Constant Coefficient Ordinary Differential Equation (ODE) is a type of linear differential equation where the coefficients of the derivatives are constants, independent of the variable $x$. These types of equations are particularly straightforward to solve because the constant coefficients allow us to use specific methods, such as the characteristic equation or the method of undetermined coefficients.\n\nThe general form of a constant coefficient ODE is:\n\n$$a_n y^{(n)} + a_{n-1} y^{(n-1)} + \\cdots + a_1 y' + a_0 y = f(x),$$\n\nwhere $a_n, a_{n-1}, \\ldots, a_0$ are constants and $f(x)$ is a known function of $x$. If $f(x) = 0$, the equation is homogeneous; otherwise, it is non-homogeneous.\n\nCharacteristic Equation\n\nThe Characteristic Equation is a crucial tool for solving linear constant coefficient ODEs, especially for homogeneous equations. It helps convert a differential equation into an algebraic equation, which is easier to solve.\n\nTo solve a homogeneous constant coefficient ODE of the form:\n\n$$a_n y^{(n)} + a_{n-1} y^{(n-1)} + \\cdots + a_1 y' + a_0 y = 0,$$\n\nis assumed a solution of the form $y = e^{rx}$, where $r$ is a constant to be determined.\n\n- Substitute $y = e^{rx}$ into the differential equation.\n- This substitution leads to the characteristic equation:\n\n$$a_n r^n + a_{n-1} r^{n-1} + \\cdots + a_1 r + a_0 = 0.$$\n\n- The roots of the characteristic equation give us the form of the general solution to the differential equation.\n\nCases of Roots:\n\n1. Distinct Real Roots: If the characteristic equation has $n$ distinct real roots $r_1, r_2, \\ldots, r_n$, the general solution is:\n\n$$y(x) = c_1 e^{r_1 x} + c_2 e^{r_2 x} + \\cdots + c_n e^{r_n x},$$\n\nwhere $c_1, c_2, \\ldots, c_n$ are constants determined by initial conditions.\n2. Repeated Roots: If a root $r$ has multiplicity $m$, the solution includes terms like:\n\n$$y(x) = (c_1 + c_2 x + \\cdots + c_m x^{m-1})e^{rx}.$$ \n\n3. Complex Roots: If the characteristic equation has complex roots of the form $r = \\alpha \\pm \\beta i$, the general solution involves sines and cosines:\n\n$$y(x) = e^{\\alpha x} (c_1 \\cos(\\beta x) + c_2 \\sin(\\beta x)).$$\n\n**Example:** For the ODE $y'' - 3y' + 2y = 0$, the characteristic equation is\n\n$$r^2 - 3r + 2 = 0,$$\n\nwhich has roots $r_1 = 1$ and $r_2 = 2$. So, the general solution is:\n\n$$y(x) = c_1 e^x + c_2 e^{2x}.$$\n\n**Undetermined Coefficient Method**\n\nThe Undetermined Coefficient Method is a technique for solving non-homogeneous linear ODEs where the non-homogeneous term $f(x)$ is a simple function like a polynomial, exponential, sine, or cosine. This method is based on the assumption that the particular solution has a similar form to $f(x)$, and is possible to determine the coefficients of this assumed form. Here some steps:\n\n1. Solve the Homogeneous Equation: First, solve the homogeneous version of the equation (i.e., set $f(x) = 0$) to find the complementary solution $y_c(x)$.\n\n2. Guess the Form of the Particular Solution: Based on the form of $f(x)$, guess a form for the particular solution $y_p(x)$. Common forms are:\n - If $f(x) = e^{ax}$, guess $y_p(x) = Ae^{ax}$.\n - If $f(x) = \\sin(bx)$ or $\\cos(bx)$, guess $y_p(x) = A \\cos(bx) + B \\sin(bx)$.\n - If $f(x)$ is a polynomial of degree $n$, guess a polynomial of degree $n$.\n\n3. Adjust for Overlap with the Homogeneous Solution: If the guessed form for $y_p(x)$ is already part of the homogeneous solution $y_c(x)$, multiply the guess by $x$ (or higher powers of $x$) until it’s independent of $y_c(x)$.\n\n4. Substitute and Solve: Substitute $y_p(x)$ into the original ODE and solve for the unknown coefficients.\n5. Form the General Solution: The general solution is:\n\n\\[ y(x) = y_c(x) + y_p(x). \\]\n\n**Example:** For the ODE \\( y'' - 3y' + 2y = e^x \\), the complementary solution is \\( y_c(x) = c_1e^x + c_2e^{2x} \\), and the guess for the particular solution is \\( y_p(x) = Axe^x \\), since \\( e^x \\) is already part of the homogeneous solution. Substituting this into the ODE gives \\( A = 1 \\), so the particular solution is \\( y_p(x) = xe^x \\). The general solution is:\n\n\\[ y(x) = c_1e^x + c_2e^{2x} + xe^x. \\]\n\n**Wronskian**\n\nThe Wronskian is a determinant used to test whether a set of solutions to a linear differential equation is linearly independent. For \\( n \\) functions \\( y_1(x), y_2(x), \\ldots, y_n(x) \\), the Wronskian \\( W(y_1, y_2, \\ldots, y_n)(x) \\) is defined as the determinant of the matrix formed by these functions and their derivatives up to order \\( n - 1 \\).\n\nFor two functions \\( y_1(x) \\) and \\( y_2(x) \\), the Wronskian is:\n\n\\[\nW(y_1, y_2)(x) = \\begin{vmatrix} y_1(x) & y_2(x) \\\\ y_1'(x) & y_2'(x) \\end{vmatrix} = y_1(x)y_2'(x) - y_1'(x)y_2(x).\n\\]\n\nFor \\( n \\) functions \\( y_1(x), y_2(x), \\ldots, y_n(x) \\), the Wronskian is:\n\n\\[\nW(y_1, y_2, \\ldots, y_n)(x) = \\det \\begin{vmatrix} y_1(x) & y_2(x) & \\cdots & y_n(x) \\\\ y_1'(x) & y_2'(x) & \\cdots & y_n'(x) \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ y_1^{(n-1)}(x) & y_2^{(n-1)}(x) & \\cdots & y_n^{(n-1)}(x) \\end{vmatrix}.\n\\]\n\nIf the Wronskian \\( W(y_1, y_2, \\ldots, y_n)(x) \\neq 0 \\) for some \\( x \\), then the functions are linearly independent and form a fundamental set of solutions to the differential equation.\n\nIf \\( W(y_1, y_2, \\ldots, y_n)(x) = 0 \\) for all \\( x \\), then the functions are linearly dependent.\n\n**Example:** For the functions \\( y_1(x) = e^x \\) and \\( y_2(x) = e^{2x} \\), the Wronskian is\n\n\\[\nW(y_1, y_2) = \\begin{vmatrix} e^x & e^{2x} \\\\ e^x & 2e^{2x} \\end{vmatrix} = e^x \\cdot 2e^{2x} - e^x \\cdot e^{2x} = e^{3x}.\n\\]\n\nSince \\( W(y_1, y_2) \\neq 0 \\), the functions are linearly independent.\n**Problem:** Find the general solution of the differential equation\n\n\\[ x'' + 4x' + 4x = 2e^{-2t}. \\]\n\nThis is a non-homogeneous linear differential equation with constant coefficients. It is possible to solve it in two parts: finding the complementary solution (solving the homogeneous equation) and finding the particular solution using the method of undetermined coefficients.\n\n1. Solve the Homogeneous Equation:\n\n The homogeneous form of the given equation is:\n\n \\[ x'' + 4x' + 4x = 0. \\]\n\n To solve this, find the characteristic equation. Assume the solution has the form \\( x(t) = e^{rt} \\), where \\( r \\) is a constant. Substituting into the homogeneous equation:\n\n \\[ r^2 e^{rt} + 4re^{rt} + 4e^{rt} = 0. \\]\n\n Dividing by \\( e^{rt} \\) (which is never zero):\n\n \\[ r^2 + 4r + 4 = 0. \\]\n\n This is a quadratic equation. Solving using the quadratic formula:\n\n \\[ r = \\frac{-4 \\pm \\sqrt{4^2 - 4(1)(4)}}{2(1)} = \\frac{-4 \\pm \\sqrt{16 - 16}}{2} = \\frac{-4 \\pm 0}{2} = -2. \\]\n\n The root \\( r = -2 \\) is a repeated root. For repeated roots, the general solution to the homogeneous equation is:\n\n \\[ x_h(t) = (c_1 + c_2t)e^{-2t}, \\]\n\n where \\( c_1 \\) and \\( c_2 \\) are constants to be determined later by initial conditions, if available.\n\n2. Find the Particular Solution:\n\n Next, it is possible to find a particular solution \\( x_p(t) \\) for the non-homogeneous equation:\n\n \\[ x'' + 4x' + 4x = 2e^{-2t}. \\]\n\n Since the right-hand side of the equation is \\( 2e^{-2t} \\), which is already part of the solution of the homogeneous equation, multiply the guess for the particular solution by \\( t \\) to avoid duplication with the homogeneous solution.\nLet the particular solution be of the form:\n\n\\[ x_p(t) = At^2e^{-2t}. \\]\n\nNow, is necessary to find the derivatives of \\( x_p(t) \\):\n\n- First derivative:\n \\[ x'_p(t) = A \\left( 2te^{-2t} + t^2(-2)e^{-2t} \\right) = Ae^{-2t} \\left( 2t - 2t^2 \\right). \\]\n\n- Second derivative:\n \\[ x''_p(t) = A \\left[ e^{-2t} (2 - 4t) - 2e^{-2t} (2t - 2t^2) \\right] = Ae^{-2t} \\left( 2 - 8t + 4t^2 \\right). \\]\n\nSubstitute \\( x_p(t) \\), \\( x'_p(t) \\), and \\( x''_p(t) \\) into the original differential equation:\n\n\\[ Ae^{-2t} \\left( 2 - 8t + 4t^2 \\right) + 4Ae^{-2t} \\left( 2t - 2t^2 \\right) + 4At^2e^{-2t} = 2e^{-2t}. \\]\n\nFactor out \\( Ae^{-2t} \\):\n\n\\[ Ae^{-2t} \\left[ 2 - 8t + 4t^2 + 8t - 8t^2 + 4t^2 \\right] = 2e^{-2t}. \\]\n\nSimplify the expression inside the brackets:\n\n\\[ Ae^{-2t} (2) = 2e^{-2t}. \\]\n\nThis gives:\n\n\\[ 2A = 2, \\]\n\nso \\( A = 1 \\). Thus, the particular solution is:\n\n\\[ x_p(t) = t^2e^{-2t}. \\]\n\n3. General Solution:\n\nSum the complementary solution \\( x_h(t) \\) and the particular solution \\( x_p(t) \\):\n\n\\[ x(t) = x_h(t) + x_p(t) = (c_1 + c_2t)e^{-2t} + t^2e^{-2t}. \\]\n\nFactoring out \\( e^{-2t} \\):\n\n\\[ x(t) = e^{-2t} \\left( c_1 + c_2t + t^2 \\right). \\]\n\nThis is the general solution to the given differential equation.\n\nReferences: Çengel, Y. A., and III, W. J. P. (2014). Differential Equations for Engineers and Scientists. ISBN 9780073385907.", "id": "./materials/957.pdf" }, { "contents": "• **Linear Systems**\n\nA linear system refers to a set of linear equations involving multiple variables. In the context of differential equations, linear systems involve a set of linear differential equations with respect to one or more dependent variables. For example, a system of first-order linear differential equations could look like:\n\n\\[\n\\frac{dx_1}{dt} = a_{11}x_1 + a_{12}x_2 + \\cdots + a_{1n}x_n + g_1(t),\n\\]\n\\[\n\\frac{dx_2}{dt} = a_{21}x_1 + a_{22}x_2 + \\cdots + a_{2n}x_n + g_2(t),\n\\]\n\\[\n\\vdots\n\\]\n\\[\n\\frac{dx_n}{dt} = a_{n1}x_1 + a_{n2}x_2 + \\cdots + a_{nn}x_n + g_n(t),\n\\]\n\nwhere the \\(a_{ij}\\)'s are constants or functions of \\(t\\), and \\(g_i(t)\\) are non-homogeneous terms. The goal is to find solutions for \\(x_1(t), x_2(t), \\ldots, x_n(t)\\).\n\n• **Elimination Method**\n\nThe elimination method is a technique used to solve systems of equations, either algebraic or differential. It involves eliminating one variable at a time by manipulating the equations. The basic idea is to combine equations in such a way that one of the variables \"cancels out\", reducing the system to fewer variables and simplifying the problem.\n\nFor example, if you have two equations:\n\n\\[\nx + y = 3,\n\\]\n\\[\n2x - y = 1,\n\\]\n\nyou can add the two equations to eliminate \\(y\\), yielding:\n\n\\[\n3x = 4 \\quad \\Rightarrow \\quad x = \\frac{4}{3}.\n\\]\n\nThen substitute this back into one of the original equations to find \\(y\\).\n\nIn differential equations, elimination works similarly but focuses on eliminating one of the dependent variables, reducing the system to a simpler set of equations.\n• **Euler’s Method**\n\nEuler’s method is a numerical technique used to approximate the solutions of ordinary differential equations (ODEs). It’s particularly useful for solving first-order ODEs. The basic idea is to approximate the solution step by step, using a finite difference approximation for the derivative.\n\nGiven an initial value problem of the form:\n\n\\[\n\\frac{dy}{dt} = f(t, y), \\quad y(t_0) = y_0,\n\\]\n\nEuler’s method computes the solution at subsequent points by using the formula:\n\n\\[\ny_{n+1} = y_n + hf(t_n, y_n),\n\\]\n\nwhere \\( h \\) is a small step size.\n\nThis method can be extended to systems of differential equations by applying it to each equation in the system.\n\n• **Integrating Factor**\n\nThe integrating factor method is a technique used to solve first-order linear differential equations of the form:\n\n\\[\n\\frac{dy}{dt} + P(t)y = Q(t).\n\\]\n\nThe idea is to multiply both sides of the equation by an integrating factor \\( \\mu(t) \\), which is chosen so that the left-hand side becomes the derivative of a product:\n\n\\[\n\\mu(t) = e^{\\int P(t)dt}.\n\\]\n\nMultiplying through by this integrating factor gives:\n\n\\[\n\\mu(t)\\frac{dy}{dt} + \\mu(t)P(t)y = \\mu(t)Q(t),\n\\]\n\nwhich simplifies to:\n\n\\[\n\\frac{d}{dt} (\\mu(t)y) = \\mu(t)Q(t).\n\\]\n\nYou can then integrate both sides to solve for \\( y(t) \\).\n• Lagrange’s Method\n\nLagrange’s method typically refers to a technique for solving higher-order linear differential equations using a systematic approach to find particular integrals (solutions). One common version of Lagrange’s method is the method of variation of parameters, used to find particular solutions to non-homogeneous linear differential equations.\n\nGiven a second-order non-homogeneous equation:\n\n\\[ y'' + P(t)y' + Q(t)y = g(t), \\]\n\nyou first solve the homogeneous equation to find the complementary solution. Then, you assume the particular solution takes the form:\n\n\\[ y_p(t) = u_1(t)y_1(t) + u_2(t)y_2(t), \\]\n\nwhere \\( y_1(t) \\) and \\( y_2(t) \\) are solutions to the homogeneous equation, and \\( u_1(t) \\), \\( u_2(t) \\) are functions to be determined. You find these functions by substituting into the original equation and solving for \\( u_1(t) \\) and \\( u_2(t) \\).\n**Problem:** Find the complete solution of the following system of differential equations\n\n\\[\n\\begin{align*}\nx'(t) &= x(t) + y(t), \\\\\ny'(t) &= -2y(t).\n\\end{align*}\n\\]\n\nSteps:\n\n1. Solve the Second Equation Separately\n\n The second equation is a first-order linear differential equation:\n\n \\[ y'(t) = -2y(t). \\]\n\n Using the integrating factor method:\n - Rewrite in standard form:\n \\[ y'(t) + 2y(t) = 0. \\]\n - Find the integrating factor:\n \\[ \\mu(t) = e^{\\int 2\\,dt} = e^{2t}. \\]\n - Multiply the equation by the integrating factor and simplifies it:\n \\[ e^{2t}y'(t) + 2e^{2t}y(t) = 0 \\Rightarrow \\frac{d}{dt}(e^{2t}y(t)) = 0. \\]\n - Integrate both sides:\n \\[ e^{2t}y(t) = C_2, \\]\n where \\( C_2 \\) is a constant.\n - Solve for \\( y(t) \\):\n \\[ y(t) = C_2e^{-2t}. \\]\n - Substitute \\( y(t) \\) into the First Equation\n \\[ x'(t) = x(t) + C_2e^{-2t}. \\]\n\n2. Solve the First Equation Using an Integrating Factor\n\n - Rewrite the equation in standard form:\n \\[ x'(t) - x(t) = C_2e^{-2t}. \\]\n• Find the integrating factor:\n\n\\[ \\mu(t) = e^{\\int -1 \\, dt} = e^{-t}. \\]\n\n• Multiply the equation by the integrating factor and simplifies it:\n\n\\[ e^{-t}x'(t) - e^{-t}x(t) = C_2 e^{-3t}. \\Rightarrow \\frac{d}{dt}(e^{-t}x(t)) = C_2 e^{-3t}. \\]\n\n• Integrate both sides:\n\n\\[ e^{-t}x(t) = \\int C_2 e^{-3t} \\, dt. \\]\n\nThe integral is:\n\n\\[ \\int C_2 e^{-3t} \\, dt = \\frac{C_2}{3} e^{-3t} + C_1, \\]\n\nso there is:\n\n\\[ e^{-t}x(t) = \\frac{C_2}{3} e^{-3t} + C_1. \\]\n\n• Solve for \\( x(t) \\):\n\nMultiply both sides by \\( e^t \\):\n\n\\[ x(t) = -\\frac{C_2}{3} e^{-2t} + C_1 e^t. \\]\n\nThus, the complete solution of the system is:\n\n\\[\n\\begin{align*}\n x(t) &= -\\frac{C_2}{3} e^{-2t} + C_1 e^t, \\\\\n y(t) &= C_2 e^{-2t}.\n\\end{align*}\n\\]\n\nReferences: Çengel, Y. A., and III, W. J. P. (2014). Differential Equations for Engineers and Scientists. ISBN 9780073385907.", "id": "./materials/958.pdf" }, { "contents": "• **Symmetric Matrix**\n\nA symmetric matrix is a square matrix that is equal to its transpose. This means that for a matrix $A$, it holds that\n\n$$A = A^T.$$ \n\nSymmetric matrices have several important properties:\n\n- **Real Eigenvalues**: All eigenvalues of a symmetric matrix are real numbers.\n- **Orthogonal Eigenvectors**: The eigenvectors corresponding to distinct eigenvalues are orthogonal to each other.\n\nMathematically, if $A$ is symmetric, then\n\n$$a_{ij} = a_{ji}$$\n\nfor all elements of the matrix, where $a_{ij}$ represents the element in the $i$-th row and $j$-th column.\n\n• **Trace of a Matrix**\n\nThe trace of a matrix, denoted as $\\text{tr}(A)$, is the sum of the diagonal elements of a square matrix. It has the following properties:\n\n- The trace is also equal to the sum of the eigenvalues of the matrix.\n\nFor a matrix\n\n$$A = \\begin{bmatrix} a_{11} & a_{12} & \\cdots \\\\ a_{21} & a_{22} & \\cdots \\\\ \\vdots & \\vdots & \\ddots \\end{bmatrix},$$\n\nthe trace is given by:\n\n$$\\text{tr}(A) = a_{11} + a_{22} + \\cdots + a_{nn}.$$\nProblems\n\na) Given the matrix\n\n\\[ A = \\begin{bmatrix} 4 & 2 & 0 \\\\ 2 & 3 & 1 \\\\ 0 & 1 & 5 \\end{bmatrix}, \\]\n\nVerify that \\( A \\) is symmetric.\n\nA matrix \\( A \\) is symmetric if \\( A = A^T \\). The transpose of matrix \\( A \\) is:\n\n\\[ A^T = \\begin{bmatrix} 4 & 2 & 0 \\\\ 2 & 3 & 1 \\\\ 0 & 1 & 5 \\end{bmatrix}. \\]\n\nSince \\( A = A^T \\), the matrix \\( A \\) is indeed symmetric.\n\nb) Given the matrix:\n\n\\[ A = \\begin{bmatrix} -3 & 1 \\\\ 1 & -3 \\end{bmatrix} \\]\n\nTo find the trace of the matrix \\( A \\):\n\n\\[ \\text{tr}(A) = -3 + (-3) = -6. \\]\n\nSo, the trace of the matrix \\( A \\) is \\(-6\\).\n\nReference: Nicholson, W. K. (2006). Elementary Linear Algebra. ISBN 85-86804-92-4.", "id": "./materials/959.pdf" }, { "contents": "Every definite integration problem begins with checking if the conditions for the Fundamental Theorem of Calculus are met or not.\n\nEvaluate \\( \\int_{\\sqrt{3}}^{2} \\frac{\\sqrt{4-x^2}}{x} \\, dx \\)\n\nConverting to form \\( R(x, \\sqrt{1-x^2}) \\)\n\nSo,\n\n\\[ I(x) = \\int \\frac{\\sqrt{4-x^2}}{x} \\, dx = \\int \\frac{\\sqrt{(4-x^2)x^2}}{x} \\, dx \\]\n\n\\[ = 2 \\int \\frac{\\sqrt{1-(\\frac{x}{2})^2}}{x} \\, dx \\]\n\nPerforming trigonometric substitution,\n\n\\( \\frac{x}{2} = \\sin(t) \\)\n\n\\( \\Rightarrow x = 2 \\sin(t) \\)\n\n\\( \\Rightarrow dx = 2 \\cos(t) \\, dt \\)\n\n\\[ = 2 \\int \\frac{\\sqrt{\\cos^2(t)}}{\\sin(t)} \\cdot 2 \\cos(t) \\, dt \\]\n\n\\[ = 2 \\int \\frac{\\cos(t)}{\\sin(t)} \\cdot \\cos(t) \\, dt \\]\n\n\\[ = 2 \\int \\frac{\\cos^2(t)}{\\sin(t)} \\, dt \\]\n\n\\[ = 2 \\int \\frac{1-\\sin^2(t)}{\\sin(t)} \\, dt \\]\n\\[ = 2 \\int \\csc(t) \\cdot \\sin(t) \\, dt \\]\n\\[ = 2 \\int \\csc(t) \\, dt - 2 \\int \\sin(t) \\, dt \\]\n\\[ = 2 \\ln |\\csc(t) - \\cot(t)| + 2 \\sin \\cos(t) + C \\]\n\nConverting to function dependent on \\( x \\):\n\n\\[ \\sin(t) = \\frac{x}{2} \\]\n\\[ \\csc(t) = \\frac{2}{x} \\]\n\\[ \\cos(t) = \\sqrt{1 - \\sin^2(t)} = \\sqrt{1 - \\frac{x^2}{4}} = \\frac{\\sqrt{4-x^2}}{2} \\]\n\\[ \\cot(t) = \\frac{\\cos(t)}{\\sin(t)} = \\frac{\\sqrt{4-x^2}}{x} \\]\n\n\\[ = 2 \\ln \\left| \\frac{2}{x} - \\frac{\\sqrt{4-x^2}}{x} \\right| + \\frac{x}{2} \\sqrt{4-x^2} + C \\]\n\\[ = 2 \\ln \\left| \\frac{2 - \\sqrt{4-x^2}}{x} \\right| + \\sqrt{4-x^2} + C \\]\nNow, \\[\n\\int_{\\sqrt{3}}^{2} \\frac{\\sqrt{4-x^2}}{x} \\, dx = \\left[ \\frac{I(x)}{\\sqrt{3}} \\right]^{2}_{\\sqrt{3}}\n\\]\n\\[\n= \\left[ 2 \\ln \\left( \\frac{2 - \\sqrt{4-x^2}}{x} + \\sqrt{4-x^2} \\right) \\right]^{2}_{\\sqrt{3}}\n\\]\n\\[\n= \\left( 2 \\ln \\left( \\frac{2 - \\sqrt{4-4}}{2} + \\sqrt{4-4} \\right) \\right) - \\left( 2 \\ln \\left( \\frac{2 - \\sqrt{4-3}}{\\sqrt{3}} + \\sqrt{4-3} \\right) \\right)\n\\]\n\\[\n= - \\left( 2 \\ln \\frac{1}{\\sqrt{3}} + 1 \\right)\n\\]\n\\[\n= -2 \\ln \\frac{1}{\\sqrt{3}} - 1\n\\]\n\\[\n= -2 \\ln (3) - 1\n\\]\n\\[\n= -2x - \\frac{1}{2} \\ln(3) - 1\n\\]\n\\[\n= \\ln(3) - 1\n\\]", "id": "./materials/96.pdf" }, { "contents": "• **Spectrum**\n\nThe spectrum of a matrix $A$, denoted by $S(A)$, is the set of all eigenvalues of $A$. In other words, it consists of all scalar values $\\lambda$ such that there exists a non-zero vector $v$ (an eigenvector) for which:\n\n$$Av = \\lambda v$$\n\nThe spectrum gives a complete picture of the behavior of the matrix’s action in terms of scaling along the directions of its eigenvectors.\n\n• **Spectral Radius**\n\nThe spectral radius of a matrix $A$, denoted $\\rho(A)$, is the largest absolute value of its eigenvalues. If $S(A)$ is the set of eigenvalues of $A$, then the spectral radius is:\n\n$$\\rho(A) = \\max\\{|\\lambda| : \\lambda \\in S(A)\\}$$\n\nIn other words, it gives the magnitude of the \"dominant\" eigenvalue, which has the greatest influence on the long-term behavior of the matrix’s powers.\n\n• **Spectral Shift**\n\nA spectral shift refers to modifying the eigenvalues of a matrix by adding a scalar multiple of the identity matrix $I$ to the original matrix $A$. If $A$ has eigenvalues $\\lambda_1, \\lambda_2, \\ldots, \\lambda_n$, then adding $cI$ to $A$ results in a new matrix $A + cI$, whose eigenvalues are shifted by $c$:\n\n$$\\text{Eigenvalues of } A + cI = \\{\\lambda_1 + c, \\lambda_2 + c, \\ldots, \\lambda_n + c\\}$$\n\nThis spectral shift preserves the relative spacing between the eigenvalues but translates them by a fixed amount.\nProblems\n\na) The set of eigenvalues of the matrix \\( A = \\begin{bmatrix} -3 & 1 \\\\ 1 & -3 \\end{bmatrix} \\) is?\n\nIs necessary find the eigenvalues, solving the characteristic equation:\n\n\\[\n\\det(A - \\lambda I) = 0\n\\]\n\nWhere \\( I \\) is the identity matrix, and \\( \\lambda \\) represents the eigenvalues. The characteristic equation becomes:\n\n\\[\n\\det \\begin{bmatrix} -3 - \\lambda & 1 \\\\ 1 & -3 - \\lambda \\end{bmatrix} = 0\n\\]\n\nThe determinant is:\n\n\\[\n(-3 - \\lambda)(-3 - \\lambda) - (1)(1) = \\lambda^2 + 6\\lambda + 9 - 1 = \\lambda^2 + 6\\lambda + 8 = 0\n\\]\n\nSolve the quadratic equation:\n\n\\[\n\\lambda^2 + 6\\lambda + 8 = 0\n\\]\n\nThus, the eigenvalues are:\n\n\\[\n\\lambda_1 = -2, \\quad \\lambda_2 = -4\n\\]\n\nThe set of eigenvalues is \\( \\{-2, -4\\} \\).\n\nb) The spectral radius of the matrix \\( A = \\begin{bmatrix} -3 & 1 \\\\ 1 & -3 \\end{bmatrix} \\) is?\n\nThe absolute values of the eigenvalues are, according to a):\n\n\\[\n|\\lambda_1| = 2, \\quad |\\lambda_2| = 4\n\\]\n\nThe spectral radius is the largest of these values:\n\n\\[\n\\rho(A) = 4\n\\]\nc) Consider the set \\{-2, 2, 5\\} of eigenvalues of the real $3 \\times 3$ matrix $A + 3I$, where $I$ denotes the identity matrix. Find the eigenvalues of $A$.\n\nThe eigenvalues of $A + 3I$ are related to the eigenvalues of $A$ by a shift of $+3$. In other words:\n\n$$\\text{Eigenvalues of } A + 3I = \\{\\lambda_1 + 3, \\lambda_2 + 3, \\lambda_3 + 3\\}$$\n\nAre given that the eigenvalues of $A + 3I$ are \\{-2, 2, 5\\}. Therefore, to find the eigenvalues of $A$, it is necessary to subtract 3 from each of these values:\n\n$$\\text{Eigenvalues of } A = \\{-2 - 3, 2 - 3, 5 - 3\\} = \\{-5, -1, 2\\}$$\n\nThus, the eigenvalues of $A$ are \\{-5, -1, 2\\}.\n\nReference: Nicholson, W. K. (2006). Elementary Linear Algebra. ISBN 85-86804-92-4.", "id": "./materials/960.pdf" }, { "contents": "• **Inverse Matrix**\n\nAn inverse matrix $A^{-1}$ of a square matrix $A$ is a matrix that, when multiplied by $A$, gives the identity matrix $I$:\n\n$$AA^{-1} = A^{-1}A = I$$\n\nFor a matrix to have an inverse, it must be non-singular (i.e., its determinant must not be zero). The eigenvalues of the inverse matrix $A^{-1}$ are the reciprocals of the eigenvalues of $A$.\n\nFor example, if $\\lambda$ is an eigenvalue of $A$, then $\\frac{1}{\\lambda}$ is an eigenvalue of $A^{-1}$.\n\n• **Orthonormal Matrix**\n\nAn orthonormal matrix (or orthogonal matrix) $Q$ is a square matrix where the rows (and columns) are orthonormal vectors, meaning:\n\n$$QQ^T = Q^TQ = I$$\n\nThis implies that the inverse of an orthonormal matrix is equal to its transpose $Q^{-1} = Q^T$. The eigenvalues of an orthonormal matrix have a modulus (absolute value) of 1, meaning they lie on the unit circle in the complex plane.\n\n• **Power Matrix**\n\nThe power of a matrix $A^n$ refers to the matrix multiplied by itself $n$ times. The eigenvalues of the matrix power $A^n$ are the eigenvalues of $A$ raised to the $n$-th power. For example, if $\\lambda$ is an eigenvalue of $A$, then $\\lambda^n$ is an eigenvalue of $A^n$. \nProblems\n\na) The eigenvalues of the matrix $A$ are $\\{1, 3\\}$. Find its $A^{-1}$\n\nTo find the eigenvalues of $A^{-1}$, take the reciprocal of each eigenvalue of $A$:\n\nFor $\\lambda_1 = 1$, \\[ \\frac{1}{\\lambda_1} = \\frac{1}{1} = 1 \\]\n\nFor $\\lambda_2 = 3$, \\[ \\frac{1}{\\lambda_2} = \\frac{1}{3} \\]\n\nThus, the eigenvalues of $A^{-1}$ are:\n\n$\\{1, \\frac{1}{3}\\}$\n\nb) The eigenvalues of the matrix $A$ are $\\{1, 3\\}$. Find its $A^2$\n\nTo find the eigenvalues of $A^2$, square each eigenvalue of $A$:\n\nFor $\\lambda_1 = 1$, \\[ \\lambda_1^2 = 1^2 = 1 \\]\n\nFor $\\lambda_2 = 3$, \\[ \\lambda_2^2 = 3^2 = 9 \\]\n\nThus, the eigenvalues of $A^2$ are:\n\n$\\{1, 9\\}$\n\nReference: Nicholson, W. K. (2006). Elementary Linear Algebra. ISBN 85-86804-92-4.", "id": "./materials/961.pdf" }, { "contents": "• Eigenvalues and Eigenvectors\n\nGiven a square matrix $A$, an eigenvector of $A$ is a non-zero vector $v$ such that when $A$ acts on $v$, the resulting vector is a scalar multiple of $v$. Mathematically, this can be expressed as:\n\n$$Av = \\lambda v$$\n\nWhere:\n- $A$ is a square matrix (i.e., $n \\times n$).\n- $v$ is a non-zero vector called an eigenvector.\n- $\\lambda$ is a scalar value called an eigenvalue.\n\nThe equation $Av = \\lambda v$ means that applying the matrix $A$ to $v$ simply stretches or compresses $v$ by the factor $\\lambda$, but does not change its direction.\n\nEigenvalues and eigenvectors are essential because they simplify matrix operations, especially when diagonalizing matrices or solving systems of differential equations. Eigenvectors provide insight into the geometric transformations represented by a matrix, while eigenvalues indicate how much vectors are stretched or compressed in the direction of eigenvectors.\n\nTo find the eigenvalues and eigenvectors of a matrix $A$, we follow these steps:\n\n1. Solve the Characteristic Equation\n\n The eigenvalues of a matrix are obtained by solving the characteristic equation:\n\n $$\\det(A - \\lambda I) = 0$$\n\n Here:\n - $I$ is the identity matrix.\n - $\\lambda$ is the scalar (the eigenvalue we are solving for).\n - $\\det(A - \\lambda I)$ is the determinant of the matrix $(A - \\lambda I)$.\n\n This equation gives us a polynomial (called the characteristic polynomial) in $\\lambda$, and the roots of this polynomial are the eigenvalues of $A$. \n2. Find the Eigenvectors\n\nOnce we have the eigenvalues $\\lambda_1, \\lambda_2, \\ldots, \\lambda_n$, we can find the corresponding eigenvectors by substituting each eigenvalue $\\lambda_i$ into the equation:\n\n$$(A - \\lambda I)v = 0$$\n\nand solving for $v$.\n\nThis system of equations typically has infinitely many solutions, so we find a basis for the set of solutions, which will give us the eigenvectors.\n\n**Geometric Interpretation of Eigenvalues and Eigenvectors**\n\nEigenvalues represent directions in which a transformation matrix $A$ acts by only stretching or compressing (not rotating or reflecting). Eigenvalues represent the scaling factor by which the eigenvector is stretched or compressed when acted upon by the matrix $A$.\n\nFor example, if $\\lambda = 2$, the matrix $A$ stretches the eigenvector by a factor of 2. If $\\lambda = -1$, the matrix $A$ reflects the eigenvector and scales it by 1.\n\n- **Eigenspaces**\n\n The eigenspace corresponding to an eigenvalue $\\lambda$ is the set of all eigenvectors associated with that eigenvalue, along with the zero vector. Formally, the eigenspace of $\\lambda$ is the null space of the matrix $(A - \\lambda I)$.\n\n $$\\text{Eigenspace of } \\lambda = \\{v \\mid (A - \\lambda I)v = 0\\}$$\n\n The dimension of the eigenspace corresponding to an eigenvalue $\\lambda$ is called the *algebraic multiplicity* of $\\lambda$. This dimension tells how many linearly independent eigenvectors are associated with $\\lambda$.\n\n Example: If $\\lambda = 0$ is an eigenvalue of $A$, the eigenspace corresponds to the null space of $A$, meaning it includes all vectors $v$ that satisfy $Av = 0$. This eigenspace is the set of all solutions to the homogeneous system $Av = 0$. \n\nProblems\n\na) The set of eigenvalues of the $3 \\times 3$ real matrix $A$ is $\\{0, 1, 2\\}$. What is the rank of $A$?\n\n1. Understanding the Eigenvalues and Eigenspaces:\n\n The eigenvalues of a matrix are the values $\\lambda$ for which the equation $Av = \\lambda v$ holds for some non-zero vector $v$, which is called an eigenvector. For each eigenvalue, the set of all eigenvectors corresponding to that eigenvalue, along with the zero vector, forms the eigenspace of $\\lambda$.\n\n **Eigenvalue 0:**\n - If $\\lambda = 0$ is an eigenvalue, it means there is a non-trivial null space. The null space is the set of vectors $v$ such that $Av = 0$. The dimension of this null space is called the nullity of the matrix.\n - The eigenspace corresponding to $\\lambda = 0$ is the null space of $A$.\n - Since $\\lambda = 0$ is an eigenvalue, the nullity of $A$ is at least 1.\n\n **Eigenvalues 1 and 2:**\n - For $\\lambda = 1$ and $\\lambda = 2$, each has a corresponding eigenspace, which contains the eigenvectors associated with these eigenvalues.\n\n2. Relationship Between Rank and Nullity\n\n The rank-nullity theorem states that for any $n \\times n$ matrix $A$:\n\n $$\\text{rank}(A) + \\text{nullity}(A) = n$$\n\n Here, $n = 3$, since $A$ is a $3 \\times 3$ matrix. So:\n\n $$\\text{rank}(A) + \\text{nullity}(A) = 3$$\n\n3. Determine the Rank of $A$ Using the rank-nullity theorem:\n\n - The matrix $A$ has three eigenvalues: 0, 1, 2. The eigenspace corresponding to $\\lambda = 0$ gives us the nullity, and the eigenspaces for $\\lambda = 1$ and $\\lambda = 2$ contribute to the rank.\n - If the nullity of $A$ is 1, then the rank of $A$ must be:\n\n $$\\text{rank}(A) = 3 - \\text{nullity}(A) = 3 - 1 = 2$$\n\n The rank of the matrix $A$ is 2.\n\nReference: Nicholson, W. K. (2006). Elementary Linear Algebra. ISBN 85-86804-92-4.", "id": "./materials/962.pdf" }, { "contents": "• **Diagonalization**\n\nDiagonalization is a process that simplifies the study of matrices by transforming a given matrix into a diagonal form, provided it meets certain conditions. The goal is to find a diagonal matrix $D$ that is similar to a given matrix $A$, meaning there is an invertible matrix $P$ such that:\n\n$$A = PDP^{-1}$$\n\nHere, $D$ is a diagonal matrix whose diagonal entries are the eigenvalues of $A$, and the columns of $P$ are the eigenvectors of $A$.\n\nConditions for diagonalization:\n\n1. A matrix $A$ can be diagonalized if and only if it has enough linearly independent eigenvectors to form the matrix $P$. Specifically, an $n \\times n$ matrix can be diagonalized if it has $n$ linearly independent eigenvectors.\n\n2. If $A$ has distinct eigenvalues, it is guaranteed to be diagonalizable.\n\nThe process of diagonalization allows to transform complex matrix operations into simpler ones. For example, powers of a diagonalizable matrix can be computed easily by diagonalizing it:\n\n$$A^k = PD^kP^{-1}$$\n\nwhere $D^k$ is easy to compute because it just involves raising the diagonal entries to the power $k$.\n\n• **Diagonalization with Orthonormal Basis**\n\nDiagonalization with an orthonormal basis (DOB) involves diagonalizing a matrix using an orthonormal set of eigenvectors. When this is possible, the matrix is orthogonally diagonalizable, meaning it can be written as:\n\n$$A = Q\\Lambda Q^T$$\n\nwhere: $Q$ is an orthogonal matrix (its columns are orthonormal eigenvectors, and $Q^{-1} = Q^T$); $\\Lambda$ is a diagonal matrix of eigenvalues.\n\nThis special diagonalization is possible when the matrix $A$ is symmetric ($n \\times n$). Symmetric matrices have real eigenvalues, and their eigenvectors can always be chosen to be orthonormal.\nOrthogonal diagonalization has several advantages, especially in numerical computation, because orthogonal matrices preserve lengths and angles, making the calculations more stable.\n\n- **Similar Matrices**\n\n Similar matrices are matrices that represent the same linear transformation under different bases. Two matrices $A$ and $B$ are similar if there exists an invertible matrix $P$ such that:\n\n $$B = P^{-1}AP$$\n\n Similar matrices have the same eigenvalues and the same characteristic polynomial, but their eigenvectors differ (since they are expressed in different bases). In practical terms, similarity means that $A$ and $B$ describe the same transformation, but in different coordinate systems.\nProblems\n\na) Let $-1, 1$ and $2$ be the eigenvalues of a real $3 \\times 3$ matrix, $A$. Find the eigenvalues of $A - 2I$, where $I$ denotes the identity matrix.\n\nAccording to the property of matrices and eigenvalues:\nSuppose $\\lambda$ is an eigenvalue of a matrix $A$, with eigenvector $v$ such that:\n\n$$Av = \\lambda v$$\n\nNow, consider the matrix $A - 2I$. We want to find its eigenvalues. Start by applying $A - 2I$ to the eigenvector $v$:\n\n$$(A - 2I)v = Av - 2Iv$$\n\nSince $Iv = v$, this simplifies to:\n\n$$(A - 2I)v = Av - 2v = \\lambda v - 2v \\Rightarrow (A - 2I)v = (\\lambda - 2)v$$\n\nThus, if $\\lambda$ is an eigenvalue of $A$, then $\\lambda - 2$ is an eigenvalue of $A - 2I$. Is possible only to subtract 2 from each eigenvalue of $A$ to find the eigenvalues of $A - 2I$.\n\nIn the given problem, the eigenvalues of $A$ are $\\{-1, 1, 2\\}$, so the eigenvalues of $A - 2I$ are:\n\n$$\\{-1 - 2, 1 - 2, 2 - 2\\} = \\{-3, -1, 0\\}$$\n\nThus, the eigenvalues of $A - 2I$ are $\\{-3, -1, 0\\}$.\n\nb) Find a Similar Matrix to $A = \\begin{bmatrix} 1 & 1 \\\\ -2 & 4 \\end{bmatrix}$\n\nTo find a matrix similar to $A$, is necessary to find a matrix that has the same eigenvalues but is diagonal.\n\nFirst, calculate the characteristic polynomial:\n\n$$\\det(A - \\lambda I) = \\det \\begin{bmatrix} 1 - \\lambda & 1 \\\\ -2 & 4 - \\lambda \\end{bmatrix}$$\n\n$$= (1 - \\lambda)(4 - \\lambda) - (-2)(1) = \\lambda^2 - 5\\lambda + 6$$\n\nSolving this quadratic equation gives eigenvalues $\\lambda_1 = 2$ and $\\lambda_2 = 3$.\n\nThus, the matrix $A$ is similar to the diagonal matrix:\n\n$$\\begin{bmatrix} 2 & 0 \\\\ 0 & 3 \\end{bmatrix}$$\n\nReference: Nicholson, W. K. (2006). Elementary Linear Algebra. ISBN 85-86804-92-4.", "id": "./materials/963.pdf" }, { "contents": "• **Characteristic Polynomial**\n\nThe characteristic polynomial of a square matrix $A$ is a polynomial that is essential in determining the eigenvalues of the matrix. For an $n \\times n$ matrix $A$, it is defined as the determinant of the matrix $A - \\lambda I$, where $\\lambda$ is a scalar (an eigenvalue) and $I$ is the identity matrix of the same dimensions as $A$. Mathematically, this can be written as:\n\n$$p(\\lambda) = \\det(A - \\lambda I)$$\n\nThe characteristic equation is formed by setting the characteristic polynomial equal to zero, i.e.,\n\n$$\\det(A - \\lambda I) = 0$$\n\nSolving this equation gives the eigenvalues of the matrix $A$.\n\n• **Determinant**\n\nThe determinant of a square matrix $A$, denoted $\\det(A)$, is a scalar value that provides important information about the matrix. For instance:\n\n- If $\\det(A) = 0$, the matrix is singular and has no inverse.\n- The determinant of a matrix is also the product of its eigenvalues. For an $n \\times n$ matrix with eigenvalues $\\lambda_1, \\lambda_2, \\ldots, \\lambda_n$, the determinant can be expressed as:\n\n$$\\det(A) = \\lambda_1 \\cdot \\lambda_2 \\cdot \\cdots \\cdot \\lambda_n$$\nProblems\n\na) Calculate the characteristic polynomial of the matrix:\n\n\\[ A = \\begin{bmatrix} 1 & 2 & -1 \\\\ 1 & 0 & 1 \\\\ 4 & -4 & 5 \\end{bmatrix} \\]\n\n1. Subtract \\( \\lambda I \\) from the matrix \\( A \\)\n\nThe identity matrix \\( I \\) is:\n\n\\[ I = \\begin{bmatrix} 1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 1 \\end{bmatrix} \\]\n\nNow subtract \\( \\lambda I \\) from matrix \\( A \\):\n\n\\[ A - \\lambda I = \\begin{bmatrix} 1 & 2 & -1 \\\\ 1 & 0 & 1 \\\\ 4 & -4 & 5 \\end{bmatrix} - \\begin{bmatrix} \\lambda & 0 & 0 \\\\ 0 & \\lambda & 0 \\\\ 0 & 0 & \\lambda \\end{bmatrix} = \\begin{bmatrix} 1 - \\lambda & 2 & -1 \\\\ 1 & -\\lambda & 1 \\\\ 4 & -4 & 5 - \\lambda \\end{bmatrix} \\]\n\n2. Find the determinant of \\( A - \\lambda I \\)\n\nNow, compute the determinant of the matrix \\( A - \\lambda I \\):\n\n\\[ \\det(A - \\lambda I) = \\det \\begin{bmatrix} 1 - \\lambda & 2 & -1 \\\\ 1 & -\\lambda & 1 \\\\ 4 & -4 & 5 - \\lambda \\end{bmatrix} \\]\n\nUse cofactor expansion along the first row. The formula for determinant expansion is:\n\n\\[ \\det(A - \\lambda I) = (1 - \\lambda) \\det \\begin{bmatrix} -\\lambda & 1 \\\\ -4 & 5 - \\lambda \\end{bmatrix} - 2 \\det \\begin{bmatrix} 1 & 1 \\\\ 4 & 5 - \\lambda \\end{bmatrix} + (-1) \\det \\begin{bmatrix} 1 & -\\lambda \\\\ 4 & -4 \\end{bmatrix} \\]\n\n3. Compute the 2×2 determinants\n\n- Compute the first 2×2 determinant:\n\n\\[ \\det \\begin{bmatrix} -\\lambda & 1 \\\\ -4 & 5 - \\lambda \\end{bmatrix} = (-\\lambda)(5 - \\lambda) - (1)(-4) = -5\\lambda + \\lambda^2 + 4 = \\lambda^2 - 5\\lambda + 4 \\]\n\n- Compute the second 2×2 determinant:\n\n\\[ \\det \\begin{bmatrix} 1 & 1 \\\\ 4 & 5 - \\lambda \\end{bmatrix} = (1)(5 - \\lambda) - (1)(4) = 5 - \\lambda - 4 = 1 - \\lambda \\]\n• Compute the third $2 \\times 2$ determinant:\n\n$$\\det \\begin{bmatrix} 1 & -\\lambda \\\\ 4 & -4 \\end{bmatrix} = (1)(-4) - (-\\lambda)(4) = -4 + 4\\lambda = 4\\lambda - 4$$\n\n4. Put it all together\n\nSubstitute the $2 \\times 2$ determinants into the cofactor expansion:\n\n$$\\det(A - \\lambda I) = (1 - \\lambda)(\\lambda^2 - 5\\lambda + 4) - 2(1 - \\lambda) + (-1)(4\\lambda - 4)$$\n\n5. Simplify the expression\n\nFirst, expand $(1 - \\lambda)(\\lambda^2 - 5\\lambda + 4)$:\n\n$$(1 - \\lambda)(\\lambda^2 - 5\\lambda + 4) = \\lambda^2 - 5\\lambda + 4 - \\lambda^3 + 5\\lambda^2 - 4\\lambda = -\\lambda^3 + 6\\lambda^2 - 9\\lambda + 4$$\n\nNow simplify the rest:\n\n$$-2(1 - \\lambda) = -2 + 2\\lambda$$\n$$-(4\\lambda - 4) = -4\\lambda + 4$$\n\nFinally, combine everything:\n\n$$\\det(A - \\lambda I) = (-\\lambda^3 + 6\\lambda^2 - 9\\lambda + 4) + (2\\lambda - 2) + (-4\\lambda + 4)$$\n\n$$= -\\lambda^3 + 6\\lambda^2 - 11\\lambda + 6$$\n\nThe characteristic polynomial is:\n\n$$p(\\lambda) = \\lambda^3 - 6\\lambda^2 + 11\\lambda - 6$$\n\nb) A real matrix $A$ ($2 \\times 2$) whose set of eigenvalues is $\\{-2, 3\\}$ has which determinant?\n\nThe determinant of a matrix is the product of its eigenvalues. Given the eigenvalues $\\{-2, 3\\}$:\n\n$$\\det(A) = (-2) \\times 3 = -6$$\n\nThus, the determinant of matrix $A$ is $-6$. \n\n3\nc) The determinant of a real matrix $A$ of dimension $3 \\times 3$, whose eigenvalues are given by the set $\\{0, 2, 3\\}$, is?\n\nAgain, the determinant is the product of the eigenvalues. Here, the eigenvalues are $\\{0, 2, 3\\}$:\n\n$$\\det(A) = 0 \\times 2 \\times 3 = 0$$\n\nSince one of the eigenvalues is zero, the determinant of matrix $A$ is 0. This indicates that $A$ is a singular matrix and has no inverse.\n\nReference: Nicholson, W. K. (2006). Elementary Linear Algebra. ISBN 85-86804-92-4.", "id": "./materials/964.pdf" }, { "contents": "• **Independent Systems**\n\nA system of equations is said to be independent if all the equations describe distinct constraints on the variables. Such a system typically has a unique solution, where no equation can be derived as a linear combination of the others.\n\nThis means:\n\n1. **Unique Constraints:** Each equation in an independent system represents a distinct constraint on the variables. None of the equations can be derived from a combination of the others.\n\n2. **Unique Solution:** Because each equation is uniquely contributing to the constraints, an independent system typically has a single solution that satisfies all equations simultaneously. This single solution represents the intersection point (or points, in higher dimensions) where all equations meet.\n\n3. **Example of an Independent System:** Consider a simple system:\n\n\\[\n\\begin{align*}\n x + y &= 5 \\\\\n 2x - y &= 1\n\\end{align*}\n\\]\n\nHere, both equations contribute unique constraints, and solving them will lead to one specific solution, making it an independent system.\n\n• **Dependent System**\n\nA system is considered dependent when one or more equations are linearly dependent on others. This means that one equation can be written as a linear combination of the others, leading to an infinite number of solutions (if the system is consistent) or no solution (if inconsistent).\n\nThis means:\n\n1. **Redundant Information:** In a dependent system, one or more equations can be formed by a linear combination of the others. This redundancy means that not all equations contribute unique constraints.\n\n2. **Infinite Solutions or No Solutions:**\n - If the system is consistent (the equations do not contradict each other), a dependent system will have infinitely many solutions, as there are “extra” degrees of freedom due to the dependent equations.\n - If the system is inconsistent (e.g., parallel but non-intersecting planes in three dimensions), it will have no solution.\n3. Example of a Dependent System:\n\n\\[\n\\begin{align*}\n x + y &= 2 \\\\\n 2x + 2y &= 4\n\\end{align*}\n\\]\n\nIn this case, the second equation is simply twice the first, so both equations represent the same line. Since they don’t offer unique constraints, this system is dependent and has infinitely many solutions along that line.\n\n**Problems**\n\na) Solve the system:\n\n\\[\n\\begin{align*}\n x + y + z &= 5 \\\\\n 5x + 3y + 2z &= 0 \\\\\n y - z &= 0\n\\end{align*}\n\\]\n\n1. Simplify the System\n\nFrom the third equation, it is possible to express \\( z \\) in terms of \\( y \\):\n\n\\[ y - z = 0 \\Rightarrow z = y. \\]\n\nSubstitute \\( z = y \\) into the first and second equations:\n\n- \\( x + y + y = 5 \\Rightarrow x + 2y = 5 \\).\n- \\( 5x + 3y + 2y = 0 \\Rightarrow 5x + 5y = 0 \\).\n\nNow, there is the following simplified system:\n\n\\[\n\\begin{align*}\n x + 2y &= 5 \\\\\n 5x + 5y &= 0\n\\end{align*}\n\\]\n\n2. Solve for \\( x \\) and \\( y \\)\n\nDivide the second equation by 5 to simplify:\n\n\\[ x + y = 0. \\]\n\nNow there are:\n\n\\[\n\\begin{align*}\n x + 2y &= 5 \\\\\n x + y &= 0\n\\end{align*}\n\\]\nSubtract the second equation from the first equation to eliminate $x$:\n\n$$(x + 2y) - (x + y) = 5 - 0 \\Rightarrow y = 5.$$ \n\nNow that is known that $y = 5$, substitute this value into the second equation:\n\n$$x + 5 = 0 \\Rightarrow x = -5.$$ \n\n3. Find $z$ using $z = y$\n\nSince $z = y$:\n\n$$z = 5.$$ \n\nThe solution to the system is:\n\n$$(x, y, z) = (-5, 5, 5).$$ \n\nThis means the equations in this system intersect at exactly one point in 3-dimensional space, so they are linearly independent and have a unique solution.\n\nb) Solve this system of linear equations:\n\n$$\\begin{align*}\n x_1 - x_2 - 3x_3 + 4x_4 &= 1 \\\\\n x_1 + x_2 + x_3 + 2x_4 &= -1 \\\\\n -x_2 - 2x_3 + x_4 &= 1 \\\\\n x_1 + 2x_2 + 3x_3 + x_4 &= -2\n\\end{align*}$$\n\n1. Eliminate $x_1$ from Equations 2 and 4. Starting with the first equation, use it to eliminate $x_1$ from the other equations.\n\nFrom Equation 1:\n\n$$x_1 = x_2 + 3x_3 - 4x_4 + 1$$\n\nSubstitute into Equation 2:\n\n$$(x_2 + 3x_3 - 4x_4 + 1) + x_2 + x_3 + 2x_4 = -1$$\n\nCombine like terms:\n\n$$2x_2 + 4x_3 - 2x_4 = -2$$\n\nSimplify:\n\n$$x_2 + 2x_3 - x_4 = -1 \\quad (\\text{Equation 5})$$\nSubstitute into Equation 4:\n\n\\[(x_2 + 3x_3 - 4x_4 + 1) + 2x_2 + 3x_3 + x_4 = -2\\]\n\nCombine like terms:\n\n\\[3x_2 + 6x_3 - 3x_4 = -3\\]\n\nSimplify:\n\n\\[x_2 + 2x_3 - x_4 = -1 \\quad \\text{(Equation 6)}\\]\n\n2. Simplify the System. Now there is a simplified system with three equations:\n\n\\[\n\\begin{align*}\n x_2 + 2x_3 - x_4 &= -1 \\quad \\text{(from Equation 5)} \\\\\n -x_2 - 2x_3 + x_4 &= 1 \\quad \\text{(from Equation 3)} \\\\\n x_2 + 2x_3 - x_4 &= -1 \\quad \\text{(from Equation 6)}\n\\end{align*}\n\\]\n\nNotice that Equations 5 and 6 are identical, so was left with just two unique equations:\n\n\\[\n\\begin{align*}\n x_2 + 2x_3 - x_4 &= -1 \\\\\n -x_2 - 2x_3 + x_4 &= 1\n\\end{align*}\n\\]\n\n3. Add the Equations to Eliminate \\(x_2\\) and \\(x_4\\).\n\nAdd these two equations:\n\n\\[(x_2 + 2x_3 - x_4) + (-x_2 - 2x_3 + x_4) = -1 + 1\\]\n\nThis simplifies to:\n\n\\[0 = 0\\]\n\nThis identity suggests that the equations are dependent, meaning there are infinitely many solutions. To find the general solution, is possible to express some variables in terms of others.\n\n4. Express \\(x_2\\) in Terms of \\(x_3\\) and \\(x_4\\).\n\nFrom Equation 5:\n\n\\[x_2 + 2x_3 - x_4 = -1\\]\n\nSolving for \\(x_2\\):\n\n\\[x_2 = -1 - 2x_3 + x_4\\]\n5. Express $x_1$ in Terms of $x_3$ and $x_4$.\n\nNow, using the expression for $x_1$ from Step 1:\n\n$$x_1 = x_2 + 3x_3 - 4x_4 + 1$$\n\nSubstitute $x_2 = -1 - 2x_3 + x_4$:\n\n$$x_1 = (-1 - 2x_3 + x_4) + 3x_3 - 4x_4 + 1$$\n\nSimplify:\n\n$$x_1 = x_3 - 3x_4$$\n\n6. The solution in terms of free variables $x_3$ and $x_4$ is:\n\n$$\\begin{cases} \n x_1 = x_3 - 3x_4 \\\\\n x_2 = -1 - 2x_3 + x_4 \\\\\n x_3 = x_3 \\\\\n x_4 = x_4 \n\\end{cases}$$\n\nThe system is dependent since has many solutions depending on two parameters:\n\n$$(x_3 - 3x_4, -1 - 2x_3 + x_4, x_3, x_4), \\forall x_3, x_4 \\in \\mathbb{R}$$\n\nReference: Nicholson, W. K. (2006). Elementary Linear Algebra. ISBN 85-86804-92-4.", "id": "./materials/965.pdf" }, { "contents": "Evaluate \\( \\int_{0}^{1} \\frac{1}{(x^2 + 1)^2} \\, dx \\)\n\n**HINT:**\n\nUse integration by trigonometric substitution:\n\nUse \\( x = \\tan t \\)\n\n**NOTE**\n\nRemember that, when \\( x = \\tan t \\) then \\( t = \\arctan x \\).", "id": "./materials/97.pdf" }, { "contents": "Grooved razor clam (*Solen marginatus*) in the mid-Atlantic Azores: unravelling ecology, phylogeny, and population biology of a new population\n\nAlberto Machado¹, Daniel Machado¹, Rita Castilho²,³,⁴, Luana S. Corona²,³, Gisela Dionísio⁵,⁶,⁷ and José Nuno Gomes-Pereira⁵,⁶,⁸\n\n¹GEO – Grupo de Estudos Oceânicos, Portimão, Portugal; ²Campus de Gambelas, Universidade do Algarve, Faro, Portugal; ³Campus de Gambelas, Centre of Marine Sciences (CCMAR/CIMAR LA), Universidade do Algarve, Faro, Portugal; ⁴Pattern Institute, Faro, Portugal; ⁵Naturalist – Science and Tourism, MARE-Startup, Horta, Portugal; ⁶AtlanticNaturalist, Monsenhor Silveira de Medeiros, Horta, Portugal; ⁷Laboratório Marítimo da Guia, Faculdade de Ciências, MARE, Marine and Environmental Sciences Centre/ARNET, Aquatic Research Network, Universidade de Lisboa, Cascais, Portugal and ⁸Institute of Marine Sciences – OKEANOS, University of the Azores, Horta, Portugal\n\nAbstract\n\nThe first insights on habitat and phylogenetic origin of a newly found population of *Solen marginatus* are provided in the mid-North Atlantic Azores archipelago, in the bay of Praia da Vitória, Terceira Island. Distribution is confined to the northern portion and most sheltered part of the bay down to 14.2 m depth. Densities with an average of 12.69 individuals/m² were found at 8.4 m depth, using 4 × 20 sqm visual transects. Sizes of shell length between 10 and 12 cm comprised 60% of collected specimens (*n* = 118), ranging between 8.79 and 15.4 cm and averaging 11.28 cm. Considering shell length, the high densities and dispersion area, a settlement period above 20 years is estimated. Greater genetic affinity was found in the Ria de Aveiro (North of Portugal) and the Asturias populations (North of Spain). The source origin remains undetermined, with intentional or non-intentional anthropogenic introduction, as well as natural dispersion remaining possible, although more unlikely. Due to the commercial value of this species, a new clam fishery is likely to develop in the area, requiring further studies and immediate conservation measures.\n\nIntroduction\n\nRazor clams are ecologically and economically important marine bivalves (Saeedi and Costello, 2019a), inhabiting sand and muddy bottoms within the lower intertidal and subtidal areas. Razor clams have been part of the pre-glaciation malacoфаuna of the mid-North Atlantic nine-island Azores archipelago (36°–40°N, 24°–31°W). *Ensis minor* (Chenu, 1843) was present on Pleistocene deposits in Santa Maria Island (Ávila et al., 2002), although, possibly due to a severe drop in sea surface temperature (Ávila et al., 2008), disappeared from the archipelago with other littoral bivalves living in fine sand. *Solen marginatus* Pulteney, 1799 was referenced for the shores of São Miguel by Drouet (1861: 47). *Solen vagina* Linnaeus, 1758 was also indicated to the Azores (0 to ~18 m) by Jeffrey (1881: 929), but it has been considered a dubious record (Ávila et al., 2000). After the submission of this manuscript, Álvaro et al. (2024) reported the presence of *S. marginatus* from beach observations at one site in Praia da Vitória, Terceira Island (Figure 1). Based on 11 individuals and opportunistic observations down to 2.5 m, the species was validated through molecular tools, advocating the existence of a new population (Álvaro et al., 2024), however, no phylogenetic analysis or information on species density and dispersion were provided.\n\nAs part of the ecological studies undertaken onboard ‘OCEANUS II’ around the Archipelago of the Azores, *S. marginatus* was identified on the 17 August 2023 in several locations in the bay of Praia da Vitória, Terceira Island. The species was verified by classic and molecular taxonomy using cytochrome oxidase I (COI) mitochondrial gene to improve the accuracy of species-level identification and allow to perform intraspecies diversity analysis in an attempt to understand level genetic divergence and population source (Hebert et al., 2003; Fernández-Tajes and Méndez, 2007; Cunha et al., 2008; Leray and Knowlton, 2015). In support of conservation measures, the species’ habitat range and population biology were further addressed for the bay of Praia da Vitória, by analysing dispersion and densities through visual surveys and size ranges, providing the first ecological insights on habitat and phylogeny of the new population of grooved razor clam in the Azores.\n\nMaterial and methods\n\nRazor clams were observed and collected on several locations in the north part of the bay of Praia da Vitória, Terceira Island on the 17 August 2023. Samples were kept in 96% ethanol for molecular studies and −20°C for subsequent analysis at the Atlantic Naturalist Collection in...\nHorta, Faial Island, Azores. Identification was conducted by morphology-based taxonomy and molecular analysis. For the molecular identification 20 razor clam individuals from Praia da Vitória, Terceira Island, together with an additional sample of five specimens obtained in Faro (south Portugal) were used to establish a control group. A fragment of the foot was dissected and preserved in 96% ethanol until DNA processing. DNA extractions were conducted on 10–20 mg of foot tissue using the DNeasy Plant Mini Kit (Qiagen, Hilden, Germany). The COI amplification of the ‘Folmer’ barcode region at the 5’ start of the cytochrome c oxidase 1 gene (COI), was done with LCO1490 (5’-GGTCAACAAATCATAAAGATATTG-3’) and HCO2198 (5’-TAAACTTCAGGGTGACCAAAAAATCA-3’) universal primers (Folmer et al., 1994). Amplifications consisted of 25 μl mix containing 1 μl (10–100 ng) of genomic DNA, 5 mM GoTaq buffer (5x), 0.2 mM dNTP (Promega, Madison, WI, USA), 1.5 mM MgCl₂, 0.2 mM of each primer, and 1 U of GoTaq DNA polymerase (Promega). The PCR cycling profile for COI amplification included an initial denaturation step at 94°C for 3 min, followed by 30 cycles of denaturation at 94°C for 30 s, annealing at 42°C for 30 s, extension at 72°C for 40 s, and a final extension at 72°C for 5 min. The PCR products were purified through ethanol/sodium acetate precipitation Field (Green and Sambrook, 2016) and subsequently sequenced using the corresponding PCR primers. Sequencing was performed on an Applied Biosystems 3130xl Genetic Analyser, utilising Sanger technology and the BigDye® Terminator v3.1 kit. The generated COI gene sequences were assembled and trimmed using the Geneious Prime (version 2020.0.3, Biomatters, New Zealand, https://www.geneious.com). The assembled sequences were BLASTn-searched in the National Centre for Biotechnology (NCBI) and compared with closely related sequences. Sequences\n\nFigure 1. Observations of Solen marginatus in Praia da Vitória bay, Terceira Island, Azores; (a) Circles indicate presence, (A, G–K), squares indicate absence, (B–F) with letters representing different sampling sites of S. marginatus in August 2023, triangle indicates the single sighting location of Álvaro et al. (2024); (b) largest and a smaller specimen collected at Site K; (c) size distribution at Site A (n = 118); (d) dorsal and (e) ventral view of shells: aam, anterior adductor muscle; avp, antero-ventral mantle projection; ant, anterior; dor, dorsal; lig, ligament; LV, left valve; mf, marginal furrow, pam, posterior adductor muscle; pl, pallial line; pos, posterior; pvp, postero-ventral mantle projection; ps, pallial sinus; rct, right cardinal tooth; lct, left cardinal tooth; RV, right valve.\nwere aligned using MUSCLE in Geneious Prime. Overall mean diversity was computed with MEGA v 11.0.13. The maximum likelihood (ML) analysis was conducted in RAxMLGUI v 2.0.10 (Tamura et al., 2021) with 100 bootstrap replicates and K81uf + gamma model. As an outgroup, we used a COI sequence from *Ensis ensis* (ACCN: HF970367).\n\nTo address habitat range, a visual assessment was conducted on the seafloor at 10 locations throughout the bay of Praia da Vitória (Figure 1). The presence of *S. marginatus* was verified using specific bioturbation burrows with an eight or keyhole shape. With a low diversity of sand bivalves, the identification was straightforward, however, caution was taken by excluding old or unclear lebensspuren (e.g. Pereyra et al., 2023). Quadrats of 0.5 x 0.5 m were used three times in each location to assess densities per m² at the various locations throughout the bay density was estimated in more detail using visual line transects conducted by two scuba divers at Site A, within the area of occurrence of the species at 8.6 m depth (Site A). Four non-overlapping random line transects of 20 square metres were conducted at this site, totalling 80 sqm.\n\nPopulation biology was assessed through size distribution by collecting 118 specimens at site A with a ‘salting method’ using a saline solution and measuring shell length from posterior to anterior furthest distance using a vernier calliper (0.1 cm). Size distribution was evaluated with a Kolmogorov–Smirnov test (*P* < 0.05).\n\n**Results**\n\n**Morphological analysis**\n\nThe species was identified from its equivalent shell, elongated and rather broad shape–shell ratio 1:6.1, with razor-like dorsal and ventral margins straight and parallel (Figure 1; Fischer et al., 1987; Carpenter and De Angelis, 2016). The shells are anterior obliquely truncated and longer at the ventral edge. There is only one cardinal tooth on each valve (with no lateral teeth). Anterior adductor muscle scar is nearly as long as the ligament. The distinct marginal furrow bordering the anterior margin of valves is very conspicuous and distinct from other species (Figure 1d; Dautzenberg, 1897; Luczak and Dewarumez, 1992). The outer colour of the shell ranges from yellowish to pale brownish growth zones.\n\n**Phylogenetic analysis**\n\nThe COI gene sequence generated from the *Solen* samples from Azores and Faro shared 100% similarities with *S. marginatus* from Ria de Aveiro, Portugal (ACCN: MK779736) and Asturias, Spain (ACCN: KJ818881). Furthermore, the Azorean samples shared 99% similarity with samples from Asturias, Spain (ACCN: KJ818881, KJ818886, KJ818884, KJ818877, KJ818890, KJ818889, KJ818882, KJ818879) and Ria de Aveiro, Portugal (ACCN: MK779734). Within the *S. marginatus* the lowest shared similarity (90%) was with a sample from Italy (ACCN: MN630857). The overall mean diversity of the *S. marginatus* clade (including *S. marginatus* and the Azorean samples) was 1%. Phylogenetic relationships of the samples from Azores with other closely related *Solen* species, as inferred by the COI gene sequences using ML analysis, showed a formation of a distinct *S. marginatus* clade (Figure 2).\n\n**Habitat range and density**\n\nThe habitat range of *Solen marginatus* in Praia da Vitória, Terceira Island from the preliminary visual census is presented in Figure 1a. The species was present on the sites encircled (average densities given from the quadrats) as follows: A (8.6 m depth; 12.4 ind/m²), I (9.6 m; 9.3 ind/m²), K (8.9 m; 10.7 ind/m²); with some of the sites of occurrence hosting densities below 1 ind/m²: G (7.1 m depth), H (7.5 m), and J (14.5 m). No burrows were observed on locations B (7.2 m depth), C (20 m), D (14.2 m), E (10 m), and F (4.6 m).\n\nThe distribution seems confined to the northern part of the bay (Figure 1). Presence was noted northward of a theoretical line between the green lighthouse (right peer as you enter the bay), and the peer south of Praia Grande (next to location F, Figure 1). Next to Prainha they were mainly observed on depths of 8 m and deeper (Figure 1, sample location K), although generally spread, with greater incidence at the bottom of the slope. At this site, only at specific locations *S. marginatus* was found up to 2.5 m depths. Densities estimated by visual transects at Site A were 12.69 individuals/m² (STDEV = 2.6; *n* = 1015 per 80 sqm). Shell length of *S. marginatus* at Site A averaged 11.3 cm (STDEV = 1.3), ranging between 8.79 and 15.4 cm. Size distribution is presented in Figure 1c. Kolmogorov–Smirnov test revealed a non-normal distribution (*D* = 0.345), with a leptokurtic positive skewness (excess kurtosis = 1.2585) around size classes between 10 and 11.99 cm length, consisting of 60% of the population. The preliminary weight measurements averaged 19.4 g (Stdev = 6.9) per specimen.\n\n**Discussion**\n\n*Solen marginatus* is widespread along the European continent, east of the Azores, from the Baltic Sea to the Mediterranean, and from the Black Sea along the African coast to down to Senegal (Saeedi and Costello, 2019b). The mid-North Atlantic Azorean malaco fauna has affinities with the Mediterranean, Portugal, and Madeira Island (Ávila et al., 2000), so the occurrence of this population is not unexpected. It is noticeable however, that such a dense aggregation of a species with high commercial value remained unreported until present. *Solen marginatus* has not been reported on other islands since the original report by Drouet (1861), and was not included in the recent check-list for the Azores (Ávila et al., 1998; Borges et al., 2010). Our findings on size distribution and preliminary habitat range of the community demonstrate that this population is well established and reaching high densities in the northern area of Praia da Vitória Bay, with no individuals observed in the south of the Bay. The settling location is one of the less hydrodynamic areas of the bay, including the recreational dock and larger sandy beaches. Considering the densities revealed and the historical presence of the solenoids in the region, the grooved razor clam may expand to other locations in the archipelago.\n\nThe dominance of reproductive sizes also indicates a well-established population, as *S. marginatus* reaches maturity between the first and the second year (ca. 4.7 cm length; Maia, 2006, in Ria de Aveiro, Portugal). Having 60% of the population with a medium age of 4 years old shows potential for reproduction and growth. While settlement occurred more than 10 years ago, judging from the existence of individuals above 14 cm length (Maia, 2006), a settlement period of over 20 years is suggested by the shallow phylogenetic divergence with other populations. The origin of this population remains undetermined, with greater genetic affinity to the Ria de Aveiro (North of Portugal) and the Asturias (North of Spain) populations. Our findings on the source populations, agree with the hypothesis of Álvaro et al. (2024), that suggested an\nanthropogenic origin, likely from Portugal or Spain, due to the 8–9 day settlement period of the razor clam larvae (Da Costa et al., 2012). Intentional or non-intentional anthropogenic sources remain possible, as well as natural dispersion. The regular arrival of boats, coupled with the availability of live specimens for purchase in local marketplaces for use as human consumption or fish bait, will more likely function as introduction vectors once the distance to the Northeast Atlantic coastal areas (at least 1800 km) makes natural dispersion more complex.\n\nFuture works should cover the entire sand bank habitats on the island as well as other potential habitats in the archipelago. The use of different genetic markers may prove useful in the search for genetic divergence and understanding the population origin. Future population studies should comprise reproductive histological analysis, toxicological and finer distribution assessments if fisheries exploitation is to be considered. These would support establishing no-take areas, closure dates, minimum sizes, or total allowable catches. Future extractive techniques may include dredging, which places this newly described and confined population and the overall habitat under imminent threat requiring urgent conservation.\n\nData. Data is kept at Atlantic Naturalist collection in Horta, Faial and can be made available under request.\n\nAcknowledgements. This study was financed by the regional ONG Atlantic Naturalist Association and the Portuguese national funds from FCT – Foundation for Science and Technology through projects UIDB/04326/2020 (DOI:10.54499/UIDB/04326/2020), UIDP/04326/2020 (DOI:10.54499/UIDP/04326/2020) and LA/P/0101/2020 (DOI:10.54499/LA/P/0101/2020). The authors wish to acknowledge Maria, Margarida, and Cristina Machado from GEO – Grupo de Estudos Oceânicos for field work assistance, Carlos Alfonso from CCMAR/Universidade do Algarve and Mafalda Albuquerque Naturalist – Science and Tourism.\n\nAuthor Contributions. A. M., D. M. and J. N. G.-P., species discovery, field work, and paper writing, G. D. data analysis and paper writing, R. C. and L. C. molecular analysis and paper writing.\n\nFinancial Support. This study was financed by the regional ONG Atlantic Naturalist Association and the Portuguese national funds from FCT – Foundation for Science and Technology through projects UIDB/04326/2020 (DOI:10.54499/UIDB/04326/2020), UIDP/04326/2020 (DOI:10.54499/UIDP/04326/2020) and LA/P/0101/2020 (DOI:10.54499/LA/P/0101/2020).\n\nCompeting interests. All authors have no conflict of interest.\n\nEthical Standards. The present survey complied with all ethical standards.\n\nReferences\nÁlvaro N, Sinigaglia L, Madeira P, Hipólito A, Melo CS, Arruda S, Fernandes JP, Baptista L and Ávila SP (2024) The razor clam Solen marginatus Pulteney, 1779: a new anthropogenic marine introduction in the Azores Archipelago. Regional Studies in Marine Science 70, 103387.\nÁvila SP, Amen RG, Azevedo J, Cachão M and García-Talavera F (2002) Checklist of the Pleistocene marine molluscs of Praiahna and Lagoinhás (Santa Maria Island, Azores). Açoreana. Revista de Estudos Açoreanos 9, 343–370.\n\nÁvila SP, Azevedo J, Gonçalves JM, Fontes J and Cardigos F (1998) Checklist of the shallow-water marine molluscs of the Azores: 1-Pico, Faial, Flores and Corvo. Açoreana. Revista de Estudos Açoreanos 8, 487–523.\n\nÁvila SP, Azevedo J, Gonçalves JM, Fontes J and Cardigos F (2000) Checklist of the shallow-water marine molluscs of the Azores: 2-São Miguel Island. Arquipélago. Life and Marine Sciences. Supplement 2, 99–131.\n\nÁvila SP, Madeira P, Mendes N, Rebelo A, Medeiros A, Gomes C, Garcia-Fernández-Tajes J and Méndez J (2008) Mass extinctions in the Azores during the last glaciation: fact or myth? Journal of Biogeography 35, 1123–1129.\n\nBorges P, Bried J, Costa A, Cunha R, Gabriela R, Gonçalves V, Martins A, Melo I, Parente M, Raposeiro P, Rodrigues P, Santos R, Silva L, Vieira P, Vieira V, Mendonça E and Boieiro M (2010) Lista de espécies para os besouros de marinhos dos Açores. Cascais, Portugal: Princípio.\n\nCarpenter KE and De Angelis N (2016) The Living Marine Resources of the Eastern Central Atlantic. Volume 2 Bivalves, Gastropods, Hagfishes, Sharks, Batoid Fishes and Chimaeras. Rome: Food & Agriculture Organization.\n\nCunha R, Tenorio M, Afonso C, Castilho R and Zardoya R (2007) Identification of the razor clam species Ensis directus (Conrad, 1843). Cahiers de biologie marine 33, 515–518.\n\nDa Costa E, Nóvoa S, Ojea J and Martínez-Patiño D (2012) Effects of algal diets and starvation on growth, survival and fatty acid composition of Solen marginatus larvae. Scientia Marina 76, 527–537.\n\nDautzenberg P (1897) Atlas de poche des coquilles des Côtes de France: Manche, Atlantique, Méditerranée, Vol. 6. Paris: P. Klincksieck, p. 54.\n\nDrouët H (1861) Éléments de la faune Açoréenne. Paris: J.B. Baiilliére & Fils, 555 pp.\n\nFischer W, Bauchot ML and Schneider M (1987) Fiches FAO d’identification des espèces pour les besoins de la pêche. Rev. 1. Méditerranée et mer Noire. Zone de pêche 37. Vol. I. Végétaux et invertébrés. Publication préparée par la FAO et la Commission des Communautés européennes (Projet GCP/INT/422/EEC) financées conjointement par ses deux organisations. Rome: FAO, 60 pp.\n\nFolmer OM, Black W, Hoeh R, Lutz R and Vrijenhoek R (1994) DNA primers for amplification of mitochondrial cytochrome c oxidase subunit I from diverse metazoan invertebrates. Molecular Marine Biology & Biotechnology 3, 294–299.\n\nGreen MR and Sambrook J (2016) Precipitation of DNA with ethanol. Cold Spring Harbor protocols 2016. doi: 10.1101/pdb.prot093377\n\nHebert PDN, Cywinska A, Ball SL and deWaard JR (2003) Biological identifications through DNA barcodes. Proceedings of the Royal Society of London. Series B: Biological Sciences 270, 313–321.\n\nJeffreys JG (1881) On the Mollusca procured during the “Lightning and Porcupine” expeditions 1868-1870. Part IV. Proceedings of the Zoological Society of London, 922–952.\n\nLeray M and Knowlton N (2015) DNA barcoding and metabarcoding of standardized samples reveal patterns of marine benthic diversity. Proceedings of the National Academy of Sciences 112, 2076–2081.\n\nLuczak C and Dewarumez JM (1992) Note on the identification of Ensis directus (Conrad, 1843). Cahiers de biologie marine 33, 515–518.\n\nMaia FMSR (2006) Estudo do ciclo reprodutor e do crescimento de Solen marginatus e Venerupis pullastra na Ria de Aveiro (Tese de Mestrado). Universidade de Aveiro.\n\nPereyra CA, Bel Haouz W and Lagnaoui A (2023) New bivalve burrows from the mid-Holocene of northeastern Buenos Aires Province (Argentina): ichnotaxonomy and ethology. Palaeoworld 32, 174–187.\n\nSaeedi H and Costello MJ (2019a) The biology, ecology, and societal importance of razor clams. Reference Module in Earth Systems and Environmental Sciences, 494–498.\n\nSaeedi H and Costello MJ (2019b) A world dataset on the geographic distributions of Solenidae razor clams (Mollusca: Bivalvia). Biodiversity Data Journal 7, e31375.\n\nTamura K, Stecher G and Kumar S (2021) MEGA11: molecular evolutionary genetics analysis version 11. Molecular Biology and Evolution 38, 3022–3027.", "id": "./materials/975.pdf" }, { "contents": "Evaluate \\( \\int_{2}^{4} \\frac{\\sqrt{4x^2 - 16}}{x} \\, dx \\)\n\n**HINT:**\n\nFirst convert to form \\( \\sqrt{f^2 - 1} \\) and perform trigonometric substitution:\n\nUse \\( f = \\sec t \\)", "id": "./materials/98.pdf" }, { "contents": "Evaluate \\[ \\int_{0}^{3} \\frac{1}{(x+1)(x-2)} \\, dx \\]\n\n* All the conditions for Fundamental theorem of calculus are met.\n\nSince, \\( m < n \\), the partial fractions should be obtained.\n\nFor \\( I(x) = \\int \\frac{1}{(x+1)(x-2)} \\, dx \\), the partial fractions are:\n\n\\[\n\\frac{1}{(x+1)(x-2)} = \\frac{A}{x+1} + \\frac{B}{x-2}\n\\]\n\n\\( \\Rightarrow \\)\n\n\\[\n1 = A(x-2) + B(x+1)\n\\]\n\n\\( \\Rightarrow \\)\n\n\\[\n1 = Ax - 2A + Bx + B\n\\]\n\n\\( \\Rightarrow \\)\n\n\\[\n1 = x(A+B) - 2A + B\n\\]\n\n\\( \\Rightarrow \\)\n\n\\[\n0 \\cdot x + 1 = x(A+B) - 2A + B\n\\]\n\nComparing coefficients of left and right hand side.\n\n\\[\nA + B = 0 \\quad \\text{(i)}\n\\]\n\n\\[\n-2A + B = 1 \\quad \\text{(ii)}\n\\]\n\nSolving eqn (i) and (ii) we get:\n\n\\[\nA = -\\frac{1}{3}\n\\]\n\n\\[\nB = \\frac{1}{3}\n\\]\n\\[ I(x) = \\int \\frac{-1}{3(x+1)} + \\frac{1}{3(x-2)} \\, dx \\]\n\n\\[ = -\\frac{1}{3} \\int \\frac{1}{x+1} \\, dx + \\frac{1}{3} \\int \\frac{1}{x-2} \\, dx \\]\n\n\\[ = -\\frac{1}{3} \\ln |x+1| + \\frac{1}{3} \\ln |x-2| + C \\]\n\nNow,\n\n\\[ \\int_{0}^{3} \\frac{1}{(x+1)(x-2)} \\, dx = \\left[ I(x) \\right]_{0}^{3} \\]\n\n\\[ = \\left[ -\\frac{1}{3} \\ln |x+1| + \\frac{1}{3} \\ln |x-2| \\right]_{0}^{3} \\]\n\n\\[ = \\left( -\\frac{1}{3} \\ln 4 + \\frac{1}{3} \\ln 1 \\right) \\]\n\n\\[ - \\left( -\\frac{1}{3} \\ln 1 + \\frac{1}{3} \\ln 1 \\right) \\]\n\n\\[ = -\\frac{1}{3} \\ln 4 + \\frac{1}{3} \\ln 1 \\]\n\n\\[ + \\frac{1}{3} \\ln 1 - \\frac{1}{3} \\ln 2 \\]\n\n\\[ = -\\frac{1}{3} \\left( \\ln 4 + \\ln 2 \\right) \\]\n\n\\[ = -\\frac{1}{3} \\ln (4 \\times 2) \\]\n\n\\[ = -\\frac{1}{3} \\ln 8 \\]\n\n\\[ = -\\frac{1}{3} \\ln (2^3) \\]\n\n\\[ = -\\frac{3}{3} \\ln 2 \\]\n\n\\[ = -\\ln 2 \\]", "id": "./materials/99.pdf" }, { "contents": "Assessment Instructions\n\nAfter logging in as a student, the “Assessment” can be visualized in the main menu.\n\nUnder the “Assessment” section, questions can be solved according to the selected topic. In this section, there are topics with and without subtopics; if subtopics exist, one of them must be selected to start the evaluation.\nThe question format is multiple choice, with 5 available options. Each question has only one correct answer and is categorized by level. If the topic is solved for the first time, level 1 will be selected by default.\n\nIf the level 1 question is answered correctly, MathE will automatically proceed to a higher-level question. If answered correctly, a subsequent higher-level question will be attributed; if answered incorrectly, the level will go backward to the previous one.\nAfter solving a set of questions on the topic, the results page will be shown, where the percentage of correct answers can be visualized. Correct answers appear in green. After checking the results, a new test can be taken by clicking “New test.”\n\nIf the selected answer is wrong, the correct answer will appear in green and the chosen wrong answer in red. The sections \"Video lessons\" and \"Teaching materials\" can automatically be used.\nUnder “Video Lessons”, can be found on the topic of the incorrectly answered question.\nUnder “Teaching Materials”, theoretical materials on the subject in PDF format are available.", "id": "./materials/Assessment.pdf" }, { "contents": "Library Instructions\n\nFinally, in the “MathE Library” section, all the theoretical and video materials associated with the desired topic and subtopic can be found.\n\nAfter selecting the topic, it is also possible to filter by the keywords of the desired subject. By selecting none or “select all,” all the materials associated with the topic will be available.\nThe “Video Collection” section contains all the videos on the topic.\n\nUnder “Teaching Materials,” all the theoretical materials on the subject are available.", "id": "./materials/Library.pdf" }, { "contents": "Privacy Policy in accordance with EU Regulation 2016/679\n\nContents\nPrivacy Policy in accordance with EU Regulation 2016/679 ................................................................. 2\nБългарски: Политика за поверителност в съответствие с Регламент на ЕС 2016/679 ........................................... 6\nČeština: Zásady ochrany osobních údajů v souladu s nařízením EU 2016/679 .......................................................... 11\nDeutsch: Datenschutzerklärung gemäß EU-Verordnung 2016/679 ........................................................................ 15\nDansk: Privatlivspolitik i henhold til EU-forordning 2016/679 .............................................................................. 19\nEesti keel : Privaatsuspoliitika vastavalt EL määrusele 2016/679 ........................................................................ 23\nEspañol: Política de Privacidad conforme al Reglamento UE 2016/679 ............................................................. 27\nSuomeksi: EU-asetuksen 2016/679 mukainen tietosuojakäytäntö ................................................................. 31\nFrançais : Politique de confidentialité conformément au Règlement UE 2016/679 .................................................. 35\nΕλληνικά: Πολιτική Απορρήτου σύμφωνα με τον Κανονισμό ΕΕ 2016/679 .............................................................. 39\nHrvatski: Politika privatnosti u skladu s Uredbom EU 2016/679 ........................................................................... 44\nMagyar: Adatvédelmi szabályzat a 2016/679 EU Rendeletnek megfelelően ......................................................... 48\nGaeilge: Beartas Príobháideachta de réir Rialachán AE 2016/679 ................................................................. 52\nItaliano: Privacy Policy ai sensi del Regolamento UE 2016/679 ........................................................................ 56\nLietuviškai: Privatumo politika pagal ES reglamentą 2016/679 ................................................................. 60\nLatviešu valodā: Privātuma politika saskaņā ar ES regulu 2016/679 ................................................................. 64\nMalti: Privac y Policy skont ir-Regolament tal-UE 2016/679 ........................................................................... 68\nNederlands: Privacybeleid in overeenstemming met EU-verordening 2016/679 .................................................... 72\nNorsk: Personvernerklæring i henhold til EU-forordning 2016/679 ................................................................. 76\nPolski: Polityka prywatności zgodna z Rozporządzeniem UE 2016/679 ............................................................ 80\nPortuguês: Política de Privacidade de acordo com o Regulamento da UE 2016/679 ............................................. 84\nRomana: Politica de confidentialitate in conformitate cu Regulamentul UE 2016/679 ........................................ 88\nSvenska: Integritetspolicy i enlighet med EU-förordning 2016/679 ................................................................. 92\nSlovenčina: Zásady ochrany osobných údajov v súlade s Nariadením EÚ 2016/679 ........................................... 96\nSlovenščina: Politika zasebnosti v skladu z Uredbo EU 2016/679 ................................................................. 100\nPrivacy Policy in accordance with EU Regulation 2016/679\n\nWho Collects Your Data\nPursuant to Article 13 of EU Regulation 2016/679 (GDPR), the data controller is the Applicant of the project, and it is responsible of collecting the data. We would like to inform you that our organization is legally bound to process the data you have provided us under the aforementioned regulation.\n\nYour data will be processed lawfully and fairly, under the provision of article 5 of EU Regulation 2016/679. Further details might be provided at a later stage.\n\nData protection officer (dpo): the presence of a possible DPO must be requested from the data controller.\n\nWhat Personal Data We Collect\nIn accordance with Article 4 of EU Regulation 2016/679:\n- “personal data” means any information relating to an identified or identifiable natural person (‘data subject’); an identifiable natural person is one who can be identified, directly or indirectly, in particular by reference to an identifier such as a name, an identification number, location data, an online identifier or to one or more factors specific to the physical, physiological, genetic, mental, economic, cultural or social identity of that natural person;\n- “processing” means any operation or set of operations which is performed on personal data or on sets of personal data, whether or not by automated means, such as collection, recording, organization, structuring, storage, adaptation or alteration, retrieval, consultation, use, disclosure by transmission, dissemination or otherwise making available, alignment or combination, restriction, erasure or destruction.\n\nWith reference to the above mentioned definitions, we underline that we collect only the information you provide us with for the purposes of your involvement in our initiatives and/or your legal relationship with our organization:\n- Personal information: name and surname of natural persons, contacts such as address, ZIP code, city, region, telephone number, email;\n- Data concerning professionals/organizations/businesses: information concerning businesses, name, fiscal address and other identifiers (fax and telephone number, tax code or VAT number).\n\nMoreover, we may collect data provided when you access our sites, through cookies and other similar technology; and when you contact us via email, social media, or similar technologies. Even though such data are not collected so as to be associated with the natural person, these online identifiers might be used and combined so as to create personal profiles. Among the online identifiers we may find IP address, browser type and plug-in details, device type (e.g. desktop, laptop, tablet, phone, etc.) operating system, local time zone. These data are used solely for the production of statistical results.\n\nWe would like to remind you that we will not be processing personal data revealing racial or\nethnic origin, political opinions, religious or philosophical beliefs, or trade union membership, and the processing of genetic data, biometric data for the purpose of uniquely identifying a natural person, data concerning health or data concerning a natural person’s sex life or sexual orientation.\n\n**Why and How We Process Your Data**\n\nWe will use your data in the following ways:\n\n1. Organize and implement initiatives in the field of education and training (e.g. training courses, conferences, European projects etc.)\n2. To produce administrative documents (e.g. invoices) in relation to the initiatives above\n3. For statistical purposes\n4. Carry out communication activities via email concerning our initiatives.\n5. Respond to requests using the forms on the site (if it is present)\n6. Allow registration for access to confidential educational content (if it is present)\n\nYour conferment is compulsory for purposes under paragraphs 1, 2, 5, 6 in order to comply with juridical obligations and EU laws and regulations; refusal to provide personal data will not allow our organization to offer you, our services.\n\nYour consent is optional for purposes under paragraph 3 and 4; we will send you marketing communication via e-mail or postal service. You can exercise your rights any time, pursuant to Article 15 and later of EU Regulation 2016/679 regarding opting out from receiving such communication or choosing other communication modalities.\n\nWe will keep your personal data collected for the purposes under all paragraphs as long as we need so as to provide you with the services offered by our organization and for up to 10 (ten) years.\n\nYou can withdraw your consent at any time.\n\nThe legal basis of the treatment consists of the commercial relationship created by the sale or purchase of goods and / or services, pre-contractual for information (article 6 paragraph b and c), and by consent for marketing activities. (article 6 paragraph a)\n\nWe will process and store your data solely for the aforementioned purposes, using digital devices and in relevant databases ensuring appropriate safeguards so as to ensure ongoing confidentiality, integrity, availability and resilience of processing systems, as set out by EU regulation 2016/679. Only subjects who have obtained access to personal data from the controller or the processor can process such information.\n\nWe do not sell, trade, or otherwise transfer to other third parties your personally identifiable information. However, we may release your information when we believe release is necessary to comply with the law, enforce our site policies, or protect ours or others' rights, property, or safety.\n\n**Profiling**\n\nYour data will not be subjected to a decision based solely on automated processing, which produces legal effects that affect it or that significantly affects its person. Cancellation and Amendment: you have the right to know, at any time, what are your data at the individual data controllers, that is at our company or at the above mentioned persons to whom we communicate them, and how they are used; they also have the right to update, supplement,\ncorrect or cancel them, request their block and oppose their treatment. For the exercise of your rights, as well as for more detailed information about the subjects or categories of subjects to whom the data are communicated or who are aware of it as managers or agents can contact the data controller or one of his managers, identified in this statement.\n\nSocial networks\nOur website may offer access to social network. The terms of service and Privacy Policy applicable to such platforms are published on their website. Pixel cannot control the way data shared on a public forum, chat or dashboard are used, being the data subject responsible of such communication.\n\nComplaints\nYou can also contact the Italian Data Protection Authority using the following link http://www.garanteprivacy.it/home/footer/contatti, or the European Data Protection Supervisor using the following link: https://edps.europa.eu/data-protection/our-role-supervisor/complaints_en\n\nCookies\nAs set out by regulation “Cookie and other tracking tools guidelines - June 10, 2021”, there are three main categories of cookies:\n\nTechnical cookies\nThese are used for the sole purpose of “transmitting communications to an electronic communication network, or to the extent strictly necessary for the provision of a service by the information company explicitly requested by the contracting party or the user in order to provide the said service” These are not used for any ulterior purposes and they are normally installed directly by the owner or the manager of the website (so-called “proprietary” or “editorial” cookies). These can be divided into: browsing or session cookies, which guarantee normal navigation and use of the website (making it possible for example, to make purchases or be authenticated in order to access reserved areas); analytics cookies assimilated by the technical cookies where they are used directly by the manager of the website to collect information, in an associated form (anonymous), about the number of users and the manner in which they visit the website; functional cookies that allow the user to navigate in relation to a series of select criteria (for example, the language or the products selected for purchase) in order to improve the service provided, provided that we inform our users as set out by article 13 EU Regulation 2016/679.\nThe prior consent of the user is not requested in order to install these cookies.\n\nAnalytics cookies\nThe site uses only google analytics, which is used to create profiles of the users and are employed for sending advertising messages according to the preferences shown by the same during their online navigation. Due to their particular invasiveness with regard to the users’ privatesphere, European and Italian regulations require that users be adequately informed about their use of the same and are thus required to express their valid consent. But in the specific case google analytics has been anonymized (IP masking) and the sharing of navigation data with google has been blocked: in this way the analytical cookie is similar to the technical cookies indicated above and does not require consent.\n\nProfiling cookies\nThis type of cookie is not used in the site.\n\nSpecific note:\nThe embedded videos of YouTube on the site do not use cookies as it has been specified \"nocookie\" the privacy-enhanced embed code for all your YouTube video embeds.\n\n**International and European data transfer**\nYour data will be processed solely in European Economic Area. Your rights with regard to the personal data we hold under EU Regulation 2016/679\n\n**Your rights**\nYou can exercise your rights any time, as set out by Article 7, par. 3, and articles 15 and following of EU regulation 2016/679:\n\n- Right to access personal data\n- Right to rectification and erasure of personal data;\n- Right to restriction of processing;\n- Right to data portability;\n- Right to object to processing of personal data\n- Right to legal claim to Italian Data Protection Authority.\n\nYou can exercise your rights by sending us an email at info@pixel-online.net or a letter addressed to Pixel, via Luigi Lanzi, 12 – 50134 – Firenze, Italy. Further information concerning data processing can be added when collecting data.\n\n12 February 2022 rev.03\nБългарски: Политика за поверителност в съответствие с Регламент на ЕС 2016/679\n\nКой събира вашите данни\nСъгласно член 13 от Регламент на ЕС 2016/679 (GDPR), администраторът на данни е заявителят на проекта и той отговаря за събирането на данните. Бихме искали да ви информираме, че нашата организация е законово задължена да обработва данните, които сте ни предоставили съгласно гореспоменатия регламент.\n\nВашите данни ще бъдат обработвани законосъобразно и справедливо, съгласно разпоредбата на член 5 от Регламент на ЕС 2016/679. Допълнителни подробности може да бъдат предоставени на по-късен етап.\n\nДлъжностно лице по защита на данните (dpo): присъствието на възможно DPO трябва да бъде поискано от администратора на данни.\n\nКакви лични данни събираме\nВ съответствие с член 4 на ЕС Регламент 2016/679 г.:\n- \"лични данни\" означава всяка информация, свързана с идентифицирано или подлежащо на идентифициране физическо лице ('данни предмет'); ан разпознаваем естествено лице е един Кой мога бъда идентифициран, директно или косвено, по-специално чрез позоваване на идентификатор като име, идентификационен номер, местоположение данни, ан онлайн идентификатор или да се един или Повече ▼ фактори специфични да се на физически, физиологичен, генетичен, психически, икономически, културни или социални идентичност на че естествено лице;\n- \"обработка\" означава всякакви операция или комплект на операции който е изпълнено На лични данни или На комплекти на лични данни, дали или не от автоматизиран означава, такъв като колекция, запис, организация, структуриране, съхранение, адаптиране или промяна, извличане, консултация, използване, разкриване чрез предаване, разпространение или предоставяне по друг начин, привеждане в съответствие или комбинация, ограничение, изтриване или унищожаване.\n\nС справка да се на по-горе споменат определения, ние подчертайте че ние събират само на информация ти предостави ни за цели на вашето участие в нашите инициативи и/или вашия законен връзка с нашите организации:\n- Лична информация: име и фамилия на физически лица, контакти като адрес, пощенски код код, град, регион, телефон номер, електронна поща;\n- Данни относно професионалисти/организации/бизнеси: информация относно фирми, име, фискален адрес и други идентификатори (факс и телефонен номер, данъчен код или ДДС номер).\n\nОсвен това, ние може да събирате данни, предоставени при достъп до нашите сайтове, чрез бисквитки и други подобна технология; и когато се свържете с нас по имейл, социални медии или подобни технологии. Въпреки че такива данни не се събират, за да бъдат свързани с физическото лице, те онлайн идентификаторите могат да се използват и комбинират за създаване на лични профили. Сред онлайн идентификатори, които\nможе да намерим IP адрес, тип браузър и подробности за приставката, тип устройство (напр. настолен компютър, лаптоп, таблет, телефон и др.) операционна система, местна часова зона. Тези данни се използват единствено за Производство на статистически резултати.\n\nБихме искали да ви напомним, че няма да обработваме лични данни, разкриващи расова или етнически произход, политически възгледи, религиозни или философски убеждения или членство в профсъюз, и обработката на генетични данни, биометрични данни с цел еднозначно идентифициране на природни човек, данни относно здраве или данни относно естествено на човек секс живот или сексуален ориентация.\n\nЗащо и как обработваме вашите данни\n\nние ще използвайте Вашият данни в следното начини:\n\n7. Организира и реализира инициативи в областта на образованието и обучението (напр. обучение курсове, конференции, европейски проекти и др.)\n8. Да се произвеждат административна документи (напр фактури) в отношение да се на инициативи по-горе\n9. За статистически цели\n10. Носи навън комуникация дейности чрез електронна поща относно нашите инициативи.\n11. Отговаряйте на заявки чрез формулярите на сайта (ако има такива)\n12. Разрешаване на регистрация за достъп до поверително образователно съдържание (ако има такова)\n\nВашето връчване е задължително за целите по параграфи 1, 2, 5, 6, за да се съобразят с юридически задължения и ЕС закони и регламенти; отказ да се предоставят лични данни ще не позволяванашите организация до оферта ти, нашите услуги.\n\nВашето съгласие не е задължително за целите по параграфи 3 и 4; ние ще ви изпратим маркетинг комуникация чрез електронна поща или пощенска услуга. Можете да упражнявате правата си по всяко време, съгласно Член 15 и по-късно от Регламент 2016/679 на ЕС относно отказ от получаване на такива комуникация или избирайки друго комуникация модалности.\n\nНие ще съхраняваме вашите лични данни, събрани за целите по всички параграфи както ни е необходимо, за да Ви предоставим предлаганите услуги от нашата организация и за до 10 (десет години).\n\nТи мога оттегли се Вашият съгласие в всякакви време.\n\nПравното основание на третирането се състои от търговското правоотношение, създадено от продажбата или покупка на стоки и/или услуги, преддоговорни за информация (член 6, параграфи б и в),и по съгласие за маркетинг дейности. ( статия 6 параграф а)\n\nНие ще обработваме и съхраняваме вашите данни единствено за гореспоменатите цели, използвайки цифрови устройства и в съответните бази данни, осигуряващи подходящи предпазни мерки, за да се гарантира непрекъсната поверителност, целостта, наличността и устойчивостта на системите за обработка, както е посочено в регламент на ЕС 2016/679. Могат само субекти, които са получили достъп до лични данни от\nадминистратора или обработващия процес такъв информация.\n\nНие не продаваме, търгуваме или по друг начин не прехвърляме на други трети лица, които можете да идентифицирате лично информация. Въпреки това, ние може да предоставим вашата информация, когато смятаме, че това е необходимо спазват с на закон, налагат се нашите сайт политики, или защитават нашите или други права, Имот, или безопасност.\n\nПрофилиране\nВашите данни няма да бъдат подложени на решение, основано единствено на автоматизирана обработка, която произвежда правни последици, които го засягат или засягат значително неговата личност. Отмяна и изменение: вие имате право да знаете по всяко време какви са вашите данни при отделните администратори на данни, т.е нашата компания или при посочените по-горе лица, на които ги съобщаваме, и как те са използвани; те също имат право да ги актуализират, допълват, коригират или отменят, да поискат тяхното блокиране и се противопоставят на тяхното лечение. За упражняване на правата си, както и за по-подробна информация относно субектите или категориите субекти, на които се съобщават данните или които са запознати от него като мениджъри или агенти могат да се съвържат с администратора на данни или един от неговите мениджъри, посочени в това изявление.\n\nСоциални мрежи\nНашият уебсайт може да предлага достъп до социалната мрежа. Приложими са условията за ползване и Политиката за поверителност към такива платформи са публикувани на уебсайта им. Pixel не може да контролира начина, по който данните се споделят на а обществено форум, чат или табло са използван, битие на данни предмет отговорен на такъв комуникация.\n\nОплаквания\nТи мога също контакт на Италиански Данни Защита Власт използвайки на следвайки връзка http://www.garanteprivacy.it/home/footer/contatti, или Европейски надзорен орган по защита на данните, използвайки следвайки връзка: https://edps.europa.eu/data-protection/our-role-supervisor/complaints_en\n\nБисквитки\nКато се т ou t от регламент „Насоки за бисквитки и други инструменти за проследяване – 10 юни 2021 г.“, т тук три основни категории на бисквитки:\n\nТехнически бисквитки\nТези са използван за на подметка предназначение на \"предавателна комуникация\" или до степента, която е строго необходима за предоставяне на услуга от страна на информационна компания, изрично поискана от договорящата страна или ползвателя с цел предоставяне споменатата услуга\" Те не се използват за никакви скрити цели и обикновено се инсталират директно от на собственик или на мениджър на на уебсайт (т.нар \"собственически\" или „редакция\" бисквитки). Тези мога бъда разделени в: съфирание или сесия бисквитки, който гаранция нормално навигация и използване на уебсайта (което прави възможно например да се правят покупки или да се удостоверени за достъп до запазени зони);\nаналитични бисквитки, усвоени от техническите бисквитки, когато се използват директно от мениджъра на уебсайта за събиране на информация, в свързана форма (анонимен), за броя на потребителите и начина, по който посещават уебсайт; функционални бисквитки, които позволяват на потребителя да навигира във връзка с поредица от критерии за избор (например езикът или продуктите, избрани за покупка), с цел подобряване на услугата при условие, при условие че информираме нашите потребители като тръгна от статия 13 ЕС Регламент 2016/679.\n\nВ преди съгласие на на потребител е не поискано в поръчка да се Инсталирай тези бисквитки.\n\nБисквитки за анализ\nСайтът използва само google analytics, който се използва за създаване на профили на потребителите и са използвани за изпращане на рекламни съобщения според предпочитанията, показани от същите по време на тяхната онлайн навигация. Поради особената им инвазивност по отношение на личните данни на потребителите сфера, европейски и Италиански регламенти изискват че потребители бъда адекватно информиран относно техен използване на на един и същ и са по този начин задължително да се експресно техен валиден съгласие. Но в конкретния случай google analytics е анонимизиран (IP маскиране) и споделянето на навигационни данни с google е блокирано: по този начин аналитичната бисквитка е подобна на техническите бисквитки, посочени по-горе и не изисква съгласие.\n\nБисквитки за профилиране\nТози тип бисквитки не се използва в сайта.\n\nСпецифична забележка:\nВградените видеоклипове на YouTube на сайта не използват бисквитки, тъй като е посочено „ nocookie “ кода за вграждане с подобрена поверителност за всички ваши вградени видеоклипове в YouTube.\n\nМеждународен и европейски трансфер на данни\nВашите данни ще бъдат обработвани единствено в Европейското икономическо пространство. Вашите права по отношение налични данни ние задръжте под ЕС Регламент 2016/679\n\nВашите права\nТи мога упражнение Вашият права всякакви време, като комплект навън от член 7, пар. 3, и статии 15 и следвайкина ЕС регулиране 2016/679 г.:\n\n- правилно влизам лични данни\n- правилно да се коригиране и изтриване на лични данни;\n- правилно да се ограничение на обработка;\n- правилно да се данни преносимост;\n- правилно да се обект да се обработка на лични данни\n- правилно да се правен иск да се Италиански Данни Защита Власт.\n\nМожете да упражните правата си, като ни изпратите имейл на info@pixel-online.net или адресирано писмо до Pixel, чрез Луиджи Ланци , 12 – 50134 – Флоренция, Италия. Допълнителна информация относно обработката на данни може да се добави кога\nсъбиране на данни.\n\n12 февруари 2022 г rev.03\nČeština: Zásady ochrany osobních údajů v souladu s nařízením EU 2016/679\n\nKdo shromažďuje vaše údaje\nV souladu s článkem 13 nařízení EU 2016/679 (GDPR) je správcem údajů žadatel projektu a je odpovědný za shromažďování údajů. Rádi bychom vás informovali, že naše organizace je ze zákona zavážána zpracovávat údaje, které jste nám poskytli podle výše uvedeného nařízení. Vaše údaje budou zpracovány zákonně a spravedlivě v souladu s ustanovením článku 5 nařízení EU 2016/679. Další podrobnosti mohou být poskytnuty později.\n\nJaké osobní údaje shromažďujeme\nv soulad s Článkem 4 z EU Nařízení 2016/679:\n- \"osobními údaji\" se rozumí jakékoli informace týkající se identifikované nebo identifikovatelné fyzické osoby ('data předmět'); an identifikovatelné přírodní osoba je jeden SZO umět být identifikovaný, přímo nebo nepřímo, zejména odkazem na identifikátor, jako je jméno, identifikační číslo, umístění data, an online identifikátor nebo na jeden nebo více faktory charakteristický na a fyzický, fyziologický, genetický, duševní, ekonomické, kulturní nebo sociální identita že přírodní osoba;\n- \"zpracovává se\" prostředek žádný úkon nebo soubor z operace který je provedeno na osobní data nebo na sady z osobní data, zda nebo ne podle Automatizovaný prostředek, takový tak jako sbírka, záznam, organizace, strukturování, ukládání, adaptace nebo změna, vyhledávání, konzultace, použití, zveřejnění přenosem, šířením nebo jiným zpřístupněním, sladěním popř kombinace, omezení, vymazání nebo zničení.\n\nS odkaz na a výše zmíněno definice, my zdůraznit že my sbírat pouze a informace vy poskytnout nám pro účely vašeho zapojení do náš iniciativy a/nebo vaše právní vztah s náš organizace:\n- Osobní údaje: jméno a příjmení fyzických osob, kontakty jako adresa, PSČ kód, město, kraj, telefon číslo, e-mail;\n- Data Pokud jde o profesionálové/organizace/podniky: informace Pokud jde o firmy, název, fiskální adresu a další identifikátory (číslo faxu a telefonu, daňový kód nebo DPH číslo).\n\nKromě toho můžeme shromažďovat údaje poskytnuté při přístupu na naše stránky prostřednictvím souborů cookie a dalších podobná technologie; a když nás kontaktujete prostřednictvím e-mailu, sociálních médií nebo podobných technologií. I když takové údaje nejsou shromažďovány tak, aby byly spojeny s fyzickou osobou, tyto online identifikátory mohou být použity a kombinovány za účelem vytvoření osobních profilů. Mezi online identifikátory, které můžeme najít IP adresu, typ prohlížeče a podrobnosti o zásuvném modulu, typ zařízení (např. notebook, tablet, telefon atd.) operační systém, místní časové pásma. Tyto údaje se používají výhradně pro produkce statistický Výsledek.\n\nRádi bychom Vám připomněli, že nebudeme zpracovávat osobní údaje odhalující rasové resp\netnický původ, politické názory, náboženské nebo filozofické přesvědčení nebo členství v odborech a zpracování genetických údajů, biometrických údajů za účelem jednoznačné identifikace přírodního osoba, data Pokud jde o zdraví nebo data Pokud jde o A přírodní osoby sex život nebo sexuální orientace.\n\nProč a jak zpracováváme vaše údaje\nMy vůle použití vaše data v následující způsoby:\n13. Organizovat a realizovat iniciativy v oblasti vzdělávání a odborné přípravy (např kurzy, konference, evropské projekty atd.)\n14. Na vyrobit správní dokumenty (např faktury) v vztah na a iniciativy výše\n15. Pro statistický účely\n16. Nést ven sdělení činnosti přes e-mailem Pokud jde o náš iniciativy.\n17. Odpovídejte na požadavky pomocí formulářů na webu (pokud jsou k dispozici)\n18. Povolit registraci pro přístup k důvěrnému vzdělávacímu obsahu (pokud existuje)\n\nVáš souhlas je povinný pro účely podle odstavců 1, 2, 5, 6, aby byly dodrženy právní závazky a EU zákony a předpisy; zamítnutí na poskytnout osobní data vůle ne dovolitnáš organizace do nabídka vy, naše služby.\nVáš souhlas je pro účely podle odstavce 3 a 4 nepovinný; zašleme vám marketing komunikace prostřednictvím e-mailu nebo pošty. Svá práva můžete uplatnit kdykoli v souladu s Článek 15 a pozdější nařízení EU 2016/679 o odstoupení od přijímání takovýchto zpráv sdělení nebo výběr jiného sdělení způsoby.\nVaše osobní údaje shromážděné pro účely uvedené ve všech odstavcích budeme uchovávat tak dlouho jak potřebujeme, abychom mohli poskytovat nabízené služby naší organizaci a to až do 10 (deset let).\nVy umět ustoupit vaše souhlas v žádný čas.\n\nPrávní základ úpravy tvoří obchodní vztah vzniklý prodejem resp nákup zboží a/nebo služeb, předsmluvně pro informaci (čl. 6 odst. b a c),a souhlasem pro marketing činnosti. (článek 6 odstavec A)\n\nVaše údaje budeme zpracovávat a uchovávat výhradně pro výše uvedené účely pomocí digitálních zařízení a v příslušných databázích zajišťujících vhodná ochranná opatření, aby byla zajištěna trvalá důvěrnost, integrita, dostupnost a odolnost systémů zpracování, jak je stanoveno nařízením EU 2016/679. Přístup k osobním údajům mohou získat pouze subjekty, které od správce nebo zpracovatele získaly proces takový informace.\n\nNeprodáváme, neobchodujeme ani jinak nepřevádíme na jiné třetí strany, které by vás osobně mohly identifikovat informace. Můžeme však uvolnit vaše informace, pokud se domníváme, že je to nutné vyhovět s a zákon, prosadit náš místo opatření, nebo chránit naše nebo ostatní práva, vlastnictví, nebo bezpečnost.\n\nProfilování\nVaše údaje nebudou předmětem rozhodnutí založeného výhradně na automatizovaném zpracování, které má právní účinky, které se jich dotýkají nebo které se významně dotýkají její osoby. Zrušení a změna: máte právo kdykoli vědět, jaké jsou vaše údaje u jednotlivých správců údajů, tedy u naší společnosti nebo u výše uvedených osob, kterým je sdělujeme, a jak jsou využívány; dále mají právo je aktualizovat, doplňovat, opravovat nebo rušit, požadovat jejich\nblokování a bránit jejich zpracování. Pro výkon vašich práv, jakož i pro podrobnější informace o subjektech nebo kategoriích subjektů, kterým jsou údaje sdělovány nebo které jsou jako manažeři nebo zástupci o nich informováni, se mohou obrátit na správce údajů nebo na některého z jeho manažerů uvedených v toto tvrzení.\n\nSociální sítě\nNaše webové stránky mohou nabízet přístup k sociální síti. Platí podmínky služby a zásady ochrany osobních údajů na tyto platformy jsou zveřejněny na jejich webových stránkách. Pixel nemůže ovládat způsob sdílení dat na a veřejnost Fórum, povídat si nebo přístrojová deska jsou použitý, bytost a data předmět odpovědný z takovým sdělením.\n\nStížnosti\nVy umět taky Kontakt a italština Data Ochrana Autorita použitím a Následující odkaz http://www.garanteprivacy.it/home/footer/contatti nebo evropského inspektora ochrany údajů pomocí Následující odkaz: https://edps.europa.eu/data-protection/our-role-supervisor/complaints_en\n\nSoubory cookie\nTak jako se t ou t podle nařízení „Pokyny pro soubory cookie a další nástroje pro sledování – 10. června 2021 “, t zde jsou tři hlavní kategorie z cookies:\n\nTechnické cookies\nTyto jsou použity pro a jediný účel z „vysílání komunikace na an elektronický komunikační sítě, nebo v rozsahu nezbytně nutném pro poskytování služby ze strany informace společnosti výslovně vyžádané smluvní stranou nebo uživatelem za účelem poskytnutí uvedená služba“ Nejsou používány pro žádné postranní účely a jsou běžně instalovány přímo podle a majitel nebo a manažer z a webová stránka (tzv \"proprietární\" nebo \"redakční\" cookies). Tyto umět být rozdělený do: procházení nebo zasedání sušenky, který záruka normální navigace a používání webových stránek (umožňující např. nákupy nebo být ověřeno pro přístup k vyhrazeným oblastem); analytické soubory cookie asimilované technickými cookies, kde je používá přímo správce webu ke shromažďování informací, v an přidružený formulář (anonymní), o počtu uživatelů a způsobu, jakým navštěvují webová stránka; funkční soubory cookie, které uživateli umožňují navigaci ve vztahu k řadě vybraných kritérií (například jazyk nebo produkty vybrané k nákupu) za účelem zlepšení služby pokud, pokud informujeme naše uživatelů tak jako stanoveno podle článk 13 EU Nařízení 2016/679. The předchozí souhlas z a uživatel je ne vyžádáno v objednat na Nainstalujte tyto cookies.\n\nAnalytické cookies\nStránka používá pouze google analytics, které se používají k vytváření profilů uživatelů a jsou slouží k zasílání reklamních zpráv podle preferencí zobrazených stejným během jejich online navigace. Kvůli jejich zvláštní invazivnosti s ohledem na soukromí uživatelů koule, evropský a italština předpisy vyžadovat že uživatelů být přiměřeně informovaný o jejich použití z a stejný a jsou tím pádem Požadované na vyjádřit jejich platný souhlas . Ale v konkrétním případě byla analytika Google anonymizována (maskování IP) a sdílení navigačních dat se společností google bylo zablokováno: analytický soubor cookie je tímto způsobem podobný technickým souborům cookie uvedeným výše a nevyžaduje souhlas.\n\nProfilovací soubory cookie\nTento typ cookie se na webu nepoužívá.\nKonkrétní poznámka:\nVložená video z YouTube na stránce nepoužívají soubory cookie, protože bylo specifikováno „nocookie“ kód pro vložení s vylepšeným soukromím pro všechna vaše vložení video na YouTube.\n\nMezinárodní a evropský přenos dat\nVaše údaje budou zpracovávány výhradně v Evropském hospodářském prostoru. Vaše práva s ohledem na osobní data my držet pod EU Nařízení 2016/679\n\nVaše práva\nVy umět cvičení vaše práv žádný čas, tak jako soubor ven podle Článek 7, odst. 3, a články 15 a Následující EU nařízení 2016/679:\n- Že jo mít přístup osobní data\n- Že jo na náprava a vymazání z osobní data;\n- Že jo na omezení z zpracovává se;\n- Že jo na data přenosnost;\n- Že jo na objekt na zpracovává se z osobní data\n- Že jo na právní Nárok na italština Data Ochrana Autorita.\n\npráva můžete uplatnit zasláním e-mailu na adresu info@pixel-online.net nebo adresného dopisu do Pixel, přes Luigi Lanzi, 12 – 50134 – Firenze, Itálie. Další informace o zpracování dat lze přidat, když sbírat data.\n\n12. února 2022 rev.03\nDeutsch: Datenschutzerklärung gemäß EU-Verordnung 2016/679\n\nWer sammelt Ihre Daten\nGemäß Artikel 13 der EU-Verordnung 2016/679 (DSGVO) ist der Datenverantwortliche der Antragsteller des Projekts und für die Erhebung der Daten verantwortlich. Wir möchten Sie darüber informieren, dass unsere Organisation gesetzlich verpflichtet ist, die Daten, die Sie uns gemäß der oben genannten Verordnung bereitgestellt haben, zu verarbeiten. Ihre Daten werden gemäß Artikel 5 der EU-Verordnung 2016/679 rechtmäßig und fair verarbeitet. Weitere Details können zu einem späteren Zeitpunkt bereitgestellt werden. Datenschutzbeauftragter (dpo): Die Anwesenheit eines möglichen Datenschutzbeauftragten muss beim für die Datenverarbeitung Verantwortlichen beantragt werden.\n\nWelche personenbezogenen Daten wir erheben\nIm Übereinstimmung mit Artikel 4 von EU Verordnung 2016/679:\n- „Personenbezogene Daten“ sind alle Informationen, die sich auf eine identifizierte oder identifizierbare natürliche Person beziehen ('Daten Fach'); ein identifizierbar natürlich Person ist ein WHO kann sein identifiziert, direkt oder indirekt, insbesondere unter Bezugnahme auf eine Kennung wie einen Namen, eine Identifikationsnummer, Lage Daten, ein online Kennung oder zu ein oder mehr Faktoren Spezifisch zu der körperlich, physiologisch, genetisch, geistig, wirtschaftlich, kulturell oder Sozial Identität von das natürlich Person;\n- \"wird bearbeitet\" bedeutet irgendein Operation oder einstellen von Operationen welcher ist durchgeführt an persönlich Daten oder an setzt von persönlich Daten, ob oder nicht durch automatisiert bedeutet, eine solche als Sammlung, Erfassung, Organisation, Strukturierung, Speicherung, Anpassung oder Veränderung, Abruf, Abfrage, Nutzung, Offenlegung durch Übertragung, Verbreitung oder anderweitige Zurverfügungstellung, Ausrichtung oder Kombination, Beschränkung, Löschen oder Zerstörung.\n\nMit Hinweis zu der Oben erwähnt Definitionen, wir unterstreichen das wir sammeln nur der Information Sie stellen Sie uns für die zur Verfügung Zwecke Ihres Engagements in unsere Initiativen und/oder Ihr Recht Beziehung mit unsere Organisation:\n- Personenbezogene Daten: Vor- und Nachname natürlicher Personen, Kontaktdaten wie Adresse, PLZ Code, Stadt, Region, Telefon Anzahl, Email;\n- Daten über Fachleute/Organisationen/Unternehmen: Information über Unternehmen, Name, Steueradresse und andere Identifikatoren (Fax- und Telefonnummer, Steuernummer oder Mehrwertsteuer Anzahl).\n\nDarüber hinaus können wir Daten sammeln, die bereitgestellt werden, wenn Sie auf unsere Websites zugreifen, durch Cookies und andere ähnliche Technologie; und wenn Sie uns per E-Mail, Social Media oder ähnlichen Technologien kontaktieren. Auch wenn solche Daten nicht erhoben werden, um sie der natürlichen Person zuzuordnen, sind diese Online-Identifikatoren können verwendet und kombiniert werden, um personenbezogene Profile zu erstellen. Unter den Online Identifikatoren, die wir möglicherweise finden, IP-Adresse, Browsertyp und Plug-\nin-Details, Gerätetyp (z. B. Desktop, Laptop, Tablet, Telefon usw.), Betriebssystem, lokale Zeitzone. Diese Daten werden ausschließlich für die Produktion von statistisch Ergebnisse.\n\nWir möchten Sie daran erinnern, dass wir keine personenbezogenen Daten verarbeiten, aus denen die Rasse oder Herkunft hervorgeht ethnische Herkunft, politische Meinungen, religiöse oder weltanschauliche Überzeugungen oder Gewerkschaftszugehörigkeit und die Verarbeitung von genetischen Daten, biometrischen Daten zum Zwecke der eindeutigen Identifizierung einer natürlichen Person Person, Daten über die Gesundheit oder Daten über ein natürlich Person Sex Leben oder sexuell Orientierung.\n\n**Warum und wie wir Ihre Daten verarbeiten**\n\nWir Wille verwenden dein Daten in die folgende Wege:\n\n19. Initiativen im Bereich der allgemeinen und beruflichen Bildung organisieren und umsetzen (zKurse, Konferenzen, europäisch Projekte etc.)\n20. Zu produzieren administrativ Unterlagen (z.B Rechnungen) in Beziehung zu der Initiativen Oben\n21. Zum statistisch Zwecke\n22. Tragen aus Kommunikation Aktivitäten über Email über unsere Initiativen.\n23. Beantworten Sie Anfragen mithilfe der Formulare auf der Website (falls vorhanden)\n24. Registrierung für den Zugriff auf vertrauliche Bildungsinhalte zulassen (sofern vorhanden)\n\nIhre Übertragung ist für Zwecke gemäß den Absätzen 1, 2, 5, 6 obligatorisch, um sie zu erfüllen rechtlich Verpflichtungen und EU Gesetze und Vorschriften; Ablehnung zu bieten persönlich Daten Wille nicht ermöglichen unsere Organisation zu Angebot Sie, Unsere Dienstleistungen.\n\nIhre Einwilligung ist für die Zwecke gemäß Absatz 3 und 4 optional; Wir senden Ihnen Marketing Kommunikation per E-Mail oder Post. Sie können Ihre Rechte jederzeit gem Artikel 15 und später der EU-Verordnung 2016/679 in Bezug auf die Ablehnung des Erhalts solcher Kommunikation oder andere wählen Kommunikation Modalitäten.\n\nWir werden Ihre personenbezogenen Daten, die für die Zwecke gemäß allen Absätzen erhoben wurden, so lange aufbewahren wie wir sie benötigen, um Ihnen die angebotenen Dienstleistungen zur Verfügung zu stellen durch unsere Organisation und bis 10 (10 Jahre. Sie kann zurückziehen dein Zustimmung bei irgendein Zeit.\n\nDie Rechtsgrundlage der Behandlung besteht in der durch den Verkauf bzw. entstandenen Geschäftsbeziehung Kauf von Waren und / oder Dienstleistungen, vorvertragliche Informationen (Artikel 6 Absatz b und c),und nach Zustimmung zum Marketing Aktivitäten. ( Artikel 6 Absatz ein)\n\nWir werden Ihre Daten ausschließlich zu den vorgenannten Zwecken mit digitalen Geräten verarbeiten und speichern und in einschlägigen Datenbanken unter Gewährleistung angemessener Sicherheitsvorkehrungen, um die fortlaufende Vertraulichkeit zu gewährleisten, Integrität, Verfügbarkeit und Belastbarkeit von Verarbeitungssystemen gemäß der EU-Verordnung 2016/679. Nur Personen, die vom Verantwortlichen oder Auftragsverarbeiter Zugang zu personenbezogenen Daten erhalten haben, können dies tun Prozess eine solche Information.\nWir verkaufen, tauschen oder übertragen Ihre persönlich identifizierbaren Daten nicht an Dritte Information. Wir können Ihre Informationen jedoch freigeben, wenn wir der Meinung sind, dass eine Freigabe erforderlich ist einhalten mit der Gesetz, erzwingen unsere Seite? Politik, oder beschützen unsere oder Andere' Rechte, Eigentum, oder Sicherheit.\n\n**Profilierung**\nIhre Daten werden keiner ausschließlich auf einer automatisierten Verarbeitung beruhenden Entscheidung unterworfen, die Rechtswirkungen entfaltet, die sie betreffen oder die ihre Person erheblich beeinträchtigen . Widerruf und Änderung: Sie haben jederzeit das Recht zu erfahren, was Ihre Daten bei den einzelnen Datenverantwortlichen sind, das heißt bei unserem Unternehmen oder bei den oben genannten Personen, denen wir sie mitteilen, und wie sie verwendet werden; Sie haben auch das Recht, sie zu aktualisieren, zu ergänzen, zu korrigieren oder zu löschen, ihre Sperrung zu verlangen und sich ihrer Behandlung zu widersetzen. Für die Ausübung Ihrer Rechte sowie für nähere Informationen zu den Subjekten oder Kategorien von Subjekten, denen die Daten mitgeteilt werden oder die davon als Manager oder Beauftragte Kenntnis haben, können Sie sich an den Datenverantwortlichen oder einen seiner Manager wenden, der in angegeben ist diese Aussage.\n\n**Soziale Netzwerke**\nUnsere Website bietet möglicherweise Zugang zu sozialen Netzwerken. Es gelten die Nutzungsbedingungen und die Datenschutzerklärung zu solchen Plattformen werden auf deren Website veröffentlicht. Pixel kann die Art und Weise, wie Daten auf a geteilt werden, nicht kontrollieren öffentlich zugänglich Forum, Plaudern oder Armaturenbrett sind benutzt, Sein der Daten Fach verantwortlich von eine solche Kommunikation.\n\n**Beschwerden**\nSie kann Auch Kontakt der Italienisch Daten Schutz Behörde verwenden der folgende Verknüpfung http://www.garanteprivacy.it/home/footer/contatti, oder der Europäische Datenschutzbeauftragte unter Verwendung der folgende Link: https://edps.europa.eu/data-protection/our-role-supervisor/complaints_en\n\n**Kekse**\nAls t _ du t durch Verordnung „Richtlinien für Cookies und andere Tracking - Tools – 10. Juni 2021“, Ich bin hier a re drei Hauptkategorien von Kekse:\n\n**Technisch Kekse**\nDiese sind benutzt zum der Sohle, einzig, alleinig Zweck von „übertragen Kommunikation zu ein elektronisch Kommunikationsnetzwerk oder in dem Umfang, der für die Bereitstellung eines Dienstes durch das unbedingt erforderlich ist Informationsgesellschaft, die ausdrücklich vom Vertragspartner oder dem Benutzer zur Bereitstellung angefordert wurde den besagten Dienst“ Diese werden nicht für andere Zwecke verwendet und sind normalerweise installiert direkt durch der Eigentümer oder der Manager von der Webseite (sogenannt „proprietär“ oder „Redaktion“ Kekse). Diese kann sein geteilt hinein: durchsuchen oder Sitzung Kekse, welcher Garantie normal Navigation und Nutzung der Website (die es beispielsweise ermöglichen, Einkäufe zu tätigen oder zu sein authentifiziert, um auf reservierte Bereiche zuzugreifen); Analyse-Cookies, die von den technischen assimiliert werden Cookies, wo sie direkt vom Verwalter der Website verwendet werden, um Informationen zu sammeln, in einem verbundenes Formular (anonym), über die Anzahl der Benutzer und die Art und Weise,\nwie sie die besuchen Webseite; funktionale Cookies, die es dem Benutzer ermöglichen, in Bezug auf eine Reihe ausgewählter Kriterien zu navigieren (zum Beispiel die Sprache oder die zum Kauf ausgewählten Produkte), um den Service zu verbessern bereitgestellt, unter der Vorraussetzung, dass wir informieren unsere Benutzer als ausgehen durch Artikel 13 EU Verordnung 2016/679.\n\nDie frühere Zustimmung von der Benutzer ist nicht angefordert in Befehl zu Installieren diese Kekse.\n\n**Analyse-Cookies**\n\nDie Website verwendet nur Google Analytics, mit dem Profile der Benutzer erstellt werden und werden verwendet, um Werbebotschaften gemäß den von ihnen angezeigten Präferenzen zu versenden während ihrer Online-Navigation. Aufgrund ihrer besonderen Invasivität in Bezug auf die Privatsphäre der Benutzer Kugel, europäisch und Italienisch Vorschriften benötigen das Benutzer sein angemessen unterrichtet Über ihr verwenden von der gleich und sind daher erforderlich zu ausdrücken ihr gültig Zustimmung . Aber in dem konkreten Fall wurde Google Analytics anonymisiert (IP-Maskierung) und die Weitergabe von Navigationsdaten an Google blockiert: Auf diese Weise ähnelt das analytische Cookie den oben angegebenen technischen Cookies und erfordert keine Zustimmung.\n\n**Profiling-Cookies**\n\nDiese Art von Cookies wird auf der Website nicht verwendet.\n\nSpezifischer Hinweis:\n\nDie eingebetteten Videos von YouTube auf der Website verwenden keine Cookies, da „nocookie “, der datenschutzerweiterte Einbettungscode für alle Ihre YouTube-Videoeinbettungen, angegeben wurde.\n\n**Internationale und europäische Datenübertragung**\n\nIhre Daten werden ausschließlich im Europäischen Wirtschaftsraum verarbeitet. Ihre Rechte bzgl.persönlich Daten wir halten unter EU Verordnung 2016/679\n\n**Deine Rechte**\n\nSie kann die Übung dein Rechte irgendein Zeit, als einstellen aus durch Artikel 7, Par. 3, und Artikel fünfzehn und folgendevon EU Verordnung 2016/679:\n\n- Rechts zugreifen persönlich Daten\n- Rechts zu Berichtigung und Löschen von persönlich Daten;\n- Rechts zu Beschränkung von wird bearbeitet;\n- Rechts zu Daten Portabilität;\n- Rechts zu Objekt zu wird bearbeitet von persönlich Daten\n- Rechts zu legal beanspruchen zu Italienisch Daten Schutz Behörde.\n\nSie können Ihre Rechte ausüben, indem Sie uns eine E-Mail an info@pixel-online.net oder einen adressierten Brief senden an Pixel, über Luigi Lanzi , 12 – 50134 – Florenz, Italien. Weitere Informationen zur Datenverarbeitung kann wann hinzugefügt werden Daten sammeln.\n\n12. Februar 2022 rev.03\nDansk: Privatlivspolitik i henhold til EU-forordning 2016/679\n\nHvem indsamler dine data\nI henhold til artikel 13 i EU-forordning 2016/679 (GDPR) er den dataansvarlige ansøger af projektet, og denne er ansvarlig for at indsamle dataene. Vi vil gerne informere dig om, at vores organisation er juridisk forpligtet til at behandle de data, du har givet os i henhold til ovennævnte forordning.\nDine data vil blive behandlet lovligt og retfærdigt i henhold til bestemmelsen i artikel 5 i EU-forordning 2016/679. Yderligere detaljer vil muligvis blive givet på et senere tidspunkt.\nDatabeskyttelsesansvarlig (dpo): tilstedeværelsen af en eventuel DPO skal anmodes om hos den dataansvarlige.\n\nHvilke personlige data vi indsamler\nI overensstemmelse med Artikel 4 af EU Regulering 2016/679:\n- \"personoplysninger\" betyder enhver information, der vedrører en identificeret eller identificerbar fysisk person ('data emne'); en identificerbar naturlig person er en WHO kan være identificeret, direkte eller indirekte, især ved henvisning til en identifikator, såsom et navn, et identifikationsnummer, Beliggenhed data, en online identifikator eller til en eller mere faktorer bestemt til det fysisk, fysiologiske, genetiske, mental, økonomisk, kulturel eller social identitet af at naturlig person;\n- \"forarbejdning\" midler nogen operation eller sæt af operationer hvilken er udført på personlig data eller på sæt af personlig data, om eller ikke ved automatiseret midler, sådan som kollektion, registrering, organisering, strukturering, opbevaring, tilpasning eller ændring, genfinding, konsultation, brug, offentliggørelse ved transmission, formidling eller på anden måde tilgængeliggørelse, tilpasning eller kombination, begrænsning, sletning eller ødelæggelse.\n\nMed reference til det over nævnte definitioner, vi understrege at vi indsamle kun det Information du give os til formål af dit engagement i vores initiativer og/eller din lovlige forhold med vores organisation:\n- Personlige oplysninger: navn og efternavn på fysiske personer, kontakter såsom adresse, postnummer kode, by, område, telefon nummer, e-mail;\n- Data vedrørende fagfolk/organisationer/virksomheder: Information vedrørende virksomheder, navn, skatteadresse og andre identifikatorer (fax- og telefonnummer, skattekode eller moms nummer).\n\nDesuden kan vi indsamle data, der leveres, når du besøger vores websteder, gennem cookies og andet lignende teknologi; og når du kontakter os via e-mail, sociale medier eller lignende teknologier. Selvom sådanne data ikke er indsamlet med henblik på at blive knyttet til den fysiske person, er disse online identifikatorer kan bruges og kombineres for at skabe personlige profiler. Blandt online identifikatorer, vi kan finde IP-adresse, browsertype og plug-in detaljer, enhedstype (f.eks. desktop, bærbar, tablet, telefon osv.) styresystem, lokal tidszone. Disse data bruges udelukkende til produktion af statistisk resultater.\nVi vil gerne minde dig om, at vi ikke vil behandle personoplysninger, der afslører race eller etnisk oprindelse, politiske holdninger, religiøse eller filosofiske overbevisninger eller fagforeningsmedlemskab, og behandlingen af genetiske data, biometriske data med det formål entydigt at identificere en naturlig person, data vedrørende sundhed eller data vedrørende -en naturlig personens køn liv eller seksuel orientering.\n\nHvorfor og hvordan vi behandler dine data\nVi vilje brug dit data i det følgende måder:\n\n25. Organiser og implementer initiativer inden for uddannelse og træning (f.eks. træningkurser, konferencer, europæisk projekter etc.)\n26. Til fremstille administrative Dokumenter (for eksempel fakturaer) i forhold til det initiativer over\n27. Til statistisk formål\n28. Bære ud meddelelse aktiviteter via e-mail vedrørende vores initiativer.\n29. Besvar forespørgsler ved hjælp af formularerne på webstedet (hvis det er til stede)\n30. Tillad registrering for adgang til fortroligt undervisningsindhold (hvis det er til stede)\n\nDin overdragelse er obligatorisk til formål i henhold til paragraf 1, 2, 5, 6 for at overholde juridiske forpligtelser og EU love og forskrifter; afslag til give personlig data vilje ikke give lov tilvores organisation til tilbud du, vores tjenester.\n\nDit samtykke er valgfrit til formål i henhold til punkt 3 og 4; vi sender dig markedsføring kommunikation via e-mail eller postvæsen. Du kan til enhver tid udøve dine rettigheder i henhold til artikel 15 og senere i EU-forordning 2016/679 vedrørende fravalg af at modtage sådanne meddelelse eller at vælge andet meddelelse modaliteter.\n\nVi vil opbevare dine personlige data indsamlet til formålene under alle paragraffer så længe som vi har brug for for at give dig de tilbudte tjenester af vores organisation og for op til 10 (ti) år.\n\nDu kan trække sig tilbage dit samtykke på nogen tid.\n\nRetsgrundlaget for behandlingen består i det kommercielle forhold, der er skabt ved salget eller køb af varer og/eller tjenesteydelser, forudgående kontraktligt til orientering (artikel 6, stk. b og c),og ved samtykke til markedsføring aktiviteter. ( artikel 6 afsnit en)\n\nVi behandler og opbevarer dine data udelukkende til ovennævnte formål ved hjælp af digitale enheder og i relevante databaser for at sikre passende sikkerhedsforanstaltninger for at sikre løbende fortrolighed, integritet, tilgængelighed og modstandsdygtighed af behandlingssystemer, som fastsat i EU-forordning 2016/679. Kun personer, der har fået adgang til personoplysninger fra den dataansvarlige eller databehandleren, kan behandle sådan Information.\n\nVi hverken sælger, handler eller på anden måde overfører dine personligt identificerbare til andre tredjeparter Information. Vi kan dog frigive dine oplysninger, når vi mener, at frigivelse er nødvendig overholde med det lov, håndhæve vores websted politikker, eller beskytte vores eller andres rettigheder, ejendom, eller sikkerhed.\n\nProfilering\nDine data vil ikke blive underlagt en beslutning, der udelukkende er baseret på automatiseret behandling, som har retsvirkninger, der påvirker dem eller som væsentligt påvirker dens\nperson. Annullering og ændring: du har ret til til enhver tid at vide, hvad dine data er hos de enkelte dataansvarlige, det vil sige hos vores virksomhed eller hos de ovennævnte personer, som vi kommunikerer dem til, og hvordan de bruges; de har også ret til at opdatere, supplerere, rette eller annullere dem, anmode om deres blokering og modsætte sig deres behandling. For at udøve dine rettigheder, samt for mere detaljerede oplysninger om de emner eller kategorier af emner, som dataene kommunikeres til, eller som er klar over det som ledere eller agenter, kan du kontakte den dataansvarlige eller en af dennes ledere, identifieret i denne udtalelse.\n\nSociale netværk\nVores hjemmeside kan tilbyde adgang til sociale netværk. Servicevilkårene og privatlivspolitikken er gældende til sådanne platforme offentliggøres på deres hjemmeside. Pixel kan ikke kontrollere, hvordan data deles på en offentlig forum, snak eller dashboard.\n\nKlager\nDu kan også kontakte det italiensk Data Beskyttelse Myndighed ved brug af det følge link http://www.garanteprivacy.it/home/footer/contatti, eller den europæiske tilsynsførende for databeskyttelse ved hjælp af følge link: https://edps.europa.eu/data-protection/our-role-supervisor/complaints_en\n\nCookies\nSom se t ud t ved re g ula t i o n \"Retningslinjer for cookies og andre sporingsværktøjer - 10. juni 2021\", t her a re tre hovedkategorier af cookies:\n\nTeknisk cookies\nDisse er Brugt til det eneste formål af \"sender kommunikation til en elektronisk kommunikationsnetværk, eller i det omfang det er strengt nødvendigt for levering af en tjeneste af informationsvirksomhed, som aftaleparten eller brugeren udtrykkeligt har anmodet om for at give den nævnte service\" Disse bruges ikke til skjulte formål, og de er normalt installeret direkte ved det ejer eller det Manager af det internet side (såkaldt \"proprietære\" eller \"redaktionel\" cookies). Disse kan være opdelt ind i: browsing eller session cookies, hvilken garanti normal navigation og brug af hjemmesiden (gør det muligt for eksempel at foretage køb eller være autentificeret for at få adgang til reserverede områder); analytics cookies assimileret af den tekniske cookies, hvor de bruges direkte af administratoren af hjemmesiden til at indsamle oplysninger, i en tilhørende formular (anonym), om antallet af brugere og måden, hvorpå de besøger internet side; funktionelle cookies, der giver brugeren mulighed for at navigere i forhold til en række udvalgte kriterier (for eksempel sproget eller de produkter, der er valgt til køb) for at forbedre servicen stillet til rådighed, forudsat at vi informerer vores brugere som satte ud ved artikel 13 EU Regulering 2016/679.\nDet tidligere samtykke af det bruger er ikke anmodet om i bestille til installere Disse cookies.\n\nAnalytics-cookies\nSiden bruger kun google analytics, som bruges til at oprette profiler af brugerne og er ansat til at sende reklamebeskedder i henhold til præferencerne vist af samme under deres online navigation. På grund af deres særige invasivitet med hensyn til brugernes private sfære, europæisk og italiensk forskrifter kræve at brugere være tilstrækkeligt informeret om deres\nbrug af det samme og er dermed påkrævet til udtrykke deres gyldig samtykke. Men i det konkrete tilfælde er google analytics blevet anonymiseret (IP-maskering), og deling af navigationsdata med google er blevet blokeret: på denne måde ligner den analytiske cookie de tekniske cookies angivet ovenfor og kræver ikke samtykke.\n\n**Profileringscookies**\nDenne type cookie bruges ikke på siden.\n\nSpecifik bemærkning:\nDe indlejrede videoer af YouTube på webstedet bruger ikke cookies, da det er blevet specificeret \" nocookie \" den fortrolighedsforbedrede indlejringskode for alle dine YouTube-videoindlejringer.\n\n**International og europæisk dataoverførsel**\nDine data vil udelukkende blive behandlet i Det Europæiske Økonomiske Samarbejdsområde. Dine rettigheder mht personlig data vi holde under EU Regulering 2016/679\n\n**Dine rettigheder**\nDu kan dyrke motion dit rettigheder nogen tid, som sæt ud ved Artikel 7, stk. 3, og artikler 15 og følgeaf EU regulering 2016/679:\n- Højre adgang personlig data\n- Højre til berigtigelse og sletning af personlig data;\n- Højre til begrænsning af forarbejdning;\n- Højre til data bæbarhed;\n- Højre til objekt til forarbejdning af personlig data\n- Højre til gyldige påstand til italiensk Data Beskyttelse Myndighed.\n\nDu kan udøve dine rettigheder ved at sende os en e-mail på info@pixel-online.net eller et adresseret brev til Pixel, via Luigi Lanzi, 12 – 50134 – Firenze, Italien. Yderligere information om databehandling kan tilføjes hvornår indsamle data.\n\n12 februar 2022 rev.03\nEesti keel : Privaatsuspoliitika vastavalt EL määrusele 2016/679\n\nKes teie andmeid kogub\nEL määruse 2016/679 (GDPR) artikli 13 kohaselt on vastutav andmetöötleja projekti taotleja, kes vastutab andmete kogumise eest. Anneme teile teada, et meie organisatsioon on juridiliselt kohustatud töötlema andmeid, mille olete meile eelnimetatud määruse alusel esitanud.\nTeie andmeid töödeldakse EL määruse 2016/679 artikli 5 alusel seaduslikult ja õiglaselt. Täpsemat teavet võidakse esitada hilisemas etapis.\nAndmekaitseametnik (dpo): võimaliku andmekaitseametniku kohalolekut tuleb taotleda vastutavalt töötlejalt.\n\nMilliseid isikuandmeid me kogume\nsisse kooskõlas koos Artikkel 4 kohta EL määruse 2016/679:\n- „isikuandmed“ – igasugune teave tuvastatud või tuvastatava füüsilise isiku kohta (‘andmed teema’); an tuvastatav loomulik isik on üks WHO saab olla tuvastatud, otse või kaudselt, eelkõige viidates identifikaatorile, nagu nimi, identifitseerimisnumber, asukoht andmed, an võrgus identifikaator või juurde üks või rohkem tegevusid spetsiifiline juurde a füüsiline, füsioloogiline, geneetiline, vaimne, majanduslik, kultuuriline või sotsiaalne identiteet et loomulik isik;\n- \"töötlemine\" tähendab ükskõik milline operatsiooni või seatud kohta operatsioonid mis on sooritatud peal isiklik andmeid või peal komplektid kohta isiklik andmed, kas või mitte kõrval automatiseritud tähendab, selline nagu kollektsooni, salvestamine, organiseerimine, struktureerimine, säilitamine, kohandamine või muutmine, otsimine, konsulteerimine, kasutamine, avalikustamine edastamise, levitamine või muul viisil kättesaadavaks tegemise teel, vastavusse viimine või kombinatsiooni, piirang, kustutamine või hävitamine.\n\nKoos viide juurde a eespool mainitud määratlused, meie joon alla et meie koguda ainult a teavet sina anda EL määruse 2016/679:\n- Isikuandmed: füüsiline isikute ees- ja perekonnanimi, kontaktid nagu aadress, ZIP kood, linn, piirkond, telefon number, email;\n- Andmed mis puudutab spetsialistid/organisatsioonid/ettevõtted: teavet mis puudutab ettevõtted, nimi, maksuaadress ja muud tunnused (faksi- ja telefoninumber, maksukood või käibemaks number).\n\nLisaks võime koguda küpsiste ja muu kaudu meie saitidele sisenemisel esitatud andmeid sarnane tehnoloogia; ja kui võtate meiega ühendust e-posti, sotsiaalmeedia või sarnaste tehnoloogiate kaudu. Kuigi selliseid andmeid ei koguta füüsiline isikuga seostamiseks, on need Isiklike profiilide loomiseks võidakse kasutada ja kombineerida veebipõhiseid identifikaatorid. Interneti hulgas identifikaatorid, mida võime leida IP-aadressi, brauseri tüübi ja pistikprogrammi üksikasjad, seadme tüübi (nt töölaud, sülearvuti, tahvelarvuti, telefoni jne) operatsioonisüsteem, kohalik ajavöönd. Neid andmeid kasutatakse ainult tootmine statistiline\ntulemused.\n\nTuletame meelde, et me ei töötle isikuandmeid, mis paljastavad rassilise või etniline päritolu, poliitilised vaated, usulised või filosoofilised veendumused või ametiühingusse kuulumine ja geneetiliste andmete, biomeetriliste andmete töötlemine loodusliku isiku unikaalse tuvastamise eesmärgil inimene, andmeid mis puudutab tervist või andmeid mis puudutab a loomulik inimese oma seks elu või seksuaalne orientatsiooni.\n\nMiks ja kuidas me teie andmeid töötleme\nMeie tahe kasutada sinu andmeid järgnev viisid:\n31. Korraldada ja viia ellu algatusi hariduse ja koolituse valdkonnas (nt koolitus kursused, konverentsid, euroopalik projektid jne.)\n32. To toota administratiivne dokumente (nt arved) sisse suhe juurde a algatused eespool\n33. Sest statistiline eesmärkidel\n34. Kanna välja suhtlemine tegevused kaudu meili mis puudutab meie algatused.\n35. Päringutele vastamine saidil olevate vormide abil (kui see on olemas)\n36. Luba registreerimine juurdepääsuks konfidentsiaalsele haridussisule (kui see on olemas)\n\nTeie kinkimine on lõigetes 1, 2, 5 ja 6 sättestatud eesmärkidel kohustuslik, et järgida juriidiline kohustusi ja EL seadused ja määrused; keeldumine juurde pakkuda isiklik andmeid tahe mitte lubamameie organisatsiooni juurde pakkuma sina, meie teenuseid.\nTeie nõusolek on lõigetes 3 ja 4 sättestatud eesmärkidel vabatahtlik; saadame teile turundust suhtlemine e-posti või posti teel. Saate oma õigusi igal ajal kasutada vastavalt EL-i määruse 2016/679 artikkel 15 ja hilisemad, mis käsitlevad sellise saamisest loobumist suhtlemine või valides muu suhtlemine viisid.\nSäilitame teie isikuandmeid, mis on kogutud kõigis lõigetes nimetatud eesmärkidel, nii kaua nagu me vajame teile pakutavate teenuste pakkumiseks meie organisatsiooni poolt ja kuni 10 (kümme aastat.\nSina saab taganema sinu nõusolekut juures ükskõik milline aega.\n\nRavi õiguslik alus seisneb müügiga tekkinud ärisuhtes või kaupade ja/või teenuste ostmine, lepingueelne teavitamine (artikli 6 lõiked b ja c), ja nõusolekul jaoks juures tegevused. (artikkel 6 lõik a)\n\nTöötleme ja säilitame teie andmeid ainult eelnimetatud eesmärkidel, kasutades digitaalseid seadmeid ja asjakohastes andmebaasides, tagades asjakohased kaitsemeetmed, et tagada pidev konfidentsiaalsus, töötlemissüsteemide terviklikkus, kättesaadavus ja vastupidavus, nagu on sättestatud EL määruses 2016/679. Ainult subjektid, kes on saanud vastutavalt töötlejalt või volitatud töötlejalt juurdepääsu isikuandmetele protsessi selline teavet.\n\nMe ei müü, ei vaheta ega anna muul viisil viisil teistele kolmandatele isikutele üle teie isikut tuvastada teavet. Siiski võime teie teavet avaldada, kui usume, et see on vajalik järgima koos a seadus, jõustada meie sait poliitika, või kaitsta meie oma või teiste oma õigused, vara, või ohutus.\n\nProfileerimine\nTeie andmete suhtes ei kohaldata üksnes automatiseeritud töötlemisel põhinevat otsust, millel on õiguslikke tagajärgi, mis mõjutavad neid või mis mõjutavad oluliselt tema isikut.\nTühistamine ja muutmine: teil on igal ajal õigus teada, millised on teie andmed vastutavate andmetöötlejate juures, st meie ettevõttes või ülalnimetatud isikute juures, kellele me neid edastame, ja kuidas neid kasutatakse; neil on ka õigus neid ajakohastada, täiendada, parandada või tühistada, taotleda nende blokeerimist ja olla vastu nende ravile. Oma õiguste kasutamiseks, samuti täpsema teabe saamiseks nende subjektide või subjektide kategooriate kohta, kellele andmeid edastatakse või kes on sellest teadlikud kui juhid või esindajad, võivad võtta ühendust vastutava töötleja või ühe tema haldajaga, kes on märgitud see väide.\n\nSotsiaalsed võrgustikud\nMeie veebisait võib pakkuda juurdepääsu sotsiaalvõrgustikule. Kohaldatavad teenusettingimused ja privaatsuspoliitika sellistele platvormidele avaldatakse nende veebisaidil. Pixel ei saa juhtida seda, kuidas andmeid jagatakse avalikfoorum, vestlus või armatuurlaud on kasutatud, olemine aandmeid teema vastutav kohta selline suhtlemine.\n\nKaebused\nSina saab samuti kontakti a itaalia keel Andmed Kaitse Asutus kasutades a järgnev link http://www.garanteprivacy.it/home/footer/contatti või Euroopa andmekaitseinspektor, kasutades järgnev link: https://edps.europa.eu/data-protection/our-role-supervisor/complaints_en\n\nKüpsised\nNagu se t ou t kõrval määrus “ Küpsiste ja muude jälgimistööriistade juhised – 10. juuni 2021” , t siin on kolm peamist kategooriat kohta küpsised:\n\nTehniline küpsised\nNeed on kasutatud jaoks a tald eesmärk kohta \"edastades side juurde an elektrooniline sidevõrgus või ulatuses, mis on vältimatult vajalik teenuse osutamiseks lepinguosaline või kasutaja selgesõnaliselt taotlenud teabefirmat nimetatud teenus” Neid ei kasutata mitte mingiks varjatud otstarbeks ja need on tavaliselt paigaldatud otse kõrval a omanik või a juht kohta a veebisait (nn \"varaline\" või \"toimetus\" küpsised). Need saab olla jagatud sisse: sirvimine või istungil küpsised, mis garantii normaalne veebisaidil navigeerimine ja kasutamine (võimaldab näiteks ostude sooritamist või viibimist autenditud, et pääseda reserveeritud aladele); tehniliste poolt samastatud analüütilised küpsised küpsised, kui veebisaidi haldur kasutab neid otse teabe kogumiseks seotud vorm (anoniümne), kasutajate arvu ja selle külalastamise viisi kohta veebisait; funktsionaalsed küpsised, mis võimaldavad kasutajal teatud kriteeriumide alusel navigeerida (näiteks keel või ostmiseks valitud tooted), et teenust täiustada tingimusel, tingimusel, et teavitame oma kasutajad nagu välja seadnud kõrval artiklit 13 EL määrus 2016/679.\n\nThe eelnev nõusolekut kohta a kasutaja on mitte taotletud sisse tellida juurde installida need küpsised.\n\nAnalyticisi küpsised\nSait kasutab ainult google analüütitakat, mille abil luuakse kasutajate ja kasutajate profiile kasutatakse reklamšöömitse saatmiseks vastavalt sama näidatud eelistustele veebis navigeerimise ajal. Nende erilise invasiivsuse tõttu kasutajate privaatsuse suhtes kera, euroopalik ja itaalia keel määrused nõuda et kasutajad olla adekvaatselt teavitatud umbes nende kasutada kohta a sama ja on seega nõutud juurde väljendada nende kehtiv nõusolek . Kuid konkreetsel juhul on google analytics anonüümseks muudetud (IP maskeerimine) ja navigeerimisandmete jagamine Google'iga blokeeritud: sel viisil sarnaneb analüütiline küpsis\nülaltoodud tehniliste küpsistega ega vaja nõusolekut.\n\n**Profileerimisküpsised**\nSeda tüüpi küpsiseid sellel saidil ei kasutata.\n\nKonkreetne märkus:\nSaidil olevad YouTube'i manustatud videod ei kasuta küpsiseid, kuna kõigile teie YouTube'i videote manustele on määratud \"nocookie\" privaatsusega täiustatud manustamiskood.\n\n**Rahvusvaheline ja Euroopa andmeedastus**\nTeie andmeid töödeldakse ainult Euroopa Majanduspiirkonnas. Teie õigused seoses isiklik andmeid meie hoia all EL määrus 2016/679\n\n**Sinu õigused**\nSina saab harjutus sinu õigused ükskõik milline aeg, nagu seatud välja kõrval Artikkel 7, par. 3, ja artiklid 15 ja järgnevkohta EL määrus 2016/679:\n\n- Õige ligi pääsema isiklik andmeid\n- Õige juurde parandamine ja kustutamine kohta isiklik andmed;\n- Õige juurde piirang kohta töötlemine;\n- Õige juurde andmeid teisaldatavus;\n- Õige juurde objektiis juurde töötlemine kohta isiklik andmeid\n- Õige juurde seaduslik väide juurde itaalia keel Andmed Kaitse Asutus.\n\nSaate oma õigusi kasutada, saates meile e- kirja aadressile info@pixel-online.net või adresseeritud kirja Pixelile Luigi Lanzi kaudu, 12 – 50134 – Firenze, Itaalia. Lisateave andmetöötluse kohta saab lisada millal andmete kogumine.\n\n12. veebruar 2022. aasta rev.03\nEspañol: Política de Privacidad conforme al Reglamento UE 2016/679\n\nQuién recopila sus datos\nDe conformidad con el artículo 13 del Reglamento de la UE 2016/679 (GDPR), el controlador de datos es el solicitante del proyecto y es responsable de recopilar los datos. Le informamos que nuestra organización está legalmente obligada a procesar los datos que nos ha proporcionado en virtud de la normativa antes mencionada.\nSus datos serán tratados de forma lícita y leal, en virtud de lo dispuesto en el artículo 5 del Reglamento UE 2016/679. Es posible que se proporcionen más detalles en una etapa posterior.\nde protección de datos (dpo): se debe solicitar al responsable del tratamiento la presencia de un posible DPO.\n\nQué datos personales recopilamos\nEn conformidad con Artículo 4 de UE Regulación 2016/679:\n- “datos personales” significa cualquier información relativa a una persona física identificada o identificable (‘datos tema’); un identificable natural persona es una quién lata ser identificado, directamente o indirectamente, en particular por referencia a un identificador como un nombre, un número de identificación, ubicación datos, un en línea identificador o para una o más factores específico para los físico, fisiológico, genético, mental, económicos, culturales o social identidad de que natural persona;\n- \"Procesando\" medio ninguna operación o colocar de operaciones cual es realizado en personal datos o en conjuntos de personal datos, ya sea o no por automatizado medio, tal como colección, grabación, organización, estructuración, almacenamiento, adaptación o alteración, recuperación, consulta, uso, divulgación por transmisión, difusión o puesta a disposición de otro modo, alineamiento o combinación, restricción, borradura o destrucción.\n\nCon referencia para los encima mencionado definiciones, nosotros subrayar que nosotros recolectar solamente los información usted proporcionarnos para el propósitos de su participación en nuestro iniciativas y/o su legal relación con nuestro organización:\n- Datos personales: nombre y apellidos de personas físicas, contactos como dirección, código postal código de ciudad, región, teléfono número, Email;\n- Datos sobre profesionales/organizaciones/empresas: información sobre negocios, nombre, dirección fiscal y otros identificadores (número de fax y teléfono, código fiscal o IVA número).\n\nAdemás, podemos recopilar datos proporcionados cuando accede a nuestros sitios, a través de cookies y otros tecnología similar; y cuando se comunica con nosotros por correo electrónico, redes sociales o tecnologías similares. Si bien dichos datos no se recopilan para asociarlos a la persona física, estos los identificadores en línea pueden usarse y combinarse para crear perfiles personales. entre los en línea identificadores que podemos encontrar dirección IP, tipo de navegador y detalles del complemento, tipo de dispositivo (por ejemplo, escritorio, computadora portátil, tableta, teléfono, etc.) sistema operativo, zona horaria local.\nEstos datos se utilizan únicamente para la producción de estadístico resultados\n\nLe recordamos que no procesaremos datos personales que revelen raza o origen étnico, opiniones políticas, creencias religiosas o filosóficas, o afiliación sindical, y el tratamiento de datos genéticos, datos biométricos con el fin de identificar de forma unívoca a un persona, datos sobre salud o datos sobre a natural de la persona sexo la vida o sexual orientación.\n\nPor qué y cómo procesamos sus datos\nNosotros voluntad utilizar tu datos en la siguiente formas:\n\n37. Organizar e implementar iniciativas en el campo de la educación y la formación (p. ej., formación cursos, conferencias, europeo proyectos etc)\n38. Para Produce administrativo documentos (p.ej facturas) en relación para los iniciativas encima\n39. Para estadístico propósitos\n40. Llevar fuera comunicación ocupaciones vía Email sobre nuestro iniciativas.\n41. Responder a las solicitudes utilizando los formularios en el sitio (si está presente)\n42. Permitir el registro para acceder a contenido educativo confidencial (si está presente)\n\nSu concesión es obligatoria a los efectos de los párrafos 1, 2, 5, 6 para cumplir con jurídico obligaciones y UE leyes y reglamentos; rechazo para proveer personal datos voluntad no permitirnuestro organización a oferta usted, Nuestros servicios.\n\nSu consentimiento es opcional a los efectos de los párrafos 3 y 4; le enviaremos marketing comunicación a través de correo electrónico o servicio postal. Puede ejercer sus derechos en cualquier momento, de conformidad con Artículo 15 y posteriores del Reglamento de la UE 2016/679 con respecto a la opción de no recibir dicha comunicación o eligiendo otro comunicación modalidades.\n\nConservaremos sus datos personales recopilados para los fines previstos en todos los párrafos mientras según sea necesario para brindarle los servicios ofrecidos por nuestra organización y hasta por 10 (diez años).\n\nTú lata retirar tu consentimiento en ninguna hora.\n\nLa base jurídica del tratamiento consiste en la relación comercial creada por la venta o compra de bienes y/o servicios, precontractuales para información (artículo 6 inciso b y c), y por consentimiento por marketing ocupaciones. (artículo 6 párrafo a)\n\nProcesaremos y almacenaremos sus datos únicamente para los fines antes mencionados, utilizando dispositivos digitales. y en bases de datos relevantes asegurando salvaguardas apropiadas para asegurar la confidencialidad continua, integridad, disponibilidad y resiliencia de los sistemas de procesamiento, tal como lo establece el reglamento de la UE 2016/679. Solo los sujetos que hayan obtenido acceso a los datos personales del controlador o del procesador pueden proceso tal información.\n\nNo vendemos, intercambiamos ni transferimos de otro modo a otros terceros su identificación personal información. Sin embargo, podemos divulgar su información cuando creamos que la divulgación es necesaria para cumplir con los ley, hacer cumplir nuestro sitio políticas, o proteger nuestro o otros’ derechos, propiedad, o la seguridad.\n\nperfilado\nSus datos no serán objeto de una decisión basada únicamente en el tratamiento automatizado, que produzca efectos jurídicos que le afecten o que afecte significativamente a su persona. Cancelación y Modificación: tiene derecho a saber, en cualquier momento, cuáles son sus datos en los controladores de datos individuales, es decir, en nuestra empresa o en las personas mencionadas anteriormente a quienes los comunicamos, y cómo se utilizan; también tienen derecho a actualizarlos, complementarlos, corregirlos o cancelarlos, solicitar su bloqueo y oponerse a su tratamiento. Para el ejercicio de sus derechos, así como para obtener información más detallada sobre los sujetos o categorías de sujetos a los que se comunican los datos o que tienen conocimiento de ellos como encargados o agentes puede dirigirse al responsable del tratamiento o a uno de sus encargados, identificado en esta declaración.\n\nRedes sociales\nNuestro sitio web puede ofrecer acceso a redes sociales. Los términos de servicio y la Política de Privacidad aplicable a dichas plataformas se publican en su sitio web. Pixel no puede controlar la forma en que se comparten los datos en un público foro, chat o tablero están usado, siendo los datos tema responsable de tal comunicación.\n\nQuejas\nTú lata además contacto los italiano Datos Proteccion Autoridad utilizando los siguiente Enlace http://www.garanteprivacy.it/home/footer/contatti, o el Supervisor Europeo de Protección de Datos utilizando el siguiente enlace: https://edps.europa.eu/data-protection/our-role-supervisor/complaints_en\n\nGalletas\nComo establecer _ fuera _ por reg ula c i ó n “ Directrices sobre cookies y otras herramientas de seguimiento - 10 de junio de 2021”, aquí _ son tres categorías principales de galletas:\n\nTécnico galletas\nEstas están usado por los único propósito de “transmitiendo comunicaciones para un electrónico red de comunicación, o en la medida estrictamente necesaria para la prestación de un servicio por parte del información de la empresa expresamente solicitada por el contratante o el usuario con el fin de proporcionar dicho servicio” No se utilizan para fines ulteriores y normalmente se instalan directamente por los dueño o los gerente de los sitio web (así llamado \"propiedad\" o \"editorial\" galletas). Estas lata ser dividido dentro: hojead o sesión galletas, cual garantía normal navegación y uso del sitio web (permitiendo, por ejemplo, realizar compras o ser autenticado para acceder a las áreas reservadas); cookies analíticas asimiladas por los técnicos cookies cuando son utilizadas directamente por el administrador del sitio web para recopilar información, en un formulario asociado (anónimo), sobre el número de usuarios y la forma en que visitan el sitio web; cookies funcionales que permiten al usuario navegar en relación a una serie de criterios de selección (por ejemplo, el idioma o los productos seleccionados para la compra) con el fin de mejorar el servicio proporcionó, siempre que informamos a nuestro usuarios como exponer por artículo 13 UE Regulación 2016/679.\nlos previo consentimiento de los usuario es no solicitado en pedido para Instalar en pc estas galletas.\n\nCookies analíticas\nEl sitio utiliza solo Google Analytics, que se utiliza para crear perfiles de los usuarios y son empleados para el envío de mensajes publicitarios de acuerdo con las preferencias mostradas por los mismos durante su navegación en línea. Por su particular invasividad con respecto a la privacidad de los usuarios esfera, europeo y italiano regulaciones exigir que usuarios ser adecuadamente informado sobre sus utilizar de los mismo y están por lo tanto requerido para Rápido sus válido consentimiento _ Pero en el caso específico, Google Analytics se ha anonimizado (enmascaramiento de IP) y se ha bloqueado el intercambio de datos de navegación con Google: de esta manera, la cookie analítica es similar a las cookies técnicas indicadas anteriormente y no requiere consentimiento.\n\n**Cookies de perfil**\nEste tipo de cookie no se utiliza en el sitio.\n\nNota específica:\nLos videos incrustados de YouTube en el sitio no utilizan cookies, ya que se ha especificado \"nocookie \", el código de incrustación de privacidad mejorada para todas sus incrustaciones de videos de YouTube.\n\n**Transferencia de datos internacional y europea**\nSus datos serán tratados únicamente en el Espacio Económico Europeo. Sus derechos con respecto a lapersonal datos nosotros sostener bajo UE Regulación 2016/679\n\n**Tus derechos**\nTú lata ejercicio tu derechos ninguna hora, como colocar fuera por Artículo 7, par. 3, y artículos 15 y siguiente de UE regulación 2016/679:\n- Derecha acceder personal datos\n- Derecha para rectificación y borradura de personal datos;\n- Derecha para restricción de Procesando;\n- Derecha para datos portabilidad;\n- Derecha para objeto para Procesando de personal datos\n- Derecha para legal afirmar para italiano Datos Proteccion Autoridad.\n\nPuede ejercer sus derechos enviándonos un correo electrónico a info@pixel-online.net o una carta dirigida a Pixel, a través de Luigi Lanzi , 12 – 50134 – Florencia, Italia. Más información sobre el tratamiento de datos Se puede agregar cuando recolectando datos.\n\n12 de febrero 2022 rev.03\nSuomeksi: EU-asetuksen 2016/679 mukainen tietosuojakäytäntö\n\nKuka kerää tietosi\nEU-asetuksen 2016/679 (GDPR) artiklan 13 mukaisesti rekisterinpitäjä on hankkeen hakija ja se vastaa tietojen keräämisestä. Haluamme ilmoittaa, että organisaatiomme on laillisesti velvollinen käsittelemään meille antamiasi tietoja edellä mainitun asetuksen mukaisesti. Tietojasi käsitellään laillisesti ja oikeudenmukaisesti EU-asetuksen 2016/679 artiklan 5 mukaisesti. Lisätietoja saatetaan toimittaa myöhemmin.\n\nTietosuojavastaava (dpo): mahdollisen DPO:n läsnäoloa on pyydetä rekisterinpitäjältä.\n\nMitä henkilötietoja keräämme\nSisäänmukaisesti kanssa Artikla 4 / EU Sääto 2016/679:\n- \"henkilötiedoilla\" tarkoitetaan kaikkia tunnistettuun tai tunnistettavissa olevaan luonnolliseen henkilöön liittyviä tietoja ('tiedot aihe'); an tunnistettavissa luonnollinen henkilö On yksi WHO voi olla tunnistettu, suoraan tai epäsuorasti, erityisesti viittaamalla tunnisteseen, kuten nimeen, tunnistenumeroon, sijainti tiedot, an verkossa tunnistetut tai kohtaan yksi tai lisää tekijät erityistä kohtaan the fyysinen, fysiologinen, geneettinen, henkinen, taloudellinen, kulttuurinen tai sosiaalinen henkilöllisyys että luonnollinen henkilö;\n- \"käsittely\" tarkoittaa minkä tahansa operaatio tai aseta / toiminnon mikä On suoritettu päällä henkilökohtainen tiedot tai päällä sarjat / henkilökohtainen tiedot, onko tai ei kirjoittaja automatisoitu tarkoittaa, sellaisia kuten kokoelma, tallennus, järjestäminen, jäsentäminen, varastointi, mukauttaminen tai muuttaminen, haku, konsultointi, käyttö, paljastaminen lähettämällä, levittämällä tai muutoin saataville, yhdistelmä, rajoitus, poistaminen tai tuhoaminen.\n\nKanssa viite kohtaan the edellä mainitsi määritelmät, me korostaa että me kerätä vain the tiedot sinä tarjota meille tietoiksi osallistumisesta meidän aloitteita ja tai sinun laillinen suhdetta kanssa meidän organisaatio:\n- Henkilötiedot: luonnollisten henkilöiden etu- ja sukunimi, yhteystiedot, kuten osoite, postinumero koodi, kaupunki, alue, puhelin määrä, sähköposti;\n- Data koskien ammattilaiset/järjestöt/yritykset: tiedot koskien yritykset, nimi, verotusosoite ja muut tunnisteet (faksi- ja puhelinnumero, verotunnus tai ALV määrä).\n\nLisäksi voimme kerätä tietoja, jotka on annettu, kun käytät sivustojamme, evästeiden ja muiden avulla samanlainen teknikka; ja kun otat meihin yhteyttä sähköpostitse, sosiaalisessa mediassa tai vastaavilla teknikoilla. Vaikka tällaisia tietoja ei kerätä liitettäväksi luonnolliseen henkilöön, nämä online-tunnisteita voidaan käyttää ja yhdistää henkilökohtaisten profiilien luomiseksi. Netin joukossa tunnisteta, joita voimme löytää IP-soittojen, selaimen tyyppin ja laajennuksen tiedot, laitetyyppin (esim. kannettava tietokone, tabletti, puhelin jne.) käyttöjärjestelmä, paikallinen aikavyöhyke. Näitä tietoja käytetään ainoastaan tuottama tilastollinen tuloksia.\n\nHaluamme muistuttaa, että emme käsittele henkilötietoja, jotka paljastavat rodun tai etninen\nalkuperä, poliittiset mielipiteet, uskonnolliset tai filosofiset vakaumukset tai ammattiliittojen jäsenyyys, ja geneettisten tietojen, biometristen tietojen käsittely luonnollisen yksilöivän yksilöllisen tunnistamiseksi henkilö, tiedot koskien terveys tai tiedot koskien a luonnollinen henkilöt seksia elämää tai seksuaalinen suuntautuminen.\n\nMiksi ja miten käsittelemme tietojasi\nMe tahtoo käyttää sinun tiedot sisään seuraavat tapoja:\n\n43. Järjestä ja toteutta koulutusalan aloitteita (esim kursseja, konferensseja, eurooppalainen hankkeita jne.)\n44. Vastaanottaja tuottaa hallinnollinen asiakirjoja (esim laskut) sisään suhde kohtaan the aloitteita edellä\n45. varten tilastollinen tarkoituksiin\n46. Kanna ulos viestintää toimintaa kautta sähköposti koskien terveys tai tiedot koskien a luonnollinen henkilöt sekä elämää tai seksuaalinen suuntautuminen.\n\nLahjoituksenne on pakollinen kohtien 1, 2, 5 ja 6 mukaisissa tarkoituksissa noudattaakseen juridinen velvoitteet ja EU lait ja määräykset; epäaminen kohtaan tarjota henkilökohtainen tiedot tahtoa ei sallia ja organisaatiolle tarjotaksemme sinulle palvelumme. Suostumuksesi on valinnainen kohtien 3 ja 4 mukaisissa tarkoituksissa; lähettämme sinulle markkinointia yhteydenpito sähköpostitse tai postitse. Voit käyttää oikeuksiasi milloin tahansa sen mukaisesti EU-asetuksen 2016/679 15 artikla ja myöhemmat kohdat, jotka koskevat tällaisten vastaanottamisen kieltäytymistä viestintää tai valita muuta viestintää menettelytavat.\n\nSäilytämme henkilötietojasi, jotka on kerätty kaikkien mukaisiin tarkoituksiin niin kauan kuin kuten tarvitsemme tarjotaksemme sinulle tarjottuja palveluita organisaatiomme toimesta ja jopa 10 (kymmenen vuotta. Sinä voi peruuttaa sinun suostumus klo minkä tahansa aika.\n\nHoidon oikeudellinen perusta on kaupalla syntyvä kauppasuhde tai tavaroiden ja/tai palvelujen ostot, tiedoksi ennen sopimusta (6 artiklan b ja c kohta),ja suostumuksella varten markkinointi toimintaa. ( artikkele 6 kohta a)\n\nKäsittelemme ja tallennamme tietojasi vain edellä mainittuihin tarkoituksiin digitaalisten laitteiden avulla ja asiaankuuluvissa tietokantoissa varmistaan asianmukaiset suojatoimet jatkuvan luottamuksellisuuden varmistamiseksi, käsittelyjärjestelmien eheys, saatavuus ja joustavuus EU-asetuksen 2016/679 mukaisesti. Vain henkilöt, jotka ovat saaneet pääsyn henkilötietoihin rekisterinpitäjältä tai käsittelijältä, voivat käsitellä asioita sellaisia tiedot.\n\nEmme myy, vaihda tai muutoin siirrä muille kolmansille osapuolille, jotka voit tunnistaa henkilökohtaisesti tiedot. Voimme kuitenkin luovuttaa tietosi, jos uskomme, että se on tarpeen noudattaa kanssa the laki, panna täytäntöön meidän sivusto politiikkaa, tai suojella meidän tai muiden oikeudet, omaisuutta, tai turvallisuutta.\n\nProfilointi\nTietoihisi ei tehdä yksinomaan automaattiseen käsitteleyyn perustuvaa päätöstä, jolla on siihen vaikuttavia tai sen henkilöön merkittävästi vaikuttavia oikeusvaikutuksia . Peruuttaminen ja muuttaminen: sinulla on milloin tahansa oikeus tietää, mitkä tietosi ovat yksittäisillä\nrekisterinpitäjillä, eli yrityksellämme tai edellä mainituilla henkilöillä, joille välitämme ne, ja miten niitä käytetään; heillä on myös oikeus päivittää, täydentää, korjata tai peruuttaa niitä, pyytää niiden estoaa ja vastustaa hoitoa. Oikeuksien mukaan käyttämiseksi sekä tarkempien tietojen saamiseksi henkilöistä tai henkilöryhmistä, joille tietoja välitetään tai jotka ovat asiasta tietoisia johtajina tai asiamiehinä, voivat ottaa yhteyttä rekisterinpitäjään tai johonkin hänen johtajistaan, joka on tunnistettu Tämä lausunto.\n\nSosiaaliset verkostot\nSivustomme voi tarjota pääsyn sosiaaliseen verkostoon. Sovellettavat palveluehdot ja tietosuojakäytäntö tällaisille alustoille julkaistaan heidän verkkosivuillaan. Pixel ei voi hallita tapaa, jolla tiedot jaetaan a julkinen foorumi, keskustella tai kojelauta ovat käytetty, oleminen the tiedot aihe vastuussa / sellaisia viestintää.\n\nValituukset\nSinä voi myös ottaa yhteyttä the italialainen Data Suojaus viranomainen käyttämällä the seurata linkki http://www.garanteprivacy.it/home/footer/contatti tai Euroopan tietosuojavaltuutettu käyttämällä seurata linkki: https://edps.europa.eu/data-protection/our-role-supervisor/complaints_en\n\nKeksit\nKuten se t sinä t kirjoittaja sääntö \" Evästeiden ja muiden seurantatyökalujen ohjeet - 10.6.2021 \", t täällä kolme pääluokka / keksit:\n\nTekninen keksit\nNämä ovat käytetty varten the pohja tarkoitus / \"lähetettä viestintää kohtaan an elektroninen viestintäverkkoon tai siinä määrin kuin se on ehdottoman välttämätöntä palvelun tarjoamiseksi tietoyohtiö, jota sopimuspuoli tai käyttäjä on nimenomaisesti pyytänyt antamaan mainittu palvelu\" Näitä ei käytetä mihinkään myöhempään tarkoituksen ja ne asennetaan normaalisti suoraan kirjoittaja the omistaja tai the johtaja / the verkkosivusto (niin sanottu \"omistusoikeus\" tai \"Pääkirjoitus\" keksit). Nämä voi olla jaettu osaksi: selailua tai istunto keksit, mikä takuu normaali navigointi ja sivuston käyttö (jotka mahdollistavat esimerkiksi ostosten tekemisen tai olemisen todennetaan varattuille alueille pääsyä varten); analytiikkaevästeet, jotka vastaavat teknisiä evästeet, joissa verkkosivuston ylläpitäjä käyttää niitä suoraan tietojen keräämiseen liittyvät lomakkeet (anonyymi), käyttäjien lukumäärästä ja tavasta, jolla he vieraillevat verkkosivusto; toiminnallisia evästeitä, joiden avulla käyttäjä voi navigoida tiettyjen kriteerien perusteella (esimerkiksi ostettavaksi valittu kieli tai tuotteet) palvelun parantamiseksi edellyttäen, edellyttäen että ilmoitamme meille käyttäjiä kuten liikkeelle kirjoittaja artikla 13 EU Säätö 2016/679. The ennen suostumus / the käyttäjä On ei pyydetty sisään Tilaus kohtaan Asentaa nämä keksit.\n\nAnalytics-evästeet\nSivusto käyttää vain google analyticsia, jota käytetään profilien lumiseen käyttäjistä ja käyttäjistä käytetään mainosviestien lähettämiseen saman osoittamien mieltymysten mukaan online-navigoinnin aikana. Niiden erityisen invasiivisuuden vuoksi käyttäjien yksityisyysen suhteenpallo, eurooppalainen ja italialainen määräyksiä vaatia että käyttäjä olla riittävästi tiedotettu noin heidän käyttää / the sama ja ovat täten edellyttetään kohtaan ilmaista heidän pätevä suostumus . Mutta kyseisessä tapauksessa google analytics on anonymisoitu (IP masking) ja navigoititietojen jakaminen googlen kanssa estetty: tällä tavalla analyyttinen eväste on samanlainen kuin yllä mainitut tekniset evästeet eikä vaadi suostumusta.\nProfilointievästeet\nTämän tyyppistä evästettä ei käytetä sivustolla.\n\nErityinen huomautus:\nSivuston upotetut YouTube-videot eivät käytä evästeitä, koska kaikille YouTube-videosi upotuksille on määritetty \"nocookie\" yksityisyyttä parantava upotuskoodi.\n\nKansainvälinen ja eurooppalainen tiedonsiirto\nTietojasi käsitellään ainoastaan Euroopan talousalueella. Sinun oikeutesi koskien henkilökohtainen tiedot pidä alla EU Sääto 2016/679\n\nSinun oikeutesi\nSinä voi Harjoittele sinun oikeuksia minkä tahansa aika, kuten aseta ulos kirjoittaja Artikla 7, par. 3, ja artikkeleita 15 ja seurata/ EU sääto 2016/679:\n- Oikein päästäkseen käsiksi henkilökohtainen tiedot\n- Oikein kohtaan oikaisua ja poistaminen / henkilökohtainen tiedot;\n- Oikein kohtaan rajoitus / käsittely;\n- Oikein kohtaan tiedot siirrettävyys;\n- Oikein kohtaan esine kohtaan käsittelyä / henkilökohtainen tiedot\n- Oikein kohtaan laillinen väittää kohtaan italialainen Data Suojaus viranomainen.\n\nVoit käyttää oikeuksiasi lähettämällä meille sähköpostin osoitteeseen info@pixel-online.net tai osoitteeseen Pixelille Luigi Lanzin kautta, 12 – 50134 – Firenze, Italia. Lisätietoja tietojen käsittelystä voidaan lisätä milloin kerätä dataa.\n\n12 helmikuuta 2022 rev.03\nFrançais : Politique de confidentialité conformément au Règlement UE 2016/679\n\nQui collecte vos données\nConformément à l'article 13 du Règlement UE 2016/679 (RGPD), le responsable du traitement est le Demandeur du projet, et il est responsable de la collecte des données. Nous tenons à vous informer que notre organisation est légalement tenue de traiter les données que vous nous avez fournies en vertu de la réglementation susmentionnée. Vos données seront traitées de manière licite et loyale, conformément aux dispositions de l'article 5 du règlement UE 2016/679. De plus amples détails pourraient être fournis ultérieurement.\n\nDélégué à la protection des données ( dpo ) : la présence d'un éventuel DPO doit être demandée au responsable du traitement.\n\nQuelles données personnelles nous collectons\nDans conformité avec Article 4 de UE Régulation 2016/679 :\n- « données personnelles » désigne toute information relative à une personne physique identifiée ou identifiable ('Les données matière'); une identifiable Naturel la personne est une OMS peut être identifié, directement ou indirectement, notamment par référence à un identifiant tel qu'un nom, un numéro d'identification, lieu Les données, une en ligne identifiant ou pour une ou Suite les facteurs spécifique pour la physique, physiologique, génétique, mental, économique, culturel ou social identité de ce Naturel la personne;\n- \"En traitement\" moyens quelconque opération ou ensemble de opérations lequel est effectué au personnel Les données ou au ensembles de personnel Les données, qu'il s'agisse ou ne pas par automatique moyens, tel comme collection, enregistrement, organisation, structuration, stockage, adaptation ou altération, récupération, consultation, utilisation, divulgation par transmission, diffusion ou autre mise à disposition, alignement ou combinaison, restriction, effacement ou destruction.\n\nAvec référence pour la au dessus mentionné définitions, nous souligner ce nous collecter seulement la information toi nous fournir pour le fins de votre implication dans notre initiatives et/ou votre juridique relation amoureuse avec notre organisation:\n- Informations personnelles : nom et prénom des personnes physiques, contacts tels que l'adresse, le code postal indicatif, ville, Région, Téléphone numéro, e-mail;\n- Données concernant professionnels/organismes/entreprises : information concernant entreprises, nom, adresse fiscale et autres identifiants (numéro de fax et de téléphone, code fiscal ou TVA numéro).\n\nDe plus, nous pouvons collecter des données fournies lorsque vous accédez à nos sites, par le biais de cookies et d'autres technologie similaire; et lorsque vous nous contactez par e-mail, réseaux sociaux ou technologies similaires. Même si ces données ne sont pas collectées pour être associées à la personne physique, ces des identifiants en ligne peuvent être utilisés et combinés afin de créer des profils personnels. Parmi les en ligne identifiants que nous pouvons trouver adresse IP, type de navigateur et détails du plug-in, type d'appareil (par exemple, ordinateur de bureau, ordinateur portable, tablette, téléphone, etc.) système\nd'exploitation, fuseau horaire local. Ces données sont utilisées uniquement pour la production de statistique résultats.\n\nNous vous rappelons que nous ne traiterons pas de données personnelles révélant des informations raciales ou l'origine ethnique, les opinions politiques, les convictions religieuses ou philosophiques ou l'appartenance à un syndicat, et le traitement de données génétiques, de données biométriques à des fins d'identification unique d'un la personne, Les données concernant santé ou Les données concernant une Naturel personne sexe la vie ou sexuel orientation.\n\n**Pourquoi et comment nous traitons vos données**\n\nNous volonté utiliser ton Les données dans ce qui suit façons:\n\n49. Organiser et mettre en œuvre des initiatives dans le domaine de l'éducation et de la formation (p.cours, conférences, européen projets etc.)\n50. Pour produire administratif documents (par exemple factures) dans relation pour la initiatives au dessus\n51. Pour statistique fins\n52. Porter en dehors la communication Activités via e-mail concernant notre initiatives.\n53. Répondre aux demandes en utilisant les formulaires présents sur le site (s'il est présent)\n54. Autoriser l'inscription pour accéder au contenu éducatif confidentiel (s'il est présent)\n\nVotre attribution est obligatoire aux fins visées aux paragraphes 1, 2, 5, 6 afin de se conformer à juridique obligations et UE lois et règlements; refus pour apporter personnel Les données volonté ne pas Autoriser notre organisation à offrir toi, Nos services.\n\nVotre consentement est facultatif aux fins des paragraphes 3 et 4 ; nous vous enverrons du marketing communication par e-mail ou service postal. Vous pouvez exercer vos droits à tout moment, conformément à Article 15 et suivants du règlement UE 2016/679 concernant l'opt-out de recevoir ces la communication ou en choisissant d'autres la communication modalités. Nous conserverons vos données personnelles collectées aux fins de tous les paragraphes aussi longtemps dont nous avons besoin pour vous fournir les services offerts par notre organisation et jusqu’à 10 (dix ans.\nToi pouvez retirer ton consentement à quelconque temps.\n\nLa base juridique du traitement est constituée par la relation commerciale créée par la vente ou achat de biens et/ou services, précontractuel pour information (article 6 paragraphe b et c),et par consentement pour commercialisation Activités. ( articles 6 paragraphe une)\n\nNous traiterons et stockerons vos données uniquement aux fins susmentionnées, en utilisant des appareils numériques et dans les bases de données pertinentes assurant des garanties appropriées afin d'assurer une confidentialité continue, l'intégrité, la disponibilité et la résilience des systèmes de traitement, telles que définies par le règlement UE 2016/679. Seuls les sujets qui ont obtenu l'accès aux données personnelles du responsable du traitement ou du sous-traitant peuvent traiter tel information.\n\nNous ne vendons, n'échangeons ni ne transférons à d'autres tiers vos données personnellement identifiables information. Cependant, nous pouvons divulguer vos informations lorsque nous pensons que la divulgation est nécessaire pour se conformer avec la droit, imposer notre placer Stratégies, ou protéger les notres ou les autres' droits, biens, ou\nsécurité.\n\n**Profilage**\nVos données ne feront pas l'objet d'une décision fondée uniquement sur un traitement automatisé, produisant des effets juridiques les affectant ou portant atteinte de manière significative à leur personne. Annulation et modification : vous avez le droit de savoir, à tout moment, quelles sont vos données chez les responsables individuels du traitement, c'est-à-dire chez nous ou chez les personnes susmentionnées auxquelles nous les communiquons, et comment elles sont utilisées ; ils ont également le droit de les mettre à jour, de les compléter, de les corriger ou de les annuler, de demander leur blocage et de s'opposer à leur traitement. Pour l'exercice de vos droits, ainsi que pour des informations plus détaillées sur les sujets ou catégories de sujets auxquels les données sont communiquées ou qui en ont connaissance en tant que responsables ou agents, vous pouvez contacter le responsable du traitement ou l'un de ses responsables, identifié dans Cette déclaration.\n\n**Réseaux sociaux**\nNotre site Internet peut proposer un accès à des réseaux sociaux. Les conditions d'utilisation et la politique de confidentialité applicables à ces plateformes sont publiées sur leur site Web. Pixel ne peut pas contrôler la manière dont les données sont partagées sur un Public forum, discuter ou tableau de bord sont utilisé, être la Les données matière responsable de tel la communication.\n\n**Plaintes**\nToi pouvez également contact la italien Données protection Autorité en utilisant la Suivant lien http://www.garanteprivacy.it/home/footer/contatti, ou le contrôleur européen de la protection des données en utilisant le Suivant lien : https://edps.europa.eu/data-protection/our-role-supervisor/complaints_en\n\n**Biscuits**\nComme régler _ out t par règlement « Directives relatives aux cookies et autres outils de suivi - 10 juin 2021 », c'est ici sont trois catégories principales de biscuits:\n\n**Technique biscuits**\nCelles-ci sont utilisé pour la Unique but de \"transmettre communication pour une électronique réseau de communication, ou dans la mesure strictement nécessaire à la fourniture d'un service par le société d'information explicitement demandée par le contractant ou l'utilisateur afin de fournir ledit service » Ceux-ci ne sont pas utilisés à des fins ultérieures et sont normalement installés directement par la propriétaire ou la directeur de la site Internet (soi-disant \"propriétaire\" ou \"éditorial\" biscuits). Celles-ci peuvent être divisé dans: navigation ou session biscuits, lequel garantie Ordinaire la navigation et l'utilisation du site (permettant par exemple d'effectuer des achats ou d'être authentifié pour accéder aux espaces réservés) ; cookies d'analyse assimilés par la technique cookies lorsqu'ils sont utilisés directement par le gestionnaire du site Web pour collecter des informations, dans un formulaire associé (anonyme), sur le nombre d'utilisateurs et la manière dont ils visitent le site Internet; cookies fonctionnels qui permettent à l'utilisateur de naviguer par rapport à une série de critères sélectionnés (par exemple, la langue ou les produits sélectionnés pour l'achat) afin d'améliorer le service à condition de, à condition que nous informons nos utilisateurs comme partir planifier par article 13 UE Régulation 2016/679.\nle avant consentement de la utilisateur est ne pas demandé dans commande pour installer celles-ci biscuits.\n\n**Cookies analytiques**\nLe site utilise uniquement Google Analytics, qui est utilisé pour créer des profils d'utilisateurs et est utilisé pour envoyer des messages publicitaires selon les préférences indiquées par le même lors de leur navigation en ligne. En raison de leur caractère particulièrement envahissant à l'égard de la vie privée des utilisateurs, européen et italien règlements exiger ce utilisateurs être adéquatement informé sur leur utiliser de la même et sont Donc obligatoire pour Express leur valide consentement. Mais dans le cas précis, google analytics a été anonymisé (masquage IP) et le partage des données de navigation avec google a été bloqué : de cette manière, le cookie analytique est similaire aux cookies techniques indiqués ci-dessus et ne nécessite pas de consentement.\n\n**Cookies de profilage**\nCe type de cookie n’est pas utilisé sur le site.\n\nRemarque spécifique :\nLes vidéos intégrées de YouTube sur le site n’utilisent pas de cookies car il a été spécifié \"nocookie \" le code d'intégration à confidentialité renforcée pour toutes vos intégrations vidéo YouTube.\n\n**Transfert de données international et européen**\nVos données seront traitées uniquement dans l'Espace économique européen. Vos droits en matière de personnel Les données nous tenir en dessous de UE Régulation 2016/679\n\n**Tes droits**\nToi pouvez exercer ton droits quelconque temps, comme ensemble en dehors par Article 7, par. 3, et des articles 15 et Suivant de UE régulation 2016/679 :\n\n- Droit accéder personnel Les données\n- Droit pour rectification et effacement de personnel Les données;\n- Droit pour restriction de En traitement;\n- Droit pour Les données portabilité;\n- Droit pour objet pour En traitement de personnel Les données\n- Droit pour légal réclamer pour italien Données protection Autorité.\n\nVous pouvez exercer vos droits en nous adressant un email à info@pixel-online.net ou un courrier adressé à Pixel, via Luigi Lanzi , 12 – 50134 – Firenze, Italie. Informations complémentaires concernant le traitement des données peut être ajouté lorsque la collecte de données.\n\n12 février 2022 rév.03\nΕλληνικά: Πολιτική Απορρήτου σύμφωνα με τον Κανονισμό ΕΕ 2016/679\n\nΠοιος συλλέγει τα δεδομένα σας\n\nΣύμφωνα με το Άρθρο 13 του Κανονισμού ΕΕ 2016/679 (GDPR), ο υπεύθυνος επεξεργασίας δεδομένων είναι ο αιτών του έργου και είναι υπεύθυνος για τη συλλογή των δεδομένων. Θα θέλαμε να σας ενημερώσουμε ότι ο οργανισμός μας δεσμεύεται νομικά να επεξεργάζεται τα δεδομένα που μας παρέχετε βάσει του προαναφερθέντος κανονισμού.\n\nΗ επεξεργασία των δεδομένων σας θα γίνει νόμιμα και δίκαια, σύμφωνα με τη διάταξη του άρθρου 5 του Κανονισμού ΕΕ 2016/679. Περισσότερες λεπτομέρειες ενδέχεται να δοθούν σε μεταγενέστερο στάδιο.\n\nΥπεύθυνος προστασίας δεδομένων (dpo): Η παρουσία πιθανού DPO πρέπει να ζητηθεί από τον υπεύθυνο επεξεργασίας δεδομένων.\n\nΠοια Προσωπικά Δεδομένα Συλλέγουμε\n\nΣε συμφωνία με Άρθρο 4 του ΕΕ Κανονισμού λειτουργίας 2016/679:\n\n- «προσωπικά δεδομένα» σημαίνει κάθε πληροφορία που σχετίζεται με ταυτοποιημένο ή ταυτοποιήσιμο φυσικό πρόσωπο (‘δεδομένα θέμα’); ένα αναγνωρίσιμο φυσικό πρόσωπο είναι ένας ο οποίος μπορεί να αναγνωριστεί, κατευθείαν ή έμμεσα, ιδίως με αναφορά σε ένα αναγνωριστικό, όπως ένα όνομα, έναν αριθμό αναγνώρισης, τοποθεσία δεδομένα, ένα αναγνωριστικό ή συνδέσμος αναγνωριστικών του ή περισσότερων παράγοντων ειδικός προς την οφειλή του φυσικού προσώπου;\n\n- \"επεξεργασία\" που σημαίνει όποιος λειτουργία ή σειρά του επιχειρήσεως οι οποίες είναι εκτελούνται επί προσωπικών δεδομένων ή επί σκηνικών του προσωπικού δεδομένα, αν ή δεν με αυτοματοποιημένη που σημαίνει, τέτοιος όπως και συλλογή, καταγραφή, οργάνωση, δόμηση, αποθήκευση, προσαρμογή ή τροποποίηση, ανάκτηση, διαβούλευση, χρήση, αποκάλυψη μέσω διαβίβασης, διάδοση ή με άλλο τρόπο διάθεση, ευθυγράμμιση ή συνδυασμός, περιορισμός, διαγραφή ή καταστροφή.\n\nΜε αναφορά προς την παρούσα πολιτική, οι πληροφορίες που συλλέγονται από τους χρήστες τους ιστοτόπους μας, μέσω cookies και άλλων παρόμοιων τεχνολογιών και όταν επικοινωνείτε μαζί μας μέσω email, μέσω κοινωνικής δικτύωσης ή παρόμοιων τεχνολογιών. Παρόλο που τέτοια δεδομένα δεν συλλέγονται ώστε να συνδέονται με το φυσικό πρόσωπο, αυτά θα διαδικτυακά αναγνωριστικά μπορούν να χρησιμοποιηθούν και να συνδυαστούν έτσι ώστε να\nδημιουργηθούν προσωπικά προφίλ. Μεταξύ των διαδικτυακών αναγνωριστικών που ενδέχεται να βρούμε διεύθυνση IP, τύπο προγράμματος περιήγησης και λεπτομέρειες προσθήκης, τύπο συσκευής (π.χ. επιτραπέζιο υπολογιστή, φορητό υπολογιστή, tablet, τηλέφωνο κ.λπ.) λειτουργικό σύστημα, τοπική ζώνη ώρας. Αυτά τα δεδομένα χρησιμοποιούνται αποκλειστικά για την παραγωγή του στατιστικός Αποτελέσματος.\n\nΘα θέλαμε να σας υπενθυμίσουμε ότι δεν θα επεξεργαζόμαστε προσωπικά δεδομένα που αποκαλύπτουν φυλετικά ή εθνική καταγωγή, πολιτικές απόψεις, θρησκευτικές ή φιλοσοφικές πεποιθήσεις ή συμμετοχή σε συνδικάτα και η επεξεργασία γενετικών δεδομένων, βιομετρικών δεδομένων με σκοπό τον μοναδικό προσδιορισμό ενός φυσικού πρόσωπο, δεδομένα σχετικά με υγεία ή δεδομένα σχετικά με ένα φυσικό του ατόμου φύλο ΖΩΗ ή σεξουαλικός προσανατολισμός.\n\nΓιατί και πώς επεξεργαζόμαστε τα δεδομένα σας\n\nΕμείς θα χρήση τα δικα σου δεδομένα σε το ακόλουθο τρόποι:\n\n55. Οργάνωση και υλοποίηση πρωτοβουλιών στον τομέα της εκπαίδευσης και της κατάρτισης (π.χ. μαθήματα, συνέδρια, ευρωπαϊκά έργα και τα λοιπά.)\n\n56. Προς την παράγω διοικητικό έγγραφα (π.χ. τιμολόγια) σε σχέση προς την ο πρωτοβουλίες πάνω από\n\n57. Για στατιστικούς σκοποί\n\n58. Μεταφέρω έξω επικοινωνία δραστηριότητες μέσω ΗΛΕΚΤΡΟΝΙΚΗ ΔΙΕΥΘΥΝΣΗ σχετικά με μας πρωτοβουλίες.\n\n59. Απαντήστε σε αιτήματα χρησιμοποιώντας τις φόρμες στον ιστότοπο (εάν υπάρχει)\n\n60. Να επιτρέπεται η εγγραφή για πρόσβαση σε εμπιστευτικό εκπαιδευτικό περιεχόμενο (εάν υπάρχει)\n\nΗ παράδοσή σας είναι υποχρεωτική για σκοπούς σύμφωνα με τις παραγράφους 1, 2, 5, 6 προκειμένου να συμμορφωθείτε με δικαστικός υποχρεώσεις και ΕΕ του νόμου και Κανονισμό; άρνηση προς την προμηθεύω προσωπικός δεδομένα θα δεν επιτρέπω μας οργάνωση προς προσφορά εσείς, Οι Υπηρεσίες μας.\n\nΗ συγκατάθεσή σας είναι προαιρετική για σκοπούς σύμφωνα με τις παραγράφους 3 και 4. θα σας στείλουμε μάρκετινγκ επικοινωνία μέσω e-mail ή ταχυδρομικής υπηρεσίας. Μπορεί τε να ασκήσετε τα δικαίωμα σας στο έως 10 (Δέκα χρόνια.)\n\nΗ νομική βάση της μεταχείρισης συνίσταται στην εμπορική σχέση που δημιουργείται από την πώληση ή αγορά αγαθών ή/και υπηρεσιών, προσυμβατικά για ενημέρωση (άρθρο 6 παράγραφοι β και γ), και με συναίνεση Για εμπορία δραστηριότητες. (άρθρο 6 παράγραφος ένα)\n\nΘα επεξεργαστούμε και θα αποθηκεύουμε τα δεδομένα σας αποκλειστικά για τους προαναφερθέντες σκοπούς, χρησιμοποιώντας ψηφιακές συσκευές και σε σχετικές βάσεις δεδομένων διασφαλίζοντας κατάλληλες διασφαλίσεις ώστε να διασφαλίζεται η συνεχής\nεμπιστευτικότητα, ακεραιότητα, διαθεσιμότητα και ανθεκτικότητα των συστημάτων επεξεργασίας, όπως ορίζεται στον κανονισμό 2016/679 της ΕΕ. Μόνο τα υποκείμενα που έχουν αποκτήσει πρόσβαση στα προσωπικά δεδομένα από τον υπεύθυνο επεξεργασίας ή τον εκτελόντα την επεξεργασία μπορούν επεξεργάζομαι, διαδικασία τέτοιου πληροφοριών.\n\nΔεν πουλάμε, εμπορευόμαστε ή με άλλο τρόπο μεταβιβάζουμε σε άλλα τρίτα μέρη τα προσωπικά σας στοιχεία πληροφορίες. Ωστόσο, ενδέχεται να κοινοποιήσουμε τις πληροφορίες σας όταν πιστεύουμε ότι η αποδέσμευση είναι απαραίτητη συμμορφώμαι με νόμο, επιβάλλω μας ιστοσελίδα πολιτικές, ή προστατεύω Δικός μας ή οι υπολοιποί δικαιώματα, ιδιοκτησία, ή ασφάλεια.\n\nΠροφίλ\nΤα δεδομένα σας δεν θα υπόκεινται σε απόφαση που βασίζεται αποκλειστικά σε αυτοματοποιημένη επεξεργασία, η οποία παράγει νομικά αποτελέσματα που τα επηρεάζουν ή που επηρεάζουν σημαντικά το πρόσωπό τους. Ακύρωση και Τροποποίηση: έχετε το δικαίωμα να γνωρίζετε, ανά πάσα στιγμή, ποια είναι τα δεδομένα σας στους μεμονωμένους υπευθύνους επεξεργασίας δεδομένων, δηλαδή στην εταιρεία μας ή στα προαναφερθέντα πρόσωπα στα οποία κοινοποιούμε και πώς χρησιμοποιούνται. Ενδέχεται να κοινοποιήσουμε τις πληροφορίες σας όταν πιστεύουμε ότι η αποδέσμευση είναι απαραίτητη συμμορφώμαι με νόμο, επιβάλλω μας ιστοσελίδα πολιτικές, ή προστατεύω Δικός μας ή οι υπολοιποί δικαιώματα, ιδιοκτησία, ή ασφάλεια.\n\nΚοινωνικά δίκτυα\nΟ ιστότοπός μας μπορεί να προσφέρει πρόσβαση στο κοινωνικό δίκτυο. Ισχύουν οι Όροι Παροχής Υπηρεσιών και η Πολιτική Απορρήτου σε τέτοιες πλατφόρμες δημοσιεύονται στον ιστότοπό τους. Το Pixel δεν μπορεί να ελέγξει τον τρόπο κοινής χρήσης των δεδομένων σε ένα δημόσιο δικαστήριο, κουβέντα ή ταμπλό είναι μεταχειρισμένος, να εισαι ο δεδομένα θέμα υπεύθυνος του τέτοιου επικοινωνία.\n\nΠαράπονα\nΕσείς μπορείτε να παρακολουθήσετε τον τρόπο κοινής χρήσης των δεδομένων σε ένα δημόσιο δικαστήριο, κουβέντα ή ταμπλό είναι μεταχειρισμένος, να εισαι ο δεδομένα θέμα υπεύθυνος του τέτοιου επικοινωνία.\n\nΜπισκότα\nΟπως και σε t ou t με re gula t i o n \" Οδηγίες για τα cookie και άλλα εργαλεία παρακολούθησης - 10 Ιουνίου 2021\", εδώ _ τρεις βασικές κατηγορίες του μπισκότα:\n\nΤεχνικός μπισκότα\nΑυτά τα είναι μεταχειρισμένος Για ο αποκλειστική σκοπό του «μεταδίδοντας διαβιβάσεις προς την ένα ηλεκτρονικό δικτύο επικοινωνίας, ή στο βαθμό που είναι απολύτως απαραίτητο για την παροχή μιας υπηρεσίας από την εταιρεία πληροφοριών που ζητείται ρητά από το συμβαλλόμενο μέρος ή τον χρήστη προκειμένου να παρέχει η εν λόγω\nυπηρεσία» Δεν χρησιμοποιούνται για κανέναν απώτερο σκοπό και εγκαθίστανται κανονικά κατευθείαν με ο ιδιοκτήτης ή ο διευθυντής του δικτυακός τόπου (το λεγόμενο \"ιδιόκτητος\" ή \"σύνταξης\" μπισκότα). Αυτά τα μπισκότα είναι διαιρεμένα σε: περιήγηση ή συνεδρίαση μπισκότα, οι οποίες εγγύηση κανονικός πλοήγηση και χρήση του ιστότοπου (καθιστώντας δυνατή, για παράδειγμα, να κάνετε αγορές ή να είστε έλεγχος ταυτότητας για πρόσβαση σε δεσμευμένες περιοχές). cookies ανάλυσης που έχουν αφομοιωθεί από την τεχνική cookies όπου χρησιμοποιούνται απευθείας από τον διαχειριστή του ιστότοπου για τη συλλογή πληροφοριών, σε μια σχετική φόρμα (ανώνυμη), σχετικά με τον αριθμό των χρηστών και τον τρόπο με τον οποίο επισκέπτονται το δικτυακό τόπο; λειτουργικά cookies που επιτρέπουν στον χρήστη να πλοηγηθεί σε σχέση με μια σειρά επιλεγμένων κριτηρίων (για παράδειγμα, τη γλώσσα ή τα προϊόντα που έχουν επιλεγεί για αγορά) προκειμένου να βελτιωθεί η υπηρεσία υπό την προϋπόθεση, υπό την προϋπόθεση ότι ενημερώνουμε μας όπως και ορίζεται με άρθρο 13 ΕΕ Κανονισμός λειτουργίας 2016/679.\n\nο πριν συγκατάθεση του χρήστης είναι δεν ζητείται σε Σειρά προς την εγκαθιστώ αυτά τα μπισκότα.\n\nCookies Analytics\nΟ ιστότοπος χρησιμοποιεί μόνο google analytics, το οποίο χρησιμοποιείται για τη δημιουργία προφίλ των χρηστών και είναι απασχολείται για την αποστολή διαφημιστικών μηνυμάτων σύμφωνα με τις προτιμήσεις που δείχνει η ίδια κατά την ηλεκτρονική πλοήγησή τους. Λόγω της ιδιαίτερης επεμβατικότητάς του σε σχέση με τα προσωπικά των χρηστών σφαίρα, ευρωπαϊκός και ιταλικός Κανονισμοί απαιτώ ότι χρήστες είναι επαρκώς προειδοποιητοί σχετικά με δικά τους χρήση του ιστότοπου ο ιδίο και είναι έτσι απαιτείται προς την εξπρές δικα τους έγκυρο συγκατάθεση. Όμως στη συγκεκριμένη περίπτωση το google analytics έχει ανωνυμοποιηθεί (απόκρυψη IP) και έχει αποκλειστεί η κοινή χρήση δεδομένων πλοήγησης με την google: με αυτόν τον τρόπο το αναλυτικό cookie είναι παρόμοιο με τα τεχνικά cookie που αναφέρονται παραπάνω και δεν απαιτεί συναίνεση.\n\nΠροφίλ των cookies\nΑυτός ο τύπος cookie δεν χρησιμοποιείται στον ιστότοπο.\n\nΕιδική σημείωση:\nΤα ενσωματωμένα βίντεο του YouTube στον ιστότοπο δεν χρησιμοποιούν cookie καθώς έχει οριστεί ως \" nocookie \" ο ενσωματωμένος κώδικας ενσωμάτωσης με βελτιωμένο απόρρητο για όλες τις ενσωματώσεις βίντεο στο YouTube.\n\nΔιεθνής και ευρωπαϊκή μεταφορά δεδομένων\nΤα δεδομένα σας θα υποβληθούν σε επεξεργασία αποκλειστικά στον Ευρωπαϊκό Οικονομικό Χώρο. Τα δικαιώματά σας σε σχέση με το προσωπικό δεδομένα εμείς Κρατήστε κάτω από ΕΕ Κανονισμός λειτουργίας 2016/679\n\nΤα δικαιώματά σας\nΕσείς μπορώ άσκηση τα δικαίωμα όποιος χρόνος, όπως και σειρά έξω με Αρθρο 7, par. 3, και άρθρα 15 και ΕΠΟΜΕΝΟτου ΕΕ κανονισμός λειτουργίας 2016/679:\n\n- σωστά για να έχω πρόσβαση σε προσωπικός δεδομένα\n- σωστά προς την διόρθωση και διαγραφή του προσωπικού δεδομένα;\n- σωστά προς την περιορισμό του επεξεργασία;\n- σωστά προς την δεδομένα φορητότητα;\n• σωστά προς την αντικείμενο προς την επεξεργασία του προσωπικός δεδομένα\n• σωστά προς την νομικός απαίτηση προς την ιταλικός Δεδομένα ΠΡΟΣΤΑΣΙΑ Εξουσία.\n\nΜπορείτε να ασκήσετε τα δικαιώματά σας στέλνοντάς μας ένα email στη διεύθυνση info@pixel-online.net ή μια επιστολή με διεύθυνση προς Pixel, μέσω Luigi Lanzi, 12 – 50134 – Firenze, Ιταλία. Περαιτέρω πληροφορίες σχετικά με την επεξεργασία δεδομένων μπορεί να προστεθεί όταν συλλογή δεδομένων.\n\n12 Φεβρουαρίου 2022 αναθ.03\nHrvatski: Politika privatnosti u skladu s Uredbom EU 2016/679\n\nTko prikuplja vaše podatke\nSukladno članku 13. EU Uredbe 2016/679 (GDPR), voditelj obrade podataka je prijavitelj projekta i odgovoran je za prikupljanje podataka. Želimo vas obavijestiti da je naša organizacija zakonski obvezna obraditi podatke koje ste nam dostavili u skladu s prethodno spomenutom uredbom.\nVaši će se podaci obrađivati zakonito i pošteno, prema odredbi članka 5. Uredbe EU-a 2016/679. Daljnji detalji mogli bi se dati u kasnijoj fazi.\nSlužbenik za zaštitu podataka (dpo): prisutnost mogućeg DPO-a mora se zatražiti od voditelja obrade podataka.\n\nKoje osobne podatke prikupljamo\nU suglasnost s Članak 4 od EU Regulativa 2016/679:\n- „osobni podaci” znači sve informacije koje se odnose na identificiranu fizičku osobu ili osobu koja se može identificirati (‘podaci predmet’); an prepoznatljivi prirodnim osoba je jedan tko limenka biti identificiran, direktno ili neizravno, posebno upućivanjem na identifikator kao što je ime, identifikacijski broj, mjesto podaci, an na liniji identifikator ili do jedan ili više čimbenici specifično do the fizički, fiziološki, genetski, mentalno, gospodarski, kulturni ili društvenim identiteta da prirodnim osoba;\n- \"obrada\" sredstva bilo koji operacija ili skupa od operacije koji je izvedena na osobnim podaci ili na skupova od osobnim podaci, da li ili ne po automatizirani sredstva, takav kao kolekcija, snimanje, organizacija, strukturiranje, pohrana, prilagodba ili izmjena, dohvat, konzultacije, korištenje, otkrivanje prijenosom, širenjem ili na drugi način stavljanjem na raspolaganje, usklađivanjem ili kombinacija, ograničenje, brisanje ili uništenje.\n\nS referenca do the iznad spomenuti definicije, mi naglasiti da mi skupljati samo the informacija vas pružiti nam za svrhe vašeg sudjelovanja u naše inicijative i/ili vaš pravni odnos s naše organizacija:\n- Osobni podaci: ime i prezime fizičkih osoba, kontakti kao što su adresa, poštanski broj kod, grad, regija, telefon broj, e-mail;\n- Podaci o profesionalci/organizacije/poslovi: informacija o tvrtke, naziv, fiskalnu adresu i druge identifikatore (broj faksa i telefona, porezni broj ili PDV-a broj).\n\nŠtoviše, možemo prikupljati podatke dostavljene kada pristupate našim stranicama, putem kolačića i ostalog slična tehnologija; i kada nas kontaktirate putem e-pošte, društvenih medija ili sličnih tehnologija. Iako se takvi podaci ne prikupljaju kako bi bili povezani s fizičkom osobom, ovi internetski identifikatori mogu se koristiti i kombinirati kako bi se stvorili osobni profili. Među online identifikatore koje možemo pronaći IP adresu, vrstu preglednika i pojedinosti o dodacima, vrstu uređaja (npr. desktop, laptop, tablet, telefon itd.) operativni sustav, lokalna vremenska zona. Ovi podaci se koriste isključivo za proizvodnja od statistički rezultate.\n\nPodsjećamo vas da nećemo obrađivati osobne podatke koji otkrivaju rasnu ili etničko\npodrijetlo, politička mišljenja, vjerska ili filozofska uvjerenja ili članstvo u sindikatu, i obrada genetskih podataka, biometrijskih podataka u svrhu jedinstvene identifikacije prirodnog osoba, podaci o zdravlje ili podaci o a prirodnim osobe seks život ili seksualni orijentacija.\n\nZašto i kako obrađujemo vaše podatke\nMi htjemo koristiti tvoj podaci u sljedeće načini:\n\n61. Organizirati i provoditi inicijative u području obrazovanja i osposobljavanja (npr tečajevi, konferencije, europski projekti itd.)\n62. Do proizvoditi upravni dokumente (npr fakture) u odnos do the inicijative iznad\n63. Za statistički svrhe\n64. Nosite van komunikacija aktivnosti preko email o naše inicijative.\n65. Odgovorite na zahtjeve koristeći obrasce na stranici (ako postoje)\n66. Omogućite registraciju za pristup povjerljivom obrazovnom sadržaju (ako je prisutan)\n\nVaše je davanje obvezne za svrhe iz stavaka 1, 2, 5, 6 kako bi se ispunili pravnim obveze i EU zakonima i propisi; odbijanje do pružiti osobnim podaci htjeti ne dopustitinaše organizacija do ponuda vas, Naše usluge.\nVaš pristanak nije obavezan za svrhe iz stavaka 3. i 4.; poslat ćemo vam marketing komunikacija putem e-pošte ili pošte. Svoja prava možete ostvariti bilo kada, u skladu s Članak 15. i kasnije Uredbe EU-a 2016/679 o isključivanju iz primanja takvih komunikacija ili birajući drugo komunikacija modaliteti.\nVaše osobne podatke ćemo čuvati prikupljene za svrhe prema svim stavcima koliko nam je potrebno kako bismo vam pružili ponuđene usluge od strane naše organizacije i to do 10 (deset godina.\nVas limenka povući tvoj pristanak na bilo koji vrijeme.\n\nPravni temelj postupanja čini trgovački odnos nastao prodajom odn kupnja robe i/ili usluga, predugovorna radi informiranja (članak 6. stavak b i c),i po pristanku za Marketing aktivnosti. ( članak 6 stavak a)\n\nVaše ćemo podatke obraditi i pohraniti isključivo u gore navedene svrhe, koristeći digitalne uređaje i u relevantnim bazama podataka osiguravajući odgovarajuće zaštitne mjere kako bi se osigurala trajna povjerljivost, integritet, dostupnost i otpornost sustava obrade, kako je navedeno u uredbi EU-a 2016/679. Mogu samo subjekti koji su od voditelja obrade ili izvršitelja obrade dobili pristup osobnim podacima postupak takav informacija.\n\nNe prodajemo, ne trgujemo ili na bilo koji drugi način prenosimo trećim stranama koje vas osobno identificiraju informacija. Međutim, možemo objaviti vaše podatke kada smatramo da je njihovo objavljivanje potrebno pridržavati se s the zakon, provoditi naše mjesto politike, ili zaštititi naše ili drugi' prava, vlasništvo, ili sigurnost.\n\nProfiliranje\nVaši podaci neće biti podvrgnuti odluci koja se temelji isključivo na automatiziranoj obradi, koja proizvodi pravne učinke koji utječu na njih ili koji značajno utječu na njegovu osobu. Otkazivanje i dopuna: imate pravo u bilo kojem trenutku znati koji su vaši podaci kod pojedinačnih voditelja obrade podataka, odnosno u našoj tvrtki ili kod gore navedenih osoba kojima ih prenosimo i kako se koriste; također imaju pravo ažurirati, dopuniti, ispraviti ili otkazati ih, zatražiti njihovu blokadu i usprotiviti se njihovom tretmanu. Za ostvarivanje vaših\nprava, kao i za detaljnije informacije o subjektima ili kategorijama subjekata kojima se podaci dostavljaju ili koji su svjesni toga kao voditelji ili agenti mogu se obratiti voditelju obrade podataka ili jednom od njegovih upravitelja, navedenim u ovu izjavu.\n\n**Društvene mreže**\nNaša web stranica može nuditi pristup društvenoj mreži. Primjenjivi uvjeti usluge i Pravila privatnosti na takve platforme objavljeni su na njihovoj web stranici. Pixel ne može kontrolirati način na koji se podaci dijele na javnost forum, razgovor ili kontrolna ploča su korišteno, biće the podaci predmet odgovoran od takav komunikacija.\n\n**Pritužbe**\nVas limenka također kontakt the talijanski Podaci Zaštita Autoritet korištenjem the slijedeći veza http://www.garanteprivacy.it/home/footer/contatti, ili Europski nadzornik za zaštitu podataka pomoću slijedeći poveznica: https://edps.europa.eu/data-protection/our-role-supervisor/complaints_en\n\n**Kolačići**\nKao se t ou t po pravilnik \"Smjernice za kolačiće i druge alate za praćenje - 10. lipnja 2021. \", t ovdje a re tri glavne kategorije od kolačići:\n\n**Tehnički kolačići**\nOve su korišteni za the jedini Svrha od “prenoseći komunikacije do an elektronička komunikacijske mreže, ili u mjeri koja je strogo potrebna za pružanje usluge od strane informacijska tvrtka koju izričito zahtijeva ugovorna strana ili korisnik radi pružanja spomenuta usluga” Oni se ne koriste u nikakve skrivene svrhe i obično se instaliraju direktno po the vlasnik ili the menadžer od the web stranica (takozvani \"vlasnički\" ili “uvodnik” kolačići). Ove limenka biti podijeljena u: pregledavanje ili sjednica kolačići, koji jamčiti normalan navigaciju i korištenje web stranice (omogućujući npr. kupnju ili autentificiran kako bi se pristupilo rezerviranim područjima); analitičke kolačiće asimilirane tehničkim kolačići gdje ih izravno koristi upravitelj web stranice za prikupljanje informacija, u povezani obrazac (anonimno), o broju korisnika i načinu na koji posjećuju web stranica; funkcionalni kolačići koji omogućuju korisniku navigaciju u odnosu na niz kriterija odabira (na primjer, jezik ili proizvodi odabrani za kupnju) kako biste poboljšali uslugu pod uvjetom, pod uvjetom da obavještavamo naše korisnika kao krenuo po članak 13 EU Regulativa 2016/679. The prije pristanak od the korisnik je ne zatraženo u narudžba do instalirati ove kolačići.\n\n**Kolačići za analitiku**\nStranica koristi samo google analytics, koja se koristi za izradu profila korisnika i su zaposleni za slanje reklamnih poruka prema preferencijama koje isti pokazuju tijekom njihove internetske navigacije. Zbog njihove posebne invazivnosti s obzirom na privatnost korisnika sfera, europski i talijanski propisi zahtijevati da korisnika biti adekvatno informirani oko njihov koristiti od the isti i su Tako potreban do izraziti njihov valjano pristanak . Ali u konkretnom slučaju google analytics je anonimiziran (IP maskiranje) i dijeljenje navigacijskih podataka s googleom je blokirano: na taj je način analitički kolačić sličan gore navedenim tehničkim kolačićima i ne zahtijeva pristanak.\n\n**Kolačići za profiliranje**\nOva vrsta kolačića se ne koristi na stranici.\nPosebna napomena:\nUgrađeni videozapisi YouTubea na web-mjestu ne koriste kolačiće jer je naveden \"nocookie\" kod za ugradnju s poboljšanom privatnošću za sve vaše ugrađene videozapise na YouTubeu.\n\nMeđunarodni i europski prijenos podataka\nVaši će se podaci obrađivati isključivo u Europskom gospodarskom prostoru. Vaša prava u vezi sosobnim podaci mi držite pod, ispod EU Regulativa 2016/679\n\nVaša prava\nVas limenka vježbanje tvoj prava bilo koji vrijeme, kao skupa van po Članak 7, par. 3, i članaka 15 i slijedećiod EU propis 2016/679:\n- Pravo pristupiti osobnim podaci\n- Pravo do ispravljanje i brisanje od osobnim podaci;\n- Pravo do ograničenje od obrada;\n- Pravo do podaci prenosivost;\n- Pravo do objekt do obrada od osobnim podaci\n- Pravo do pravnim zahtjev do talijanski Podaci Zaštita Autoritet.\n\nprava možete ostvariti slanjem e-pošte na info@pixel-online.net ili pismom na adresu na Pixel, preko Luigija Lanzija , 12 – 50134 – Firenze, Italija. Dodatne informacije o obradi podataka može se dodati kada prikupljanje podataka.\n\n12 veljače 2022 rev.03\nMagyar: Adatvédelmi szabályzat a 2016/679 EU Rendeletnek megfelelően\n\nKi gyűjti az Ön adatait\nA 2016/679 EU Rendelet (GDPR) 13. cikke értelmében az adatkezelő a projekt Pályázója, aki az adatok gyűjtéséért felelős. Tájékoztatjuk, hogy szervezetünk jogilag köteles feldolgozni az Ön által részünkre megadott adatokat a fent említett szabályozás értelmében.\nAz Ön adatait jogszorúan és tisztességesen kezeljük a 2016/679 EU-rendelet 5. cikkének rendelkezései szerint. További részleteket egy későbbi szakaszban lehet közölni.\nAdatvédelmi tisztviselő (dpo): az esetleges adatvédelmi tisztviselő jelenlétét az adatkezelőtől kell kérni.\n\nMilyen személyes adatokat gyűjtünk\nBan ben összhangban val vel Cikk 4 nak,-nek EU Szabályozás 2016/679:\n- „személyes adat”: bármely azonosított vagy azonosítható természetes személyre vonatkozó információ ('adat tantárgy'); an azonosítható természetes személy van egy ki tud lenni azonosított, közvetlenül vagy közvetetten, különösen egy azonosítóra, például névre, azonosító számra hivatkozva, elhelyezkedés adat, an online azonosító vagy nak nek egy vagy több tényezőket különleges nak nek az fizikai, fiziológiai, genetikai, szellemi, gazdasági, kulturális vagy társadalmi identitása hogy természetes személy;\n- \"feldolgozás\" eszközök Bármily művelet vagy készlet nak,-nek tevékenységek melyik van teljesített tovább személyes adat vagy tovább készletek nak,-nek személyes adat, vajon vagy nem által automatizált eszközök, ilyen mint Gyűjtemény, rögzítés, rendszerezés, strukturálás, tárolás, adaptálás vagy megváltoztatás, visszakeresés, konzultáció, felhasználás, nyilvánosságra hozatal továbbítással, terjesztéssel vagy más módon elérhetővé tételel, összehangolással vagy kombináció, korlátozás, törlés vagy megsemmisítés.\n\nVal vel referencia nak nek az felett említett definíciók, mi aláhúzás hogy mi gyűjti csak az információ Ön biztosítson nekünk a célokra az ön részvételéről a miénk kezdeményezések és/vagy a jogi kapcsolatot val a miénk szervezet:\n- Személyes adatok: természetes személyek vezeték- és vezetékneve, elérhetőségek, például cím, irányítószám kód, város, vidék, telefon szám, email;\n- Adat vonatkozó szakemberek/szervezetek/vállalkozások: információ vonatkozó vállalkozások, név, adózási cím és egyéb azonosítók (fax- és telefonszám, adószám vagy ÁFA szám).\n\nEzen túlmenően adatokat gyűjthetünk, amikor Ön eléri oldalainkat, cookie-k és egyéb módon hasonló technológia; és amikor kapcsolatba lép velünk e-mailben, közösségi médián vagy hasonló technológiákon keresztül. Annak ellenére, hogy az ilyen adatokat nem úgy gyűjti, hogy azokat a természetes személyel társításak, ezek online azonosítók használhatók és kombinálhatók személyes profilok létrehozásához. Az online között azonosítókat találhatunk IP-címet, böngészőtípust és beépülő modul adatait, eszköztípust (pl. asztali, laptop, tablet, telefon stb.) operációs rendszer, helyi időzóna. Ezeket az adatokat kizárólag a termelése statisztikai eredmények.\nSzeretnénk emlékeztetni, hogy nem dolgozunk fel faji, ill. etnikai származás, politikai vélemény, vallási vagy filozófiai meggyőződés vagy szakszervezeti tagság, és genetikai adatok, biometrikus adatok feldolgozása természetes egyedi azonosítás céljából személy, adat vonatkozó Egészség vagy adat vonatkozó a természetes személyé szex élet vagy szexuális orientáció.\n\n**Miért és hogyan dolgozzuk fel az Ön adatait**\n\nMi akarat használat a te adatban ben a következő módokon:\n\n67. Kezdeményezések szervezése és végrehajtása az oktatás és képzés területén (pl. tanfolyamok, konferenciák, európai projektek stb.)\n68. Nak nek előállítani közigazgatási dokumentumokat (például számlák) ban ben kapcsolat nak nek az kezdeményezések felett\n69. Mert statisztikai célokra\n70. Vízsi ki kommunikáció tevékenysége keresztül email vonatkozó a miénk kezdeményezések.\n71. Válaszoljon a kérésekre a webhelyen található űrlapok segítségével (ha van)\n72. Regisztráció engedélyezése a bizalmas oktatási tartalomhoz való hozzáféréshez (ha van ilyen)\n\nAz Ön adományozása kötelező az (1), (2), (5) és (6) bekezdésben meghatározott célokra annak érdekében, hogy megfeleljen a bírói kötelezettségeket és EU törvényeket és előírások; elutasítás nak nek biztosítani személyes adat akarat nem lehetővé teszia miénk szervezethez ajánlat Önn, Szolgáltatásaink.\n\nAz Ön hozzájárulása a (3) és (4) bekezdés szerinti célokra nem kötelező; marketinget küldünk Önnnek kommunikáció e-mailben vagy postai úton. alapján bármikor gyakorolhatja jogait A 2016/679 EU-rendelet 15. és későbbi cikkei az ilyen átvételtől való lemondásra vonatkozóan kommunikáció vagy más választása kommunikáció módozatait.\n\nAz összes bekezdés szerinti célból gyűjtött személyes adatait mindaddig megőrizzük amennyire szükségünk van a kínált szolgáltatások biztosításához szervezetünk által és 10-ig (tiz év.\nÖn tud visszavonulni a te beleegyezés nál nél Bármifől.\n\nA kezelés jogalapját az értékesítéssel létrejött kereskedelmi kapcsolat, ill. áruk és/vagy szolgáltatások vásárlása, szerződéskötést megelőző tájékoztatás céljából (6. cikk b) és c) pontja, és beleegyezésével számára marketing tevékenységek. (cikk 6 bekezdés a)\n\nAdatait kizárólag a fent említett célokra, digitális eszközök segítségével dolgozzuk fel és tároljuk és a megfelelő adatbázisokban megfelelő biztosítékokat biztosítva a folyamatos titoktartás biztosítása érdekében, a feldolgozó rendszerek integritása, rendelkezésre állása és rugalmassága, a 2016/679 EU-rendeletnek megfelelően. Csak azok az alanyok férhetnek hozzá a személyes adatokhoz az adatkezelőtől vagy az adatfeldolgozótól folyamat ilyen információ.\n\nNem adjuk el, nem kereskedünk vagy más módon nem adjuk át más harmadik félnek az Ön személyazonosságát információ. Azonban kiadhatjuk az Ön adatait, ha úgy gondoljuk, hogy ez szükséges betartani val vel az törvény, érvényesíteni a miénk webhely irányelvek, vagy védeni a miénk vagy másoké jogok, ingatlan, vagy biztonság.\nProfilalkotás\nAz Ön adataira nem vonatkozik kizárólag automatizált adatkezelésen alapuló olyan döntés, amely őt érintő vagy személyét jelentősen érintő joghatásokat vált ki. Lemondás és módosítás: Önnek bármikor joga van megtudni, hogy az egyes adatkezelőknél, azaz cégünknel vagy a fent említett személyeknél melyek az Ön adatai, és hogyan használják fel azokat; jogukban áll azokat aktualizálni, kiegészíteni, javítani vagy törleni, kérni tiltásukat és ellenezni kezelésüket. Az Ön jogainak gyakorlásához, valamint azokról az alanyokról vagy alanyok kategóriáiról szóló részletes tájékoztatásért, akikkel az adatot közölni, vagy akikről vezetőként vagy megbízottként tudomásuk van, fordulhatnak az adatkezelőhöz vagy valamelyik kezelőjéhez, a ez az állítás.\n\nKözösségi hálózatok\nWeboldalunk hozzáférést kínálhat a közösségi hálózatokhoz. Az érvényes szolgáltatási feltételek és adatvédelmi szabályzat az ilyen platformokon közzéteszik a weboldalukon. A Pixel nem tudja szabályozni az adatok megosztásának módját a nyilvános fórum, csevegés vagy Irányítópult vannak használt, lény az adat tantárgy felelősnek, ilyen kommunikáció.\n\nPanaszok\nÖn tud is kapcsolatba lépni az olasz Adat Védelem Hatóság segítségével az következő link http://www.garanteprivacy.it/home/footer/contatti, vagy az európai adatvédelmi biztos a következő link: https://edps.europa.eu/data-protection/our-role-supervisor/complaints_en\n\nCookie-k\nMint se t ou t által „A cookie- kra és más nyomkövető eszközökre vonatkozó irányelvek – 2021. június 10. “ rendelet, t itt három fő kategória nak,-nek sütik:\n\nMűszaki sütik\nEzek vannak használt számára az egyetlen célja nak,-nek „közvetítés kommunikáció nak nek an elektronikus kommunikációs hálózat, vagy a szolgáltatás nyújtásához feltétlenül szükséges mértékben információs társaságot a szerződő fél vagy a felhasználó kifejezetten kért annak biztosítása érdekében az említett szolgáltatás” Ezeket nem használják külső célokra, és általában telepítve vannak közvetlenül által az tulajdonos vagy az menedzer nak,-nek az weboldal (ügynevezett \"szabadalmazott\" vagy \"szerkesztőségi\" sütik). Ezek tud lenni megosztott ba: bongészés vagy ülés sütik, melyik garancia Normál navigáció és a weboldal használata (lehetővé téve például a vásárlást vagy a tartózkodást hitelesítve a fenntartott területek elérése érdekében); a technikai által asszimilált analitikai cookie-k cookie-k, ahol azokat közvetlenül a weboldal kezelője használja fel információgyűjtésre, egy kapcsolódó űrlap (anonim), a felhasználók számáról és a látogatási módról weboldal; funkcionális cookie-k, amelyek lehetővé teszik a felhasználó számára, hogy egy sor kiválasztott kritérium alapján navigáljon (például a nyelv vagy a vásárlásra kiválasztott termékek) a szolgáltatás javítása érdekében biztosítani, feltéve, hogy tájékoztatjuk magunkat felhasználókat mint elindult által cikk 13 EU Szabályozás 2016/679.\nAz előzetes beleegyezés nak,-nek az felhasználó van nem kérte ban ben rendelés nak nek telepítés ezek sütiket.\n\nAnalitikai cookie-k\nAz oldal kizárólag a google analytics-t használja, amely a felhasználók profiljának létrehozására szolgál reklámüzenetek küldésére alkalmazzák, az ugyanazon feltüntetett preferenciák szerint\nonline navigációjuk során. A felhasználók magánéletére vonatkozó különleges invazivitásuk miatt gömb, európai és olasz előírások igényelnek hogy felhasználókat lenni megfelelően tájékozott ról ról az övök használat nak,-nek az azonos és vannak így kívánt nak nek Expressz az övök érvényes beleegyezés . A konkrét esetben azonban a google analytics anonimizálása (IP maszkolás) és a navigációs adatok google-lel való megosztása blokkolva lett: így az analitikai süti hasonló a fent jelzett technikai cookie-khoz, és nem igényel hozzájárulást.\n\nProfilozó cookie-k\nEz a fajta süti nem használatos az oldalon.\n\nKülön megjegyzés:\nA webhelyen található YouTube beágyazott videói nem használnak cookie-kat, mivel a „nocookie” adatvédelemmel megerősített beágyazási kódot határoztak meg az összes YouTube-videó beágyazásához.\n\nNemzetközi és európai adatátvitel\nAz Ön adatait kizárólag az Európai Gazdasági Térség területén kezeljük. Az Ön jogai a személyes adat mi tart alatt EU Szabályozás 2016/679\n\nAz Ön jogai\nÖn tud gyakorlat a te jogokat Bármi idő, mint készlet ki által Cikk 7, par. 3, és cikkeket 15 és következőnak,-nek EU szabályozás 2016/679:\n\n- Jobb hozzáférni személyes adat\n- Jobb nak nek helyesbítés és törlés nak,-nek személyes adat;\n- Jobb nak nek korlátozás nak,-nek feldolgozás;\n- Jobb nak nek adat hordozhatóság;\n- Jobb nak nek tárgy nak nek feldolgozás nak,-nek személyes adat\n- Jobb nak nek jogi követelés nak nek olasz Adat Védelem Hatóság.\n\ninfo@pixel-online.net címre küldött e-mail vagy a címzett levél útján gyakorolhatja. Pixel, Luigi Lanzi 12 – 50134 – Firenze, Olaszország útján. További információ az adatkezelésről mikor lehet hozzáadni adatgyűjtés.\n\nfebruár 12 2022 rev.03\nGaeilge: Beartas Príobháideachta de réir Rialachán AE 2016/679\n\nCé a Bhailíonn Do Shonraí\nDe bhun Airteagal 13 de Rialachán AE 2016/679 (GDPR), is é an rialaitheoir sonraí larratasóir an tionscadail, agus tá sé freagrach as na sonraí a bhalliú. Ba mhaith linn a chur in iúl duit go bhfuil ceangal dlíthiúil ar ár n-eagraíocht na sonraí a chuir tú ar fáil dúinn faoin rialachán thuasluaite a phróiseáil.\n\nPróiseálfar do shonraí go dleathach agus go cothrom, faoi fhoráil airteagal 5 de Rialachán AE 2016/679. D’fhéadfaí sonraí breise a sholáthar ag céim níos déanaí.\n\nOifigeach um chosaint sonraí (dpo) : ní mór a iarraidh ar an rialaitheoir sonraí go bhféadfadh OCS a bheith i láthair.\n\nCad iad na Sonraí Pearsanta a Bhailímid\nI de réir le Airteagal 4 de AE Rialachán 2016/679:\n- ciallaíonn “sonraí pearsanta” aon fhaisnéis a bhaineann le duine nádúrtha sainaitheanta nó inaithheanta (‘sonraí ábhar’); an inaithheanta nádúrtha duine tá aon EDS féidir bheith aitheanta, go díreach nó go hindíreach, go háirithe trí thagairt d’aithteantóir amhail ainm, uimhir aitheantais, suíomh sonraí, an ar líne aitheantóir nó chun aon nó níos mó fachtóirí sonraí chun an fisiciúil, fiseolaíoch, géiniteach, meabhrach, eacnamaíoch, cultúrtha nó sóisialta céannacht go nádúrtha duine;\n- “próiseáil” acmhainn ar bith oibríocht nó leagtha de oibríochtaí a tá léirithe ar pearsanta sonraí nó ar tacair de pearsanta sonraí, cibé acu nó ní le uathoibrithe acmhainn, den sórt sin mar bhalliú, taifeadadh, eagrú, struchtúrú, stóráil, oiriúnú nó athrú, aisghabháil, comhairliúcháin, úsáid, nochtadh trí tharchur, scaipeadh nó cur ar fáil ar shlí eile, aillíú nó teaglaim, srian, sriosadh nó scrios.\n\nLe tagairt chun an os cionn luaite sainmhínithe, againn béim go agaínn bhalliú amháin an eolas leat sholáthar duínn le haghaidh an críocha de do pháirt i ár tionscnaimh agus/nó do dlíthiúil caidreamh le ár eagraíocht:\n- Faisnéis phearsanta: ainm agus sloinne daoine nádúrtha, teagmhálacha ar nós seoladh, ZIP cód, cathair, réigiún, teileafón uimhir, riomhphost;\n- Sonraí maidir le gairmithe/eagraíochtaí/gnóthai: eolas maidir le gnólachtaí, ainm, seoladh fioscach agus aitheantóirí eile (facs agus uimhir theileafóin, cód cánach nó CBL uimhir).\n\nIna theannta sin, féadfaímid sonraí a chuirtear ar fáil a bhalliú nuair a dhéanann tú rochtain ar ár suíomhanna, trí fhianáin agus eile teicneolaíocht den chineál céanna; agus nuair a dhéanann tú teagmháil linn trí riomhphost, na meáin shóisialta nó teicneolaíocht comhchosúla. Cé nach mbailítear sonraí den sórt sin sa chaoi is go mbaineann siad leis an duine nádúrtha, beidh siad seo d’fhéadfaí aitheantóirí ar líne a úsáid agus a chomhcheangal chun próifíil pearsanta a chruthú. I measc na ar líne aitheantóirí is féidir linn a fháil seoladh IP, cineál brabhsálaí agus sonraí breiseán, cineál gléis (m.sh. deasc, riomhaire glúine, táibléad, fón, srl.) córas oibríucháin, crios ama áitiúil. Úsáidtear na sonraí seo le haghaidh an tairgeadh de staitistiúil torthaí.\nBa mhaith linn a mheabhrú duit nach mbeidh sonraí pearsanta á bpróiseáil agaínn a thaispeáann ciníochas nó bunús eitneach, tuairimí polaitiúla, creidimh reiligíúnaacha nó fealsúnacha, nó ballráíocht ceardchumainn, agus próiseáil sonraí géiniteacha, sonraí bithmhéadacha chun críche nádúrtha a shainaithint go huathúil duine, sonraí maidir le sláinte nó sonraí maidir le a nádúrtha duine gnéas saol nó gnéasach treoshuíomh.\n\nCén Fáth agus Conas a Phróiseálaímid Do Shonraí\nmuid beidh úsáid do sonraí ísteach méid seo a leanas bealaí:\n\n73. Tionscnaimh i réimse an oideachais agus na hoiliúna a eagrú agus a chur i bhfeidhm (m.sh. oiliúint cúrsaí, comhdhálacha, Eorpach tionscadail srl.)\n74. Chun tháirgeadh riaracháin doiciméid (m.sh sonraisc) ísteach caidreamh chun an tionscnaimh os cionn\n75. Le haghaidh staitistiúil críocha\n76. Iompar amach cumarsáide gniomhaíochtaí tríd riomhphost maidir le an tionscnaimh.\n77. Freagair iarraitais ag baint úsáide as na foirméach ar an suíomh (má tá sé i láthair)\n78. Clárú a cheadú chun rochtain a fháil ar ábhar oideachais rúnda (má tá sé i láthair)\n\ndo bhronnadh éigeantach chun críocha faoi mhíreanna 1, 2, 5, 6 chun cloí le dlíniúil oibleagáidí agus AE dlíthe agus rialacháin; diúltú chun sholáthar pearsanta sonraí beidh ní ceadaigh eagraíocht eagraíochta chun tairiscint tusa, Ár Seirbhísí.\n\nTá do thoilíú roghnach chun críocha faoi mhír 3 agus 4; seolfaimid margaocht chugat cumarsáid trí r-phost nó seirbhís poist. Féadfaidh tú do chearta a fheidhmiú am ar bith, dá réir Airtseagal 15 agus níos déanaí de Rialachán AE 2016/679 maidir le tarraingt siar as a leithéid a fháil cumarsáide nó ag roghnú eile cumarsáide módúlachtaí.\n\nCoinneoidim do shonraí pearsanta a bhailítear chun na gcrioch faoi gach alt chomh fada mar is gá dúnna chun na seirbhísí a chur ar fáil duit ag ár n-eagraíocht agus ar feadh suas le 10 gcinn (deich) bliana.\n\ntu féidir tarraingt siar do toiliú ag ar bith am.\n\nIs éard atá i mbunús dlí na cóireála an caidreamh tráchtála a chruthaigh an díolachán nó ceannach earráí agus/nó seirbhísí, réamhchonarthach mar fhaisnéis (Airtseagal 6 mír b agus c), agus le toiliú le haghaidh margaochta gniomhaíochtaí. ( alt 6 alt a)\n\nDéanfaimid do shonraí a phróiseáil agus a stóráil chun na críocha thuasluaite amháin, ag baint úsáide as gléasanna digiteacha agus i mbunachair shonraí ábhartha cosaintí iomchuí a áirithiú chun rúndacht leanúnach a áirithiú, sláine, infhaighteacht agus athléimneacht na gcóras próiseála, mar atá leagtha amach i Rialachán AE 2016/679. Ní féidir ach le hábhair a bhfuil rochtain faighe agu ar shonraí pearsanta ón rialaitheoir nó ón bpróiseálaí próiseas den sорт sin eolas.\n\nNí chuirimid dhiol, trádáil, nó ar shlí eile a aistriú chuig tríú páirtithe eile do inaitheanta pearsanta eolas. Mar sin féin, féadfaidim do chuid faisnéise a scaoileadh nuair a chreidimid go bhfuil gá le scaoileadh géilleadh le an Dlí, fhorfheidhmiú ár suíomh polasaithe, nó chosaint linne nó daoine eile cearta, maoin, nó sábháilteacht.\n\nPróifíliú\nNí bheidh do shonraí faoi réir cinnidh atá bunaithe go hiomlán ar phróiseáil uathoibrithe, rud\na tháirgeann éifeachtaí dlíthiúla a dhéanann difear dóibh nó a dhéanann difear suntasach dá duine. Cealú agus Leasú: tá sé de cheart agat fios a bheith, ag am ar bith, cad iad do shonraí ag rialaitheoirí sonraí aonair, atá ag ár gcuideachta nó ag na daoine thuasluaite a gcuirimid in iúl dóibh iad, agus conas a úsáidtear iad; tá sé de cheart acu freisin iad a nuashonrú, a fhlorionadh, a cheartú nó a chealú, a bhloc a iarraidh agus cur i gcóireála. Chun do chearta a hheidhmiú, agus chun fainseáis níos mionsonraithe a fháil faoi na hábhair nó na catagóirí ábhar a gcuirtear na sonraí in iúl dóibh nó a bhfuil eolas acu orthu mar is féidir le bainisteoirí nó gniomhairí dul i dteagmháil leis an rialaitheoirí nó le duine dá bhainisteoirí, arna sainaithint i. an ráiteas seo.\n\nLíonraí sóisialta\nFéadfaidh ár suíomh Gréasáin rochtain ar líonra sóisialta a thairiscint. Na téarmaí seirbhíse agus an Beartas Príobháideachta is infheidhme chuig ardáin den sórt sin a fhoilsiú ar a láithreán gréasáin. Ní féidir le picteilíní an bealach a roinntear sonraí ar a poiblí fóram, comhrá nó deais bhfuil úsáidtear, á an sonraí ábhar freagrach de den sórt sin cumarsáide.\n\nGearáin\ntu féidir freisin teagmháil an Iodálach Sonraí Cosaint Údarás ag baint úsáide as an leanas nasc http://www.garanteprivacy.it/home/footer/contatti, nó an Maoirseoir Eorpach ar Chosain t Sonraí a úsáideann an leanas nasc: https://edps.europa.eu/data-protection/our-role-supervisor/complaints_ga\n\nFianáin\nMar se t u t le rialachán “ Treoirlínte fianán agus uirlisí rianaithe eile - 10 Meitheamh, 2021 ”, t anseo a maidir le trí príomhchatagóir de fianáin:\n\nTeicniúil fianáin\niad seo bhfuil úsáidtear le haghaidh an amháin cuspóir de “tarchur cumarsáide chun an leictreonach liónra cumarsáide, nó a mhéid is fiorghá chun seirbhís a sholáthar ag an cuideachta fainseáise arna hiarradh go sainráite ag an bpáirtí conarthach nó ag an úsáideoir chun é a sholáthar an tseirbhís sin” Nó úsáidtear iad seo chun criocha ar bith eile agus is gnách iad a shuiteáil go direach le an úinéir nó an bainisteoir de an láithreán gréasáin (mar a thugtar air “dílseánach” nó “eagarthóireacht” fianáin). iad seo féidir bheith roinnte isteach: ag brabháil nó seisiún fianáin, a ráthaíocht gnáth nascleanúint agus úsáid an tsuímh Ghréasáin (a fhágann gur féidir, mar shampla, ceannacháin a dhéanamh nó a bheith fiordheimhnithe d’fhonn rochtain a fháil ar limistéir fhorchóimeádta); fianáin anailíse arna gcomhshamhlú ag an teicniúil fianáin ina n-úsáideann bainisteoir an tsuímh Ghréasáin iad go direach chun fainseáis a bhailiú, i bhfoirm foirm ghaolmhar (gan ainm), faoi lón na n-úsáideoirí agus faoin mbealach a dtugann siad cuairt ar an suíomh Gréasáin; fianáin fheidhmiúla a ligeann don úsáideoir nascleanúint a dhéanamh maidir le sraith critéar roghnaithe (mar shampla, an teanga nó na táirgí a roghnaiodh le ceannach) chun an tseirbhís a fheabhsú ar choinnioll, ar choinnioll go cuirimid ár úsáideoirí mar leagan Amach le alt 13 AE Rialachán 2016/679. Tá an roimhe toilíú de an úsáideoir tá ní iartha isteach ordú chun shuiteáil iad seo fianáin.\n\nFianáin Analytics\nNí úsáideann an suíomh ach google Analytics, a úsáidtear chun próifílí na n-úsáideoirí agus iad a chruthú fostaithe chun teachtaireachtaí fógraíochta a sheoladh de réir na sainroghanna a léirionn an céanna le linn a gcuid loingseoireachta ar line. Mar gheall ar a invasiveness ar leith maidir le príobháideachta na n-úsáideoiríSféar, Eorpach agus Iodálaí rialachán cheangal go\núsáideoirí bheith go leordhóthanach eolasach faoi a gcuid úsáid de an céanna agus bhfuil mar sin ag teastáil chun sainráite a gcuid bailí toiliú. Ach sa chás sonrach tá Google Analytics gan ainm (cascadh IP) agus cuireadh bac ar roinnt sonraí loingseoireachta le google: ar an mbealach seo tá an fianán anailíseach cosuíl leis na fianáin theicniúla a luaitear thuas agus ní gá toiliú a fháil.\n\n**Fianáin a phróifíliú**\nNí úsáidtear fianán den chineál seo ar an suíomh.\n\nNóta sonrach:\nNí úsáideann na físeáin leabaithe de YouTube ar an suíomh fianán mar go bhfuil \"nocookie\" sonraíthe ann, an cód leabaithe priobháideachais le haghaidh do leabaithe físeáin YouTube go léir.\n\n**Aistriú sonraí idirnáisiúnta agus Eorpacha**\nDéanfar do shonraí a phróiseáil sa Limistéar Eorpach Eacnamaíoch amháin. Do chearta maidir leis an pearsanta sonraí agaín shealbhú faoi AE Rialachán 2016/679\n\n**Do chearta**\ntu féidir aclaíocht do cearta ar bith am, mar leagtha amach le Airteagal 7, par. 3, agus ailt 15 agus leanasde AE rialachán 2016/679:\n- Ceart a rochtain pearsanta sonraí\n- Ceart chun ceartúchán agus scriosadh de pearsanta sonraí;\n- Ceart chun srian de próiseáil;\n- Ceart chun sonraí iniomparthacht ;\n- Ceart chun réad chun próiseáil de pearsanta sonraí\n- Ceart chun dlíthiúil éileamh chun Iodálach Sonraí Cosaint Údarás.\n\nIs féidir leat do chearta a fheidhmiú trí ríomhphost a sheoladh chugainn ag info@pixel-online.net nó litir seolta go Pixel, trí Luigi Lanzi , 12 - 50134 - Firenze, an Iodáil. Tuilleadh faisnéise maidir le próiseáil sonraí is féidir a chur leis nuair a ag bailiú sonraí.\n\n12 Feabhra 2022 ath.03\nItaliano: Privacy Policy ai sensi del Regolamento UE 2016/679\n\nChi raccoglie i tuoi dati\nAi sensi dell'articolo 13 del Regolamento UE 2016/679 (GDPR), il titolare del trattamento è il Richiedente del progetto, ed è responsabile della raccolta dei dati. La informiamo che la nostra organizzazione è legalmente obbligata a trattare i dati che ci ha fornito ai sensi del suddetto regolamento.\nI suoi dati saranno trattati in modo lecito e secondo correttezza, ai sensi dell’articolo 5 del Regolamento UE 2016/679. Ulteriori dettagli potrebbero essere forniti in una fase successiva. Responsabile della protezione dei dati (dpo): la presenza di un eventuale DPO deve essere richiesta al titolare del trattamento.\n\nQuali dati personali raccogliamo\nAi sensi dell’articolo 4 del Regolamento UE 2016/679:\n- \"dati personali\": qualsiasi informazione relativa a una persona fisica identificata o identificabile (\"interessato\"); una persona fisica identificabile è una persona che può essere identificata, direttamente o indirettamente, in particolare mediante riferimento a un identificatore come un nome, un numero di identificazione, dati relativi all’ubicazione, un identificatore online o a uno o più fattori specifici dell’aspetto fisico, fisiologico, identità genetica, mentale, economica, culturale o sociale di quella persona fisica;\n- \"trattamento\" indica qualsiasi operazione o insieme di operazioni che viene eseguito su dati personali o su insiemi di dati personali, anche con mezzi automatizzati, come la raccolta, la registrazione, l’organizzazione, la strutturazione, la conservazione, l’adattamento o la modifica, il reperimento, la consultazione, uso, divulgazione mediante trasmissione, diffusione o altrimenti messa a disposizione, allineamento o combinazione, restrizione, cancellazione o distruzione.\nInsieme a riferimento a il sopra menzionato definizioni, noi sottolineare Quello noi raccogliere solo il informazione tu fornirci per il scopi del tuo coinvolgimento in nostro iniziative e/o il tuo legale relazione insieme a nostro organizzazione:\n- Dati personali: nome e cognome delle persone fisiche, contatti quali indirizzo, CAP codice, città, regione, telefono numero, e-mail;\n- Dati concernente professionisti/organizzazioni/imprese: informazione concernente attività, nome, indirizzo fiscale e altri identificatori (fax e numero di telefono, codice fiscale o IVA numero).\n\nInoltre, è possibile richiedere i dati forniti quando si accede ai nostri siti, cookie e altra tecnologia similitudine e quando ci contatti tramite e-mail, social media o tecnologie simili. Anche se tali dati non sono raccolti per essere associati alla persona fisica, questi identificatori online potrebbero essere utilizzati e combinati in modo da creare profili personali. Tra gli identificatori on line è possibile trovare indirizzo IP, tipo di browser e dettagli del plug-in, tipo di dispositivo (es. desktop, laptop, tablet, telefono, ecc.) sistema operativo, fuso orario locale. Questi dati vengono utilizzati esclusivamente per la produzione di risultati statistici.\nRicordiamo che non tratteremo dati personali che rivelino la razza o origine etnica, opinioni politiche, convinzioni religiose o filosofiche o appartenenza sindacale, e il trattamento di dati genetici, dati biometrici al fine di definire univocamente un naturale persona, dati concernente Salute o dati concernente un naturale di persona sesso vita o orientamento sessuale.\n\n**Perché e come trattiamo i tuoi dati**\n\nUtilizzeremo i tuoi dati nei seguenti modi:\n\n1. Organizzare e realizzare iniziative nel campo dell'istruzione e della formazione (es. corsi di formazione, conferenze, progetti europei ecc.)\n2. Produrre documenti amministrativi (es. fatture) in relazione alle iniziative di cui sopra\n3. A fini statistici\n4. Svolgere attività di comunicazione via e-mail in merito alle nostre iniziative.\n5. Rispondere alle richieste utilizzando i moduli presenti nel sito (se presente)\n6. Consentire la registrazione per l'accesso a contenuti didattici riservati (se presenti)\n\nIl Suo conferimento è obbligatorio per le finalità di cui ai commi 1, 2, 5, 6 al fine di ottemperare agli obblighi giuridici e alle leggi e ai regolamenti comunitari; il rifiuto a fornire i dati personali non consentirà alla nostra organizzazione di offrirvi i nostri servizi.\n\nIl Suo consenso è facoltativo per le finalità di cui ai paragrafi 3 e 4; ti invieremo comunicazioni di marketing via e-mail o servizio postale. In ogni momento Lei potrà esercitare i Suoi diritti, ai sensi degli artt. 15 e ss. Regolamento UE 2016/679, in merito alla rinuncia a ricevere tale comunicazione o alla scelta di altre modalità di comunicazione.\n\nConserveremo i tuoi dati personali raccolti per le finalità di cui a tutti i paragrafi per tutto il tempo necessario per fornirti i servizi offerti dalla nostra organizzazione e per un massimo di 10 (dieci) anni.\n\nPuoi revocare il tuo consenso in qualsiasi momento.\n\nLa base giuridica del trattamento è costituita dal rapporto commerciale instaurato dalla compravendita di beni e/o servizi, precontrattuale a titolo informativo (art. 6 comma b e c), e dal consenso per attività di marketing. (articolo 6 comma a)\n\nTratteremo e conserveremo i tuoi dati esclusivamente per le finalità di cui sopra, utilizzando dispositivi digitali e in pertinenti banche dati assicurando adeguate salvaguardie in modo da garantire la continua riservatezza, integrità, disponibilità e resilienza dei sistemi di trattamento, come previsto dal regolamento UE 2016/679. Solo i soggetti che hanno avuto accesso ai dati personali dal titolare o dal responsabile del trattamento possono trattare tali informazioni.\n\nNon vendiamo, scambiamo o altrimenti trasferiamo ad altre terze parti le tue informazioni di identificazione personale. Tuttavia, potremmo divulgare le tue informazioni quando riteniamo che il rilascio sia necessario per rispettare la legge, far rispettare le nostre politiche del sito o proteggere i nostri o altri diritti, proprietà o sicurezza.\n\n**Profilazione**\n\nI suoi dati non saranno oggetto di una decisione basata unicamente sul trattamento automatizzato, che produca effetti giuridici che lo riguardano o che incida in modo significativo sulla sua persona. Cancellazione e Rettifica: hai il diritto di conoscere, in ogni momento, quali\nsono i tuoi dati presso i singoli titolari del trattamento, ovvero presso la nostra società o presso i suddetti soggetti a cui li comuniciamo, e come vengono utilizzati; hanno inoltre il diritto di aggiornarli, integrarli, rettificarli o cancellarli, chiederne il blocco ed opporsi al loro trattamento. Per l'esercizio dei Suoi diritti, nonché per informazioni più dettagliate circa i soggetti o le categorie di soggetti ai quali i dati sono comunicati o che ne hanno conoscenza in qualità di responsabili o incaricati possono rivolgersi al titolare del trattamento o ad un suo responsabile, individuati in questa dichiarazione.\n\nSocial networks\nIl nostro sito Web può offrire l'accesso al social network. I termini di servizio e la Privacy Policy applicabili a tali piattaforme sono pubblicati sul loro sito web. Pixel non può controllare il modo in cui vengono utilizzati i dati condivisi su un forum pubblico, una chat o una dashboard, essendo l'interessato responsabile di tale comunicazione.\n\nDenunce, contestazioni\nÈ inoltre possibile contattare il Garante per la protezione dei dati personali italiano utilizzando il seguente link http://www.garanteprivacy.it/home/footer/contatti, o il Garante europeo della protezione dei dati utilizzando il a seguire link: https://edps.europa.eu/data-protection/our-role-supervisor/complaints_en\n\nCookies\nCome previsto dal regolamento “Linee guida sui cookie e altri strumenti di tracciamento - 10 giugno 2021”, si distinguono tre categorie principali di cookies:\n\nCookies tecnici\nSono utilizzati al solo fine di “effettuare la trasmissione di comunicazioni su una rete di comunicazione elettronica, o nella misura strettamente necessaria all'erogazione di un servizio da parte della società dell'informazione esplicitamente richiesto dal contraente o dall'utente al fine di erogare tale servizio”. Questi non vengono utilizzati per scopi ulteriori e sono normalmente installati direttamente dal titolare o gestore del sito web (c.d. cookie “proprietari” o “editoriali”). Questi possono essere suddivisi in: cookie di navigazione o di sessione, che garantiscono la normale navigazione e fruizione del sito web (permettendo, ad esempio, di realizzare un acquisto o autenticarsi per accedere ad aree riservate); cookie analytics assimilati ai cookie tecnici laddove utilizzati direttamente dal gestore del sito per raccogliere informazioni, in forma associata (anonima), sul numero degli utenti e su come questi visitano il sito stesso; cookie di funzionalità che consentono all'utente la navigazione in relazione ad una serie di criteri selezionati (ad esempio, la lingua o i prodotti selezionati per l'acquisto) al fine di migliorare il servizio reso allo stesso, a condizione che si informi gli utenti come previsto dall'articolo 13 UE Regolamento 2016/679.\nPer installare questi cookie non è richiesto il preventivo consenso dell'utente.\n\nCookies analitici\nIl sito utilizza esclusivamente google analytics, che vengono utilizzati al fine di creare profili relativi agli utenti e vengono utilizzati al fine di inviare messaggi pubblicitari in linea con le preferenze manifestate dallo stesso nell'ambito della navigazione in rete. Per la loro particolare invasività rispetto alla sfera privata degli utenti, le normative europee e italiane richiedono che gli utenti siano adeguatamente informati circa il loro utilizzo degli stessi e sono quindi tenuti ad esprimere il loro valido consenso. Ma nel caso specifico google analytics è stato anonimizzato\n(mascheramento IP) ed è stata bloccata la condivisione dei dati di navigazione con google: in questo modo il cookie analitico è assimilabile ai cookie tecnici sopra indicati e non necessita di consenso.\n\n**Cookies di profilazione**\nQuesto tipo di cookies non viene utilizzato nel sito.\n\nNota specifica:\nI video incorporati di YouTube sul sito non utilizzano i cookies in quanto è stato specificato \"nocookies\" il codice di incorporamento ottimizzato per la privacy per tutti i tuoi incorporamenti video di YouTube.\n\n**Trasferimento dati internazionale ed europeo**\nI tuoi dati saranno trattati esclusivamente nello Spazio Economico Europeo. I tuoi diritti in relazione ai dati personali in nostro possesso ai sensi del Regolamento UE 2016/679\n\n**I tuoi diritti**\nIn ogni momento Lei potrà esercitare i Suoi diritti, come previsto dall'art. 7, par. 3, e artt. 15 e seguenti del Regolamento UE 2016/679:\n- Diritto di accesso ai dati personali\n- Diritto di rettifica e cancellazione dei dati personali;\n- Diritto alla limitazione del trattamento;\n- Diritto alla portabilità dei dati;\n- Diritto di opposizione al trattamento dei dati personali\n\nDiritto di azione legale al Garante per la protezione dei dati personali italiano.\nPuoi esercitare i tuoi diritti inviandoci una e-mail all'indirizzo info@pixel-online.net o una lettera indirizzata a Pixel, via Luigi Lanzi, 12 – 50134 – Firenze, Italia. Ulteriori informazioni sul trattamento dei dati può essere aggiunto quando raccolta dati.\n\n12 febbraio 2022 rev.03\nLietuviškai: Privatumo politika pagal ES reglamentą 2016/679\n\nKas renka jūsų duomenis\nPagal ES reglamento 2016/679 (BDAR) 13 straipsnį duomenų valdytojas yra projekto pareiškėjas, kuris yra atsakingas už duomenų rinkimą. Informuojame, kad mūsų organizacija yra teisiškai įpareigota tvarkyti duomenis, kuriuos mums pateikėte pagal pirmiau minėtą reglamentą.\n\nJūsų duomenys bus tvarkomi teisėtai ir sąžiningai pagal ES reglamento 2016/679 5 straipsnio nuostatą. Išsamesnė informacija gali būti pateikta vėliau.\n\nDuomenų apsaugos pareigūnas (dpo): galimo DAP dalyvavimas turi būti paprašytas duomenų valdytojo.\n\nKokius asmens duomenis renkame\nĮ atitikimą su Straipsnis 4 apie ES reglamentas 2016/679:\n- „asmens duomenys“ – tai bet kokia informacija, susijusi su fiziniu asmeniu, kurio tapatybė nustatyta arba gali būti nustatyta („duomenys tema“); an identifikuoti natūralus asmuo yra vienas PSO gali būti identifikuotas, tiesiogiai arba netiesiogiai, visų pirma nurodant identifikatorius, pvz., vardą, identifikavimo numerį, vieta duomenys, an prisiungęs identifikatorius arba į vienas arba daugiau faktoriai specifinis į jį fizinis, fiziologinis, genetinė, protinis, ekonominis, kultūrinis arba socialiniai tapatybę kad natūralus asmuo;\n- \"apdorojimas\" reiškia bet koks operacija arba rinkinys apie operacijos kurios yra atliktą įjungta Asmeninis duomenis arba įjungta rinkiniai apie Asmeninis duomenys, arba ne pateikė automatizuotas reiškia, toks kaip kolekcija, įrašymas, organizavimas, struktūrizavimas, saugojimas, pritaikymas ar keitimas, paieška, konsultavimas, naudojimas, atskleidimas perduodant, platinant ar kitaip padarant prieinamą, suderinant arba derinys, apribojimas, trynimas arba sunaikinimas.\n\nSu nuoroda į jį aukščiau paminėta apibrėžimai, mes pabraukti kad mes rinkti tik į informacija tu suteikite mums už tikslai apie jūsų dalyvavimą mūsų iniciatyvas ir/arba jūsų teisėtas santykiai su mūsų organizacija:\n- Asmeninė informacija: fizinio asmenio vardas ir pavardė, kontaktai, tokie kaip adresas, ZIP kodas, miestas, telefonas numeris, paštas;\n- Duomenys susijusius profesionalai / organizacijos / įmonės: informacija susijusius įmonės, pavadinimas, fiskalinis adresas ir kiti identifikatoriai (fakso ir telefono numeris, mokesčių kodas arba PVM skaičius).\n\nBe to, mes galime rinkti duomenis, pateiktus jums prisiungęs prie mūsų svetainių, naudodami slapukus ir kt panaši technologiją; ir kai susisiekiate su mumis el. paštu, socialiniuose tinkluose ar panašiomis technologijomis. Net jei tokie duomenys nėra renkami taip, kad būtų siejami su fizinio asmenio, šie internetiniai identifikatoriai gali būti naudojami ir derinami siekiant sukurti asmeninius profilius. Tarp internetinių identifikatoriai, kuriuos galime rasti IP adresas, naršyklės tipas ir papildinio informacija, įrenginio tipas (pvz., darbalaukis, nešiojamasis kompiuteris, planšetinis kompiuteris, telefonas ir kt.) operacinė sistema, vietinė laiko juosta. Šie duomenys naudojami tik gamyba statistiniai rezultatus.\nNorime priminti, kad netvarkysime asmens duomenų, atskleidžiančių rasinę ar etninę kilmę, politinės pažiūros, religinių ar filosofinių įsitikinimų arba narystę profesinėse sąjungose, ir genetinių duomenų, biometrinių duomenų tvarkymas siekiant unikaliai identifikuoti natūralų asmuo, duomenis susijusius sveikata arba duomenis susijusius a natūralus asmens seksas gyvenimą arba seksualinis orientacija.\n\nKodėl ir kaip apdorojame jūsų duomenis\nMes valios naudoti tavo duomenis in Sekantis būdai:\n\n79. Organizuoti ir įgyvendinti iniciatyvas švietimo ir mokymo srityje (pvz., mokymas kursai, konferencijos, Europos projektus ir tt)\n80. Į gaminti administracinis dokumentus (pvz sąskaitos faktūros) in santykį jį iniciatyvas aukščiau\n81. Dėl statistiniai tikslai\n82. Nešioti išeiti bendravimas veikla per paštu susijusius mūsų iniciatyvas.\n83. Atsakykite į užklausas naudodami svetainėje esančias formas (jei jos yra)\n84. Leisti užsiregistruoti, kad gautumėte prieigą prie konfidencialaus švietimo turinio (jei jis yra)\n\nJūsų suteikimas yra privalomas 1, 2, 5 ir 6 dalyse nurodytais tikslais, kad būtų laikomasi jurinės įsipareigojimų ir ES įstatymai ir reglamentas; atsisakymas jį teikti Asmeninis duomenis valios ne leistimūs organizacijai pasiūlymas tu, Mūsų Paslaugos. Jūsų sutikimas yra neprivalomas 3 ir 4 dalyse nurodytais tikslais; atsisakyme jums rinkodarą bendravimas elektroniniu paštu arba pašto paslaugomis. Jūs galite bet kada pasinaudoti savo teisėmis pagal ES reglamento 2016/679 15 ir vėlesniais straipsniais dėl atsisakymo tokių gauti bendravimas arba pasirenkant kitą bendravimas būdus.\n\nMes saugosime jūsų asmens duomenis, surinktus visose pastraipose nurodytais tikslais, naudodami skaitmeninius įrenginius ir atitinkamose duomenų bazėse, užtikrinant tinkamas apsaugos priemones, kad būtų užtikrintas nuolatinis konfidencialumas, apdorojimo sistemų vientisumas, prieinamumas ir atsparumas, kaip nustatyta ES reglamente 2016/679. Gali tik tie subjektai, kurie iš duomenų valdytojo arba duomenų tvarkytojo gavo prieigą prie asmens duomenų procesas toks informacija.\n\nMes neparduodame, neprekiaujame ar kitaip neperduodame kitoms trečiosioms šalims, kurias galite identifikuoti informacija. Tačiau mes galime paskelbti jūsų informaciją, kai manome, kad tai būtina laikytis su jį įstatymas, vykdyti mūsų svetainė politika, arba apsaugoti mūsų arba kitų teises, nuosavybė, arba saugumo.\n\nProfiliavimas\nJūsų duomenims nebus taikomas sprendimas, pagrįstas tik automatizuotu tvarkymu, dėl kurio atsiranda teisinių padarinių, turinčių jiems įtakos arba reikšmingą poveikį jo asmeniui. Panaikinimas ir pakeitimas: Jūs turite teisę bet kuriuo metu sužinoti, kokie Jūsų duomenys yra pas atskirus duomenų valdytojus, tai yra mūsų įmonėje arba pas aukščiau nurodytus nurodytus asmenis, kuriems juos perduodame, ir kaip jie naudojami; jie taip pat turi teisę juos atnaujinti, papildyti, taisyti ar panaikinti, prašyti blokuoti ir prieštarauti jų gydymui. Dėl savo teisių įgyvendinimo, taip pat dėl išsamesnės informacijos apie subjektus ar subjektų kategorijas, kuriems perduodami duomenys arba kurie apie tai žino kaip valdytojai ar agentai, gali kreiptis į duomenų valdytoją arba vieną iš jo vadovų, nurodytus šis teiginys.\n\nSocialiniai tinklai\nMūsų svetainė gali pasiūlyti prieigą prie socialinio tinklo. Taikomos paslaugų teikimo sąlygos ir privatumo politika tokioms platformoms yra paskelbti jų svetainėje. „Pixel“ negali valdyti duomenų bendrinimo būdo a viešas forumas, pokalbis arba prietaisų skydelis yra naudotas, esamas į duomenis tema atsakingas apie toks bendravimas.\n\nSkundai\nTu gali taip pat kontaktas į italų Duomenys Apsauga Valdžia naudojant į sekantis nuoroda http://www.garanteprivacy.it/home/footer/contatti_arba Europos duomenų apsaugos priežiūros pareigūnas, naudodamas sekantis nuoroda: https://edps.europa.eu/data-protection/our-role-supervisor/complaints_en\n\nSlapukai\nKaip se t tu t pateikė reglamentas „ Slapukų ir kitų sekimo įrankių gairės – 2021 m. birželio 10 d. “, t čia yra trys pagrindinės kategorijos apie slapukai:\n\nTechninė sausainiai\nŠie yra naudojamas dėl į padas tikslas apie „perdavimas komunikacijos į an elektroninis ryšio tinkle arba tiek, kiek tai griežtai būtina paslaugai teikti informacijos įmonė, kurios susitariančioji šalis arba vartotojas aškiai paprašė pateikti minėtą paslaugą“ Jie nenaudojami jokiems išoriniams tikslams ir paprastai įrengiami tiesiogiai pateikė į savininkas arba į vadovas apie į Interneto svetainė (vadinamas \"patentuotas\" arba \"redakcinis\" slapukus). Šie gali būti padalintas į: naudomasis arba sesija sausainiai, kurios garantija normalus našymas ir naudojimasis svetaine (suteikia galimybę, pavyzdžiui, pirkėjui ar būti autentifikuotas, kad būtų galima pasiekti rezervuotas sritis); analitiniai slapukai, asimiliuojami techninių slapukai, kai juos tiesiogiai naudoja svetainės valdytojas informacijai rinkti susijusi forma (anoniminė), apie vartotojų skaičių ir būdą, kuriuo jie lankosi Interneto svetainė; funkciniai slapukai, leidžiantys vartotojui našyti pagal pasirinktų kriterijų seriją (pavyzdžiui, kalba ar perkami produktai), kad būtų patobulinta paslauga jeigu, su sąlyga, kad informuojame savo vartotojų kaip išdėstyti pateikė straipsnis 13 ES reglamentas 2016/679. The anksčiau sutikimas apie į Vartotojas yra ne paprašė ir įsakymas į diegti šie sausainiai.\n\nAnalitikos slapukai\nSvetainėje naudojama tik google analytics, kuri naudojama vartotojų profiliams kurti ir yra naudojami reklaminiams pranešimams siųsti pagal to paties nurodytas nuostatas našydami internete. Dėl jų ypatingo invaziškumo vartotojų privataus atžvilgiu sfera, Europos ir italų reglamentas reikalauja kad vartotojų būti adekvačiai informuotas apie jų naudoti apie į tas pats ir yra taigi reikalaujama į išreikšti jų galioja sutikimas . Bet konkrečiu atveju google\nanalytics buvo anonimizuotas (IP maskavimas) ir užblokuotas naršymo duomenų bendrinimas su google: tokiu būdu analitinis slapukas yra panašus į aukščiau nurodytus techninius slapukus ir jam sutikimo nereikia.\n\n**Profiliavimo slapukai**\nŠio tipo slapukai svetainėje nenaudojami.\n\nKonkreti pastaba:\nSvetainėje įterptuose „YouTube“ vaizdo įrašuose slapukai nenaudojami, nes buvo nurodytas „nocookie“ – privatumo patobulintas įterpimo kodas visiems jūsų „YouTube“ vaizdo įrašų įterpimams.\n\n**Tarptautinis ir Europos duomenų perdavimas**\nJūsų duomenys bus tvarkomi tik Europos ekonominėje erdvėje. Jūsų teisės, susijusios su Asmeninis duomenis mes laikykite pagal ES reglamentas 2016/679\n\n**Jūsų teisės**\nTu gali pratimas tavo teises bet koks laikas, kaip rinkinys išeiti pateikė Straipsnis 7, par. 3, ir straipsnius 15 ir sekantis apie ES reglamentas 2016/679:\n- Teisingai prieiti Asmeninis duomenis\n- Teisingai į ištaisymas ir trynimas apie Asmeninis duomenys;\n- Teisingai į apribojimas apie apdorojimas;\n- Teisingai į duomenis perkeliamumas;\n- Teisingai į objektas į apdorojimas apie Asmeninis duomenis\n- Teisingai į legalus pretenzija į italų Duomenys Apsauga Valdžia.\n\nSavo teisėmis galite pasinaudoti atsiųsdami mums el. laišką adresu info@pixel-online.net arba laišką adresu į Pixel, per Luigi Lanzi, 12 – 50134 – Firenze, Italija. Daugiau informacijos apie duomenų tvarkymą galima pridėti kada renkant duomenis.\n\nvasario 12 d 2022 m rev.03\nLatviešu valodā: Privātuma politika saskaņā ar ES regulu 2016/679\n\nKas vāc jūsu datus\nSaskaņā ar ES Regulas 2016/679 (VDAR) 13. pantu datu pārzinis ir projekta iesniedzējs, un tas ir atbildīgs par datu vākšanu. Vēlamies jūs informēt, ka mūsu organizācijai ir juridiskas saistības apstrādāt datus, ko esat mums sniedzis saskaņā ar iepriekšminēto regulu.\n\nJūsu dati tiks apstrādāti likumīgi un godīgi saskaņā ar ES Regulas 2016/679 5. panta nosacījumu. Sīkāka informācija var tikt sniegta vēlāk.\n\nDatu aizsardzības inspektors (dpo): iespējamā DAI kārtībā ir jāpieprasa datu pārzinim.\n\nKādus personas datus mēs apkopojam\nIn saskaņā ar Raksts 4 no ES regula 2016/679:\n- \"personas dati\" ir jebkura informācija, kas attiecas uz identificētu vai identificējamu fizisku personu (\"dati priekšmets\"); an identificējami dabīks persona ir viens PVO var būt identificēts, tieši vai netieši, jo īpaši atsaucoties uz identifikatoru, piemēram, vārdu, identifikācijas numuru, atrašanās vieta dati, an tiešsaistē identifikators vai uz viens vai vairāk faktoriem specifisks uz uz fiziska, fizioloģiska, ģenētiska, garīgs, ekonomikas, kultūras vai sociālā identitāte ka dabīks persona;\n- \"apstrāde\" nozīmē jebkura darbība vai komplekts no operācijas kuras ir veikta ieslēgts personisks datus vai ieslēgts komplekti no personisks dati, vai vai nē autors automatizēti nozīmē, tāds kā kolekcija, ierakstīšana, organizēšana, strukturēšana, uzglabāšana, pielāgošana vai pārveidošana, izguve, konsultācijas, izmantošana, izpaušana, pārraidot, izplatot vai citādi padarot pieejamu, saskaņojot vai kombinācija, ierobežojums, dzēšana vai iznīcināšana.\n\nAr atsauce uz uz virs minēts definīcijas, mēs pasvītrot ka mēs savākt tikai uz informāciju tu sniedz mums par mērķiem par jūsu iesaistīšanos mūsu iniciatīvas un/vai jūsu juridiskais attiecības ar mūsu organizāciju:\n- Personas informācija: fizisko personu vārds un uzvārds, kontakti, piemēram, adrese, pasta indekss kods, pilsēta, novads, telefons numurs, e-pasts;\n- Dati kas attiecas uz profesionāļi/organizācijas/uzņēmumi: informāciju kas attiecas uz uzņēmumi, nosaukums, fiskālā adrese un citi identifikatori (faksa un tālruņa numurs, nodokļu kods vai PVN numurs).\n\nTurklāt mēs varam apkopot datus, kas tiek sniegti, kad pieklūstat mūsu vietnēm, izmantojot sīkfailus un citus līdzīgu tehnoloģiju; un sazinoties ar mums, izmantojot e-pastu, sociālos medijus vai līdzīgas tehnoloģijas. Pat ja šādi dati netiek vākti tā, lai tie būtu saistīti ar fizisko personu, šie tiešsaistes identifikatorus var izmantot un kombinēt, lai izveidotu personiskus profilus. Starp tiešsaistes identifikatorus kurus mēs varam atrast IP adresi, pārlūkprogrammas veidu un spraudņa informāciju, ierīces veidu (piemēram, darbvīrsmu, klēpjodators, planšetdators, tālrunis utt.) operētājsistēma, vietējā laika josla. Šie dati tiek izmantoti tikai ražošana statistikas rezultātus.\n\nAtgādinām, ka mēs neapstrādāsim personas datus, kas atklāj rases vai etniskā izcelsme,\npolitiskie uzskati, reliģiskā vai filozofiskā pārliecība vai dalība arodbiedrībās, un ģenētisko datu, biometrisko datu apstrādi, lai unikāli identificētu dabisku persona, datus kas attiecas uz veselība vai datus kas attiecas uz a dabisks personas sekss dzīvi vai seksuāla orientācija.\n\nKāpēc un kā mēs apstrādājam jūsu datus\nMēs gribu izmantot jūsu datus iekšā sekojošais veidi:\n\n85. Organizēt un īstenot iniciatīvas izglītības un apmācības jomā (piemēram, apmācībakursi, konferences, Eiropas projekts utt.)\n86. Uz ražot administratīvā dokumentus (piemērē, attiecības uz iniciatīvas virs\n87. Priekš statistikas mērķiem\n88. Nēsājiet ārā komunikācija aktivitātes caur e-pasts kas attiecas uz mūsu iniciatīvas.\n89. Atbildiet uz pieprasījumiem, izmantojot vietnē esošās veidlapas (ja tādas ir)\n90. Atļaut registrešties, lai pieklūtu konfidenciālam izglītības saturam (ja tāds ir)\n\nJūsu dāvināšana ir obligāta 1., 2., 5. un 6. punktā minētajiem mērķiem, lai ievērotu juridiskās saistības un ES likumus un noteikumi; atteikums uz nodrošināt personiskās datus gribu nē Atļautmūsu organizācijai piedāvājums tu, Mūsu pakalpojumi. Jūsu piekrišana nav obligāta 3. un 4. punktā minētajiem mērķiem; mēs nosūtīsim jums mārketingu saziņu pa e-pastu vai pasta pakalpojumu. Jūs varat izmantot savas tiesības jebkura laikā saskaņā ar ES Regulas 2016/679 15. pants un jaunāki noteikumi par atteikšanos no šādas saņemšanas komunikācijas vai izvēloties citu komunikācijas modalitātes. Mēs glabāsim jūsu personas datus, kas savākti visos punktos minētajiem mērķiem, tik ilgi, kāmēr cilvēks nepieciešams, lai sniegtu jums piedāvātos pakalpojumus mūsu organizācijā un līdz 10 (desmit) gadi. Tu var atsaukt jūsu piekrišanu plkst jebkura laiks.\n\nApstrādes tiesisko pamatu veido komercattiecības, kas izveidotas, pārdodot vai preču un/vai pakalpojumu iegāde, pirms līguma noslēgšanas informācijas nolūkos (6. panta b un c punkts), un pēc piekrišanas priekš mārketinga aktivitātes. ( raksts 6 paragrāfs a)\n\nMēs nepārdodam, netirgojam vai kā citādi nenododam citām trešajām personām, kurās jūs varētu identificēt informāciju. Tomēr mēs varam izpaust jūsu informāciju, ja uzskatām, ka tā ir nepieciešama ievērot ar uz likums, īstenot mūsu vietnes politikas, vai aizsargāt mūsu tiesības, vai drošību.\n\nProfilēšana\nUz jūsu datiem netiks attiecināts lēmums, kas balstīts tikai uz automatizētu apstrādi, kas rada tiesiskas sekas, kas to ietekmē vai būtiski ietekmē tās personu. Anulēšana un grozīšana: jums ir tiesības jebkura laikā uzzināt, kādi ir jūsu dati pie individuālajiem datu pārziņiem, tas ir, mūsu uzņēmumā vai pie iepriekš minētajām personām, kurām mēs tos nododam, un kā tie tiek izmantoti; viņiem ir arī tiesības to atjaunināt, papildināt, labot vai atcelt, pieprasīt to bloķēšanu un iebilst pret to ārstēšanu. Lai īstenotu savas tiesības, kā arī lai iegūtu sīkāku\ninformāciju par subjektiem vai subjektu kategorijām, kuriem dati tiek nodoti vai kuri par to ir informēti kā vadītāji vai agenti, var sazināties ar datu pārziņu vai kādu no viņa pārvaldītājiem, kas norādīti šo paziņojumu.\n\n**Sociālie tīkli**\nMūsu vietne var piedāvāt piekļuvi sociālajam tīklam. Piemērojamie pakalpojumu sniegšanas noteikumi un privātuma politika platformām ir publicētas to tīmekļa vietnē. Pixel nevar kontrolēt veidu, kā dati tiek kopīgoti a publiski forums, tērzēšana vai mērinstrumentu panelis ir lietots, būtne uz datu priekšmets atbildīgs no tās komunikācijā.\n\n**Sūdzības**\nTu var arī kontaktpersona uz itāļu valodu Dati Aizsardzība Autoritāte izmantojot uz sekojošās saite: http://www.garanteprivacy.it/home/footer/contatti vai Eiropas Datu aizsardzības uzraudzītājs, izmantojot sekojošās saite: https://edps.europa.eu/data-protection/our-role-supervisor/complaints_en\n\n**Sīkdatnes**\nKā se t ou t autors noteikums “ Sīkfailu un citu izsekošanas rīku vadlīnijas — 2021. gada 10. jūnijs”, t šeit ir trīs galvenās kategorijas no cepumi:\n\n**Tehnisks cepumi**\nŠīs ir lietots priekš uz zole mērķis no \"pārraidīšana komunikācijas uz an elektroniski sakaru tīklā vai tiktāl, cik tas noteikti nepieciešams pakalpojuma sniegšanai informācijas uznēmums, ko līgumslēdzēja puse vai lietotājs ir skaidri pieprasījis, lai sniegtu minētās pakalpojums” Tie netiek izmantoti nekādiem slēptiem nolūkiem un parasti tiek uzstādīti tieši autors uz īpašnieks vai uz vadītājs no uz vietne (tā sauktais \"ipašums\" vai \"redakcionāls\" cepumi). Šīs var būt sadalīts iekšā: pārlūkošanu vai sesija cepumi, kuras garantija normāli navigācija un vietnes lietošana (dodot iespēju, piemēram, veikt pirkumus vai būt autentificēts, lai piekļūtu rezervētajām zonām); analītikas sīkfaili, ko pielīdzina tehniskie sīkdatnes, ja tās tieši izmanto vietnes pārvaldnieks, lai apkopotu informācijā saistīto veidlapu (anonīmu), par lietotāju skaitu un veidu, kādā viņi apmeklē tīmekļa vietne; funkcionālās sīkdatnes, kas lauj lietotājam pārvietoties saistībā ar virknī atslēguma kritēriju (piemēram, valoda vai iegādei izvēlētie produkti), lai uzlabotu pakalpojumu ar nosacījumu, ka informējam mūsu lietotājiem kā izklāstīts autors rakstā 13 ES regula 2016/679. The iepriekš piekrišanu no uz lietotājs ir nē pieprasīts iekšā pasūtījums uz uzstādīt šie cepumi.\n\n**Analytics sīkfaili**\nVietne izmanto tikai google analytics, kas tiek izmantota, lai izveidotu lietotāju profilus un ir izmanto reklāmas ziņojumu sūtīšanai atbilstoši tā norādītajām preferencēm tiešsaistes navigācijas laikā. Sakarā ar to īpašo invazitīti attiecībā uz lietotāju privāto sfēru, Eiropas un itāļu valoda noteikumi pieprasīt ka lietotājiem būt adekvāti informēts par viņu izmantot no uz tas pats un ir tātad nepieciešams uz izteikt viņu derīgs piekrišana . Taču konkrētajā gadījumā google analytics ir anonimizēts (IP maskēšana) un ir bloķēta navigācijas datu koplietošana ar Google: tādā veidā analītiskais sīkfails ir līdzīgs iepriekš norādītajām tehniskajām sīkdatnēm un tai nav nepieciešama piekrišana.\n\n**Profilēšanas sīkfaili**\nŠīs sīkfailu veids vietnē netiek izmantots.\nĪpaša piezīme:\nVietnē iegultajos YouTube videoklipos netiek izmantoti sīkfaili, jo visiem jūsu YouTube video iegulumiem ir norādīts \"nocookie\" kā konfidencialitātes uzlabots iegulšanas kods.\n\nStarptautiskā un Eiropas datu pārraide\nJūsu dati tiks apstrādāti tikai Eiropas Ekonomikas zonā. Jūsu tiesības attiecībā uz personisks datus mēs turiet zem ES regula 2016/679\n\nJūsu tiesības\nTu var vingrinājums jūsu tiesības jebkura laiks, kā komplekts ārā autors Raksts 7, pag. 3, un rakstus 15 un sekojošno ES regulējums 2016/679:\n\n- Taisnība piekļūt personisks datus\n- Taisnība uz labošana un dzēšana no personisks dati;\n- Taisnība uz ierobežojums no apstrāde;\n- Taisnība uz datus pārnesamība;\n- Taisnība uz objektu uz apstrāde no personisks datus\n- Taisnība uz juridiski prasība uz itāļu valodu Dati Aizsardzība Autoritāte.\n\nJūs varat īstenot savas tiesības, nosūtot mums e-pastu uz info@pixel-online.net vai adresētu vēstuli uz Pixel, izmantojot Luigi Lanzi , 12 – 50134 – Firenze, Itālija. Sīkāka informācija par datu apstrādi var pievienot, kad vācot datus.\n\n12. februāris 2022. gads rev.03\nMalti: Privacy Policy skont ir-Regolament tal-UE 2016/679\n\nMin Jiġbor id-Dejta Tiegħek\nSkont l-Artikolu 13 tar-Regolament tal-UE 2016/679 (GDPR), il-kontrollur tad-dejta huwa l-Applikant tal-proġett, u huwa responsabbli li jiġbor id-dejta. Nixtiequ ninfurmak li l-organizzazzjoni tagħna hija legalment marbuta li tipproċessa d-dejta li tkun ipprovdejtna taħt ir-regolament imsemmi hawn fuq.\nId-dejta tiegħek tiġi pproċessata b’mod legali u ġust, skont id-dispożizzjoni tal-artikolu 5 tar-Regolament tal-UE 2016/679. Aktar dettalji jistgħu jiġu pprovduti fi stadju aktar tard.\nUffiċjal tal-protezzjoni tad-dejta (dpo): il-preżenza ta’ DPO possibbli għandha tintalab mingħand il-kontrollur tad-dejta.\n\nLiema Data Personali Niġbru\nFil skond bil Artikolu 4 ta UE Regolament 2016/679:\n- “data personali” tfisser kwalunkwe informazzjoni relatata ma’ persuna fiżika identifikata jew identifikabbli (‘data suġġett’); an identifikabbli naturali persuna huwa waħda min jista tkun identifikati, direttament jew indirettament, b’mod partikolari b’referenza għal identifikatur bħal isem, numru ta’ identifikazzjoni, post data, an online identifikatur jew biex waħda jew aktar fatturi speċifiċi biex il fiżiku, fiżjoloġiku, ġenetiku, mentali, ekonomiku, kulturali jew soċjali identità ta’ dawk naturali persuna;\n- \"ipproċessar\" mezzi kwalunkwe operazzjoni jew sett ta operazzjonijiet li huwa mwettqa fuq personali data jew fuq settijiet ta personali data, jekk jew le minn awtomatizzat tfisser, tali kif ġbir, registrazzjoni, organizzazzjoni, strutturar, ħażna, adattament jew alterazzjoni, irkupru, konsultazzjoni, użu, żvelar bi trażmissjoni, disseminazzjoni jew b’xi mod ieħor jagħmlu disponibbli, allinjament jew kombinazzjoni, restrizzjoni, thassir jew qerda.\n\nBil referenza biex il hawn fuq imsemmi definizzjonijiet, aħna enfasizza dak aħna jiġbru biss il informazzjoni int aqgħna għall- skopijiet tal-involvejment tiegħek fi tagħna inizjattivi u/jew legali tiegħek relazzjoni ta’ tagħna organizzazzjoni:\n- Informazzjoni personali: isem u kunjom ta’ persuni fiżiċi, kuntatti bħall-indirizz, ZIP kodici, belt, reġjun, telefon numru, email;\n- Data tikkonċerni professionisti/organizzazzjonijiet/negozji: informazzjoni tikkonċerni negozji, isem, indirizz fiskali u identifikaturi oħra (numru tal-fax u tat-telefon, kodici tat-taxxa jew VAT numru).\n\nBarra minn hekk, nistgħu niġbru data pprovduta meta taċċessa s-siti tagħna, permezz tal-cookies u oħrajn teknoloġija simili; u meta tikkuntattjana permezz ta’ email, midja soċjali, jew teknoloġiji simili. Anke jekk tali data ma tingħabax sabiex tkun assoċjata mal-persuna fiżika, dawn identifikaturi onlajn jiġi jintużaw u kkombinati sabiex jinħolqu profili personali. Fost l-online identifikaturi li nistgħu nsibu indirizz IP, tip ta’ browser u dettalji tal-plug-in, tip ta’ apparat (eż. desktop, laptop, tablet, telefon, eċċ.) sistema operattiva, żona tal-ħin lokali. Din id-dejta tintuża biss għall- produzzjoni ta statistika riżultati.\nNixtiequ nfakkruk li mhux se nkunu qed nipproċessaw data personali li tiżvela razzjali jew origini etnika, opinjonijiet politiċi, twemmin religjuż jew filosofiku, jew shubija fi trejdjunjins, u l-ipproċessar ta ‘data ġenetika, data bijometrika għall-iskop ta’ identifikazzjoni unika naturali persuna, data tikkonċerna is-sahħa jew data tikkonċerna a naturali tal-persuna sess ħajja jew sesswali orjentazzjoni.\n\nGħaliex u Kif Nipproċessaw id-Dejta Tiegħek\nAħna se użu tiegħek data fi dawn li ġejjin modi:\n\n91. Torganizza u timplimenta inizjattivi fil-qasam tal-edukazzjoni u t-taħriġ (eż. tahriġkorsijiet, konferenzi, Ewropea proġetti eċċ.)\n92. Biex jiproduċi amministrattiva dokumenti (eż fatturi) fi relazzjoni biex il inizjattivi hawn fuq\n93. Għal statistika skopijiet\n94. Ġorru barra komunikazzjoni attivitajiet via email tikkonċerna tagħna inizjattivi.\n95. Irrispondi għal talbiet billi tuża l-formoli fuq is-sit (jekk ikun preżenti)\n96.Ħalli r-registrazzjoni għall-accēss għal kontenut edukattiv kunfidenzjali (jekk ikun preżenti)\n\nkonferiment tiegħek huwa obbligatorju għall-finijiet taħt il-paragrafi 1, 2, 5, 6 sabiex tikkonforma mal-ġuridiċi obbligi u UE liġijiet u regolamenti; rifjut biex jiprovdu personali data se le jippermettu tagħna organizzazzjoni biex offerta int, is-servizzi tagħna.\nIl-kunsens tiegħek huwa fakultattiv għal skopijiet taħt il-paragrafi 3 u 4; aħna nibagħtulek marketing komunikazzjoni permezz tal-posta elettronika jew servizz postali. Tista' teżercita d-drittijiet tiegħek kwalunkwe ħin, skont L-Artikolu 15 u aktar tard tar-Regolament tal-UE 2016/679 dwar l-ghażla li ma tirċevix tali komunikazzjoni jew għażla oħra komunikazzjoni modalitajiet.\nAħna se nżommu d-dejta personali tiegħek miġbura għall-iskopijiet taħt il-paragrafi kollha sakemm kif għandna bżonn sabiex nipprovdulek is-servizzi offruti mill-organizzazzjoni tagħna u sa 10 (ghaxar) snin.\nInti jista jirtira tiegħek kunsens fi kwalunkwe ħin.\n\nIl-bażi legali tat-trattament tikkonsisti fir-relazzjoni kummerċjali maħluqa mill-bejgh jew xiri ta’ oggetti u/jew servizzi, prekuntrattwali għall-informazzjoni (l-artikolu 6 paragrafu b u c),u bil-kunsens għal marketing attivitajiet. (artiklu 6 paragrafu a)\n\nAħna nipproċessaw u naħżnu d-dejta tiegħek biss għall-iskopijiet imsemmija hawn fuq, bl-użu ta’ apparat digitali u f’databases rilevanti li jiżguraw salvagwardji xierqa sabiex tigi żgurata kunfidenzjalità kontinwa, l-integrità, id-disponibbiltà u r-reżiljenza tas-sistemi tal-ipproċessar, kif stabbilit mir-regolament tal-UE 2016/679. Is-suġġetti biss li jkunu kisbu aċċess għal data personali mingħand il-kontrollur jew il-processur jistgħu process tali informazzjoni.\n\nAħna ma nbighux, innegozjaw, jew b’xi mod ieħor nittrasferixxu lil partijiet terzi oħra identifikabbli personalment tiegħek informazzjoni. Madankollu, nistgħu nirrilaxxaw l-informazzjoni tiegħek meta nemmnu li r-rilax huwa meħtieġ tikkonforma bil il-liġi, jinforza tagħna sit politiki, jew jipprotegi tagħna jew oħrajn drittijiet, proprjetà, jew sigurtà.\n\nProfiling\nId-dejta tiegħek mhux se tkun sogġetta għal deċiżjoni bbażata biss fuq ipproċessar\nawtomatizzat, li jiproduċi effetti legali li jaffettawha jew li taffettwa b’mod sinifikanti l’-persuna tagħha. Kanċellazzjoni u Emenda: ghandek id-dritt li tkun taf, fi kwalunkwe ħin, x’inh d-dejta tieghek fil-kontrolluri tad-dejta individwali, jiġifieri fil-kumpanija tagħna jew fil-persuni msemmija hawn fuq li lihom nikkomunikawhom, u kif tintuża; ghandhom ukoll id-dritt li jaġġornawhom, jissupplementaw, jikkoreġi jew tkun normali “editorjali” installati tipprovdi kumpanija ta’ informazzjoni mitluba b’mod espliċitu mill-netwerk ta’ komunikazzjoni, jew sal-ħin li huma użu tal-“editorjali” cookies assimilati mill-kumpanija tagħhom. Għall-eżercizzju tad-drittijiet tieghek, kif ukoll għal informazzjoni aktar dettaljata dwar is-sugġetti jew kategoriji ta’ sugġetti li lihom tiġi kkomunikata d-dejta jew li huma konxji minnha bhala maniġers jew ġġenti jistgħu jikkuntattjaw lill-kontrollur tad-dejta jew lil wieħed mill-maniġers tiegħu, identifikati f’ din id-dikjarazzjoni.\n\n**Netwerks soċjali**\nIl-websajt tagħna tista’ toffri aċċess għan-netwerk soċjali. It-termini tas-servizz u l-Politika ta’ Privatezza applikabbli għal pjattaformi bħal dawn huma ppubblikati fuq il-websajt tagħhom. Pixel ma jistax jikkontrolla l-mod kif id-data kondivizja fuq a pubbliku Forum, chat jew daxxbord huma użat, qed il data sugġett responsabbli ta’ tali komunikazzjoni.\n\n**Ilmenti**\nInti jista ukoll kuntatt il Taljan Data Protezzjoni Awtorità bl-użu il wara link http://www.garanteprivacy.it/home/footer/contatti, jew il-Kontrollur Ewropew għall-Protezzjoni tad-Data billi juża l- wara link: https://edps.europa.eu/data-protection/our-role-supervisor/complaints_en\n\n**Cookies**\nKif se t ou t minn re g ula t i o n “Linji gwida dwar ghodod ta’ traċċar tal-cookies u ohrajn - 10 ta’ Ġunju 2021”, t hawn a re tliet kategoriji ewlenin ta cookies:\n\n**Tekniku cookies**\nDawn huma użati għal il lingwata għan ta “tittrasmetti komunikazzjonijiet biex an elettronici netwerk ta’ komunikazzjoni, jew sal-punt strettament meħtieġ għall-provvista ta’ servizz mill-kumpanija ta’ informazzjoni mitluba b’mod espliċitu mill-parti kontraenti jew mill-utent sabiex tipprovdi l-imsemmi servizz” Dawn ma jintużawx għal skopijiet ulterjuri u huma normalment installati direttament minn il sid jew il maniġer ta il websajt (hekk imsejħa \"proprietarju\" jew \"editorjali\" cookies). Dawn jista tkun maqsuma fi: browsing jew sessjoni cookies, li garanzija normali in-navigazzjoni u l-użu tal-websajt (li tagħmilha possibbli pereżempju, li tagħmel xiri jew tkun awtentikati sabiex jaċċessaw żoni riżervati); cookies analitiċi assimilati mill-tekniki cookies fejn huma użati direttament mill-maniġer tal-websajt biex jiġbru informazzjoni, f’an formola assoċjata (anonima), dwar in-numru ta’ utenti u l-mod li bih iżuru l- websajt; cookies funzjonali li jippermettu lill-utent jinnaviga fir-rigward ta’ serje ta’ kriterji magħżula (pereżempju, il-lingwa jew il-prodotti magħżula għax-xiri) sabiex jittejjeb is-servizz ipprovdu, basta li ninfurma tagħna utenti kif stabbiliti minn artikolu 13 UE Regolament 2016/679. Il- qabel kunsens ta il utent huwa le mitluba fi ordni biex tinstalla dawn cookies.\n\n**Cookies analitiċi**\nIs-sit juża biss google analytics, li jintuża biex jinħolqu profili tal-utenti u huma impjegati biex jintbaghtu messaġġi ta’ reklamar skont il-preferenzi murija mill-istess waqt in-navigazzjoni onlajn tagħhom. Minħabba l-invażività partikolari tagħhom fir-rigward tal-privat tal-utenti sfera, Ewropea u Taljan regolamenti jeħtieġu dak utenti tkun adegwatament infurmat dwar tagħhom użu ta il l-istess u huma għalhekk meħtieġ biex jesprimu tagħhom validu kunsens.\nIżda fil-każ speċifiku google analytics ġiet anonimizzata (IP masking) u l-kondivizjoni tad-dejta tan-navigazzjoni ma 'google ġiet imblukkata: b'dan il-mod il-cookie analitika hija simili għall-cookies tekniċi indikati hawn fuq u ma teħtieġx kunsens.\n\n**Cookies tal-profili**\nDan it-tip ta’ cookie ma jintużax fis-sit.\n\nNota speċifika:\nIl-vidjows inkorporati ta 'YouTube fis-sit ma jużawx cookies peress li ġie speċifikat \" nocookie \" il-kodiċi ta' inkorporazzjoni msahħa bil-privatezza għall-inkorporazzjonijiet kollha tal-vidjow YouTube tiegħek.\n\n**Trasferiment ta' data internazzjonali u Ewropew**\nId-dejta tiegħek tiġi pproċessata biss fiż-Żona Ekonomika Ewropea. Id-drittijiet tiegħek fir-rigward tal-personali data ahna żomm taħt UE Regolament 2016/679\n\n**Id-drittijiet tiegħek**\nInti jista eżerċizzju tiegħek drittijiet kwalunkwe ħin, kif sett barra minn Artikolu 7, par. 3, u artikoli 15 u warata UE regolamentazzjoni 2016/679:\n- Dritt għall-aċċess personali data\n- Dritt biex rettifika u thassir ta personali data;\n- Dritt biex restrizzjoni ta l-ipproċessar;\n- Dritt biex data portabbiltà;\n- Dritt biex oġġett biex ipproċessar ta personali data\n- Dritt biex legali talba biex Taljan Data Protezzjoni Awtorità.\n\nTista' teżerċita d-drittijiet tiegħek billi tibgħatilna email fuq info@pixel-online.net jew ittra indirizzata lil Pixel, via Luigi Lanzi , 12 – 50134 – Firenze, l-Italja. Aktar informazzjoni dwar l-ipproċessar tad-data jistgħu jiġu miżjuda meta ġbir ta' data.\n\n12 ta' Frar 2022 rev.03\nNederlands: Privacybeleid in overeenstemming met EU-verordening 2016/679\n\nWie verzamelt uw gegevens\nOvereenkomstig artikel 13 van EU-verordening 2016/679 (AVG) is de gegevensbeheerder de aanvrager van het project en is hij verantwoordelijk voor het verzamelen van de gegevens. Graag willen wij u erop wijzen dat onze organisatie wettelijk verplicht is om de gegevens die u ons heeft verstrekt op grond van voornoemde regeling te verwerken. Uw gegevens worden rechtmatig en eerlijk verwerkt, in overeenstemming met artikel 5 van EU-verordening 2016/679. Meer details kunnen in een later stadium worden verstrekt.\n\nvoor gegevensbescherming (dpo): de aanwezigheid van een eventuele DPO moet worden aangevraagd bij de verwerkingsverantwoordelijke.\n\nWelke persoonlijke gegevens we verzamelen?\nIn overeenstemming met Artikel 4 van EU Regulatie 2016/679:\n- \"persoonlijke gegevens\" betekent alle informatie met betrekking tot een geïdentificeerde of identificeerbare natuurlijke persoon ('gegevens onderwerp'); een identificeerbaar natuurlijk persoon is een WHO kan zijn geïdentificeerd, direct of indirect, met name door te verwijzen naar een identificator zoals een naam, een identificatienummer, plaats gegevens, een online identificatie of naar een of meer factoren specifiek naar de fysiek, fysiologisch, genetisch, mentaal, economisch, cultureel of sociaal identiteit van Dat natuurlijk persoon;\n- \"verwerken\" middelen elk operatie of set van activiteiten die is uitgevoerd Aan persoonlijk gegevens of Aan sets van persoonlijk gegevens, of of niet door geautomatiseerd middelen, zo een zoals verzameling, vastleggen, ordenen, structureren, opslaan, aanpassen of wijzigen, opvragen, raadplegen, gebruik, openbaarmaking door verzending, verspreiding of anderszins beschikbaar stellen, afstemmen of combinatie, beperking, wissen of verwoesting.\n\nMet verwijzing naar de bovenstaand genoemd definities, wij onderstrepen Dat wij verzamelen alleen de informatie u geef ons voor de doeleinden van uw betrokkenheid bij ons initiatieven en/of uw wettelijke relatie met ons organisatie:\n- Persoonlijke informatie: voor- en achternaam van natuurlijke personen, contacten zoals adres, ZIP code, stad, regio, telefoon nummer, e-mail;\n- Gegevens betreft professionals/organisaties/bedrijven: informatie betreft bedrijven, naam, fiscaal adres en andere identificatiegegevens (fax- en telefoonnummer, belastingcode) of btw nummer).\n\nBovendien kunnen we gegevens verzamelen die worden verstrekt wanneer u onze sites bezoekt, via cookies en andere vergelijkbare technologie; en wanneer u contact met ons opneemt via e-mail, sociale media of vergelijkbare technologieën. Hoewel dergelijke gegevens niet worden verzameld om in verband te worden gebracht met de natuurlijke persoon, zijn deze online identifiers kunnen worden gebruikt en gecombineerd om persoonlijke profielen te\ncreëren. Onder de online ID's kunnen we IP-adres, browsertype en plug-indetails, apparaattype (bijv. desktop, laptop, tablet, telefoon, enz.) besturingssysteem, lokale tijdzone. Deze gegevens worden uitsluitend gebruikt voor de productie van statistisch resultaten.\n\nWe willen u eraan herinneren dat we geen persoonlijke gegevens zullen verwerken waaruit ras of etnische afkomst, politieke opvattingen, religieuze of filosofische overtuigingen of lidmaatschap van een vakbond, en de verwerking van genetische gegevens, biometrische gegevens met het oog op de unieke identificatie van een natuurlijke persoon, gegevens betreft Gezondheid of gegevens betreft een natuurlijk personen seks leven of seksueel oriëntatie.\n\n**Waarom en hoe we uw gegevens verwerken**\n\nWe zullen gebruiken uw gegevens in het volgende manieren:\n\n97. Initiatieven op het gebied van onderwijs en training organiseren en uitvoeren (bijv. cursussen, congressen, Europese projecten enzovoort.)\n98. Naar produceren administratief documenten (bijv facturen) in relatie naar de initiatieven bovenstaand\n99. Voor statistisch doeleinden\n100. Dragen uit communicatie activiteiten via e-mail betreft ons initiatieven.\n101. Reageren op verzoeken met behulp van de formulieren op de site (indien aanwezig)\n102. Sta registratie toe voor toegang tot vertrouwelijke educatieve inhoud (indien aanwezig)\n\nUw conferentie is verplicht voor doeleinden onder de paragrafen 1, 2, 5, 6 om te voldoen aan: juridisch verplichtingen en EU wetten en regelgeving; weigering naar voorzien in persoonlijk gegevens zullen niet toestaan ons organisatie om aanbieding u, Onze diensten. Uw toestemming is optioneel voor de doeleinden onder paragraaf 3 en 4; wij sturen u marketing communicatie via e-mail of post. U kunt uw rechten te allen tijde uitoefenen op grond van: Artikel 15 en later van EU-verordening 2016/679 met betrekking tot opt-out voor het ontvangen van dergelijke communicatie of andere kiezen communicatie modaliteiten. We bewaren uw persoonlijke gegevens die zijn verzameld voor de doeleinden onder alle paragrafen zolang zoals we nodig hebben om u de aangeboden diensten te bieden door onze organisatie en tot 10 (tien jaar). Jij kan terugtrekken uw toestemming Bij elk tijd.\n\nDe juridische basis van de behandeling bestaat uit de commerciële relatie die is ontstaan door de verkoop of aankoop van goederen en/of diensten, precontractueel ter informatie (artikel 6 lid b en c),en met toestemming voor marketing activiteiten. ( artikel 6 paragraaf een)\n\nWe verwerken en bewaren uw gegevens uitsluitend voor de bovengenoemde doeleinden, met behulp van digitale apparaten en in relevante databases die passende waarborgen bieden om voortdurende vertrouwelijkheid te waarborgen, integriteit, beschikbaarheid en veerkracht van verwerkingssystemen, zoals uiteengezet in EU-verordening 2016/679. Alleen personen die toegang hebben gekregen tot persoonsgegevens van de verwerkingsverantwoordelijke of de verwerker kunnen Verwerken zo een informatie.\n\nWij verkopen, verhandelen of dragen uw persoonlijk identificeerbare informatie. We kunnen\nuw informatie echter vrijgeven wanneer we van mening zijn dat vrijgave noodzakelijk is om: voldoen aan met de wet, afdwingen ons site beleid, of beschermen De onze of anderen rechten, eigenschap, of veiligheid.\n\n**Profilering**\nUw gegevens worden niet onderworpen aan een uitsluitend op geautomatiseerde verwerking gebaseerde beslissing die rechtsgevolgen heeft die van invloed zijn op hen of die haar persoon in aanzienlijke mate treffen. Annulering en wijziging: u hebt het recht om te allen tijde te weten wat uw gegevens zijn bij de individuele verwerkingsverantwoordelijken, dat wil zeggen bij ons bedrijf of bij de bovengenoemde personen aan wie we ze doorgeven, en hoe ze worden gebruikt; ze hebben ook het recht om ze bij te werken, aan te vullen, te corrigeren of te annuleren, hun blokking aan te vragen en zich tegen hun behandeling te verzetten. Voor de uitoefening van uw rechten, evenals voor meer gedetailleerde informatie over de onderwerpen of categorieën van onderwerpen aan wie de gegevens worden gecommuniceerd of die hiervan op de hoogte zijn, kunnen als managers of agenten contact opnemen met de gegevensbeheerder of een van zijn managers, geïdentificeerd in deze verklaring.\n\n**Sociale netwerken**\nOnze website kan toegang bieden tot sociale netwerken. De servicevoorwaarden en het privacybeleid zijn van toepassing naar dergelijke platforms worden gepubliceerd op hun website. Pixel heeft geen controle over de manier waarop gegevens worden gedeeld op een openbaar forum, chatten of dashboard zijn gebruikt, wezen de gegevens onderwerp verantwoordelijk van zo een communicatie.\n\n**Klachten**\nJij kan ook contact de Italiaans Gegevens Bescherming Autoriteit gebruik makend van de volgend koppeling http://www.garanteprivacy.it/home/footer/contatti, of de Europese Toezichthouder voor gegevensbescherming met behulp van de volgend link: https://edps.europa.eu/data-protection/our-role-supervisor/complaints_en\n\n**Koekjes**\nZoals kijk naar je moet door re g ula t i e \"Richtlijnen voor cookies en andere trackingtools - 10 juni 2021\", t hier er zijn drie hoofdcategorieën van koekjes:\n\n**Technisch koekjes**\nDeze zijn gebruikt voor de zool doel van \"zenden\" communicatie naar een elektronisch communicatienetwerk, of voor zover strikt noodzakelijk voor het verlenen van een dienst door de informatiebedrijf uitdrukkelijk gevraagd door de contractant of de gebruiker om de genoemde service\" Deze worden niet gebruikt voor andere doeleinden en worden normaal gesproken geïnstalleerd direct door de eigenaar of de manager van de website (zogenaamde \"eigendom\" of \"redactioneel\" koekjes). Deze kan zijn verdeeld naar binnen: browsen of sessie koekjes, die garantie normaal navigatie en gebruik van de website (waardoor het bijvoorbeeld mogelijk is om aankopen te doen of geauthenticeerd om toegang te krijgen tot gereserveerde gebieden); analytische cookies geassimileerd door de technische cookies waarbij ze direct door de beheerder van de website worden gebruikt om informatie te verzamelen, in een bijbehorende formulier (anoniem), over het aantal gebruikers en de manier waarop zij de website bezoeken website; functionele cookies waarmee de gebruiker kan navigeren in relatie\ntot een reeks geselecteerde criteria (bijvoorbeeld de taal of de producten die voor aankoop zijn geselecteerd) om de service te verbeteren mits, mits wij informeren onze gebruikers zoals op weg gaan door artikel 13 EU Regulatie 2016/679. De voorafgaand toestemming van de gebruiker is niet aangevraagd in volgorde naar installeren deze koekjes.\n\n**Analytics-cookies**\nDe site gebruikt alleen google analytics, die wordt gebruikt om profielen van de gebruikers te maken en zijn gebruikt voor het verzenden van reclameboodschappen volgens de door hem getoonde voorkeuren tijdens hun online navigatie. Vanwege hun bijzondere invasiviteit met betrekking tot de privégebied, Europese en Italiaans voorschriften vereisen Dat gebruikers zijn adequaat op de hoogte over hun gebruiken van de dezelfde en zijn dus vereist naar nadrukkelijk hun Geldig toestemming. Maar in het specifieke geval is google analytics geanonimiseerd (IP-masking) en is het delen van navigatiegegevens met google geblokkeerd: op deze manier lijkt de analytische cookie op de hierboven aangegeven technische cookies en is geen toestemming vereist.\n\n**Profileringscookies**\nDit type cookie wordt niet gebruikt op de site.\n\nSpecifieke opmerking:\nDe ingesloten video's van YouTube op de site maken geen gebruik van cookies omdat er \"nocookie \" is gespecificeerd als de privacy-verbeterde insluitcode voor al uw ingesloten YouTube-video's.\n\n**Internationale en Europese gegevensoverdracht**\nUw gegevens worden uitsluitend verwerkt in de Europese Economische Ruimte. Uw rechten met betrekking tot depersoonlijk gegevens wij houden onder EU Regulatie 2016/679\n\n**Jou rechten**\nJij kan oefening uw rechten elk tijd, zoals set uit door Artikel 7, par. 3, en Lidwoord 15 en volgendvan EU regulatie 2016/679:\n\n- Rechts toegang krijgen tot persoonlijk gegevens\n- Rechts naar rectificatie en wissen van persoonlijk gegevens;\n- Rechts naar beperking van verwerken;\n- Rechts naar gegevens draagbaarheid;\n- Rechts naar voorwerp naar verwerken van persoonlijk gegevens\n- Rechts naar legaal claim naar Italiaans Gegevens Bescherming Autoriteit.\n\nU kunt uw rechten uitoefenen door ons een e-mail te sturen op info@pixel-online.net of een geadresseerde brief naar Pixel, via Luigi Lanzi , 12 – 50134 – Firenze, Italië. Meer informatie over gegevensverwerking kan worden toegevoegd wanneer gegevens verzamelen.\n\n12 februari 2022 rev.03\nNorsk: Personvernerklæring i henhold til EU-forordning 2016/679\n\nHvem samler inn dataene dine\nI henhold til artikkel 13 i EU-forordning 2016/679 (GDPR), er behandlingsansvarlig søkeren av prosjektet, og den er ansvarlig for å samle inn dataene. Vi vil gjerne informere deg om at vår organisasjon er juridisk forpliktet til å behandle dataene du har gitt oss i henhold til nevnte regelverk.\n\nDine data vil bli behandlet lovlig og rettferdig, i henhold til bestemmelsen i artikkel 5 i EU-forordning 2016/679. Ytterligere detaljer kan bli gitt på et senere tidspunkt.\n\nDatabeskyttelsesansvarlig (dpo): tilstedeværelsen av en eventuell databeskyttelsesansvarlig må bes om fra behandlingsansvarlig.\n\nHvilke personopplysninger vi samler inn\nI samsvar med Artikkel 4 av EU Regulering 2016/679:\n- «personopplysninger» betyr all informasjon knyttet til en identifisert eller identifiserbar fysisk person ('data Emne'); an identifiserbar naturlig person er en hvem kan være identifisert, direkte eller indirekte, spesielt ved henvisning til en identifikator som et navn, et identifikasjonsnummer, plassering data, an på nett identifikator eller til en eller mer faktorer spesifikk til de fysisk, fysiologiske, genetisk, mental, økonomisk, kulturell eller sosial identiteten til at naturlig person;\n- \"behandling\" midler noen operasjon eller sett av operasjoner hvilken er utført på personlig data eller på settene av personlig data, om eller ikke av automatisert midler, slik som samling, registrering, organisering, strukturering, lagring, tilpasning eller endring, gjenfinning, konsultasjon, bruk, avsløring ved overføring, formidling eller på annen måte tilgjengeliggjøring, justering eller kombinasjon, begrensning, sletting eller ødeleggelse.\n\nMed referanse til de ovenfor nevnt definisjoner, vi understreke at vi samle inn kun de informasjon du gi oss for formål av ditt engasjement i vår initiativ og/eller din lovlige forhold med vår organisasjon:\n- Personopplysninger: navn og etternavn på fysiske personer, kontakter som adresse, postnummer kode, by, region, telefon Nummer, e-post;\n- Data angående fagpersoner/organisasjoner/bedrifter: informasjon angående bedrifter, navn, skatteadresse og andre identifikatorer (faks og telefonnummer, skattekode eller mva Nummer).\n\nDessuten kan vi samle inn data som oppgis når du går inn på nettsidene våre, gjennom informasjonskapsler og annet lignende teknologi; og når du kontakter oss via e-post, sosiale medier eller lignende teknologier. Selv om slike data ikke samles inn for å knyttes til den fysiske personen, er disse Internett-identifikatorer kan brukes og kombineres for å lage personlige profiler. Blant online identifikatorer vi kan finne IP-adresse, nettlesertype og plugin-detaljer, enhetstype (f.eks. skrivebord, bærbar PC, nettbrett, telefon osv.) operativsystem, lokal tidssone. Disse dataene brukes utelukkende til produksjon av statistisk resultater.\nVi vil minne deg på at vi ikke vil behandle personopplysninger som avslører rase eller etnisk opprinnelse, politiske meninger, religiøse eller filosofiske overbevisninger, eller fagforeningsmedlemskap, og behandlingen av genetiske data, biometriske data med det formål å unikt identifisere en naturlig person, data angående Helse eller data angående en naturlig personens kjønn liv eller seksuell orientering.\n\nHvorfor og hvordan vi behandler dataene dine\nVi vil bruk din data i følgende måter:\n\n103. Organisere og implementere tiltak innen utdanning og opplæring (f.eks. opplæringkurs, konferanser, europeisk prosjekter etc.)\n104. Til produsere administrativt dokumenter (f.eks fakturaer) i forhold til de initiativ ovenfor\n105. Til statistisk formål\n106. Bære ute kommunikasjon aktiviteter via e-post angående vår initiativ.\n107. Svar på forespørsler ved å bruke skjemaene på nettstedet (hvis det finnes)\n108. Tillat registrering for tilgang til konfidensielt pedagogisk innhold (hvis det finnes)\n\nDin erkjennelse er obligatorisk for formål under paragraf 1, 2, 5, 6 for å overholde juridiske forpliktelser og EU lover og forskrifter; avslag til gi personlig data vil ikke tillate vår organisasjon til by på du, Våre tjenester.\nDitt samtykke er valgfritt for formål under paragraf 3 og 4; vi sender deg markedsføring kommunikasjon via e-post eller posttjeneste. Du kan utøve dine rettigheter når som helst, iht Artikkel 15 og senere i EU-forordning 2016/679 om å velge bort å motta slike kommunikasjon eller velge andre kommunikasjon modaliteter.\nVi vil beholde dine personopplysninger som samles inn for formålene under alle paragrafer så lenge som vi trenger for å gi deg tjenestene som tilbys av vår organisasjon og for opptil 10 (ti) år.\nDu kan ta ut din samtykke på noen tid.\n\nDet rettslige grunnlaget for behandlingen består av det kommersielle forholdet som er opprettet ved salget eller kjøp av varer og/eller tjenester, pre-kontraktuell for informasjon (artikkel 6 ledd b og c), og ved samtykke til markedsføring aktiviteter. ( artikkel 6 avsnitt en)\n\nVi vil behandle og lagre dataene dine utelukkende for de nevnte formålene, ved bruk av digitale enheter og i relevante databaser for å sikre passende sikkerhetstiltak for å sikre kontinuerlig konfidentialitet, integritet, tilgjengelighet og motstandskraft til behandlingssystemer, som fastsatt i EU-forordning 2016/679. Kun forsøkspersoner som har fått tilgang til personopplysninger fra behandlingsansvarlig eller databehandler kan prosess slik informasjon.\n\nVi selger ikke, bytter eller på annen måte overfører dine personlig identifiserbare til andre tredjeparter informasjon. Vi kan imidlertid frigi informasjonen din når vi mener det er nødvendig etterkomme med de lov, håndheve vår nettstedet retningslinjer, eller beskytte vårt eller andres rettigheter, eiendom, eller sikkerhet.\n\nProfilering\nDine data vil ikke bli gjenstand for en avgjørelse basert utelukkende på automatisert behandling, som gir rettsvirkninger som påvirker dem eller som i betydelig grad påvirker\npersonens person. Kansellering og endring: du har rett til å vite, når som helst, hva som er dine data hos de individuelle behandlingsansvarlige, det vil si hos vårt selskap eller hos de ovenfor nevnte personene som vi kommuniserer dem til, og hvordan de brukes; de har også rett til å oppdatere, supplere, korrigere eller kansellere dem, be om blokkering og motsette seg behandlingen. For å utøve rettighetene dine, samt for mer detaljert informasjon om subjektene eller kategoriene av subjekter som dataene formidles til eller som er klar over det som ledere eller agenter, kan du kontakte behandlingsansvarlig eller en av hans ledere, identifisert i dette utsagnet.\n\nSosiale nettverk\nVår nettside kan tilby tilgang til sosiale nettverk. Vilkårene for bruk og personvernerklæringen gjelder til slike plattformer publiseres på deres nettside. Pixel kan ikke kontrollere måten data deles på offentlig forum, chatte eller dashbord er brukt, å være de data Emne ansvarlig av slik kommunikasjon.\n\nKlager\nDu kan også ta kontakt med de italiensk Data Beskyttelse Autoritet ved hjelp av de følgende link http://www.garanteprivacy.it/home/footer/contatti, eller European Data Protection Supervisor ved å bruke følgende lenke: https://edps.europa.eu/data-protection/our-role-supervisor/complaints_en\n\nInformasjonskapsler\nSom se t du t av re g ula t i o n \"Retningslinjer for informasjonskapsler og andre sporingsverktøy - 10. juni 2021\", t her a re tre hovedkategorier av informasjonskapsler:\n\nTeknisk informasjonskapsler\nDisse er brukt til de såle hensikt av «sender kommunikasjon til en elektronisk kommunikasjonsnettverk, eller i den grad det er strengt nødvendig for levering av en tjeneste av informasjonsselskap som er eksplisitt bedt om av avtaleparten eller brukeren for å gi nevnte tjeneste” Disse brukes ikke til andre formål, og de er normalt installert direkte av de Eieren eller de sjef av de nettsted (såkalt \"proprietær\" eller \"redaksjonell\" informasjonskapsler). Disse kan være delt inn i: surfing eller økt informasjonskapsler, hvilken garanti normal navigering og bruk av nettstedet (gjør det mulig for eksempel å foreta kjøp eller være autentisert for å få tilgang til reserverte områder); analytiske informasjonskapsler assimilert av den tekniske informasjonskapsler hvor de brukes direkte av administratoren av nettstedet for å samle informasjon, i en tilknyttet skjema (anonym), om antall brukere og måten de besøker nettsted; funksjonelle informasjonskapsler som lar brukeren navigere i forhold til en rekke utvalgte kriterier (for eksempel språket eller produktene som er valgt for kjøp) for å forbedre tjenesten sørget for, forutsatt at vi informerer vår brukere som sette ut av artikkel 1. 3 EU Regulering 2016/679.\nDe i forkant samtykke av de bruker er ikke Forespurt i rekkefølge til installere disse informasjonskapsler.\n\nAnalytics-informasjonskapsler\nSiden bruker kun google analytics, som brukes til å lage profiler av brukerne og er ansatt for å sende reklamemeldinger i henhold til preferansene vist av samme under deres online navigasjon. På grunn av deres spesielle invasivitet med hensyn til brukernes private sfære, europeisk og italiensk forskrifter krever at brukere være tilstrekkelig informert Om deres bruk av de samme og er og dermed nødvendig til uttrykke deres gyldig samtykke . Men i det\nspesifikk tilfellet er google analytics blitt anonymisert (IP-maskering) og deling av navigasjonsdata med google har blitt blokkert: på denne måten ligner den analytiske informasjonskapselen de tekniske informasjonskapslene som er angitt ovenfor og krever ikke samtykke.\n\n**Profileringsinformasjonskapsler**\nDenne typen informasjonskapsler brukes ikke på nettstedet.\n\nSpesifikk merknad:\nDe innebygde videoene til YouTube på nettstedet bruker ikke informasjonskapsler, da det er spesifisert \" nocookie \" den personvernforbedrede innebyggingskoden for alle YouTube-videoinnbyggingene dine.\n\n**Internasjonal og europeisk dataoverføring**\nDine data vil bli behandlet utelukkende i det europeiske økonomiske samarbeidsområdet. Dine rettigheter mht personlig data vi holde under EU Regulering 2016/679\n\n**Dine rettigheter**\nDu kan trening din rettigheter noen tid, som sett ute av Artikkel 7, par. 3, og artikler 15 og følgende av EU regulering 2016/679:\n- Ikke sant å få tilgang personlig data\n- Ikke sant til utbedring og sletting av personlig data;\n- Ikke sant til begrensning av behandling;\n- Ikke sant til data bæbarhet;\n- Ikke sant til gjenstand til behandling av personlig data\n- Ikke sant til lovlig krav til italiensk Data Beskyttelse Autoritet.\n\nDu kan utøve dine rettigheter ved å sende oss en e-post på info@pixel-online.net eller et brev adressert til Pixel, via Luigi Lanzi , 12 – 50134 – Firenze, Italia. Ytterligere informasjon om databehandling kan legges til når samle data.\n\n12 februar 2022 rev.03\nPolski: Polityka prywatności zgodna z Rozporządzeniem UE 2016/679\n\nKto zbiera Twoje dane\nZgodnie z art. 13 Rozporządzenia UE 2016/679 (RODO) administratorem danych jest Wnioskodawca projektu i to on odpowiada za zbieranie danych. Informujemy, że nasza organizacja jest prawnie zobowiązana do przetwarzania danych, które nam przekazałeś na podstawie w/w rozporządzenia.\nTwoje dane będą przetwarzane zgodnie z prawem i rzetelnie, na podstawie przepisu art. 5 Rozporządzenia UE 2016/679. Dalsze szczegóły mogą zostać podane na późniejszym etapie.\nInspektor ochrony danych (IOD): o obecność ewentualnego inspektora ochrony danych należy zwrócić się do administratora danych.\n\nJakie dane osobowe zbieramy\nw zgodność z Artykuł 4 z UE Rozporządzenie 2016/679:\n- „dane osobowe” oznaczają wszelkie informacje dotyczące zidentyfikowanej lub możliwej do zidentyfikowania osoby fizycznej (‘dane Przedmiot’); jakiś rozpoznawalny naturalny osoba jest jeden WHO mogą być zidentyfikowane, bezpośrednio lub pośrednio, w szczególności przez odniesienie do identyfikatora takiego jak imię i nazwisko, numer identyfikacyjny, Lokalizacja dane, jakiś online identyfikator lub do jeden lub jeszcze czynniki konkretny do ten fizyczny, fizjologiczny, genetyczny, psychiczny, ekonomiczne, kulturalne lub społeczny tożsamość że naturalny osoba;\n- \"przetwarzanie\" oznacza każdy operacja lub ustawić z operacje który jest wykonywane na osobisty dane lub na zestawy z osobisty dane, czy lub nie przez zautomatyzowany oznacza, taki Jak kolekcja, nagrywanie, organizacja, strukturyzacja, przechowywanie, adaptacja lub zmiana, odzyskiwanie, konsultacja, wykorzystanie poprzez transmisję, rozpowszechnianie lub udostępnianie w inny sposób, dostosowanie lub połączenie, ograniczenie, skasowanie lub zniszczenie.\n\nZ sprawdzenie do ten nad wspomniany definicje, my podkreślać że my zebrać tylko ten Informacja Ty daj nam na cele Twojego zaangażowania w nasz inicjatywy i/lub twój legalny relacja z nasz organizacja:\n- Dane osobowe: imię i nazwisko osób fizycznych, kontakty takie jak adres, ZIP kod, miasto, region, telefon numer, e-mail;\n- Dane dotyczący profesjonalistów/organizacje/przedsiębiorstwa: Informacja dotyczący firmy, imię i nazwisko, adres fiskalny i inne identyfikatory (numer faksu i telefonu, kod podatkowy) lub VAT numer).\n\nPonadto możemy zbierać dane podane podczas uzyskiwania dostępu do naszych witryn, za pośrednictwem plików cookie i innych podobna technologia; oraz gdy kontaktujesz się z nami za pośrednictwem poczty elektronicznej, mediów społecznościowych lub podobnych technologii. Chociaż takie dane nie są gromadzone w celu powiązania z osobą fizyczną, to jednak identyfikatory internetowe mogą być używane i łączone w celu tworzenia profili osobistych. Wśród internetowych identyfikatory, które możemy znaleźć adres IP, typ przeglądarki i dane wtyczki, typ urządzenia (np. komputer stacjonarny, laptop, tablet, telefon\nitp.) system operacyjny, lokalna strefa czasowa. Dane te są wykorzystywane wyłącznie do produkcja statystyczny wyniki.\n\nPrzypominamy, że nie będziemy przetwarzać danych osobowych ujawniających rasę lub pochodzenie etniczne, poglądy polityczne, przekonania religijne lub filozoficzne lub przynależność do związków zawodowych oraz przetwarzanie danych genetycznych, danych biometrycznych w celu jednoznacznej identyfikacji osobnika naturalnego osoba, dane dotyczący zdrowie lub dane dotyczący a naturalny osoby seks życie lub seksualny orientacja.\n\nDlaczego i jak przetwarzamy Twoje dane\n\nMy będzie stosowanie Twój dane w następujące sposoby:\n\n109. Organizować i realizować inicjatywy z zakresu edukacji i szkoleń (np. szkoleniakursy, konferencje, europejski projektowanie itp.)\n110. W celu produkować Administracyjny dokumenty (np faktury) w relacja do ten inicjatywy nad\n111. Do statystyczny cele\n112. Nosić na zewnątrz Komunikacja zajęcia przez e-mail dotyczący nasz inicjatywy.\n113. Odpowiadać na prośby za pomocą formularzy na stronie (jeśli jest obecny)\n114. Zezwól na rejestrację w celu uzyskania dostępu do poufnych treści edukacyjnych (jeśli są obecne)\n\nTwoje przyznanie jest obowiązkowe do celów określonych w paragrafach 1, 2, 5, 6 w celu spełnienia prawniczy obowiązki oraz UE prawa oraz przepisy prawne; odmowa do zapewniać osobisty dane będzie nie dopuszczać nasz organizacja do oferta Ty, Nasze Usługi. Twoja zgoda jest opcjonalna dla celów określonych w ust. 3 i 4; wyślemy Ci marketing komunikacja za pośrednictwem poczty elektronicznej lub poczty. W każdej chwili możesz skorzystać ze swoich praw, zgodnie z: Artykuł 15 i późniejsze Rozporządzenia UE 2016/679 dotyczące rezygnacji z otrzymywania takich Komunikacja lub wybierając inne Komunikacja modalności.\n\nBędziemy przechowywać Twoje dane osobowe zebrane dla celów określonych we wszystkich paragrafach tak długo, jak jak potrzebujemy, aby świadczyć Ci oferowane usługi przez naszą organizację i do 10 (dziesięć lat.)\n\nPodstawą prawną leczenia jest stosunek handlowy stworzony przez sprzedaż lub zakup towarów i/lub usług, przedumowny o informację (art. 6 ust. b i c), oraz za zgodą dla marketing zajęcia. (artykuł 6 ustęp a)\n\nTwoje dane będziemy przetwarzać i przechowywać wyłącznie w wyżej wymienionych celach, korzystając z urządzeń cyfrowych oraz w odpowiednich bazach danych zapewniających odpowiednie zabezpieczenia, aby zapewnić ciągłą poufność, integralność, dostępność i odporność systemów przetwarzania, zgodnie z rozporządzeniem UE 2016/679. Tylko podmioty, które uzyskały dostęp do danych osobowych od administratora lub podmiotu przetwarzającego mogą: proces taki Informacja.\n\nNie sprzedajemy, nie handlujemy ani w żaden inny sposób nie przekazujemy innym osobom trzecim Twoich danych osobowych Informacja. Możemy jednak ujawnić Twoje dane, jeśli uważamy, że jest to konieczne, aby zastosować się z ten prawo, egzekwować nasz teren\npolityki, lub chronić nasz lub inni' prawa, własność, lub bezpieczeństwo.\n\nProfilowy\nTwoje dane nie będą poddane decyzji opartej wyłącznie na zautomatyzowanym przetwarzaniu, która wywołuje skutki prawne wpływające na nie lub istotnie wpływa na jej osobę. Anulowanie i zmiana: masz prawo w dowolnym momencie wiedzieć, jakie są Twoje dane u poszczególnych administratorów danych, tj. w naszej firmie lub u ww. osób, którym je przekazujemy, i w jaki sposób są one wykorzystywane; mają również prawo do ich aktualizacji, uzupełniania, poprawiania lub anulowania, żądania ich zablokowania i sprzeciwienia się ich przetwarzaniu. W celu wykonania swoich praw, a także w celu uzyskania bardziej szczegółowych informacji na temat podmiotów lub kategorii podmiotów, którym dane są przekazywane lub które są tego świadome, jako menedżerowie lub agenci mogą skontaktować się z administratorem danych lub jednym z jego menedżerów, wskazanym w to oświadczenie.\n\nPortale społecznościowe\nNasza strona internetowa może oferować dostęp do sieci społecznościowych. Obowiązują warunki świadczenia usług i Polityka prywatności na takie platformy są publikowane na ich stronie internetowej. Pixel nie może kontrolować sposobu, w jaki dane są udostępniane na publiczny forum, czat lub deska rozdzielcza są używany, istnienie ten dane Przedmiot odpowiedzialny z taki Komunikacja.\n\nUskarżanie się\nty mogą Również kontakt ten Włoski Dane Ochrona Autorytet za pomocą ten Następny połączyć http://www.garanteprivacy.it/home/footer/contatti lub Europejskiego Inspektora Ochrony Danych za pomocą Następny link: https://edps.europa.eu/data-protection/our-role-supervisor/complaints_en\n\nCiasteczka\nTak jak se t jesteś _ przez regulamin „Wytyczne dotyczące plików cookie i innych narzędzi śledzących – 10 czerwca 2021 r. ” t tutaj są trzy główne kategorie z ciasteczka:\n\nTechniczny ciasteczka\nTe są używany dla ten podeszwa cel, powód z „nadawanie komunikacja do jakiś elektroniczny sieci komunikacyjnej lub w zakresie ściśle niezbędnym do świadczenia usługi przez firma informacyjna na wyraźne żądanie kontrahenta lub użytkownika w celu dostarczenia; wspomniana usługa” Nie są one wykorzystywane do żadnych ukrytych celów i są zwykle instalowane bezpośrednio przez ten właściciel lub ten menedżer z ten stronie internetowej (tak zwane \"prawnie zastrzeżony\" lub \"redakcyjny\" ciasteczka). Te mogą być podzielony do: przeglądanie lub sesja ciasteczka, który gwarancja normalna nawigacja i korzystanie z serwisu (umożliwiające np. dokonywanie zakupów lub uwierzytelnione w celu uzyskania dostępu do zastrzeżonych obszarów); analityczne pliki cookie asimilowane przez techniczne pliki cookies, w których są wykorzystywane bezpośrednio przez administratora serwisu do zbierania informacji, w tym: formularz skojarzony (anonimowy), o liczbie użytkowników i sposobie, w jaki odwiedzają stronę internetowej; funkcjonalne pliki cookie, które umożliwiają użytkownikowi nawigację w odniesieniu do szeregu wybranych kryteriów (na przykład język lub produkty wybrane do zakupu) w celu ulepszenia usługi pod warunkiem, że, pod warunkiem że informujemy nasz użytkownicy Jak wyruszać przez artykuł 13 UE Rozporządzenie 2016/679.\nten wcześniejszy zgoda z ten użytkownik jest nie wymagany w zamówienie do zainstalować te ciasteczka.\n\n**Analityczne pliki cookie**\nStrona korzysta wyłącznie z Google Analytics, która służy do tworzenia profili użytkowników i jest wykorzystywane do wysyłania komunikatów reklamowych zgodnie z preferencjami wskazanymi przez niego podczas nawigacji online. Ze względu na ich szczególną inwazyjność w odniesieniu do prywatności użytkowników kula, europejski oraz Włoski przepisy prawne wymagać że użytkownicy być odpowiednio powiadomiony o ich stosowanie z ten to samo oraz są zatem wymagany do wyrazić ich ważny zgoda. Jednak w konkretnym przypadku Google Analytics został zanonimizowany (maskowanie adresu IP), a udostępnianie danych nawigacyjnych Google zostało zablokowane: w ten sposób analityczny plik cookie jest podobny do technicznych plików cookie wskazanych powyżej i nie wymaga zgody.\n\n**Profilujące pliki cookie**\nTen rodzaj plików cookie nie jest używany w witrynie.\n\nUwaga szczegółowa:\nfilmy YouTube w witrynie nie używają plików cookie, ponieważ określono \" nocookie \" kod osadzania o zwiększonej prywatności dla wszystkich osadzonych filmów YouTube.\n\n**Międzynarodowy i europejski transfer danych**\nTwoje dane będą przetwarzane wyłącznie na terenie Europejskiego Obszaru Gospodarczego. Twoje prawa w odniesieniu doosobisty dane my utrzymać pod UE Rozporządzenie 2016/679\n\n**Twoje prawa**\ny mogą ćwiczenie Twój prawa każdy czas, Jak ustawić na zewnątrz przez Artykuł 7, par. 3, oraz artykuły 15 oraz Następny UE rozporządzenie 2016/679:\n\n- Prawidłowy mieć dostęp osobisty dane\n- Prawidłowy do sprostowanie oraz skasowanie z osobisty dane;\n- Prawidłowy do ograniczenie z przetwarzanie;\n- Prawidłowy do dane ruchliwość;\n- Prawidłowy do obiekt do przetwarzanie z osobisty dane\n- Prawidłowy do prawnego prawo do Włoski Dane Ochrona Autorytet.\n\nMożesz skorzystać ze swoich praw, wysyłając do nas wiadomość e-mail na adres info@pixel-online.net lub list zaadresowany do Pixel, via Luigi Lanzi , 12 – 50134 – Firenze, Włochy. Dalsze informacje dotyczące przetwarzania danych można dodać, gdy zbieranie danych.\n\n12 lutego 2022 rew.03\nPortuguês: Política de Privacidade de acordo com o Regulamento da UE 2016/679\n\nQuem coleta seus dados\nNos termos do artigo 13.º do Regulamento da UE 2016/679 (GDPR), o responsável pelo tratamento é o Requerente do projeto, sendo este o responsável pela recolha dos dados. Gostaríamos de informar que nossa organização está legalmente obrigada a processar os dados que você nos forneceu de acordo com o regulamento mencionado. Os seus dados serão processados de forma lícita e justa, nos termos do artigo 5.º do Regulamento da UE 2016/679. Mais detalhes podem ser fornecidos em um estágio posterior. Encarregado de proteção de dados (dpo): a presença de um possível DPO deve ser solicitada ao controlador de dados.\n\nQuais dados pessoais coletamos\nDentro conformidade com Artigo 4 do eu Regulamento 2016/679:\n- “dados pessoais” significa qualquer informação relativa a uma pessoa singular identificada ou identificável (‘dados sujeito’); a identificável natural pessoa é 1 Who posso estar identificado, diretamente ou indiretamente, em particular por referência a um identificador, como um nome, um número de identificação, localização dados, a on-line identificador ou para 1 ou mais fatores específico para a físico, fisiológico, genético, mental, econômico, cultural ou social identidade de aquele natural pessoa;\n- \"em processamento\" meios algum Operação ou definir do operações que é realizado em pessoal dados ou em conjuntos do pessoal dados, se ou não de automatizado meios, tal Como coleção, registro, organização, estruturação, armazenamento, adaptação ou alteração, recuperação, consulta, uso, divulgação por transmissão, disseminação ou disponibilização, alinhamento ou combinação, restrição, apagamento ou destruição.\n\nCom referência para a acima de mencionado definições, nós sublinhado aquele nós coletar só a em formação tu nos fornecer para propósitos do seu envolvimento nosso iniciativas e/ou seu legal relação com nosso organização:\n- Informações pessoais: nome e sobrenome de pessoas físicas, contatos como endereço, CEP código, cidade, região, Telefone número, o email;\n- Dados relativo profissionais/organizações/empresas: em formação relativo empresas, nome, endereço fiscal e outros identificadores (fax e número de telefone, código fiscal ou IVA número).\n\nAlém disso, podemos coletar dados fornecidos quando você acessa nossos sites, por meio de cookies e outros tecnologia semelhante; e quando você entrar em contato conosco por e-mail, mídia social ou tecnologias semelhantes. Ainda que tais dados não sejam recolhidos de forma a serem associados à pessoa singular, estes identificadores online podem ser usados e combinados para criar perfis pessoais. Entre os on-line identificadores podemos encontrar endereço IP, tipo de navegador e detalhes do plug-in, tipo de dispositivo (por exemplo, desktop, laptop, tablet, telefone, etc.) sistema operacional, fuso horário local. Esses dados são usados exclusivamente para a produção de estatística resultados.\nGostaríamos de lembrá-lo de que não processaremos dados pessoais que revelem informações raciais ou origem étnica, opiniões políticas, crenças religiosas ou filosóficas ou filiação sindical, e o tratamento de dados genéticos, dados biométricos com o objetivo de identificar de forma inequívoca um pessoa, dados relativo saúde ou dados relativo uma natural pessoas sexo vida ou sexual orientação.\n\n**Por que e como processamos seus dados**\n\nNós vontade usar seu dados dentro a seguir maneiras:\n\n115. Organizar e implementar iniciativas no domínio da educação e formação (por exemplo, formação cursos, congressos, europeu projetos etc.)\n116. Para produzir administrativo documentos (por exemplo: faturas) dentro relação para a iniciativas acima de\n117. Para estatística propósitos\n118. Carregar Fora comunicação Atividades através da o email relativo nosso iniciativas.\n119. Responder às solicitações usando os formulários do site (se houver)\n120. Permitir registro para acesso a conteúdo educacional confidencial (se houver)\n\nSua concessão é obrigatória para os fins dos parágrafos 1, 2, 5, 6, a fim de cumprir jurídico obrigações e eu leis e regulamentos; recusa para providenciar pessoal dados vontade não permitirnosso organização para oferta tu, Nossos serviços.\n\nO seu consentimento é opcional para os fins previstos nos parágrafos 3 e 4; nós lhe enviaremos marketing comunicação via e-mail ou serviço postal. Você pode exercer seus direitos a qualquer momento, de acordo com Artigo 15 e posterior do Regulamento da UE 2016/679 sobre a opção de não receber tais comunicação ou escolhendo outro comunicação modalidades.\n\nManteremos seus dados pessoais coletados para os fins de todos os parágrafos, desde que como precisamos para lhe fornecer os serviços oferecidos pela nossa organização e até 10 (dez anos).\n\nVocê posso retirar o seu consentimento no algum Tempo.\n\nA base legal do tratamento consiste na relação comercial criada pela venda ou aquisição de bens e/ou serviços, pré-contratual para informação (artigo 6º alíneas b e c),e por consentimento para marketing Atividades. ( artigo 6 parágrafo a)\n\nProcessaremos e armazenaremos seus dados exclusivamente para os fins acima mencionados, usando dispositivos digitais e em bancos de dados relevantes, garantindo as salvaguardas apropriadas para garantir a confidencialidade contínua, integridade, disponibilidade e resiliência dos sistemas de processamento, conforme estabelecido pelo regulamento da UE 2016/679. Somente os sujeitos que obtiveram acesso aos dados pessoais do controlador ou do processador podem processar tal em formação.\n\nNão vendemos, trocamos ou transferimos para terceiros seus dados pessoais identificáveis em formação. No entanto, podemos divulgar suas informações quando acreditarmos que a divulgação é necessária para cumprir com a lei, impor nosso local políticas, ou proteger nosso ou outras' direitos, propriedade, ou segurança.\n\n**Perfil**\n\nOs seus dados não serão sujeitos a uma decisão baseada exclusivamente no tratamento\nautomatizado, que produza efeitos jurídicos que os afetem ou que afetem significativamente a sua pessoa. Cancelamento e Alteração: você tem o direito de saber, a qualquer momento, quais são seus dados nos controladores de dados individuais, ou seja, em nossa empresa ou nas pessoas acima mencionadas a quem os comunicamos e como são usados; também têm o direito de atualizá-los, complementá-los, corrigi-los ou cancelá-los, solicitar seu bloqueio e opor-se ao seu tratamento. Para o exercício dos seus direitos, bem como para informação mais detalhada sobre os sujeitos ou categorias de sujeitos a quem os dados são comunicados ou que deles tenham conhecimento como gestores ou agentes podem contactar o responsável pelo tratamento ou um dos seus gestores, identificado no esta afirmação.\n\nRedes sociais\nNosso site pode oferecer acesso à rede social. Os termos de serviço e a Política de Privacidade aplicáveis para essas plataformas são publicados em seu site. Pixel não pode controlar a maneira como os dados são compartilhados em um público fórum, bater papo ou painel de controle está usado, ser a dados sujeito responsável do tal comunicação.\n\nReclamações\nVocê posso Além disso contato a italiano Dados Proteção Autoridade usando a Segue link http://www.garanteprivacy.it/home/footer/contatti, ou o Supervisor Europeu de Proteção de Dados usando o Segue link: https://edps.europa.eu/data-protection/our-role-supervisor/complaints_en\n\nBiscoitos\nComo definir _ fora _ de regulamento “ Diretrizes de cookies e outras ferramentas de rastreamento - 10 de junho de 2021 ”, t aqui são três categorias principais do biscoitos:\n\nTécnico biscoitos\nEsses está usado para a único objetivo do “transmitindo comunicações para a eletrônico rede de comunicação, ou na medida estritamente necessária para a prestação de um serviço pelo empresa de informações explicitamente solicitada pelo contratante ou pelo usuário para fornecer o referido serviço” Estes não são utilizados para fins ulteriores e são normalmente instalados diretamente de a proprietário ou a Gerente do a local na rede Internet (chamado “proprietário” ou \"editorial\" biscoitos). Esses posso estar dividido para dentro: navegando ou sessão biscoitos, que garantia normal navegação e uso do site (possibilitando, por exemplo, fazer compras ou ser autenticado para acessar as áreas reservadas); cookies analíticos assimilados pelo técnico cookies onde eles são usados diretamente pelo gerente do site para coletar informações, de forma formulário associado (anônimo), sobre o número de usuários e a forma como visitam o local na rede Internet; cookies funcionais que permitem ao usuário navegar em relação a uma série de critérios selecionados (por exemplo, o idioma ou os produtos selecionados para compra) para melhorar o serviço forneceu, providenciou que informamos nossos Comercial Como estabelecer de artigo 13 eu Regulamento 2016/679. O anterior consentimento do a do utilizador é não Requeridos dentro pedido para instalar esses biscoitos.\n\nCookies de análise\nO site usa apenas o google analytics, que é usado para criar perfis dos usuários e são empregado para o envio de mensagens publicitárias de acordo com as preferências apresentadas pelo mesmo durante sua navegação online. Devido à sua particular invasividade em relação à privacidade dos usuários, esfera, europeu e italiano regulamentos exigir aquele\nComercial estar adequadamente informado cerca de deles usar do a mesmo e está portanto requeridos para expressar deles válido consentimento. Mas, no caso específico, o google analytics foi anonimizado (mascaramento de IP) e o compartilhamento de dados de navegação com o google foi bloqueado: desta forma, o cookie analítico é semelhante aos cookies técnicos indicados acima e não requer consentimento.\n\n**Cookies de perfil**\nEste tipo de cookie não é usado no site.\n\nNota específica:\nOs vídeos incorporados do YouTube no site não usam cookies, pois foi especificado \" nocookie \" o código de incorporação de privacidade aprimorada para todas as incorporações de vídeos do YouTube.\n\n**Transferência de dados internacionais e europeias**\nSeus dados serão processados exclusivamente no Espaço Econômico Europeu. Seus direitos em relação ao pessoal dados nós espera sob eu Regulamento 2016/679\n\n**Seus direitos**\nVocês posso exercício seu direitos algum Tempo, Como definir Fora de Artigo 7, par. 3, e artigos 15 e Seguedo eu regulamento 2016/679:\n\n- Certo acessar pessoal dados\n- Certo para retificação e apagamento do pessoal dados;\n- Certo para restrição do em processamento;\n- Certo para dados portabilidade;\n- Certo para objeto para em processamento do pessoal dados\n- Certo para jurídico alegar para italiano Dados Proteção Autoridade.\n\nVocê pode exercer seus direitos enviando-nos um e-mail para info@pixel-online.net ou uma carta endereçada para Pixel, via Luigi Lanzi , 12 – 50134 – Firenze, Itália. Mais informações sobre o processamento de dados pode ser adicionado quando coletando dados.\n\n12 de fevereiro 2022 rev.03\nRomana: Politica de confidentialitate în conformitate cu Regulamentul UE 2016/679\n\nCine vă colectează datele\nÎn conformitate cu articolul 13 din Regulamentul UE 2016/679 (GDPR), operatorul de date este Solicitantul proiectului și este responsabil de colectarea datelor. Dorim să vă informăm că organizația noastră este obligată legal să prelucreze datele pe care ni le-ați furnizat în temeiul regulamentului menționat mai sus.\nDatele dumneavoastră vor fi prelucrate în mod legal și corect, în conformitate cu prevederile articolului 5 din Regulamentul UE 2016/679. Mai multe detalii ar putea fi furnizate într-o etapă ulterioară.\nResponsabil cu protecția datelor ( dpo ): prezența unui posibil DPO trebuie solicitată de la operatorul de date.\n\nCe date personale colectăm\nÎn conformitate cu Articol 4 de UE Regulament 2016/679:\n- „date cu caracter personal” înseamnă orice informație referitoare la o persoană fizică identificată sau identificabilă ('date subiect'); un identificabile natural persoană este unu care poate sa fi identificat, direct sau indirect, în special prin referire la un identificator, cum ar fi un nume, un număr de identificare, Locație date, un pe net identificator sau la unu sau Mai mult factori specific la cel fizic, fiziologic, genetic, mental, economic, cultural sau social identitate a acea natural persoană;\n- \"prelucrare\" mijloace orice Operațiune sau a stabilit de operațiuni care este efectuat pe personal date sau pe seturi de personal date, dacă sau nu de automatizate mijloace, astfel de la fel de Colectie, înregistrare, organizare, structurare, stocare, adaptare sau modificare, regăsire, consultare, utilizarea, dezvăluirea prin transmitere, diseminare sau punerea la dispoziție în alt mod, aliniere sau combinație, restricție, ștergere sau distrugere.\n\nCu referință la cel de mai sus menționat definiții, noi subliniază acea noi colectarea numai cel informație tu oferiți-ne pentru scopuri a implicării dumneavoastră în al nostru inițiative și/sau legala ta relație cu al nostru organizare:\n- Informații personale: numele și prenumele persoanelor fizice, contacte precum adresa, ZIP cod, oraș, regiune, telefon număr, e-mail;\n- Date cu privire la profesionist/organizatii/afaceri: informație cu privire la afaceri, nume, adresa fiscală și alți identificatori (număr de fax și telefon, cod fiscal sau TVA număr).\n\nMai mult, este posibil să colectăm date furnizate atunci când accesați site-urile noastre, prin cookie-uri și altele tehnologie similară; și când ne contactați prin e-mail, rețele sociale sau tehnologii similare. Chiar dacă astfel de date nu sunt colectate astfel încât să fie asociate cu persoana fizică, acestea identificatorii online pot fi utilizați și combinați pentru a crea profiluri personale. Printre online identificatorii putem găsi adresa IP, tipul browserului și detaliile plug-in-ului, tipul dispozitivului (de exemplu desktop, laptop, tableță, telefon etc.) sistem de operare, fus orar local. Aceste date sunt folosite exclusiv pentru producție de statistic\nrezultate.\n\nDorim să vă reamintim că nu vom prelucra date cu caracter personal care dezvăluie rasa sau origine etnică, opinii politice, convingeri religioase sau filozofice sau apartenență la sindicate și prelucrarea datelor genetice, a datelor biometrice în scopul identificării în mod unic a unui natural persoană, date cu privire la sănătate sau date cu privire la A natural ale persoanei sex viață sau sexual orientare.\n\nDe ce și cum prelucrăm datele dvs\nNoi voi utilizare ta date în următoarele moduri:\n\n121. Organizarea și implementarea inițiativelor în domeniul educației și formării (ex. formare cursuri, conferințe, european proiecte etc.)\n122. La legume și fructe administrativ documente (de exemplu facturi) în relație la cel inițiative de mai sus\n123. Pentru statistic scopuri\n124. Transporta afară comunicare Activități prin intermediul e-mail cu privire la al nostru inițiative.\n125. Raspunde la solicitari folosind formularele de pe site (daca este prezent)\n126. Permite înregistrarea pentru acces la conținut educațional confidențial (dacă este prezent)\n\nConferirea dumneavoastră este obligatorie în scopurile prevăzute la alin. 1, 2, 5, 6 pentru a respecta juridic obligații și UE legi și reguli; refuz la oferi personal date voi nu permite al nostru organizație să oferi tu, Serviciile noastre.\n\nConsimțământul dumneavoastră este opțional în scopurile prevăzute la paragrafele 3 și 4; vă vom trimite marketing comunicare prin e-mail sau serviciu poștal. Vă puteți exercita drepturile în orice moment, în conformitate cu Articolul 15 și ulterior din Regulamentul UE 2016/679 privind renunțarea de la primirea acestor comunicare sau alegand altele comunicare modalități.\n\nVom păstra datele dumneavoastră personale colectate în scopurile prevăzute în toate paragrafele atât timp cât așa cum avem nevoie pentru a vă oferi serviciile oferite de către organizația noastră și până la 10 (zece ani).\n\nTu poate sa retrage ta consimțământ la orice timp.\n\nTemeiul juridic al tratamentului constă în relația comercială creată prin vânzare sau achiziționarea de bunuri și/sau servicii, precontractuală cu titlu informativ (art. 6 alin. b și c), și prin consimțământ pentru marketing Activități. ( articol 6 paragraf A)\n\nVom prelucra și stoca datele dumneavoastră exclusiv în scopurile menționate mai sus, folosind dispozitive digitale și în bazele de date relevante, asigurând garanții adecvate pentru a asigura confidențialitatea continuă; integritatea, disponibilitatea și rezistența sistemelor de prelucrare, astfel cum sunt prevăzute de Regulamentul UE 2016/679. Numai subiecții care au obținut acces la datele cu caracter personal de la operator sau de la operator poate proces astfel de informație.\n\nNu văd, tranzacționăm sau transferăm în alt mod altor terți elementele dvs. de identificare personală informație. Cu toate acestea, este posibil să eliberăm informațiile dvs. atunci când considerăm că este necesară se conformează cu cel lege, impune al nostru site-ul politici, sau\nproteja a noastra sau alții' drepturi, proprietate, sau Siguranță.\n\nProfilare\nDatele dumneavoastră nu vor fi supuse unei decizii bazate exclusiv pe prelucrare automată, care produce efecte juridice care le afectează sau care îi afectează semnificativ persoana. Anulare și Modificare: aveți dreptul de a afla, în orice moment, care sunt datele dumneavoastră la operatorii individuali, adică la compania noastră sau la persoanele menționate mai sus cărora le comunicăm, și cum sunt utilizate; au, de asemenea, dreptul de a le actualiza, completa, corecta sau anula, solicita blocarea acestora si se opune tratarii acestora. Pentru exercitarea drepturilor dumneavoastră, precum și pentru informații mai detaliate despre subiecții sau categoriile de subiecți cărora le sunt comunicate datele sau unul dintre responsabilii acestuia, identificat în aceasta afirmatie.\n\nRetele sociale\nSite-ul nostru web poate oferi acces la rețeaua socială. Termenii și condițiile și politica de confidențialitate aplicabile pentru astfel de platforme sunt publicate pe site-ul lor. Pixel nu poate controla modul în care datele partajate pe a public forum, conversație sau bord sunt folosit, fiind cel date subiect responsabil de astfel de comunicare.\n\nReclamații\nTu poate sa de asemenea a lua legatura cu Autoritatea Europeană pentru Protecția Datelor folosind ca urmare a link: https://edps.europa.eu/data-protection/our-role-supervisor/complaints_en\n\nCookie-uri\nLa fel de se t ou t de regulament „Instrucțiuni privind cookie - urile și alte instrumente de urmărire - 10 iunie 2021 ”, aici _ sunt trei categorii principale de cookie-uri:\n\nTehnic cookie-uri\nAceste sunteti folosit pentru cel unic scop de „transmite comunicatii la un electronic rețeaua de comunicații, sau în măsura strict necesară pentru furnizarea unui serviciu de către companie de informații solicitate în mod explicit de către partea contractantă sau utilizator în vederea furnizării serviciului menționat” Acestea nu sunt utilizate în scopuri ulterioare și sunt instalate în mod normal direct de cel proprietar sau cel administrator de cel site-ul web (așa-zisul \"proprietate\" sau \"editorial\" cookie-uri). Aceste poate sa fi împărțit în: navigare sau sesiune cookie-uri, care garanție normal navigarea și utilizarea site-ului web (făcând posibil, de exemplu, efectuarea de achiziții sau a fi autentificat pentru a accesa zonele rezervate); cookie-uri analitice asimilate de cele tehnice cookie-uri unde sunt folosite direct de către administratorul site-ului web pentru a colecta informații, într-un formularul asociat (anonom), despre numărul de utilizatori și modul în care aceștia vizitează site-ul web; cookie-uri funcționale care permit utilizatorului să navigheze în raport cu o serie de criterii selectate (de exemplu, limba sau produsele selectate pentru achiziție) pentru a îmbunătăți serviciul cu condiția, cu conditia ca informăm noastre utilizatorii la fel de a pornit de articol 13 UE Regulament 2016/679.\n\nThe anterior consimțământ de cel utilizator este nu solicitat în Ordin la instalare aceste cookie-uri.\n**Cookie-uri de analiză**\nSite-ul folosește doar google analytics, care este folosit pentru a crea profiluri ale utilizatorilor și sunt angajat pentru transmiterea de mesaje publicitare conform preferintelor manifestate de acesta în timpul navigării lor online. Datorită invazivității lor deosebite în ceea ce privește privatitatea utilizatorilor sferă, european și italiană reguli cere acea utilizatorii fi adecvat informat despre al lor utilizare de cel la fel și sunteți prin urmare necesar la expres al lor valabil consimțământ. Dar în cazul specific google analytics a fost anonimizat (mascare IP) și partajarea datelor de navigare cu google a fost blocată: în acest fel cookie-ul analitic este similar cu cookie-urile tehnice indicate mai sus și nu necesită consimțământ.\n\n**Cookie-uri de profilare**\nAcest tip de cookie nu este folosit pe site.\n\nNotă specifică:\nVideoclipurile încorporate de pe YouTube pe site nu folosesc cookie-uri, deoarece a fost specificat „ nocookie ” codul de încorporare îmbunătățit pentru confidențialitate pentru toate încorporarile video YouTube.\n\n**Transfer de date internațional și european**\nDatele dumneavoastră vor fi prelucrate exclusiv în Spațiul Economic European. Drepturile dumneavoastră cu privire la personal date noi tine sub UE Regulament 2016/679\n\n**Drepturile tale**\nTu poate sa exerciți ta drepturi orice timp, la fel de a stabili afară de Articol 7, alin. 3, și articole 15 și ca urmare ade UE regulament 2016/679:\n\n- Dreapta a accesa personal date\n- Dreapta la rectificare și ștergere de personal date;\n- Dreapta la restricție de prelucrare;\n- Dreapta la date portabilitate;\n- Dreapta la obiect la prelucrare de personal date\n- Dreapta la legale Revendicare la Italiană Date Protecție Autoritate.\n\nVă puteți exercita drepturile trimițându-ne un e-mail la info@pixel-online.net sau o scrisoare adresată către Pixel, via Luigi Lanzi , 12 – 50134 – Firenze, Italia. Informații suplimentare privind prelucrarea datelor poate fi adăugat când colectarea datelor.\n\n12 februarie 2022 rev.03\nSvenska: Integritetspolicy i enlighet med EU-förordning 2016/679\n\nVem samlar in dina uppgifter\nEnligt artikel 13 i EU-förordning 2016/679 (GDPR) är den personuppgiftsansvarige den sökande till projektet och den ansvarar för att samla in uppgifterna. Vi vill informera dig om att vår organisation är juridiskt bunden att behandla de uppgifter du har tillhandahållit oss enligt ovanämnda förordning.\n\nDina uppgifter kommer att behandlas lagligt och rättvist, enligt bestämmelserna i artikel 5 i EU-förordning 2016/679. Ytterligare information kan komma att ges i ett senare skede.\n\n(dpo): närvaron av en eventuell dataskyddsombud måste begäras från den personuppgiftsansvarige.\n\nVilka personuppgifter vi samlar in\nI överensstämmelse med Artikel 4 av EU förordning 2016/679:\n- \"personuppgifter\" betyder all information som hänför sig till en identifierad eller identifierbar fysisk person (‘data ämne’); en identifierbar naturlig person är ett vem burk vara identifierade, direkt eller indirekt, särskilt genom hänvisning till en identifierare såsom ett namn, ett identifikationsnummer, plats data, en uppkopplad identifierare eller till ett eller Mer faktorer specifika till de fysisk, fysiologisk, genetisk, mental, ekonomisk, kulturell eller social identitet av den där naturlig person;\n- \"bearbetning\" innebär att några drift eller uppsättning av operationer som är genomförde på personlig data eller på set av personlig data, huruvida eller inte förbi automatiserad innebär att, sådan som samling, registrering, organisation, strukturering, lagring, anpassning eller ändring, hämtning, konsultation, användning, avslöjande genom överföring, spridning eller på annat sätt tillgängliggörande, anpassning eller kombination, restriktion, radering eller förstörelse.\n\nMed referens till de ovan nämns definitioner, vi Understrykning den där vi samla endast de information du förse oss med för syften av ditt engagemang i vår initiativ och/eller din lagliga relation med vår organisation:\n- Personuppgifter: namn och efternamn på fysiska personer, kontakter som adress, postnummer kod, stad, område, telefon siffra, e-post;\n- Data rörande proffs/organisationer/företag: information rörande företag, namn, skatteadress och andra identifierare (fax och telefonnummer, skattekod eller moms siffra).\n\nDessutom kan vi samla in data som tillhandahålls när du besöker våra webbplatser, genom cookies och annat liknande teknik; och när du kontaktar oss via e-post, sociala medier eller liknande tekniker. Även om sådana uppgifter inte samlas in för att kunna kopplas till den fysiska personen, är dessa onlineidentifierare kan användas och kombineras för att skapa personliga profiler. Bland online identifierare som vi kan hitta IP-adress, webbläsartyp och plugin-detaljer, enhetstyp (t.ex. skrivbord, bärbar dator, surfplatta, telefon etc.) operativsystem, lokal tidszon. Dessa uppgifter används endast för produktion av statistisk resultat.\nVi vill påminna dig om att vi inte kommer att behandla personuppgifter som avslöjar ras eller etniskt ursprung, politiska åsikter, religiös eller filosofisk övertygelse eller medlemskap i fackförening, och bearbetning av genetiska data, biometriska data i syfte att unikt identifiera en naturlig person, data rörande hälsa eller data rörande en naturlig personens sex liv eller sexuell orientering.\n\n**Varför och hur vi behandlar dina uppgifter**\n\nVi kommer använda sig av din data i det följande sätt:\n\n127. Organisera och genomföra initiativ inom utbildningsområdet (t.ex. utbildningskurser, konferenser, Europeiska projekt etc.)\n128. Till producera administrativ dokument (t.ex fakturor) i relation till de initiativ ovan\n129. För statistisk syften\n130. Bära ut kommunikation aktiviteter via e-post rörande vår initiativ.\n131. Svara på förfrågningar med hjälp av formulären på webbplatsen (om det finns)\n132. Tillåt registrering för åtkomst till konfidentiellt utbildningsinnehåll (om det finns)\n\nDitt erkännande är obligatoriskt för ändamål enligt punkterna 1, 2, 5, 6 för att uppfylla juridisk skyldighet och EU lagar och föreskrifter; vägran till förse personlig data kommer inte tillåtavår organisation till erbjudande du, Våra tjänster.\n\nDitt samtycke är valfritt för ändamål enligt punkt 3 och 4; vi skickar marknadsföring till dig kommunikation via e-post eller post. Du kan utöva dina rättigheter när som helst, enligt Artikel 15 och senare i EU-förordning 2016/679 om att välja bort att ta emot sådana kommunikation eller att välja andra kommunikation modaliteter.\n\nVi kommer att behålla dina personuppgifter som samlats in för ändamålen under alla paragrafer så länge som vi behöver för att förse dig med de tjänster som erbjuds av vår organisation och för upp till 10 (tio år).\n\nDu burk dra tillbaka din samtycke på några tid.\n\nDen rättsliga grunden för behandlingen utgörs av det kommersiella förhållandet som skapas genom försäljningen eller köp av varor och/eller tjänster, pre-kontraktuella för information (artikel 6 punkt b och c),och med samtycke för marknadsföring aktiviteter. ( artikel 6 paragraf a)\n\nVi kommer att behandla och lagra dina uppgifter enbart för ovan nämnda ändamål, med hjälp av digitala enheter och i relevanta databaser för att säkerställa lämpliga skyddsåtgärder för att säkerställa fortlöpande konfidentialitet, bearbetningssystems integritet, tillgänglighet och motståndskraft, enligt EU-förordning 2016/679. Endast försökspersoner som har fått tillgång till personuppgifter från den personuppgiftsansvarige eller personuppgiftsbiträdet kan bearbeta sådan information.\n\nVi säljer, handlar inte med eller på annat sätt överför dina personligt identifierbara till andra tredje parter information. Däremot kan vi lämna ut din information när vi anser att det är nödvändigt att lämna ut din information följa med de lag, förstärka vår webbplats politik, eller skydda vår eller andras rättigheter, fast egendom, eller säkerhet.\n\n**Profilering**\n\nDina uppgifter kommer inte att bli föremål för ett beslut baserat enbart på automatiserad\nbehandling, vilket ger rättsverkningar som påverkar dem eller som väsentligt påverkar dess person. Uppsägning och ändring: du har rätt att när som helst veta vilka uppgifter du har hos de enskilda personuppgiftsansvariga, det vill säga på vårt företag eller hos ovan nämnda personer som vi kommunicerar dem till, och hur de används; de har också rätt att uppdatera, komplettera, korrigera eller avbryta dem, begära deras blockering och motsätta sig deras behandling. För att utöva dina rättigheter, samt för mer detaljerad information om de subjekt eller kategorier av subjekt till vilka uppgifterna kommuniceras eller som är medvetna om det som chefer eller agenter kan kontakta den personuppgiftsansvarige eller en av dennes chefer, identifierade i Detta påstående.\n\nSociala nätverk\nVår webbplats kan erbjuda tillgång till sociala nätverk. Användarvillkoren och sekretesspolicyn gäller till sådana plattformar publiceras på deras webbplats. Pixel kan inte styra hur data delas på en offentlig forum, chatt eller instrumentbräda är Begagnade, varelse de data ämne ansvarig av sådan kommunikation.\n\nKlagomål\nDu burk också Kontakt de italienska Data Skydd Auktoritet använder sig av de följande länk http://www.garanteprivacy.it/home/footer/contatti, eller den europeiska datatillsynsmannen med hjälp av följande länk: https://edps.europa.eu/data-protection/our-role-supervisor/complaints_en\n\nSmåkakor\nSom se t du t förbi re g ula t i o n \"Riktlinjer för cookies och andra spårningsverktyg - 10 juni 2021\", t här a re tre huvudkategorier av småkakor:\n\nTeknisk småkakor\nDessa är Begagnade för de enda ändamål av \"överföring kommunikation till en elektronisk kommunikationsnät, eller i den utsträckning som är strikt nödvändig för tillhandahållande av en tjänst av informationsföretag som uttryckligen begärts av avtalsparten eller användaren för att tillhandahålla nämnda tjänst\" Dessa används inte för några andra syften och de är normalt installerade direkt förbi de ägare eller de chef av de hemsida (så kallade \"Proprietär\" eller \"redaktionell\" småkakor). Dessa burk vara dividerat in i: bläddring eller session småkakor, som garanti vanligt navigering och användning av webbplatsen (gör det möjligt att till exempel göra köp eller vara autentiserad för att komma åt reserverade områden); analyookies assimilerade av den tekniska cookies där de används direkt av chefen för webbplatsen för att samla in information, i en tillhörande form (anonym), om antalet användare och det sätt på vilket de besöker hemsida; funktionella cookies som låter användaren navigera i förhållande till en rad utvalda kriterier (till exempel språket eller de produkter som valts för köp) för att förbättra tjänsten försett, förutsatt att vi informerar vår användare som ställa ut förbi artikel 13 EU förordning 2016/679.\nDe tidigare samtycke av de användare är inte begärda i beställa till Installera dessa småkakor.\n\nAnalytics-cookies\nWebbplatsen använder endast google analytics, som används för att skapa profiler för användarna och är används för att skicka reklameddelanden enligt de preferenser som visas av densamma under deras onlinenavigering. På grund av deras speciella invasivitet med hänsyn till användarnas privatsfär, Europeiska och italienska föreskrifter behöva den där användare vara tillräckligt informerad handla om deras använda sig av de samma och är\nSåledes nödvändig till uttrycka deras giltig samtycke. Men i det specifika fallet har google analytics anonymiserats (IP-maskering) och delning av navigeringsdata med google har blockerats: på detta sätt liknar den analytiska cookien de tekniska cookies som anges ovan och kräver inget samtycke.\n\n**Profileringscookies**\nDenna typ av cookie används inte på webbplatsen.\n\nSpecifik anmärkning:\nDe inbäddade videorna från YouTube på webbplatsen använder inte cookies eftersom det har specificerats \" nocookie \", den integritetsförstärkta inbäddningskoden för alla dina YouTube-videoinbäddningar.\n\n**Internationell och europeisk dataöverföring**\nDina uppgifter kommer att behandlas enbart inom Europeiska ekonomiska samarbetsområdet. Dina rättigheter med avseende på personlig data vi håll under EU förordning 2016/679\n\n**Dina rättigheter**\nDu burk övning din rättigheter några tid, som uppsättning ut förbi Artikel 7, par. 3, och artiklar 15 och följande av EU reglering 2016/679:\n\n- Höger få tillgång till personlig data\n- Höger till rättelse och radering av personlig data;\n- Höger till restriktion av bearbetning;\n- Höger till data bärbarhet;\n- Höger till objekt till bearbetning av personlig data\n- Höger till Rättslig krav till italienska Data Skydd Auktoritet.\n\nDu kan utöva dina rättigheter genom att skicka oss ett e-postmeddelande på info@pixel-online.net eller ett adresserat brev till Pixel, via Luigi Lanzi , 12 – 50134 – Firenze, Italien. Ytterligare information om databehandling kan läggas till när samla in data.\n\n12 februari 2022 rev.03\nSlovenčina: Zásady ochrany osobných údajov v súlade s Nariadením EÚ 2016/679\n\nKto zhromažďuje vaše údaje\nPodľa článku 13 nariadenia EÚ 2016/679 (GDPR) je prevádzkovateľom údajov Žiadateľ projektu a je zodpovedný za zhromažďovanie údajov. Radi by sme Vás informovali, že naša organizácia je zo zákona zaviazaná spracúvať údaje, ktoré ste nám poskytli podľa vyššie uvedeného nariadenia.\nVaše údaje budú spracované zákonne a spravodlivo v súlade s ustanovením článku 5 nariadenia EÚ 2016/679. Ďalšie podrobnosti môžu byť poskytnuté neskôr.\nÚradník pre ochranu údajov (DPO): o prítomnosti možného DPO je potrebné požiadať správcu údajov.\n\nAké osobné údaje zhromažďujeme\nIn súlade s Článkom 4 z EÚ nariadenia 2016/679:\n- „osobné údaje“ sú akékoľvek informácie týkajúce sa identifikovanej alebo identifikovateľnej fyzickej osoby („údaje predmet“); an identifikovateľné prírodzené osoba je jeden SZO môcť byť identifikovaný, priamo alebo nepriamo, najmä odkazom na identifikátor, ako je meno, identifikačné číslo, umiestnenie údaje, an online identifikátor alebo do jeden alebo viac faktory špecifické do a fyzický, fyziologický, genetický, duševný, ekonomické, kultúrne alebo sociálna identita že prírodzené osoba;\n- \"spracovanie\" znamená akýkoľvek prevádzka alebo nastaviť z operácií ktorý je vykonané na osobné údaje alebo na súpravy z osobné údaje, či alebo nie podľa automatizované znamená, taký ako zbierka, zaznamenávanie, organizovanie, štruktúrovanie, ukladanie, prispôsobovanie alebo pozmenenie, vyhľadávanie, konzultácie, použitie, sprístupnenie prenosom, šírením alebo iným sprístupnením, zosúladením resp kombinácia, obmedzenie, vymazanie alebo zničenie.\n\nS odkaz do a vyššie spomínané definície, my zdôrazníť že my zbierať iba a informácie vy poskytnúť nám za účely vašej účasti v náš iniciatív a/alebo vaše zákonné vzťah s náš Organizácia:\n- Osobné údaje: meno a priezvisko fyzických osôb, kontakty ako adresa, PSČ kód, mesto, región, telefón číslo, email;\n- Údaje týkajúci sa profesionáli/organizácie/podniky: informácie týkajúci sa firmy, názov, fiškálna adresa a ďalšie identifikátor (číslo faxu a telefónu, daňový kód alebo DPH číslo).\n\nOkrem toho môžeme zhromažďovať údaje, ktoré nám poskytnete, keď pristupujete na naše stránky, prostredníctvom súborov cookie a iných podobná technológia; a keď nás kontaktujete prostredníctvom e-mailu, sociálnych médií alebo podobných technológií. Aj keď sa takéto údaje nezhromažďujú tak, aby boli spájané s fyzickou osobou, tieto online identifikátor môžu byť použité a kombinované na vytvorenie osobných profilov. Medzi online identifikátor môžeme nájst IP adresu, typ prehliadača a podrobnosti o zásuvnom module, typ zariadenia (napr. laptop, tablet, telefon atď.) operačný systém, miestne časové pásma. Tieto údaje sa používajú výhradne na výroba štatistické výsledky.\nPripomíname, že nebudeme spracúvať osobné údaje odhaľujúce rasové resp. etnický pôvod, politické názory, náboženské alebo filozofické presvedčenie alebo členstvo v odboroch a spracovanie genetických údajov, biometrických údajov za účelom jednoznačnej identifikácie prírodnej osoby osoba, údajov týkajúci sa zdravie alebo údajov týkajúci sa a prírodzené osoby sex života alebo sexuálne orientácia.\n\nPrečo a ako spracúvame vaše údaje\nmy bude použitie tvoj údajov v nasledujúci spôsoby:\n133. Organizujte a implementujte iniciatívy v oblasti vzdelávania a odbornej prípravy (napr. kurzy, konferencie, Európsky projektov atď.)\n134. Komu produkovať administratívne Dokumenty (napr. faktúry) v vzťah do a iniciatív vyšie\n135. Pre štatistické účely\n136. Prenášať von komunikácia činnosti cez email týkajúci sa náš iniciatív.\n137. Odpovedajte na žiadosti pomocou formulárov na stránke (ak sú k dispozícii)\n138. Povoliť registráciu na prístup k dôvernému vzdelávaciemu obsahu (ak existuje)\n\nVaše pridelenie je povinné na účely podľa odsekov 1, 2, 5, 6, aby ste vyhoveli právne povinnosti a EÚ zákonov a predpisov; odmietnutie do poskytnúť osobné údajov bude nie dovolitná organizácie do ponuka ty Naše služby.\n\nVaše údaje budeme spracovávať a uchovávať výhradne na vyššie uvedené účely pomocou digitálnych zariadení a v príslušných databázach zabezpečujúcich primerané záruky, aby sa zabezpečila trvalá dôvernosť, integrita, dostupnosť a odolnosť systémov spracovania, ako je stanovené v nariadení EÚ 2016/679. Prístup k osobným údajom môžu získať iba subjekty, ktoré od prevádzkovateľa alebo spracovateľa získali proces taký informácie.\n\nVaše údaje budeme spracovávať a uchovávať výhradne na vyššie uvedené účely pomocou digitálnych zariadení a v príslušných databázach zabezpečujúcich primerané záruky, aby sa zabezpečila trvalá dôvernosť, integrita, dostupnosť a odolnosť systémov spracovania, ako je stanovené v nariadení EÚ 2016/679. Prístup k osobným údajom môžu získať iba subjekty, ktoré od prevádzkovateľa alebo spracovateľa získali proces taký informácie.\n\nNepredávame, neobchodujeme ani inak neprevádzame na iné tretie strany, ktoré by vás osobne mohli identifikovať informácie. Vaše informácie však môžeme zverejniť, keď sa domnievame, že je to nevyhnutné vyhoviet s a zákon, presadiť náš stránky postupy, alebo chrániť náš alebo ostatných práva, nehnuteľnosť, alebo bezpečnosť.\n\nProfilovanie\nVaše údaje nebudú predmetom rozhodnutia založeného výlučne na automatizovanom spracúvaní, ktoré má právne účinky, ktoré sa ich dotýkajú alebo ktoré sa jej osoby významne\ndotýkajú. Zrušenie a zmena: máte právo kedykoľvek vedieť, aké sú vaše údaje u jednotlivých prevádzkovateľov údajov, teda u našej spoločnosti alebo u vyššie uvedených osôb, ktorým ich oznamujeme, a ako sa používajú; majú tiež právo ich aktualizovať, doplniť, opravovať alebo rušiť, požiadať o ich zablokovanie a namieteť voči ich spracovaniu. Pre uplatnenie Vašich práv, ako aj pre podrobnejšie informácie o subjektoch alebo kategóriách subjektov, ktorým sa údaje oznamujú alebo ktorí o tom vedia ako manažéri alebo zástupcovia, sa môžu obrátiť na prevádzkovateľa údajov alebo niektorého z jeho manažérov uvedených v toto vyhlásenie.\n\nSociálne siete\nNaša webová stránka môže ponúkať prístup k sociálnej sieti. Platné podmienky služby a zásady ochrany osobných údajov na takéto platformy sú zverejnené na ich webovej stránke. Pixel nemôže ovládať spôsob zdieľania údajov na a verejnosti fórum, chatovať alebo prístrojová doska sú použité, bytie a údajov predmet zodpovedný z taký komunikácia.\n\nSťažnosti\nvy môžte tiež kontakt a taliansky Údaje Ochrana autorita použitím a nasledujúce odkaz http://www.garanteprivacy.it/home/footer/contatti alebo európskeho dozorného úradníka pre ochranu údajov pomocou nasledujúce odkaz: https://edps.europa.eu/data-protection/our-role-supervisor/complaints_en\n\nCookies\nAko t_ ou t podľa regula „ Pokyny pre súbory cookie a iné nástroje sledovania – 10. júna 2021“, t tu sú tri hlavné kategórie z cookies:\n\nTechnické cookies\nTito sú použité pre a jediným účel z „vysielame komunikácie do an elektronické komunikačnej siete, alebo v rozsahu nevyhnutne potrebnom na poskytnutie služby zo strany informácie, ktoré si zmluvná strana alebo užívateľ výslovne vyžiada za účelom poskytnutia uvedená služba“ Nepoužívajú sa na žiadne vedľajšie účely a bežne sa inštalujú priamo podľa a vlastník alebo a manažér z a webové stránky (tzv \"vlastnený\" alebo \"redakčný\" cookies). Tito môžu byť rozdelený do: prehliadanie alebo relácia sušienky, ktorý záruka normálne navigácia a používanie webovej stránky (umožňujúce napr. nakupovať alebo byť overené na prístup k vyhradeným oblastiam); analytické cookies asimilované technickými cookies, kde ich používa priamo správca webovej stránky na zhromažďovanie informácií, v an pridružený formulár (anonymný), o počte používateľov a spôsobe, akým navštevujú webové stránky; funkčné súbory cookie, ktoré používateľovi umožňujú navigáciu vo vzťahu k sérii vybraných kritérií (napríklad jazyk alebo produkty vybrané na nákup) s cieľom zlepšiť službu za predpokladu, že informujeme našich používateľov ako stanovené podľa článok 13 EÚ nariadenia 2016/679. The predchádzajúci súhlas z a užívateľ je nie vyžiadané v objednať do Inštalácia tito cookies.\n\nSúbory cookie služby Analytics\nStránka používa iba analytiku Google, ktorá sa používa na vytváranie profilov používateľov a sú sa používa na odosielanie reklamných správ podľa preferencií, ktoré zobrazuje počas ich online navigácie. Vzhľadom na ich osobitnú invazívnosť s ohľadom na súkromie používateľov guľa, Európsky a taliansky predpisov vyžadovať že používateľov byť primerane informovaný o ich použitie z a rovnaký a sú teda požadovaný do expresné ich platné súhlas . V konkrétnom prípade však bola analytika Google anonymizovaná (maskovanie IP) a zdieľanie navigačných\núdajov so spoločnosťou Google bolo zablokované: analytický súbor cookie je týmto spôsobom podobný technickým súborom cookie uvedeným vyššie a nevyžaduje súhlas.\n\n**Profilovanie cookies**\nTento typ cookies sa na stránke nepoužíva.\n\nKonkrétna poznámka:\nVložené videá YouTube na stránke nepoužívajú súbory cookie, pretože bolo špecifikované „nocookie“ ako kód na vloženie s vylepšeným súkromím pre všetky vaše vložené videá na YouTube.\n\n**Medzinárodný a európsky prenos údajov**\nVaše údaje budú spracované výlučne v Európskom hospodárskom priestore. Vaše práva v súvislosti s osobnými údajmi držať pod EÚ nariadenia 2016/679\n\n**Vaše práva**\nvy môžete cvičenie tvoj práva akýkoľvek čas, ako nastaviť von podľa Článok 7, ods. 3, a články 15 a nasledujúce EÚ regulácia 2016/679:\n\n- Správny pristúpiť osobné údaje\n- Správny do náprava a vymazanie z osobné údaje;\n- Správny do obmedzenie z spracovanie;\n- Správny do údajov prenosnosť;\n- Správny do objekt do spracovanie z osobné údajov\n- Správny do legálne nárokoť si do taliansky Údaje Ochrana autorita.\n\nSvoje práva si môžete uplatniť zaslaním e-mailu na adresu info@pixel-online.net alebo adresného listu do Pixel, cez Luigi Lanzi, 12 – 50134 – Firenze, Taliansko. Ďalšie informácie o spracovaní údajov možno pridať, keď zbieranie údajov.\n\n12. februára 2022 rev.03\nSlovenščina: Politika zasebnosti v skladu z Uredbo EU 2016/679\n\nKdo zbira vaše podatke\nV skladu s 13. členom Uredbe EU 2016/679 (GDPR) je upravljavec podatkov prijavitelj projekta in je odgovoren za zbiranje podatkov. Obveščamo vas, da je naša organizacija pravno zavezana k obdelavi podatkov, ki ste nam jih posredovali v skladu z zgoraj omenjeno uredbo. Vaši podatki bodo obdelani zakonito in pošteno v skladu z določbo 5. člena Uredbe EU 2016/679. Dodatne podrobnosti bodo morda posredovane pozneje.\nPooblaščena oseba za varstvo podatkov (dpo): prisotnost morebitne DPO je treba zahtevati od upravljavca podatkov.\n\nKatere osebne podatke zbiramo\nV skladnost z Članek 4 od EU Uredba 2016/679:\n- »osebni podatki« pomenijo vse informacije, ki se nanašajo na določeno ali določljivo fizično osebo (\"podatki predmet\"); an prepoznavno naravno oseba je eno WHO lahko biti identificiran, neposredno oz posredno, zlasti s sklicevanjem na identifikator, kot so ime, identifikacijska številka, lokacija podatki, an na spletu identifikator oz do eno oz več dejavniki specifične do the fizično, fiziološki, genetski, duševno, gospodarski, kulturni oz socialno identiteteto to naravno oseba;\n- \"obravnavati\" pomeni kaj delovanje oz set od operacije ki je izvajal na osebno podatkov oz na kompleti od osebno podatki, ali oz ne od avtomatiziran pomeni, takšen kot zbirka, snemanje, organizacija, strukturiranje, shranjevanje, prilagajanje ali spreminjanje, iskanje, posvetovanje, uporaba, razkritje s prenosom, razširjanjem ali kako drugače dajanjem na voljo, uskladitvijo oz kombinacija, omejitev, izbris oz uničenje.\n\nZ referenca do the zgoraj omenjeno definicije, mi podčrtaj to mi zbirati samo the informacije ti nam zagotovite za namene vaše vpletenosti v naše pobude in/ali vaš pravni odnos z naše organizacija:\n- Osebni podatki: ime in priimek fizične osebe, kontakti kot so naslov, ZIP koda, mesto, regija, telefon številka, E-naslov;\n- Podatki glede strokovnjaki/organizacije/podjetja: informacije glede podjetja, ime, davčni naslov in drugi identifikatorji (faks in telefonska številka, davčna številka ali DDV številka).\n\nPoleg tega lahko zbiramo podatke, ki jih posredujete, ko dostopate do naših spletnih mest, prek piškotkov in drugega podobna tehnologija; in ko nas kontaktirate preko e-pošte, družbenih medijev ali podobnih tehnologij. Čeprav se ti podatki ne zbirajo zato, da bi bili povezani s fizično osebo, ti spletni identifikatorji se lahko uporabljajo in kombinirajo za ustvarjanje osebnih profilov. Med spletnimi identifikatorji, ki jih lahko najdemo IP naslov, vrsto brskalnika in podrobnosti vtičnika, vrsto naprave (npr. namizje, prenosni računalnik, tablica, telefon itd.) operacijski sistem, lokalni časovni pas. Ti podatki se uporabljajo izključno za proizvodnja statistično rezultate.\n\nOpozarjamo vas, da ne bomo obdelovali osebnih podatkov, ki bi razkrivali rasno oz etnično\nporeklo, politična mnenja, verska ali filozofska prepričanja ali članstvo v sindikatu, in obdelava genetskih podatkov, biometričnih podatkov za namen edinstvene identifikacije naravnega oseba, podatkov glede zdravje oz podatkov glede a naravno osebe seks življenje oz spolne orientacijo.\n\nZakaj in kako obdelujemo vaše podatke\nmi volja uporaba tvoj podatkov v naslednji načini:\n\n139. Organizirati in izvajati pobude na področju izobraževanja in usposabljanja (npr tečaji, konference, evropski projekti itd.)\n140. Za pridelati upravnih dokumenti (npr računi) v razmerje do the pobude zgoraj\n141. Za statistično namene\n142. Nositi ven komunikacijo dejavnosti preko E-naslov glede naše pobude.\n143. Odgovorite na zahteve z uporabo obrazcev na spletnem mestu (če so prisotni)\n144. Dovoli registracijo za dostop do zaupnih izobraževalnih vsebin (če je prisotna)\n\nVaša podelitev je obvezna za namene iz odstavkov 1, 2, 5, 6, da bi izpolnili pravni obveznosti in EU zakoni in predpisi; zavrnitev do zagotoviti osebno podatkov volja ne dovoli naše organizacijo do ponudbo ti, naše storitve.\n\nVaše soglasje ni obvezno za namene iz odstavkov 3 in 4; poslali vam bomo trženje komunikacijo preko e-pošte ali poštne storitve. Svoje pravice lahko uveljavljate kadar koli v skladu z 15. člen in pozneje Uredbe EU 2016/679 o izključitvi prejema takih komunikacijo oz izbira drugega komunikacijo modalitete.\n\nVaše osebne podatke bomo hranili, zbrane za namene kot potrebujemo, da vam lahko zagotovimo ponujene storitve s strani naše organizacije in do 10 (deset) let. ti lahko dvigniti tvoj soglasje pri kaj čas.\n\nPravno podlago obravnave sestavljajo s prodajo nastalo poslovno razmerje oz nakup blaga in/ali storitev, predpogodbeni v vednost (b in c odstavka 6. člena), in po soglasju za trženje dejavnosti. (članek 6 odstavek a)\n\nVaše podatke bomo obdelovali in hranili izključno za zgoraj omenjene namene z uporabo digitalnih naprav in v ustreznih zbirkah podatkov, ki zagotavljajo ustrezne zaščitne ukrepe za zagotovitev stalne zaupnosti, celovitost, razpoložljivost in odpornost sistemov obdelave, kot določa uredba EU 2016/679. Lahko le subjekti, ki so pridobili dostop do osebnih podatkov od upravljavca ali obdelovalca proces takšen informacije.\n\nNe prodajamo, ne trgujemo ali kako drugače prenašamo na druge tretje osebe, ki jih je mogoče prepoznati informacije. Vendar pa lahko vaše podatke objavimo, če menimo, da je to potrebno izpolnjevati z the zakon, uveljavljati naše spletne mesto pravila, oz zaščititi naše oz drugi pravice, lastnina, oz varnost.\n\nProfiliranje\nVaši podatki ne bodo predmet odločitve, ki temelji izključno na avtomatizirani obdelavi, ki ima pravne učinke, ki vplivajo nanje ali bistveno vplivajo na njegovo osebo. Preklic in sprememba: kadarkoli imate pravico vedeti, kateri so vaši podatki pri posameznih upravljavcih podatkov, to je v našem podjetju ali pri zgoraj navedenih osebah, ki jim jih sporočamo, in kako se uporabljajo; imajo tudi pravico, da jih posodobijo, dopolnijo, popravijo ali prekličejo, zahtevajo njihovo blokiranje in nasprotujejo njihovi obravnavi. Za uveljavljanje svojih pravic,\npa tudi za podrobnejše informacije o subjektih ali kategorijah subjektov, ki so jim podatki posredovani ali ki so z njimi seznanjeni kot upravljavci ali zastopniki, se lahko obrnete na upravljavca podatkov ali enega od njegovih upravljavcev, opredeljenih v to izjavo.\n\n**Socialna omrežja**\nNaše spletno mesto lahko ponuja dostop do družbenega omrežja. Veljajo pogoji storitve in politika zasebnosti do takšnih platform so objavljeni na njihovi spletni strani. Pixel ne more nadzorovati načina skupne rabe podatkov na javnosti forum, klepetati oz armaturna plošča so uporabljeno, biti the podatkov predmet odgovoren od takšen komunikacijo.\n\n**Pritožbe**\nTi lahko tudi stik the italijanski Podatki Zaščita Oblast z uporabo the sledijo povezava http://www.garanteprivacy.it/home/footer/contatti, ali evropski nadzornik za varstvo podatkov, ki uporablja sledijo povezava: https://edps.europa.eu/data-protection/our-role-supervisor/complaints_en\n\n**Piškotki**\nKot se t ou t od pravilnik \" Smernice za piškotke in druga orodja za sledenje - 10. junij 2021 \", t tukaj so tri glavne kategorije od piškotki:\n\n**Tehnični piškotki**\nTe so uporabljali za the podplat namen od \"oddaja komunikacije do an elektronski komunikacijskega omrežja ali v obsegu, ki je nujno potreben za zagotavljanje storitve s strani informacijsko podjetje, ki ga izrecno zahteva pogodbena stranka ali uporabnik, da bi zagotovili omenjeno storitev.” Ti se ne uporabljajo za nobene skrite namene in so običajno nameščeni neposredno od the lastnik oz the upravitelj od the Spletna stran (ti \"lastniška\" oz \"uvodnik\" piškotki). Te lahko biti razdeljeno v: brskanje oz seja piškotki, ki garancija normalno navigacijo in uporabo spletnega mesta (ki omogoča na primer nakupovanje ali overjen za dostop do rezerviranih območij); analitične piškotke, ki jih asimilirajo tehnični piškotki, kjer jih upravljavec spletnega mesta uporablja neposredno za zbiranje informacij, v an povezanem obrazcu (anomimno), o številu uporabnikov in načinu obiska Spletna stran; funkcionalne piškotke, ki uporabniku omogočajo navigacijo glede na vrsto izbranih kriterijev (na primer jezik ali izdelki, izbrani za nakup), da bi izboljšali storitev pod pogojem, pod pogojem, da obveščamo naše uporabnikov kot odpraviti se od Članek 13 EU Uredba 2016/679. The prej soglasje od the uporabnik je ne zahteval v naročilo do namestite te piškotki.\n\n**Analitični piškotki**\nStran uporablja samo google analytics, ki se uporablja za ustvarjanje profilov uporabnikov in so uporabljali za pošiljanje oglasnih sporočil v skladu s preferencami, ki jih pokaže isti med njihovo spletno navigacijo. Zaradi njihove posebne invazivnosti glede na zasebnost uporabnikov krogla, evropski in italijanski predpisi zahtevajo to uporabnikov biti ustrezno obveščeni približno njihov uporaba od the enako in so torej zahtevano do ekspresno njihov veljaven soglasje . Toda v konkretnem primeru je bila google analytics anonimizirana (IP maskiranje) in deljenje navigacijskih podatkov z Googlom je blokirano: na ta način je analitični piškotek podoben zgoraj navedenim tehničnim piškotkom in ne zahteva privolitve.\n\n**Piškotki za profiliranje**\nTa vrsta piškotkov se na spletne mestu ne uporablja.\nPosebna opomba:\nVdelani videoposnetki YouTuba na spletnem mestu ne uporabljajo piškotkov, saj je bila določena \"nocookie\" koda za vdelavo z izboljšano zasebnostjo za vse vaše vdelane videoposnetke v YouTubu.\n\nMednarodni in evropski prenos podatkov\nVaši podatki bodo obdelani izključno v Evropskem gospodarskem prostoru. Vaše pravice v zvezi zosebno podatkov mi drži Spodaj EU Uredba 2016/679\n\nVaše pravice\nti lahko vadbo tvoj pravice kaj čas, kot set ven od Članek 7, par. 3, in člankov 15 in sledijo od EU ureditev 2016/679:\n\n- Prav dostopati osebno podatkov\n- Prav do popravek in izbris od osebno podatki;\n- Prav do omejitev od obravnavati;\n- Prav do podatkov prenosljivost;\n- Prav do predmet do obravnavati od osebno podatkov\n- Prav do pravni trdijo do italijanski Podatki Zaščita Oblast.\n\nSvoje pravice lahko uveljavljate tako, da nam pošljete e-pošto na info@pixel-online.net ali naslovljeno pismo na Pixel, preko Luigija Lanzija , 12 – 50134 – Firenze, Italija. Dodatne informacije o obdelavi podatkov se lahko doda kdaj zbiranje podatkov.\n\n12. februarja 2022 rev.03", "id": "./materials/europe.pdf" }, { "contents": "BOLOGNINI\n- Montare content slider\n- Nuovo testo in homepage\n\nDRILLGEO\n\n| Progetto | Attività | Scadenza |\n|------------|--------------------------------------------------------------------------|----------|\n| School&Work| Video Gallery | 30 luglio|\n| | Materiale secondo meeting | |\n| | Inserire “Association” in target groups della dissemination | |\n| | Caricare le immagini nello sliding show nella home | |\n| Goerudio | Goerudio 10 | 30 luglio|\n| NOFP | Home multilingua | 30 luglio|\n| | Posizione titoli e testi | |\n| | Su Initiatives: | |\n| | - PDF della scheda in lingua nazionale | |\n| | - Other documents | |\n| | Inserire “Association” in target groups della dissemination | |", "id": "./materials/1.pdf" }, { "contents": "CONFRONTO LOCALE DI FUNZIONI\n\nGM\n\nSia $c \\in \\mathbb{R}$. Denotiamo con $\\mathcal{F}_c$ l’insieme delle funzioni definite in un intorno bucato del punto $c$, che può essere diverso per ogni funzione.\n\nDefinizione 1. Date due funzioni $f, g \\in \\mathcal{F}_c$ diremo che\n\n(i) $f$ è trascurabile rispetto a $g$ per $x \\to c$ se esiste una funzione $\\omega \\in \\mathcal{F}_c$ tale che $\\lim_{x \\to c} \\omega(x) = 0$ e\n\n$$f(x) = g(x)\\omega(x)$$\n\nper ogni $x$ in un intorno bucato di $c$;\n\n(ii) $f$ è asintotica a $g$ per $x \\to c$ se esiste una funzione $h \\in \\mathcal{F}_c$ tale che $\\lim_{x \\to c} h(x) = 1$ e\n\n$$f(x) = g(x)h(x)$$\n\nper ogni $x$ in un intorno bucato di $c$.\n\nNotazione. Se $f$ è trascurabile rispetto a $g$ per $x \\to c$ scriveremo $f \\ll_c g$ oppure $f \\ll g$ per $x \\to c$. Se $f$ è asintotica rispetto a $g$ scriveremo $f \\sim_c g$ oppure $f \\sim g$ per $x \\to c$.\n\nSe $g \\in \\mathcal{F}_c$ l’insieme $\\{f \\in \\mathcal{F}_c : f \\ll g, \\text{per } x \\to c\\}$ si denota con $o_c(g)$ oppure semplicemente con $o(g)$ quando è chiaro che stiamo considerando il comportamento per $x \\to c$ (leggi “o-piccolo” di $g$ per $x$ tendente a $c$). Quindi la notazione $f \\ll_c g$ è equivalente a $f \\in o_c(g)$. Più spesso, con un abuso di notazione, si scrive $f = o_c(g)$.\n\nProposizione 2. Supponiamo che $g$ sia diversa da zero in un intorno di $c$. Allora\n\n(i) $f \\ll_c g$ se e solo se $\\lim_{x \\to c} \\frac{f(x)}{g(x)} = 0$.\n\n(ii) $f \\sim_c g$ se e solo se $\\lim_{x \\to c} \\frac{f(x)}{g(x)} = 1$.\n\nDimostrazione. Esercizio. □\nEsercizio. Verificare che sussistano le seguenti relazioni:\n\n\\[ 1 - \\cos x \\ll_0 x; \\quad x - \\sin x \\ll_0 x^2; \\quad x^3 \\ll_0 x; \\]\n\\[ x \\ll_{+\\infty} x^3; \\quad x^{10} \\ll_{+\\infty} e^x; \\quad e^x \\ll_{-\\infty} x^{-10} \\]\n\\[ \\sin x \\sim_0 x; \\quad 1 - \\cos x \\sim_o \\frac{x^2}{2}; \\quad x - \\sin x \\sim_0 \\frac{x^3}{6}; \\]\n\\[ \\ln(1 + x) \\sim_0 x; \\quad \\sqrt{1 + 2x^3} \\sim_{+\\infty} \\sqrt{2}x^{3/2}; \\quad 3x^4 - x \\cos \\sim_{-\\infty} 3x^4. \\]\n\nSe \\( f \\) e \\( g \\) sono entrambe infinitesime per \\( x \\to c \\) e \\( f \\ll_c g \\) si dice anche che \\( f \\) è infinitesima di ordine superiore a \\( g \\) per \\( x \\to c \\) oppure che \\( f \\) tende a zero più velocemente di \\( g \\) per \\( x \\to c \\). Se \\( f \\) e \\( g \\) sono entrambe infinite per \\( x \\to c \\) e \\( f \\ll_c g \\) si dice anche che \\( g \\) è infinita di ordine superiore a \\( f \\) per \\( x \\to c \\) oppure che \\( g \\) tende a infinito più velocemente di \\( f \\) per \\( x \\to c \\).\n\nIl seguente risultato risulta molto utile nel calcolo dei limiti delle forme indeterminate.\n\n**Teorema 3.** Siano \\( f, g, f_1, g_1 \\) funzioni in \\( F_c \\) e supponiamo che \\( g(x) \\neq 0 \\) in un intorno bucato di \\( c \\).\n\n(i) Se \\( f_1 \\ll_c f \\) e \\( g_1 \\ll_c g \\) allora\n\\[\n\\lim_{x \\to c} \\frac{f(x) + f_1(x)}{g(x) + g_1(x)} = \\lim_{x \\to c} \\frac{f(x)}{g(x)}.\n\\]\n\n(ii) Se \\( f_1 \\sim_c f \\) e \\( g_1 \\sim_c g \\) allora\n\\[\n\\lim_{x \\to c} \\frac{f(x)}{g(x)} = \\lim_{x \\to c} \\frac{f_1(x)}{g_1(x)}.\n\\]\n\n**Dimostrazione.** Se \\( f_1 \\ll_c f \\) e \\( g_1 \\ll_c g \\) allora \\( f_1 = f\\omega \\) e \\( g_1 = g\\eta \\) con \\( \\omega \\) e \\( \\eta \\) infinitesime per \\( x \\to c \\). Quindi\n\\[\n\\lim_{x \\to c} \\frac{f(x) + f_1(x)}{g(x) + g_1(x)} = \\lim_{x \\to c} \\frac{f(x)}{g(x)} \\lim_{x \\to c} \\frac{1 + \\omega(x)}{1 + \\eta(x)} = \\lim_{x \\to c} \\frac{f(x)}{g(x)}.\n\\]\nSe \\( f_1 \\sim_c f \\) e \\( g_1 \\sim_c g \\) allora \\( f_1 = fh \\) e \\( g_1 = gk \\) con \\( h, k \\to 1 \\) per \\( x \\to c \\). Quindi\n\\[\n\\lim_{x \\to c} \\frac{f(x)}{g(x)} = \\lim_{x \\to c} \\frac{f_1(x)}{g_1(x)} \\lim_{x \\to c} \\frac{h(x)}{k(x)} = \\lim_{x \\to c} \\frac{f_1(x)}{g_1(x)}.\n\\]\n\n\\( \\square \\)\n\nCi riferiremo a (i) come al **principio di eliminazione** dei termini trascurabili e alla (ii) come al **principio di sostituzione** dei termini asintotici. Vediamo\nin un esempio come si possa applicare queste principi nel calcolo dei limiti di forme indeterminate. Supponiamo di voler calcolare il limite\n\\[\n\\lim_{x \\to 0} \\frac{\\sin(3x^2) + x \\ln(1 + x^2)e^x}{1 - \\cos x + e^{x^3} - 1},\n\\]\nche si presenta nella forma indeterminata 0/0. Si può verificare facilmente, usando l'Hôpital oppure i limiti notevoli, che per \\( x \\to 0 \\)\n\\[\nx \\ln(1 + x^2)e^x \\ll \\sin(3x^2) \\sim 3x^2\n\\]\n\\[\ne^{x^3} - 1 \\ll 1 - \\cos x \\sim x^2/2\n\\]\nApplicando dapprima il principio di eliminazione e, successivamente, il principio di sostituzione, si ottiene\n\\[\n\\lim_{x \\to 0} \\sin(3x^2) + x \\ln(1 + x^2)e^x = \\lim_{x \\to 0} \\sin(3x^2) = \\lim_{x \\to 0} \\frac{3x^2}{x^2/2} = 6.\n\\]\nIl problema che si pone a questo punto è: come riconoscere i termini trascurabili e come trovare dei termini che siano asintotici a quelli presenti nel limite e che siano più semplici? Per rispondere a queste domande è utile soffermarci a considerare alcune proprietà delle relazioni \\( \\ll \\) e \\( \\sim \\).\n\n**Proposizione 4.** La relazione \\( \\ll_c \\) è transitiva su \\( \\mathcal{F}_c \\):\n\\[\nf \\ll_c g, \\quad g \\ll_c h \\quad \\Rightarrow \\quad f \\ll_c h \\quad \\forall f, g, h \\in \\mathcal{F}_c.\n\\]\nLa relazione \\( \\sim_c \\) è una relazione di equivalenza su \\( \\mathcal{F}_c \\), cioè\n\\[\n\\begin{align*}\n(a) & \\quad \\text{è riflessiva: } f \\sim_c f; \\\\\n(b) & \\quad \\text{è simmetrica: } f \\sim_c g \\quad \\Rightarrow \\quad g \\sim_c f; \\\\\n(b) & \\quad \\text{è transitiva: } f \\sim_c g, \\quad g \\sim_c h \\quad \\Rightarrow \\quad f \\sim_c h.\n\\end{align*}\n\\]\n\n**Dimostrazione.** Esercizio. \\( \\square \\)\n\n**Osservazione.** È facile verificare che \\( f \\ll_c f \\) se e solo se \\( f = 0 \\) in qualche intorno di \\( c \\). Indichiamo con \\( \\mathcal{Z}_c \\) il sottoinsieme di \\( \\mathcal{F}_c \\) delle funzioni nulle in qualche intorno bucato di \\( c \\). Allora la relazione \\( \\ll_c \\) è transitiva e non-riflessiva su \\( \\mathcal{F}_c \\setminus \\mathcal{Z}_c \\), cioè è una relazione di ordine (parziale) stretto. Parziale qui si riferisce al fatto che non tutte le funzioni in \\( \\mathcal{F}_c \\) sono confrontabili. Ad esempio le funzioni \\( f(x) = x \\) e \\( g(x) = x \\sin(1/x) \\) non sono confrontabili in 0.\n\n**Proposizione 5.** Le relazioni \\( \\ll_c \\) e \\( \\sim_c \\) soddisfano le seguenti proprietà\n\\[\n\\begin{align*}\n(a) & \\quad f_1 \\ll_c g \\quad \\text{e} \\quad f_2 \\ll_c g \\quad \\Rightarrow \\quad f_1 + f_2 \\ll_c g; \\\\\n(b) & \\quad f_1 \\sim_c \\alpha g \\quad \\text{e} \\quad f_2 \\sim_c \\beta g \\quad \\text{con} \\quad \\alpha, \\beta \\in \\mathbb{R}, \\quad \\alpha + \\beta \\neq 0 \\quad \\Rightarrow \\quad f_1 + f_2 \\sim_c (\\alpha + \\beta)g;\n\\end{align*}\n\\]\n(c) \\( f_1 \\ll_c g_1 \\) e \\( f_2 \\ll_c g_2 \\) \\( \\implies \\) \\( f_1 f_2 \\ll_c g_1 g_2 \\);\n\n(d) \\( f_1 \\sim_c g_1 \\) e \\( f_2 \\sim_c g_2 \\) \\( \\implies \\) \\( f_1 f_2 \\sim_c g_1 g_2 \\);\n\n(e) \\( f \\ll_c g \\) e \\( g \\sim_c h \\) \\( \\implies \\) \\( f \\ll_c h \\);\n\n(f) \\( f \\sim_c g \\) e \\( g \\ll_c h \\) \\( \\implies \\) \\( f \\ll_c h \\).\n\n(g) Sia \\( \\phi \\in \\mathcal{F}_p \\) tale che \\( \\phi \\to c \\) per \\( x \\to p \\) e \\( \\phi(y) \\neq c \\) per ogni \\( y \\) in un intorno bucato di \\( p \\). Allora\n\n\\[\n\\begin{align*}\n(g.1) & \\quad f \\ll_c g \\implies f \\circ \\phi \\ll_p g \\circ \\phi; \\\\\n(g.2) & \\quad f \\sim_c g \\implies f \\circ \\phi \\sim_p g \\circ \\phi.\n\\end{align*}\n\\]\n\nDimostrazione. Sono tutte facili verifiche, a partire dalle definizioni di \\( \\ll_c \\) e \\( \\sim_c \\). Ad esempio verifichiamo la (b). Se \\( f_1 = \\alpha gh_1 \\) e \\( f_2 = \\beta gh_2 \\) con \\( h_1, h_2 \\to 1 \\) allora \\( f_1 f_2 = (\\alpha + \\beta) gh \\) dove \\( h = (\\alpha h_1 + \\beta h_2)/(\\alpha + \\beta) \\to 1 \\). La (g) utilizza il teorema sul limite della funzione composta. \\( \\square \\)\n\nOsservazione. È importante osservare che le seguenti implicazioni, apparentemente plausibili, sono invece FALSE\n\n\\[\n\\begin{align*}\n(F.1) & \\quad f_1 \\ll_c g_1 \\) e \\( f_2 \\ll_c g_2 \\) \\( \\implies \\) \\( f_1 + f_2 \\ll_c g_1 + g_2 \\).\n\\end{align*}\n\\]\n\n\\[\n\\begin{align*}\n(F.2) & \\quad f_1 \\sim_c g_1 \\) e \\( f_2 \\sim_c g_2 \\) \\( \\implies \\) \\( f_1 + f_2 \\sim_c g_1 + g_2 \\).\n\\end{align*}\n\\]\n\n\\[\n\\begin{align*}\n(F.3) & \\quad f \\sim_c g \\implies \\phi \\circ f \\sim_c \\phi \\circ g.\n\\end{align*}\n\\]\n\nIl lettore verifichi che un controesempio di (F.1) per \\( x \\to 0 \\) è dato dalle funzioni \\( f_1 = f_2 = x^2 \\), \\( g_1 = x \\), \\( g_2 = -\\sin(x) \\), un controesempio di (F.2) per \\( x \\to 0 \\) è dato dalle funzioni \\( f_1 = x + x^3 \\), \\( f_2 = -\\sin x \\), \\( g_1 = x \\), \\( g_2 = -x \\) e un controesempio di (F.3) per \\( x \\to +\\infty \\) è dato dalle funzioni \\( f(x) = x^2 \\), \\( g(x) = x^2 - x \\) e \\( \\phi(y) = e^y \\).\n\nTutte le considerazioni precedenti si applicano, con le ovvie modifiche, a funzioni che sono definite in un intorno destro o sinistro di \\( c \\). In particolare i principi di eliminazione e di sostituzione valgono anche per limiti sinistri o destri.\n\nApplicazione al calcolo dei limiti. Supponiamo di voler calcolare un limite della forma\n\n\\[\n\\lim_{x \\to c} \\frac{f_1 + f_2 + \\cdots + f_k}{g_1 + g_2 + \\cdots + g_m},\n\\]\n\ndove \\( f_1, \\ldots, f_k, g_1, \\ldots, g_m \\) sono funzioni in \\( \\mathcal{F}_c \\). La procedura da seguire è\n(1) determinare nel numeratore e nel denominatore separatamente i termini trascurabili e quelli dominanti.\n\n(2) Se vi è un unico termine dominante, tutti i termini trascurabili rispetto ad esso possono essere eliminati, per il principio di eliminazione.\n\n(3) Se i termini dominanti sono più d’uno, occorre considerare la loro somma come un unico termine e verificare se tale somma non sia trascurabile rispetto agli altri termini.\n\n(4) Una volta individuati i termini dominanti al numeratore e al denominatore, sostituirli con termini più semplici ad essi asintotici.\n\nConsideriamo ad esempio il seguente limite\n\n\\[ \\lim_{x \\to 0^+} \\frac{\\sin x + 1 - \\cos x + \\ln(1 + x^3)}{x + e^{x^2} - 1 - \\sin x}. \\]\n\nAl numeratore abbiamo che per \\( x \\to 0^+ \\)\n\n\\[ \\sin x \\sim x, \\quad 1 - \\cos x \\sim \\frac{x^2}{2}, \\quad \\ln(1 + x^3) \\sim x^3. \\]\n\nPertanto c’è un unico termine dominante, che è \\( \\sin(x) \\). Possiamo quindi eliminare i termini \\( 1 - \\cos x \\) e \\( \\ln(1 + x^3) \\) e sostituire \\( \\sin x \\) con \\( x \\).\n\nAl denominatore abbiamo\n\n\\[ x, \\quad e^{x^2} - 1 \\sim x^2, \\quad -\\sin x \\sim -x. \\]\n\nI termini dominanti sono due, \\( x \\) e \\( -\\sin x \\), la cui somma deve quindi essere considerata come un unico termine, \\( x - \\sin x \\). Poiché\n\n\\[ x - \\sin x \\sim \\frac{x^3}{6} \\ll x^2 \\sim e^{x^2} - 1, \\]\n\nil termine dominante al denominatore risulta essere in realtà \\( e^{x^2} - 1 \\sim x^2 \\).\n\nIn conclusione si ha che\n\n\\[ \\lim_{x \\to 0^+} \\frac{\\sin x + 1 - \\cos x + \\ln(1 + x^3)}{x + e^{x^2} - 1 - \\sin x} = \\lim_{x \\to 0^+} \\frac{x}{x^2} = +\\infty. \\]\n\nConsideriamo ancora un esempio\n\n\\[ \\lim_{x \\to 0^+} \\frac{\\cos(2x) \\sin(3x^3) - \\sin(\\pi x^2) \\sqrt{1 - \\cos(\\pi x)}}{\\sqrt{x^3 + 3x \\arctan(3x) - x \\ln^2(2\\sqrt{\\sin x + 1})}}. \\]\n\nEsaminiamo i termini al numeratore: per \\( x \\to 0^+ \\)\n\n\\[ \\cos(2x) \\sim 1, \\quad \\sin(3x^3) \\sim 3x^3, \\quad -\\sin(\\pi x^2) \\sim -\\pi x^2, \\quad \\sqrt{1 - \\cos(\\pi x)} \\sim \\frac{\\pi x}{\\sqrt{2}} \\]\n\nQuindi\n\n\\[ \\cos(2x) \\sin(3x^3) \\sim 3x^3; \\quad -\\sin(\\pi x^2) \\sqrt{1 - \\cos(\\pi x)} \\sim -\\frac{\\pi^2 x^3}{\\sqrt{2}}. \\]\nPoiché $3 - \\frac{\\pi^2}{\\sqrt{2}} \\neq 0$, per il punto (b) della Proposizione 5 possiamo concludere che\n\n$$\\cos(2x) \\sin(3x^3) - \\sin(\\pi x^2) \\sqrt{1 - \\cos(\\pi x)} \\sim \\left(3 - \\frac{\\pi^2}{\\sqrt{2}}\\right) x^3$$\n\nEsaminiamo ora il denominatore\n\n$$\\sqrt{x^2 + 3x} \\sim \\sqrt{3x}, \\quad \\arctan(3x) \\sim 3x, \\quad \\ln^2(2\\sqrt{\\sin x + 1}) \\sim \\ln^2(2).$$\n\nAllora\n\n$$\\sqrt{x^2 + 3x} \\arctan(3x) \\sim 3\\sqrt{3} x^{3/2}, \\quad x \\ln^2(2\\sqrt{\\sin x + 1}) \\sim \\ln^2(2) x.$$\n\nQuest’ultimo termine è quello dominante e quindi\n\n$$\\lim_{x \\to 0^+} \\frac{\\cos(2x) \\sin(3x^3) - \\sin(\\pi x^2) \\sqrt{1 - \\cos(\\pi x)}}{\\sqrt{x^2 + 3x} \\arctan(3x) - x \\ln^2(2\\sqrt{\\sin x + 1})} = \\lim_{x \\to 0^+} \\frac{3 - \\frac{\\pi^2}{\\sqrt{2}}}{\\ln^2(2)x} = 0.$$\n\n**Ordini di infinitesimo e di infinito, parti principali.**\n\nPer talune applicazioni (studio della convergenza di integrali impropri e di serie) è utile introdurre una misura della velocità con cui una funzione infinitesima tende a zero o una funzione infinita tende a infinito. A tale fine occorre introdurre una funzione campione, che svolge il ruolo di unità di misura. Cominciamo con il considerare il caso degli infinitesimi. Fissato un punto $c \\in \\mathbb{R}$ scegliamo una funzione $u$, definita e positiva in un intorno bucato di $c$, infinitesima per $x \\to c$, che svolgerà il ruolo di infinitesimo campione.\n\n**Definizione 6.** Si dice che una funzione $f \\in \\mathcal{F}_c$ è infinitesima di ordine $\\alpha \\in \\mathbb{R}$ per $x \\to c$ (rispetto a $u$) se esiste $\\lambda \\in \\mathbb{R}$, $\\lambda \\neq 0$ tale che\n\n$$f(x) \\sim \\lambda u(x)^\\alpha \\quad \\text{per} \\quad x \\to c.$$ \n\nIn tal caso la funzione $\\lambda u(x)^\\alpha$ è detta la *parte principale* di $f$ per $x \\to c$, (rispetto all’infinitesimo campione $u$).\n\nOsserviamo che $u^\\alpha$ è ben definita, perché abbiamo supposto che $u$ fosse strettamente positiva in un intorno bucato di $c$. Inoltre, poiché $u$ è positiva, la (1) equivale a\n\n$$\\lim_{x \\to c} \\frac{f(x)}{u(x)^\\alpha} = \\lambda \\neq 0.$$\n\nTalvolta accade che il limite non esista, ma esistano separatamente il limite destro e sinistro\n\n$$\\lim_{x \\to c^+} \\frac{f(x)}{u(x)^\\alpha} = \\lambda_+, \\quad \\lim_{x \\to c^-} \\frac{f(x)}{u(x)^\\alpha} = \\lambda_-,$$\n\ncon $\\lambda_-$ e $\\lambda_+$ numeri reali diversi da 0. Anche in tal caso diremo che $f$ è infinitesima di ordine $\\alpha$ per $x \\to c$. \nUsualmente, come infinitesimo campione si sceglie la funzione\n\\[ u(x) = \\begin{cases} |x - c| & \\text{se } c \\in \\mathbb{R}, \\\\ \\frac{1}{|x|} & \\text{se } c = \\pm \\infty. \\end{cases} \\]\n\n**Esempi.**\n\n(1) La funzione \\( 1 - \\cos x \\) è infinitesima di ordine 2 per \\( x \\to 0 \\), perché\n\\[ \\lim_{x \\to 0} \\frac{1 - \\cos x}{x^2} = \\frac{1}{2}. \\]\nLa parte principale di \\( 1 - \\cos x \\) per \\( x \\to 0 \\) è \\( x^2/2 \\).\n\n(2) La funzione \\( x - \\sin x \\) è infinitesima di ordine 3 per \\( x \\to 0 \\), perché\n\\[ \\lim_{x \\to 0} \\frac{x - \\sin x}{|x|^3} = -\\frac{1}{6}, \\quad \\lim_{x \\to 0^+} \\frac{x - \\sin x}{|x|^3} = \\frac{1}{6} \\]\nLa parte principale di \\( x - \\sin x \\) per \\( x \\to 0 \\) è \\( x^3/6 \\).\n\n(3) Non tutte le funzioni infinitesime in un punto ammettono ordine rispetto al campione scelto. Ad esempio, la funzione \\( x \\ln(x) \\) non ha ordine rispetto a \\( u(x) = |x| \\) per \\( x \\to 0^+ \\). Infatti\n\\[ \\lim_{x \\to 0^+} \\frac{x \\ln x}{x^\\alpha} = \\begin{cases} 0 & \\text{se } \\alpha < 1 \\\\ -\\infty, & \\text{se } \\alpha \\geq 1. \\end{cases} \\]\nPertanto non esiste \\( \\alpha \\in \\mathbb{R} \\) per cui il limite esiste finito e diverso da 0.\n\nLe definizione di ordine di infinito è analoga. Fissato un infinito campione \\( u \\) per \\( x \\to c \\), si dice che una funzione \\( f \\in \\mathcal{F}_c \\) è infinita di ordine \\( \\alpha \\) per \\( x \\to c \\), rispetto a \\( u \\), se esiste \\( \\lambda \\in \\mathbb{R}, \\lambda \\neq 0 \\) tale che\n\\[ f(x) \\sim \\lambda u(x)^\\alpha \\quad \\text{per } x \\to c. \\]\nIn tal caso \\( \\lambda u^\\alpha \\) si dice parte principale dell’infinito \\( f \\) per \\( x \\to c \\). Anche in questo caso la relazione (2) equivale a\n\\[ \\lim_{x \\to c} \\frac{f(x)}{u(x)^\\alpha} = \\lambda \\neq 0 \\]\ne valgono le considerazioni sui limiti destro e sinistro quando il limite non esiste.\n\nGli infiniti campione di uso più comune sono\n\\[ u(x) = \\begin{cases} \\frac{1}{|x - c|} & \\text{se } c \\in \\mathbb{R}, \\\\ |x| & \\text{se } c = \\pm \\infty. \\end{cases} \\]\nEsempi.\n\n(1) La funzione $\\sqrt{1 + x^3}$ è infinita di ordine $3/2$ per $x \\to +\\infty$, perché\n\n$$\\lim_{x \\to +\\infty} \\frac{\\sqrt{1 + x^3}}{x^{3/2}} = 1.$$ \n\nLa sua parte principale per $x \\to +\\infty$ è $x^{3/2}$.\n\n(2) La funzione $1/(1 - \\sin x)$ è infinita di ordine $2$ per $x \\to \\pi/2$, perché\n\n$$\\lim_{x \\to \\pi/2} \\frac{1}{1 - \\sin x} \\frac{1}{(x - \\pi/2)^2} = 2$$\n\nLa sua parte principale di $x - \\sin x$ per $x \\to \\pi/2$ è $2/(x - \\pi/2)^2$. ", "id": "./materials/10.pdf" }, { "contents": "Evaluate \\( \\int \\frac{x-4}{x^2-5x+6} \\, dx \\)\n\n* All the conditions for Fundamental theorem of calculus are met.\n\nsince, man, the partial fractions should be obtained.\n\nAt first, factorize the denominator,\n\n\\[ x^2-5x+6 = (x-3)(x-2) \\]\n\n\\[ \\text{Quadratic formula can be used} \\]\n\n\\[ \\rightarrow \\text{Proceed to partial fractions} \\]\n\n\\[ \\frac{x-4}{x^2-5x+6} = \\frac{A}{x-3} + \\frac{B}{x-2} \\]\n\n\\[ \\Rightarrow x-4 = A(x-2) + B(x-3) \\]\n\n\\[ \\Rightarrow x-4 = Ax-2A + Bx-3B \\]\n\n\\[ \\Rightarrow x-4 = x(A+B) - 2A - 3B \\]\n\nComparing coefficients on left and right side\n\n\\[ \\begin{cases} A + B = 1 \\\\ -2A - 3B = -4 \\end{cases} \\]\n\n\\[ \\Rightarrow \\begin{cases} A = 1 - B \\\\ -2 + 2B - 3B = -4 \\end{cases} \\]\n\n\\[ \\Rightarrow \\begin{cases} A = 1 - B \\\\ B = 2 \\end{cases} \\]\n\n\\[ \\Rightarrow \\begin{cases} A = -1 \\\\ B = 2 \\end{cases} \\]\nSo,\n\\[\n\\frac{x - 4}{x^2 - 5x + 6} = \\frac{-1}{x - 3} + \\frac{2}{x - 2}\n\\]\n\nNow,\n\\[\n\\int_0^1 \\frac{x - 4}{x^2 - 5x + 6} \\, dx = \\int_0^1 \\left( \\frac{-1}{x - 3} + \\frac{2}{x - 2} \\right) \\, dx\n\\]\n\\[\n= \\int_0^1 \\frac{-1}{x - 3} \\, dx + 2 \\int_0^1 \\frac{1}{x - 2} \\, dx\n\\]\n\\[\n= \\left[ -\\ln |x - 3| \\right]_0^1 + \\left[ 2 \\ln |x - 2| \\right]_0^1\n\\]\n\\[\n= (-\\ln 1 - 2) + (2 \\ln 1 - 2 \\ln 2)\n\\]\n\\[\n= -\\ln 2 + \\ln 3 + 2 \\ln 1 - 2 \\ln 2\n\\]\n\\[\n= \\ln 3 - \\ln 2 + 2 \\ln 1 - 2 \\ln 2\n\\]\n\\[\n= \\ln 3 - \\ln 2 - 2 \\ln 2\n\\]\n\\[\n= \\ln 3 - 3 \\ln 2\n\\]\n\\[\n= \\ln 3 - \\ln 2^3\n\\]\n\\[\n= \\ln 3 - \\ln 8\n\\]\n\\[\n= \\ln \\left( \\frac{3}{8} \\right)\n\\]", "id": "./materials/100.pdf" }, { "contents": "Evaluate \\( \\int_{1}^{2} \\frac{x^2-1}{x(x^2+1)} \\, dx \\)\n\n* All the conditions for Fundamental theorem of calculus are met.\n\nSince, \\( m < n \\), the partial fractions should be obtained.\n\nFor, \\( I(x) = \\int \\frac{x^2-1}{x(x^2+1)} \\, dx \\), the partial fractions are,\n\n\\[\n\\frac{x^2-1}{x(x^2+1)} = \\frac{A}{x} + \\frac{Cx+D}{x^2+1}\n\\]\n\n\\[\n(\\Rightarrow) \\quad x^2-1 = A(x^2+1) + (Cx+D)x\n\\]\n\n\\[\n(\\Rightarrow) \\quad x^2-1 = Ax^2 + A + Cx^2 + Dx\n\\]\n\n\\[\n(\\Rightarrow) \\quad x^2-1 = (A+C)x^2 + Dx + A\n\\]\n\n\\[\n(\\Rightarrow) \\quad x^2 + 0 \\cdot x - 1 = (A+C)x^2 + Dx + A\n\\]\n\nComparing coefficients of left and right hand side,\n\n\\[\n\\begin{align*}\nA &= -1 \\\\\nD &= 0 \\\\\nA+C &= 1\n\\end{align*}\n\\]\n\n\\[\n\\begin{align*}\nA &= -1 \\\\\nD &= 0 \\\\\n-1+C &= 1\n\\end{align*}\n\\]\n\n\\[\n\\begin{align*}\nA &= -1 \\\\\nD &= 0 \\\\\nC &= 2\n\\end{align*}\n\\]\nSo, \\( \\frac{x^2 - 1}{x(x^2 + 1)} = -\\frac{1}{x} + \\frac{2x}{x^2 + 1} \\)\n\n\\[ I(x) = \\int -\\frac{1}{x} \\, dx + \\int \\frac{2x}{x^2 + 1} \\, dx \\]\n\n\\[ = -\\ln |x| + \\ln |x^2 + 1| + C \\]\n\nNow, \\( \\int \\frac{x^2 - 1}{x(x^2 + 1)} \\, dx = \\left[ I(x) \\right]^2 \\)\n\n\\[ = \\left[ -\\ln |x| + \\ln |x^2 + 1| \\right]^2 \\]\n\n\\[ = (-\\ln(2) + \\ln(5)) - (\\ln(11) + \\ln(12)) \\]\n\n\\[ = -\\ln(2) + \\ln(5) - \\ln(2) \\]\n\n\\[ = \\ln(5) - 2\\ln(2) \\]\n\n\\[ = \\ln(5) - \\ln(2^2) \\]\n\n\\[ = \\ln(5) - \\ln(4) \\]\n\n\\[ = \\ln\\left(\\frac{5}{4}\\right) \\]", "id": "./materials/101.pdf" }, { "contents": "Evaluate \\( \\int_{-2}^{1} \\frac{4-x^2}{x-2} \\, dx \\)\n\n* Looking at the question, it seems like we have to perform long division followed by partial fraction decomposition. But, there is an easier way.\n\n* All conditions to use Fundamental theorem of calculus are met.\n\n\\[\nF(x) = \\int \\frac{4-x^2}{x-2} \\, dx\n\\]\n\n\\[\n= \\int \\frac{(2-x)(2+x)}{x-2} \\, dx\n\\]\n\n\\[\n= \\int \\frac{-x-x^2}{x-2} \\, dx\n\\]\n\n\\[\n= -\\int (2+x) \\, dx\n\\]\n\n\\[\n= -2x - \\frac{x^2}{2} + C\n\\]\n\nand, \\[\n\\int_{-2}^{1} \\frac{4-x^2}{x-2} \\, dx = \\left[ F(x) \\right]_{-2}^{1} = \\left[ -2x - \\frac{x^2}{2} \\right]_{-2}^{1}\n\\]\n\n\\[\n= -2 - \\frac{1}{2} - (4 - 2)\n\\]\n\n\\[\n= -\\frac{9}{2}\n\\]", "id": "./materials/102.pdf" }, { "contents": "$x = 5$ is a straight line.\n$y = 2x$ is a straight line.\n$y = x$ is a straight line.\n\nTo find the point of intersection of the straight lines, we solve the equations simultaneously.\n\nFor example; solving $y = 2x$ and $x = 5$, we get $y = 10$\n\nSimilarly, solving $y = x$ and $x = 5$, we get $y = 5$.\n\n**Remember that**, Area bounded by the curves is given by,\n\n$$\\text{Area} = \\int_{a}^{b} f(x) - g(x) \\, dx,$$\n\nwhere $f(x)$ is the upper curve and $g(x)$ is the lower curve and $x \\in [a, b]$.\n\nIn this case, the upper function is $f(x) = 2x$ and lower function is $g(x) = x$ and $x \\in [0, 5]$. \nArea = \\int_{a}^{b} f(x) - g(x) \\, dx\n\n= \\int_{0}^{5} 2x - x \\, dx\n\n= \\int_{0}^{5} x \\, dx\n\n= \\left[ \\frac{x^2}{2} \\right]_{0}^{5}\n\n= \\left( \\frac{5^2}{2} \\right) - \\left( \\frac{0}{2} \\right)\n\n= \\frac{25}{2} \\text{ square units}", "id": "./materials/103.pdf" }, { "contents": "We want to obtain the area of the region between the parabola and the $x$–axis.\n\n$y = 4x - x^2$ is a parabola facing\n\nSolving $4x - x^2 = 0$ we conclude that the parabola intersects the $x$–axis on points $(0,0)$ and $(4,0)$.\n\n**Remember that**, Area bounded by two curves is given by,\n\n$$\\text{Area} = \\int_{a}^{b} f(x) - g(x) \\, dx,$$\n\nwhere $f(x)$ is the upper curve and $g(x)$ is the lower curve and $x \\in [a, b]$.\n\nIn this case, the upper function is $f(x) = 4x - x^2$ and the lower function is $g(x) = 0$ and $x \\in [0, 4]$.\n\n$$\\text{Area} = \\int_{0}^{4} 4x - x^2 \\, dx$$\n\n$$= \\left[ 2x^2 - \\frac{x^3}{3} \\right]_{0}^{4} = \\frac{32}{3} \\text{ square units}$$", "id": "./materials/104.pdf" }, { "contents": "$y = x^2 - 4$ is a parabola opening upwards with vertex (0,-4).\n\nThe parabola intersects the $x-$axis on points (-2,0) and (2,0).\n\nTo find the point of intersection of the parabola and the line $x = 5$, we solve the equations simultaneously.\n\nFor example; solving $y = x^2 - 4$ and $x = 5$, we get $y = 21$.\n\n**Remember that**, Area bounded by the curves is given by,\n\n$$\\text{Area} = \\int_{a}^{b} f(x) - g(x) \\, dx,$$\n\nwhere $f(x)$ is the upper curve and $g(x)$ is the lower curve and $x \\in [a, b]$. \nIn this case, the upper function is $f(x) = x^2 - 4$ and lower function is $g(x) = 0$ and $x \\in [2, 5]$.\n\n\\[\n\\text{Area} = \\int_{a}^{b} f(x) - g(x) \\, dx \\\\\n= \\int_{2}^{5} x^2 - 4 \\, dx \\\\\n= \\left[ \\frac{x^3}{3} - 4x \\right]_{2}^{5} = 27 \\text{ square units}\n\\]", "id": "./materials/105.pdf" }, { "contents": "By: Amulya Baniya\n\n\\[ y = x^2 \\] is a parabola opening upwards with vertex (0,0).\n\n**Remember that**, Area bounded by the curves is given by,\n\n\\[\n\\text{Area} = \\int_{a}^{b} f(x) - g(x) \\, dx,\n\\]\n\nwhere \\( f(x) \\) is the upper curve and \\( g(x) \\) is the lower curve and \\( x \\in [a, b] \\).\n\nIn this case, there are two upper functions and one lower function. Therefore, it is necessary to split the region \\( R \\) into two regions (\\( R_1 \\) and \\( R_2 \\)) such that there’s only one upper function and only one lower function.\nFor $R_1$, the upper function is $f(x) = x^2$ and lower function is $h(x) = 0$ and $x \\in [0, 1]$.\n\n\\[\nA_1 = \\int_a^b f(x) - h(x) \\, dx \\\\\n= \\int_0^1 x^2 \\, dx \\\\\n= \\left[ \\frac{x^3}{3} \\right]_0^1 = \\frac{1}{3} \\text{ square units}\n\\]\n\nFor $R_2$, the upper function is $g(x) = 2 - x$ and lower function is $h(x) = 0$ and $x \\in [1, 2]$.\n\n\\[\nA_2 = \\int_a^b g(x) - h(x) \\, dx \\\\\n= \\int_1^2 2 - x \\, dx \\\\\n= \\left[ 2x - \\frac{x^2}{2} \\right]_1^2 = \\frac{1}{2} \\text{ square units}\n\\]\n\nThe total area enclosed by region $R = A_1 + A_2$\n\n\\[\n= \\frac{1}{3} + \\frac{1}{2} \\\\\n= \\frac{5}{6} \\text{ square units}\n\\]\nAlternate method (Integrating with respect to y)\n\nWhen we take $dx$, we found that we need to divide the region $R$ into 2 sub-regions because we had two different upper functions. However, when we take $dy$, i.e. treating $x$ as a function of $y$, it fixes the problem. To understand further, click this link.\n\nWhen integrating with respect to $y$, Area $= \\int_{a}^{b} f(y) - g(y) \\, dy$, where $f(y)$ is the curve on the right side and $g(x)$ is the curve on the left side and $y \\in [a, b]$.\n\nNow, rewrite the functions in function on $y$. We have $g(y) = 2 - y$ and $f(y) = \\sqrt{y}$\n\n(Note that, we don’t need $f(y) = -\\sqrt{y}$ here.)\nIn this case, the function on the right is \\( g(y) = 2 - y \\) and function of the left is \\( f(y) = \\sqrt{y} \\) and \\( y \\in [0, 1] \\).\n\n\\[\n\\text{Area} = \\int_0^1 \\text{right - left } dy \\\\\n= \\int_0^1 2 - y - \\sqrt{y} \\, dx \\\\\n= \\left[ 2y - \\frac{y^2}{2} - \\frac{2}{3}y^{\\frac{3}{2}} \\right]_0^1 = \\frac{5}{6} \\text{ square units}\n\\]", "id": "./materials/106.pdf" }, { "contents": "$y = 2x$ is a straight line which passes through the origin.\n\n$y = \\frac{8}{x}$ is a curve with both vertical and horizontal asymptote.\n\n$x = 4$ is a straight line.\n\nThe straight line $y = 2x$ intersects the curve $y = \\frac{8}{x}$ on $x = 2$ and $x = -2$\n\n**Remember that**, Area bounded by the curves is given by,\n\n$$\\text{Area} = \\int_{a}^{b} f(x) - g(x) \\, dx,$$\n\nwhere $f(x)$ is the upper curve and $g(x)$ is the lower curve and $x \\in [a, b]$. \nIn this case, the upper function is \\( f(x) = 2x \\) and lower function is \\( g(x) = \\frac{8}{x} \\) and \\( x \\in [2, 4] \\).\n\n\\[\n\\text{Area} = \\int_{a}^{b} f(x) - g(x) \\, dx \\\\\n= \\int_{2}^{4} 2x - \\frac{8}{x} \\, dx \\\\\n= \\left[ x^2 - 8 \\ln(x) \\right]_{2}^{4} \\\\\n= \\left[ x^2 \\right]_{2}^{4} - \\left[ 8 \\ln(x) \\right]_{2}^{4} \\\\\n= (16 - 4) - 8(\\ln(4) - \\ln(2)) \\\\\n= 4 - 8 \\ln(2) \\text{ square units}\n\\]", "id": "./materials/107.pdf" }, { "contents": "For R1, upper function is $f(x) = 0$ and lower function is $g(x) = x^2 - 1$ and $x \\in [0, 1]$\n\nFor R2, upper function is $f(x) = x^2 - 1$ and lower function is $g(x) = 0$ and $x \\in [1, 2]$. ", "id": "./materials/108.pdf" }, { "contents": "Hint:\n\nThe region bounded by the curves is highlighted in blue.", "id": "./materials/109.pdf" }, { "contents": "LA FUNZIONE ESPONENZIALE E IL LOGARITMO\n\nAPPUNTI PER IL CORSO\nDI ANALISI MATEMATICA I\n\nG. MAUCERI\n\nINDICE\n\n1. Introduzione 1\n2. La funzione esponenziale 2\n3. Il numero $e$ di Nepero 9\n4. L’irrazionalità di $e$ 12\n5. L’esponenziale e il logaritmo nel campo complesso 13\n6. Le funzioni trigonometriche nel campo complesso 16\n\n1. INTRODUZIONE\n\nLa funzione esponenziale è una delle funzioni più importanti in matematica. Essa gioca un ruolo cruciale nella teoria delle equazioni differenziali, in analisi armonica, in probabilità, in matematica applicata e in fisica. Vi sono molti modi di definire la funzione esponenziale (utilizzando la densità dei razionali nei reali, come somma di una serie di potenze, come soluzione di un’equazione differenziale, come inversa del logaritmo naturale...). In queste note abbiamo scelto di introdurre la funzione esponenziale a partire dalla legge degli esponenti, cioè dalla proprietà di trasformare somme in prodotti. Questa proprietà si esprime in termini più tecnici dicendo che la funzione esponenziale è un omomorfismo della struttura additiva dei numeri reali nella struttura moltiplicativa dei numeri reali positivi. Nella prima sezione mostreremo che questa proprietà caratterizza completamente la funzione esponenziale sui numeri razionali. Per caratterizzare l’esponenziale sui reali occorre aggiungere qualche ipotesi di regolarità, ad esempio che la funzione sia continua. Dimostreremo quindi che la funzione esponenziale in base $a > 0$ ha un unico prolungamento continuo da $\\mathbb{Q}$ a $\\mathbb{R}$ che soddisfa la legge degli esponenti e che nel punto 1 vale $a$. Nella terza sezione, dopo aver introdotto il numero $e$ di Nepero come limite di una successione monotona,\ncalcoliamo i limiti notevoli dell’esponenziale e del logaritmo, che sono alla base del calcolo differenziale di queste funzioni. Nella quarta sezione dimostriamo l’irrazionalità del numero $e$ di Nepero, base dei logaritmi naturali. Nelle ultime due sezioni estendiamo le funzioni esponenziale e logaritmo al campo complesso e dimostriamo le identità di Eulero che legano la funzione esponenziale alle funzioni trigonometriche. Infine utilizziamo le identità di Eulero per estendere anche le funzioni trigonometriche al campo complesso.\n\n2. La funzione esponenziale\n\nSiano $a$ un numero reale e $n$ un intero $\\geq 1$. La potenza di base $a$ e esponente $n$ è definita ricorsivamente:\n\n$$a^1 = a, \\quad a^{n+1} = a \\cdot a^n.$$ \n\nÈ facile dimostrare per induzione che\n\n$$(2.1) \\quad a^{m+n} = a^m a^n$$\n\nper ogni coppia di numeri naturali $m, n \\geq 1$. L’identità (2.1) è nota come legge degli esponenti.\n\nIl seguente teorema mostra che, se la base $a$ è positiva, c’è un unico modo di estendere la funzione $n \\mapsto a^n$ da $\\mathbb{Z}_+$ a $\\mathbb{Q}$ preservando la validità della legge degli esponenti.\n\n**Teorema 2.1.** Sia $f$ una funzione definita su $\\mathbb{Q}$ tale che $f(1) = a$ e $f(r+s) = f(r)f(s)$, per ogni coppia di numeri razionali $r$ e $s$. Allora per ogni razionale $r$ si ha\n\n$$f(r) = \\begin{cases} \\sqrt[n]{a^m}, & \\text{se } r = m/n, \\text{ con } m, n \\in \\mathbb{Z}_+ \\\\ 1, & \\text{se } r = 0 \\\\ \\frac{1}{\\sqrt[n]{a^m}}, & \\text{se } r = -m/n, \\text{ con } m, n \\in \\mathbb{Z}_+. \\end{cases}$$\n\n**Dimostrazione.** Osserviamo innanzitutto che $f(0) = 1$ perché\n\n$$a = f(1) = f(0 + 1) = f(0)f(1) = f(0)a.$$ \n\nSe $m \\in \\mathbb{Z}_+$, poiché $m = 1 + \\ldots + 1$ ($m$ volte), si ha che\n\n$$f(m) = f(1 + \\ldots + 1) = f(1) \\cdots f(1) = a^m.$$ \n\nSe $r = m/n$ con $m, n \\in \\mathbb{Z}_+$, poiché $m = r + \\ldots + r$ ($n$ volte), si ha che\n\n$$a^m = f(m) = f(r + \\cdots + r) = f(r) \\cdots f(r) = f(r)^n.$$ \n\nQuindi $f(r) = \\sqrt[n]{a^m}$.\n\nSe $r = -m/n$ la conclusione segue dall’osservazione che\n\n$$1 = f(0) = f(r + (-r)) = f(r)f(-r) = f(r)\\sqrt[n]{a^{-m}}.$$ \n\nQuindi $f(r) = 1/\\sqrt[n]{a^m}$. \\qed\nIl Teorema 2.1 motiva la seguente definizione.\n\n**Definizione.** Definiamo la funzione esponenziale in base $a > 0$ su $\\mathbb{Q}$ ponendo\n\n$$a^r = \\begin{cases} \\sqrt[n]{a^m}, & \\text{se } r = m/n, \\text{ con } m, n \\in \\mathbb{Z}_+ \\\\ 1, & \\text{se } r = 0 \\\\ \\frac{1}{\\sqrt[n]{a^m}}, & \\text{se } r = -m/n, \\text{ con } m, n \\in \\mathbb{Z}_+ . \\end{cases}$$\n\nUtilizzando le proprietà della funzione radice $n$-esima è facile verificare che la funzione $r \\mapsto a^r$ così definita soddisfa la legge degli esponenti; cioè\n\n$$a^{r+s} = a^r a^s \\quad \\forall r, s \\in \\mathbb{Q}.$$\n\n**Osservazione.** La restrizione $a > 0$ è necessaria se vogliamo che la definizione sia indipendente dalla rappresentazione dei razionali come frazione. Ad esempio, $1/2 = 2/4$, ma mentre $(-3)^{2/4} = \\sqrt[4]{(-3)^2} = \\sqrt{9}$ è ben definito, $(-3)^{1/2} = \\sqrt{-3}$ non lo è.\n\nLasciamo al lettore, come esercizio, il compito di dimostrare che la funzione esponenziale su $\\mathbb{Q}$ è sempre positiva, strettamente crescente se $a > 1$, strettamente decrescente se $a < 1$, costante uguale a 1 se $a = 1$. Inoltre $(a^r)^s = a^{rs}$, per ogni coppia di numeri razionali $r$ e $s$. Se $a$ e $b$ sono due numeri reali positivi e $r \\in \\mathbb{Q}$ allora $a^r b^r = (ab)^r$.\n\nLa legge degli esponenti, da sola, non è sufficiente a caratterizzare l’estensione della funzione esponenziale su $\\mathbb{R}$. Infatti si può dimostrare che esistono infinite funzioni su $\\mathbb{R}$, che soddisfano la legge degli esponenti e che su $\\mathbb{Q}$ coincidono con la funzione $r \\mapsto a^r$ (la dimostrazione non è elementare e si basa sull’assioma della scelta, quindi è non costruttiva). Tuttavia si può ottenere l’unicità dell’estensione della funzione esponenziale imponendo una condizione di regolarità ulteriore. A questo scopo dimostriamo due risultati preliminari.\n\n**Lemma 2.2.** Siano $f$ e $g$ due funzioni continue su $\\mathbb{R}$ le cui restrizioni a $\\mathbb{Q}$ coincidono. Allora $f = g$.\n\n**Dimostrazione.** Sia $x$ un numero reale e $(r_n)$ una successione di numeri razionali che converge a $x$. Allora $f(x) = \\lim_{n \\to +\\infty} f(r_n) = \\lim_{n \\to +\\infty} g(r_n) = g(x)$. □\n\n**Lemma 2.3.** Se $a > 0$ si ha che $\\lim_{n \\to +\\infty} a^{1/n} = 1$.\n\n**Dimostrazione.** Se $a = 1$ allora $a^{1/n} = \\sqrt[n]{1} = 1$ per ogni $n$ e la conclusione è ovvia. Supponiamo ora $a > 1$. Poniamo $b_n = a^{1/n} - 1$. Allora $b_n \\geq 0$ e\n\n$$a = (1 + b_n)^n > n b_n \\quad \\forall n \\in \\mathbb{N},$$\n\nper la disuguaglianza di Bernoulli. Quindi $0 < b_n < a/n$ per ogni $n \\geq 1$. Per il teorema del confronto\n\n$$\\lim_{n} a^{1/n} - 1 = \\lim_{n} b_n = 0.$$\nQuesto prova che \\( \\lim_{n \\to +\\infty} a^{1/n} = 1 \\) se \\( a > 1 \\). Se \\( 0 < a < 1 \\) basta osservare che\n\\[\n\\lim_{n \\to +\\infty} a^{1/n} = \\lim_{n \\to +\\infty} \\left(\\frac{1}{a}\\right)^{1/n} = 1.\n\\]\n\n\\[\\square\\]\n\n**Teorema 2.4.** Sia \\( a \\) un numero reale positivo. Esiste un’unica funzione continua \\( \\exp_a \\) su \\( \\mathbb{R} \\) tale \\( \\exp_a(1) = a \\) e \\( \\exp_a(x + y) = \\exp_a(x) \\exp_a(y) \\) per ogni coppia di numeri reali \\( x \\) e \\( y \\).\n\n**Dimostrazione.** Dimostriamo prima l’unicità. Se \\( f \\) e \\( g \\) sono due funzioni che soddisfano le ipotesi del teorema, esse coincidono sui razionali, per il Teorema 2.1. Quindi \\( f = g \\) per il Lemma 2.2.\n\nDimostriamo ora l’esistenza. Se \\( a = 1 \\) osserviamo che \\( 1^r = 1 \\) per ogni \\( r \\) in \\( \\mathbb{Q} \\). Quindi la funzione costante che assume il valore 1 in tutti i punti di \\( \\mathbb{R} \\) è un prolungamento continuo della funzione esponenziale in base 1 sui razionali e soddisfa la legge degli esponenti.\n\nSupponiamo ora \\( a > 1 \\) e dimostriamo che per ogni \\( x \\) in \\( \\mathbb{R} \\) esiste\n\\[\n\\lim_{r \\to x} a^r\n\\]\nPoiché la funzione \\( r \\mapsto a^r \\) è strettamente crescente su \\( \\mathbb{Q} \\), per il teorema sul limite delle funzioni monotone, esistono\n\\[\n\\lim_{r \\to x^-} a^r = \\sup\\{a^r : r \\in \\mathbb{Q}, r < x\\},\n\\]\n(2.2)\n\\[\n\\lim_{r \\to x^+} a^r = \\inf\\{a^r : r \\in \\mathbb{Q}, x < r\\},\n\\]\ne si ha \\( \\lim_{r \\to x^-} a^r \\leq \\lim_{r \\to x^+} a^r \\). Per dimostrare che i due limiti sono uguali consideriamo due successioni di numeri razionali, \\( (r_n) \\) e \\( (s_n) \\), tali che \\( (r_n) \\) tende crescendo a \\( x \\), \\( (s_n) \\) tende decrescendo a \\( x \\) e \\( s_n - r_n < 1/n \\), per ogni \\( n \\in \\mathbb{N}_+ \\). Allora, per la (2.2),\n\\[\na^{r_n} \\leq \\lim_{r \\to x^-} a^r \\leq \\lim_{r \\to x^+} a^r \\leq a^{s_n}.\n\\]\nPertanto\n\\[\n1 \\leq \\lim_{r \\to x^+} a^r \\leq \\lim_{r \\to x^-} a^r \\leq a^{s_n - r_n} \\leq a^{1/n}.\n\\]\nPer il Lemma 2.3 e il teorema sul confronto dei limiti ne segue che \\( \\lim_{r \\to x^-} a^r = \\lim_{r \\to x^+} a^r \\). Pertanto per ogni \\( x \\) in \\( \\mathbb{R} \\) esiste \\( \\lim_{r \\to x} a^r \\). Definiamo\n\\[\n\\exp_a x = \\lim_{r \\to x} a^r.\n\\]\nLa funzione esponenziale\n\nDimostriamo ora che \\( \\exp_a r = a^r \\) per ogni \\( r \\) in \\( \\mathbb{Q} \\). Poiché \\( (r + 1/n) \\) è una successione di numeri razionali diversi da \\( r \\) che converge a \\( r \\),\n\n\\[\n\\exp_a r = \\lim_{n \\to \\infty} a^{r+1/n} = \\lim_{n \\to \\infty} a^r a^{1/n} = a^r.\n\\]\n\nQuindi la funzione \\( \\exp_a \\) è un’estensione a \\( \\mathbb{R} \\) della funzione \\( \\mathbb{Q} \\ni r \\mapsto a^r \\). In particolare \\( \\exp_a 1 = a \\). Osserviamo che, se \\( (r_n) \\) è una qualunque successione di numeri razionali che converge a \\( x \\), si ha\n\n\\[\n(2.3) \\quad \\exp_a x = \\lim_{n \\to +\\infty} a^{r_n}.\n\\]\n\nInfatti, poiché \\( \\exp_a x = \\lim_{r \\to x} a^r \\), per la caratterizzazione del limite mediante successioni, la (2.3) è certamente vera se \\( r_n \\neq x \\) per ogni \\( n \\). Inoltre la restrizione \\( r_n \\neq x \\) può essere rimossa, perché se \\( x \\) è razionale \\( \\exp_a x = a^x \\).\n\nPer dimostrare che la funzione \\( \\exp_a \\) soddisfa l’identità\n\n\\[\n\\exp_a (x + y) = \\exp_a x \\exp_a y \\quad \\forall x, y \\in \\mathbb{R},\n\\]\n\nconsideriamo due successioni di numeri razionali \\( (r_n) \\) e \\( (s_n) \\) che convergono a \\( x \\) e a \\( y \\) rispettivamente. Allora \\( (r_n + s_n) \\) è una successione di numeri razionali che converge a \\( x + y \\) e\n\n\\[\n\\exp_a (x + y) = \\lim_{n \\to +\\infty} a^{r_n+s_n} = \\lim_{n \\to +\\infty} a^{r_n} \\lim_{n \\to +\\infty} a^{s_n} = \\exp_a x \\exp_a y.\n\\]\n\nDimostriamo ora che la funzione \\( x \\mapsto \\exp_a x \\) è positiva e strettamente crescente su \\( \\mathbb{R} \\). Infatti, per la (2.2), si ha\n\n\\[\n(2.4) \\quad \\exp_a x = \\sup\\{a^r : r \\in \\mathbb{Q}, r < x\\} = \\inf\\{a^r : r \\in \\mathbb{Q}, x < r\\}.\n\\]\n\nIn particolare \\( \\exp_a x > 0 \\) per ogni \\( x \\) in \\( \\mathbb{R} \\). Inoltre, se \\( x \\) e \\( y \\) in \\( \\mathbb{R} \\) e \\( x < y \\), presi due numeri razionali \\( r \\) e \\( s \\) tali che \\( x < r < s < y \\), per la (2.4) si ha \\( \\exp_a x \\leq a^r < a^s \\leq \\exp_a y \\).\n\nInfine, per dimostrare la continuità di \\( \\exp_a \\) su \\( \\mathbb{R} \\), proviamo che \\( \\lim_{h \\to 0} \\exp_a (x + h) = \\exp_a x \\), per ogni \\( x \\) in \\( \\mathbb{R} \\). Poiché \\( \\exp_a (x + h) = \\exp_a x \\exp_a h \\), è sufficiente provare che \\( \\lim_{h \\to 0} \\exp_a h = 1 \\). Poiché la funzione \\( \\exp_a \\) è crescente esiste il limite per \\( h \\to 0^+ \\) e, per determinarlo, è sufficiente calcolarlo su una successione di numeri razionali positivi che tende a zero, ad esempio \\( (1/n) \\), per cui si ha\n\n\\[\n\\lim_{n \\to +\\infty} \\exp_a (1/n) = \\lim_{n \\to +\\infty} a^{1/n} = 1.\n\\]\n\nIn modo analogo, utilizzando la successione \\( (-1/n) \\), si dimostra che anche anche il limite di \\( \\exp_a h \\) per \\( h \\to 0^- \\) esiste e vale 1. Quindi \\( \\lim_{h \\to 0} \\exp_a h = 1 \\).\n\nIl teorema è così dimostrato quando la base \\( a \\) è maggiore o uguale a 1. Se \\( 0 < a < 1 \\) è facile verificare che la funzione \\( \\exp_a = 1/\\exp_{1/a} \\), che è ben definita perché \\( 1/a > 1 \\), soddisfa la tesi. □\nDefinizione. La funzione \\( \\exp_a \\) si dice funzione esponenziale in base \\( a \\) su \\( \\mathbb{R} \\). Se \\( a \\) è un numero reale positivo e \\( x \\) è un numero reale qualsiasi, la potenza \\( a^x \\) è, per definizione, il numero reale \\( \\exp_a x \\).\n\nIl seguente teorema riassume alcune delle proprietà della funzione esponenziale su \\( \\mathbb{R} \\).\n\nTeorema 2.5. La funzione \\( x \\mapsto a^x \\) è positiva e strettamente crescente se \\( a > 1 \\), strettamente decrescente se \\( 0 < a < 1 \\), costante uguale a 1 se \\( a = 1 \\). Inoltre valgono le identità\n\n\\[\n\\begin{align*}\n a^{x+y} &= a^x a^y \\\\\n a^{-x} &= \\frac{1}{a^x} \\\\\n (ab)^x &= a^x b^x \\\\\n (a^x)^y &= a^{xy}\n\\end{align*}\n\\]\n\nper ogni \\( x, y \\) e \\( b \\) in \\( \\mathbb{R} \\), \\( b > 0 \\).\n\nDimostrazione. Le prime proprietà, fino alle legge degli esponenti inclusa, sono già state dimostrate nel Teorema 2.4. L’identità \\( a^{-x} = 1/a^x \\) segue dalla legge degli esponenti, perché \\( a^x a^{-x} = a^0 = 1 \\). Per il Lemma 2.2, per dimostrare la terza identità è sufficiente osservare che essa vale per \\( x \\) razionale. Per dimostrare l’ultima identità cominciamo con l’osservare che essa è vera se \\( y \\) è razionale. Infatti, se \\( y = m/n \\) con \\( m, n \\) interi e \\( n > 0 \\), si ha che\n\n\\[\n(a^x)^{m/n} = \\sqrt[n]{(a^x)^m} = \\sqrt[n]{a^{xm}} = (a^{xm})^{1/n} = a^{xm/n}.\n\\]\n\nLa conclusione segue dal Lemma 2.2 perché le funzioni \\( y \\mapsto (a^x)^y \\) e \\( y \\mapsto a^{xy} \\) sono continue su \\( \\mathbb{R} \\) e coincidono su \\( \\mathbb{Q} \\). \\( \\square \\)\n\nEsaminiamo ora i limiti della funzione esponenziale all’infinito.\n\nProposizione 2.6. Se \\( a > 1 \\)\n\n\\[\n\\begin{align*}\n \\lim_{x \\to +\\infty} a^x &= +\\infty, & \\lim_{x \\to -\\infty} a^x &= 0.\n\\end{align*}\n\\]\n\nSe \\( 0 < a < 1 \\)\n\n\\[\n\\begin{align*}\n \\lim_{x \\to +\\infty} a^x &= 0, & \\lim_{x \\to -\\infty} a^x &= +\\infty.\n\\end{align*}\n\\]\n\nDimostrazione. Supponiamo che \\( a > 1 \\). Poiché la funzione è crescente i limiti esistono e si ha\n\n\\[\n\\lim_{x \\to +\\infty} a^x = \\lim_{n \\to +\\infty} a^n, \\quad \\lim_{x \\to -\\infty} a^x = \\lim_{n \\to -\\infty} a^{-n}.\n\\]\nScriviamo $a = 1 + b$, con $b > 0$. Allora, per ogni intero positivo $n$,\n\n$$a^n = (1 + b)^n \\geq 1 + nb, \\quad a^{-n} = \\frac{1}{a^n} \\leq \\frac{1}{1 + nb}.$$ \n\nPertanto $\\lim_{n \\to +\\infty} a^n = +\\infty$ e $\\lim_{n \\to +\\infty} a^{-n} = 0$ per il teorema del confronto. Questo prova la (2.5). Per dimostrare la (2.6) basta osservare che, se $0 < a < 1$, allora $1/a > 1$ e $a^x = (1/a)^{-x}$. □\n\nLa proposizione seguente, che è una generalizzazione della Proposizione 2.6, mostra che, se la base $a$ è maggiore di 1, la funzione esponenziale tende all’infinito per $x \\to \\infty$ e tende a 0 per $x \\to -\\infty$ “più rapidamente” di qualunque potenza di $x$.\n\n**Proposizione 2.7.** Sia $k \\in \\mathbb{N}_+$. Se $a > 1$\n\n(2.7) $\\lim_{x \\to +\\infty} a^x x^k = +\\infty$, $\\lim_{x \\to -\\infty} x^k a^x = 0$.\n\nSe $0 < a < 1$\n\n(2.8) $\\lim_{x \\to +\\infty} x^k a^x = 0$, $\\lim_{x \\to -\\infty} \\frac{a^x}{x^k} = \\begin{cases} +\\infty, & \\text{se } k \\text{ è pari} \\\\ -\\infty, & \\text{se } k \\text{ è dispari}. \\end{cases}$\n\n**Dimostrazione.** Supponiamo $a > 1$. Proviamo dapprima che\n\n(2.9) $\\lim_{n \\to +\\infty} \\frac{a^n}{n + 1} = +\\infty$.\n\nInfatti, scrivendo $a = 1 + b$, con $b > 0$, e utilizzando la formula del binomio di Newton, si ha\n\n$$a^n = (1 + b)^n \\geq 1 + nb + n(n - 1)b^2 \\geq n(n - 1)b^2.$$ \n\nQuindi, per il teorema del confronto,\n\n$$\\lim_{n \\to +\\infty} \\frac{a^n}{n + 1} = \\lim_{n \\to +\\infty} \\frac{a^n}{n} \\frac{n}{n + 1} \\geq \\lim_{n \\to +\\infty} (n - 1)b^2 \\lim_{n \\to +\\infty} \\frac{n}{n + 1} = +\\infty.$$ \n\nQuesto dimostra la (2.9). Pertanto, per la definizione di limite, fissato $y \\in \\mathbb{R}$ esiste un naturale $n_y$ tale che $a^n/(n + 1) > y$ se $n \\geq n_y$. Allora, se $x > n_y + 1$ e $n = \\lfloor x \\rfloor$ (la parte intera di $x$)\n\n$$\\frac{a^x}{x} > \\frac{a^n}{n + 1} > y,$$\n\ncioè\n\n(2.10) $\\lim_{x \\to +\\infty} \\frac{a^x}{x} = +\\infty$.\n\nIl primo limite in (2.7) si ottiene da (2.10) con il cambiamento di variabile $x = ky$. Infatti\n\n$$\\lim_{x \\to +\\infty} \\frac{a^x}{x^k} = \\lim_{y \\to +\\infty} \\frac{a^{ky}}{(ky)^k} = \\frac{1}{k^k} \\lim_{y \\to +\\infty} \\left(\\frac{a^y}{y}\\right)^k = +\\infty.$$\nIl secondo limite segue dal primo con il cambiamento di variabile \\( y = -x \\). I limiti per \\( 0 < a < 1 \\) si ottengono dai precedenti osservando che \\( a^x = b^{-x} \\) con \\( b = 1/a > 1 \\).\n\n**Corollario 2.8.** Se \\( a \\neq 1 \\) l’immagine della funzione esponenziale in base \\( a \\) è \\( \\mathbb{R}_+ \\).\n\n**Dimostrazione.** Sia \\( y > 0 \\) fissato. Per la Proposizione 2.6 esistono due numeri reali \\( t \\) e \\( z \\) tali che \\( a^t < y \\) e \\( a^z > y \\). Per il teorema dei valori intermedi per le funzioni continue esiste un numero reale \\( x \\) compreso tra \\( t \\) e \\( z \\) tale che \\( a^x = y \\).\n\nSe \\( a \\neq 1 \\) la funzione esponenziale in base \\( a \\) è strettamente monotona. Pertanto essa è invertibile. La sua inversa è definita su \\( \\mathbb{R}_+ \\) e ha come immagine \\( \\mathbb{R} \\).\n\n**Definizione.** L’inversa della funzione esponenziale in base \\( a \\) si dice *logaritmo in base \\( a \\)* e si denota con il simbolo \\( \\log_a \\).\n\n**Teorema 2.9.** La funzione \\( \\log_a \\) è continua su \\( \\mathbb{R}_+ \\), strettamente crescente se \\( a > 1 \\), strettamente decrescente se \\( 0 < a < 1 \\). Essa soddisfa le seguenti identità\n\n\\[\n\\log_a(xy) = \\log_a x + \\log_a y \\\\\n\\log_a x^y = y \\log_a x \\\\\n\\log_b x = \\log_b a \\cdot \\log_a x.\n\\]\n\n**Dimostrazione.** Le proprietà di continuità e di monotonia del logaritmo seguono dai teoremi di continuità e monotonia della funzione inversa. La prima identità è una conseguenza immediata della legge degli esponenti. Per dimostrarla basta applicare il \\( \\log_a \\) al primo e all’ultimo termine dell’identità \\( a^{\\log_a x + \\log_a y} = a^{\\log_a x} a^{\\log_a y} = xy \\). Per dimostrare la seconda si applica il \\( \\log_a \\) al primo e all’ultimo termine dell’identità \\( a^{y \\log_a x} = (a^{\\log_a x})^y = x^y \\). Per dimostrare l’ultima identità si applichi \\( \\log_b \\) al primo e all’ultimo termine dell’identità \\( x = (b^{\\log_b a})^{\\log_a x} = b^{\\log_b a \\log_a x} \\).\n\n**Corollario 2.10.** Siano \\( a \\) e \\( b \\) numeri reali positivi. Allora \\( b^x = a^{x \\log_a b} \\) per ogni \\( x \\) in \\( \\mathbb{R} \\).\n\n**Dimostrazione.** È una conseguenza immediata delle identità \\( a^{x \\log_a b} = (a^{\\log_a b})^x = b^x \\).\n\n**Proposizione 2.11.** Se \\( a > 1 \\)\n\n\\[\n\\lim_{x \\to 0^+} \\log_a x = -\\infty, \\quad \\lim_{x \\to +\\infty} \\log_a x = +\\infty.\n\\]\n\nSe \\( 0 < a < 1 \\)\n\n\\[\n\\lim_{x \\to 0^+} \\log_a x = +\\infty, \\quad \\lim_{x \\to +\\infty} \\log_a x = -\\infty.\n\\]\n**Dimostrazione.** Proviamo il secondo limite nel caso $a > 1$. Fissato $y \\in \\mathbb{R}$, poiché il logaritmo in base $a$ è crescente, se $x > a^y$ si ha che $\\log_a(x) > y$. Questo prova che $\\lim_{x \\to +\\infty} \\log_a x = +\\infty$. Alternativamente, possiamo usare il teorema sul limite della funzione composta:\n\n$$\\lim_{x \\to +\\infty} \\log_a(x) = \\lim_{y \\to +\\infty} \\log_a(a^y) = \\lim_{y \\to +\\infty} y = +\\infty.$$ \n\nLasciamo al lettore come esercizio il compito di provare gli altri casi. □\n\nPoiché abbiamo definito la potenza con esponente reale possiamo ora considerare, per ogni esponente reale $\\alpha$, la funzione potenza $x \\mapsto x^\\alpha$, definita per $x$ in $\\mathbb{R}_+$. \n\n**Proposizione 2.12.** La funzione potenza $x \\mapsto x^\\alpha$ è continua su $\\mathbb{R}_+$, positiva e strettamente crescente se $\\alpha > 0$, strettamente decrescente se $\\alpha < 0$.\n\n**Dimostrazione.** Sia $a$ un numero reale maggiore di 1. Allora, per il Teorema 2.9, $x^\\alpha = a^{\\alpha \\log_a x}$ è una funzione continua perché composta di funzioni continue. Se $\\alpha > 0$ è strettamente crescente perché composta di funzioni strettamente crescenti. Se $\\alpha < 0$ la funzione è strettamente decrescente, perché $x^\\alpha = 1/x^{-\\alpha}$. □\n\n### 3. Il numero $e$ di Nepero\n\nIn questo paragrafo calcoleremo alcuni limiti notevoli che sono alla base delle formule del calcolo differenziale per le funzioni esponenziali e logaritmiche. Per prima cosa definiamo il numero $e$ di Nepero, come limite di una particolare successione limitata. Siano $(a_n)$ e $(b_n)$ le successioni definite da\n\n$$a_n = \\left(1 + \\frac{1}{n}\\right)^n, \\quad b_n = \\left(1 + \\frac{1}{n}\\right)^{n+1}$$\n\n**Lemma 3.1.** La successione $(a_n)$ è crescente, la successione $(b_n)$ è decrescente, e $a_n < b_n$ per ogni intero $n \\geq 1$.\n\n**Dimostrazione.** Utilizzeremo la disuguaglianza di Bernoulli\n\n$$(1 + x)^n \\geq 1 + nx,$$\n\nvalida per ogni $n \\in \\mathbb{N}$ e per $x \\geq -1$, la cui dimostrazione si ottiene facilmente per induzione. Per provare che la successione $(a_n)$ è crescente basta dimostrare che $a_n \\geq a_{n-1}$, per ogni $n \\geq 2$. Questa disuguaglianza equivale a\n\n$$\\left(\\frac{n+1}{n}\\right)^n \\geq \\left(\\frac{n}{n-1}\\right)^n \\left(\\frac{n-1}{n}\\right),$$\ncioè ancora a\n\\[\n\\left( \\frac{n+1}{n} \\right)^n \\left( \\frac{n-1}{n} \\right)^n = \\left( 1 - \\frac{1}{n^2} \\right)^n \\geq 1 - \\frac{1}{n}\n\\]\nQuest’ultima disuguaglianza segue dalla disuguaglianza di Bernoulli, ponendo \\( x = -1/n^2 \\). Per provare la decrescenza della successione \\((b_n)\\) basta dimostrare che \\( b_n < b_{n-1} \\), per ogni \\( n \\leq 2 \\). Con ragionamenti analoghi a quelli utilizzati in precedenza si vede che questo equivale a provare la disuguaglianza\n\\[\n\\left( 1 + \\frac{1}{n^2 - 1} \\right)^n > 1 + \\frac{1}{n},\n\\]\nche segue dalla disuguaglianza di Bernoulli ponendo \\( x = 1/(n^2 - 1) \\). Infine, per provare che \\( a_n < b_n \\), basta osservare che \\( b_n = \\left( 1 + \\frac{1}{n} \\right) a_n > a_n \\). □\n\n**Proposizione 3.2.** Le successione \\((a_n)\\) e \\((b_n)\\) convergono allo stesso limite.\n\n**Dimostrazione.** Poiché \\( a_1 \\leq a_n < b_n \\leq b_1 \\), le due successioni sono limitate. Quindi esse convergono a dei limiti finiti per il teorema sul limite delle successioni monotone. I limiti coincidono perché \\( b_n = \\left( 1 + \\frac{1}{n} \\right) a_n \\). □\n\nIl limite delle due successioni \\((a_n)\\) e \\((b_n)\\) si dice **numero di Nepero o di Eulero** e si denota con la lettera \\( e \\). Successive approssimazione del numero \\( e \\) si possono ottenere utilizzando le stime \\( a_n < e < b_n \\) per ogni \\( n \\geq 1 \\). Una approssimazione di \\( e \\) con 12 cifre decimali è 2.718281828459. Il numero \\( e \\) è irrazionale. La dimostrazione di questo fatto verrà data nella sezione 4. Si può dimostrare anche che il numero \\( e \\) non è algebrico, cioè non è radice di alcun polinomio a coefficienti razionali. Per questo motivo si dice che \\( e \\) è un numero **trascendente**. Se si sceglie \\( e \\) come base per i logaritmi si ottengono i **logaritmi Neperiani o naturali**, così detti perché la scelta di \\( e \\) come base porta a formule particolarmente semplici nel calcolo differenziale. La funzione logaritmo naturale si denota con il simbolo \\( \\ln \\) o anche \\( \\log \\), omettendo l’indicazione della base. Utilizzeremo ora la formula\n\n\\[\n(3.1) \\quad \\lim_{n \\to +\\infty} \\left( 1 + \\frac{1}{n} \\right)^n = \\lim_{n \\to +\\infty} \\left( 1 + \\frac{1}{n} \\right)^{n+1} = e\n\\]\n\nper calcolare alcuni limiti notevoli. Consideriamo la funzione \\( f(x) = \\left( 1 + \\frac{1}{x} \\right)^x \\), definita quando la base è positiva, cioè per \\( x \\in (-\\infty, -1) \\cup (0, +\\infty) \\).\n\n**Proposizione 3.3.** Si ha\n\\[\n\\lim_{x \\to -\\infty} \\left( 1 + \\frac{1}{x} \\right)^x = \\lim_{x \\to +\\infty} \\left( 1 + \\frac{1}{x} \\right)^x = e.\n\\]\nDimostrazione. Consideriamo dapprima il limite a $+\\infty$. Se $n \\leq x < n + 1$ e $n > 0$ si ha\n\n$$1 + \\frac{1}{n + 1} < 1 + \\frac{1}{x} \\leq 1 + \\frac{1}{n}.$$ \n\nPer le proprietà di monotonia delle potenze sia rispetto alla base che rispetto all’esponente, si ha\n\n$$\\left(1 + \\frac{1}{n + 1}\\right)^n < \\left(1 + \\frac{1}{x}\\right)^n \\leq \\left(1 + \\frac{1}{n}\\right)^x \\leq \\left(1 + \\frac{1}{n}\\right)^{n+1}.$$ \n\nIl primo e l’ultimo termine in questa catena di disuguaglianze tendono a $e$, per la (3.1). Infatti\n\n$$\\lim_{n \\to +\\infty} \\left(1 + \\frac{1}{n + 1}\\right)^n = \\lim_{k \\to +\\infty} \\left(1 + \\frac{1}{k}\\right)^k \\left(1 + \\frac{1}{k}\\right)^{-1} = e,$$\n\n$$\\lim_{n \\to +\\infty} \\left(1 + \\frac{1}{n}\\right)^{n+1} = \\lim_{n \\to +\\infty} \\left(1 + \\frac{1}{n}\\right)^n \\left(1 + \\frac{1}{n}\\right) = e.$$ \n\nPertanto, per il teorema del confronto dei limiti,\n\n$$\\lim_{x \\to +\\infty} \\left(1 + \\frac{1}{x}\\right)^x = e.$$ \n\nPer calcolare il limite a $-\\infty$ basta effettuare la sostituzione $x = -t - 1$. Si ottiene così\n\n$$\\lim_{x \\to -\\infty} \\left(1 + \\frac{1}{x}\\right)^x = \\lim_{t \\to +\\infty} \\left(1 + \\frac{1}{t}\\right) \\left(1 + \\frac{1}{t}\\right)^t = e.$$ \n\nI seguenti limiti notevoli giocheranno un ruolo importante nel calcolo differenziale del logaritmo e dell’esponenziale.\n\nProposizione 3.4. Si ha\n\n$$\\lim_{x \\to 0} \\frac{\\ln(1 + x)}{x} = 1$$\n\n$$\\lim_{x \\to 0} \\frac{e^x - 1}{x} = 1.$$ \n\nDimostrazione. Effettuando il cambiamento di variabile $x = 1/y$, si ottiene che\n\n$$\\lim_{x \\to 0^+} \\frac{\\ln(1 + x)}{x} = \\lim_{x \\to 0^+} \\ln(1 + x)^{1/x}$$\n\n$$= \\lim_{y \\to +\\infty} \\ln(1 + \\frac{1}{y})^y$$\n\n$$= \\ln e = 1.$$\nIl limite per $x \\to 0^-$ si calcola in modo simile. Il secondo limite si riconduce al primo, mediante il cambiamento di variabile $e^x - 1 = y$.\n\n**Osservazione.** Poiché i limiti notevoli dell’esponenziale e del logaritmo assumono una forma più semplice quando la base è $e$, in analisi si preferisce lavorare quasi sempre con questa base, riconducendo l’esponenziale e il logaritmo in base diversa a quelli in base $e$ mediante le formule\n\n$$a^x = e^{x \\ln a}, \\quad \\log_a x = \\ln x / \\ln a.$$ \n\n**Corollario 3.5.** Sia $a > 0$, $a \\neq 1$. Allora\n\n$$\\lim_{x \\to 0} \\frac{\\log_a (1 + x)}{x} = \\frac{1}{\\log a},$$\n\n$$\\lim_{x \\to 0} \\frac{a^x - 1}{x} = \\log a.$$ \n\n**Dimostrazione.** Utilizzando la formula di cambiamento di base\n\n$$\\log_a (1 + x) = \\frac{\\ln(1 + x)}{\\ln a},$$\n\nsi ottiene che\n\n$$\\lim_{x \\to 0^+} \\frac{\\log_a (1 + x)}{x} = \\frac{1}{\\ln a} \\lim_{x \\to 0^+} \\frac{\\ln(1 + x)}{x} = \\frac{1}{\\ln a}.$$ \n\nPer calcolare il secondo limite si usa la formula $a^x = e^{x \\ln a}$ e il cambiamento di variabile $y = x \\ln a$.\n\n4. **L’irrazionalità di $e$**\n\nIn questo paragrafo dimostriamo che il numero $e$ è irrazionale utilizzando la formula di Taylor dell’esponenziale.\n\n**Teorema 4.1.** Il numero $e$ è irrazionale.\n\n**Dimostrazione.** Sia $T_n$ il polinomio di Mac Laurin di ordine $n$ della funzione $x \\mapsto e^x$. Dalla formula del resto di Lagrange abbiamo che esiste un numero reale $c$, $0 < c < 1$, tale che\n\n$$e - T_n(1) = \\frac{e^c}{(n + 1)!}.$$ \n\nPoiché $1 < e^c < e < 3$, si ha che\n\n$$0 < e - \\sum_{k=0}^{n} \\frac{1}{k!} < \\frac{3}{(n + 1)!}.$$\nMoltiplicando per $n!$ si ottiene\n\n$$0 < en! - n! \\sum_{k=0}^{n} \\frac{1}{k!} < \\frac{3}{(n+1)}.$$ \n\nSe $e$ fosse razionale, cioè $e = p/q$ con $p$ e $q$ in $N_+$, scegliendo per $n$ un multiplo di $q$ maggiore di 3, si avrebbe che il numero intero $en! - n! \\sum_{k=0}^{n} \\frac{1}{k!}$ sarebbe minore di $3/4$. Questo è assurdo. □\n\n**Osservazione.** La dimostrazione del Teorema 4.1 fornisce un'altra espressione del numero $e$. Infatti si ha che\n\n$$e = \\lim_{n \\to +\\infty} \\sum_{k=0}^{n} \\frac{1}{k!}.$$ \n\nL'errore commesso approssimando $e$ mediante questa formula tende a zero molto rapidamente al crescere di $n$.\n\n5. **L'esponenziale e il logaritmo nel campo complesso**\n\nVogliamo ora estendere la funzione esponenziale al campo dei numeri complessi $\\mathbb{C}$ in modo tale che che continui a valere la legge degli esponenti $a^{z+w} = a^z a^w$ per ogni coppia di numeri complessi $z$ e $w$. Naturalmente vogliamo che la restrizione dell’esponenziale ai numeri reali coincida con la funzione $x \\mapsto a^x$ già definita. Consideriamo dapprima il caso in cui la base è $e$. La definizione di $e^z$ che daremo è motivata dalle seguenti considerazioni euristiche.\n\nSe $z = x + iy$, con $x$ e $y$ reali, per la legge degli esponenti dovremo avere che\n\n$$e^{x+iy} = e^x e^{iy}.$$ \n\nPoiché $e^x$ è già definito per $x$ reale, è sufficiente definire l’esponenziale di un numero immaginario puro $iy$. Imponiamo la condizione che la funzione $y \\mapsto e^{iy}$ soddisfi l’identità\n\n$$\\frac{d}{dy} e^{iy} = i e^{iy}. \\tag{5.1}$$ \n\nÈ immediato verificare che la funzione $f(y) = \\cos y + i \\sin(y)$ soddisfa l’equazione\n\n$$f' = if \\tag{5.2}$$ \n\nInfatti\n\n$$f'(y) = -\\sin y + i \\cos y = i(\\cos y + i \\sin y) = if(y).$$ \n\nInoltre $f(0) = 1 = e^{iy}$. Essa è anche l’unica funzione che soddisfi queste due condizioni, come mostra il lemma seguente.\n\n\\footnote{Si può dimostrare che una funzione continua $f : \\mathbb{R} \\to \\mathbb{C}$ soddisfa le legge degli esponenti se e solo è derivabile ed esiste un numero complesso $w$ tale tale che $f'(y) = w f(y)$, per ogni $y$ in $\\mathbb{R}$. Pertanto è naturale richiedere che la funzione $y \\mapsto e^{iy}$ soddisfi la (5.1).}\nLemma 5.1. Sia $g : \\mathbb{R} \\to \\mathbb{C}$ una funzione derivabile che soddisfa l’equazione (5.2) e la condizione iniziale $g(0) = 1$. Allora $g(y) = \\cos y + i \\sin y$.\n\nDimostrazione. Questo risultato è una conseguenza del teorema generale di unicità della soluzione delle equazioni differenziali lineari a coefficienti costanti. Tuttavia ne diamo una dimostrazione semplice, che non utilizza la teoria delle equazioni differenziali. Consideriamo il rapporto $g(y)/f(y)$. (Possiamo dividere per $f(y)$, perché $|f(y)|^2 = \\cos^2 y + \\sin^2 y = 1 \\neq 0$). Allora\n\n$$\\frac{d}{dy} \\frac{g(y)}{f(y)} = \\frac{f(y)g'(y) - g(y)f'(y)}{f(y)^2} = 0.$$ \n\nQuindi $g(y) = cf(y)$ e $c = 1$ perché $g(0) = f(0) = 1$. □\n\nLa discussione precedente motiva la seguente definizione\n\nDefinizione. L’esponenziale del numero complesso $z = x + iy$, con $x$ e $y$ reali, è definito da\n\n$$e^z = e^x (\\cos y + i \\sin y).$$\n\nProposizione 5.2. La funzione esponenziale complessa gode delle seguenti proprietà:\n\n(5.3) $e^{z+w} = e^z e^w$\n(5.4) $e^{-z} = \\frac{1}{e^z}$\n(5.5) $e^z = \\overline{e^z}$\n(5.6) $|e^z| = e^{\\Re z}$\n(5.7) $e^{z+2\\pi i} = e^z$,\n\nper ogni $z$ e $w$ in $\\mathbb{C}$. Inoltre la funzione $x \\mapsto e^{zx}$ è derivabile su $\\mathbb{R}$ e\n\n$$\\frac{d}{dz} e^{zx} = ze^{zx}.$$ \n\nLa verifica è semplice e viene lasciata la lettore. Se $z = x + iy$ è un numero complesso la cui rappresentazione polare è $z = r(\\cos \\theta + i \\sin \\theta)$, con $r = |z|$ e $\\theta \\in \\mathbb{R}$, possiamo scrivere $z = re^{i\\theta}$. Si noti che un numero complesso ha modulo 1 se e solo se esso ha la forma $z = e^{i\\theta}$, per qualche $\\theta \\in \\mathbb{R}$. L’argomento $\\theta$ è individuato a meno di multipli interi di $2\\pi$. Si chiama argomento principale del numero complesso $z$ la funzione $\\text{Arg} : \\mathbb{C} \\setminus \\{0\\} \\to (-\\pi, \\pi]$ che associa ad ad ogni numero complesso $z \\neq 0$ il suo argomento compreso tra $-\\pi$ e $\\pi$.\n\nConsideriamo ora la funzione esponenziale come funzione definita su $\\mathbb{C}$ a valori in $\\mathbb{C}$. \nProposizione 5.3. La funzione esponenziale ha come immagine \\( \\mathbb{C} \\setminus \\{0\\} \\). Essa è periodica di periodo \\( 2\\pi i \\). La sua restrizione alla striscia orizzontale\n\\[\nS = \\{ z : -\\pi < \\Im z \\leq \\pi \\}\n\\]\ne iniettiva.\n\nDimostrazione. Poiché \\( |e^z| = e^{\\Re z} > 0 \\) per ogni numero complesso \\( z \\), 0 non appartiene all’immagine dell’esponenziale. Sia \\( w \\in \\mathbb{C} \\setminus \\{0\\} \\) di modulo \\( r \\neq 0 \\) e argomento \\( \\theta \\). Allora, se \\( z = \\log r + i\\theta \\), si ha \\( e^z = re^{i\\theta} = w \\). Questo dimostra che l’immagine dell’esponenziale è \\( \\mathbb{C} \\setminus \\{0\\} \\). La (5.7) mostra che l’esponenziale è periodica di periodo \\( 2\\pi i \\). Per concludere la dimostrazione del teorema basta provare che essa è iniettiva sulla striscia \\( S \\). A questo scopo osserviamo che la controimmagine di 1 è \\( \\{2k\\pi i : k \\in \\mathbb{Z}\\} \\). Infatti se \\( e^z = 1 \\), con \\( z = x + iy \\), si ha \\( e^x = |e^z| = 1 \\) e \\( \\cos y + i\\sin y = 1 \\). Quindi \\( x = 0 \\) e \\( y = 2k\\pi \\), \\( k \\in \\mathbb{Z} \\). Ora, se \\( e^z = e^w \\), si ha \\( e^{z-w} = 1 \\) e quindi \\( z - w = 2k\\pi i \\), \\( k \\in \\mathbb{Z} \\). La costante \\( k \\) è 0 se \\( z \\) e \\( w \\in S \\). □\n\nLa restrizione della funzione \\( \\exp : S \\to \\mathbb{C} \\) è quindi invertibile.\n\nDefinizione. Si chiama determinazione principale del logaritmo complesso la funzione \\( \\log : \\mathbb{C} \\setminus \\{0\\} \\to S \\), inversa della restrizione dell’esponenziale alla striscia \\( S \\).\n\nProposizione 5.4. La determinazione principale del logaritmo soddisfa l’identità\n\\[\n\\log z = \\log |z| + i \\Arg z,\n\\]\nper ogni \\( z \\neq 0 \\).\n\nSostituendo la striscia \\( S \\) con una qualunque striscia orizzontale di ampiezza \\( 2\\pi \\) si ottengono altre determinazioni del logaritmo complesso.\n\nProposizione 5.5. Se \\( z \\) e \\( w \\) sono due numeri complessi tali che \\( zw \\neq 0 \\), allora\n\\[\n\\log(zw) = \\log z + \\log w + 2\\pi i \\ n(z, w),\n\\]\ndove\n\\[\n(5.8) \\quad n(z, w) = \\begin{cases} \n0 & \\text{se } -\\pi < \\Arg(z) + \\Arg(w) \\leq \\pi, \\\\\n1 & \\text{se } -2\\pi < \\Arg(z) + \\Arg(w) \\leq -\\pi, \\\\\n-1 & \\text{se } \\pi < \\Arg(z) + \\Arg(w) \\leq 2\\pi,\n\\end{cases}\n\\]\n\nDimostrazione. Scriviamo \\( z = re^{i\\theta} \\), \\( w = \\rho e^{i\\phi} \\), con \\( \\theta = \\Arg z \\) e \\( \\phi = \\Arg w \\). Allora \\( zw = r\\rho e^{i(\\theta+\\phi)} \\), con \\( -2\\pi < \\theta + \\phi \\leq 2\\pi \\). Pertanto, se \\( n(z, w) \\) è l’intero definito da (5.8), \\( \\Arg(zw) = \\Arg z + \\Arg w + 2\\pi n(z, w) \\). La conclusione segue dalla Proposizione 5.4. □\nPossiamo ora definire le potenze complesse di un numero complesso diverso da 0.\n\n**Definizione.** Se $w$ e $z$ sono due numeri complessi e $w \\neq 0$, definiamo\n\n$$w^z = e^{z \\log w}.$$ \n\nIl lettore può facilmente verificare che le potenze complesse soddisfano le seguenti regole.\n\n**Proposizione 5.6.** Siano $w$, $z$ e $\\zeta$ numeri complessi. Allora\n\n$$w^{z+\\zeta} = w^z w^\\zeta, \\quad \\text{se } w \\neq 0,$$\n\n$$(z\\zeta)^w = z^w \\zeta^w e^{2\\pi i w n(z,\\zeta)}, \\quad \\text{se } z\\zeta \\neq 0,$$\n\ndove $n(z,\\zeta)$ è l’intero definito nella Proposizione 5.5.\n\n6. **Le funzioni trigonometriche nel campo complesso**\n\nL’identità $e^{i\\theta} = \\cos \\theta + i \\sin \\theta$ stabilisce una relazione tra l’esponenziale e le funzioni trigonometriche. Considerando separatamente la parte reale e la parte immaginaria di $e^{i\\theta}$ si ottengono le sorprendenti identità di Eulero\n\n$$\\cos \\theta = \\frac{e^{i\\theta} + e^{-i\\theta}}{2},$$\n\n$$\\sin \\theta = \\frac{e^{i\\theta} - e^{-i\\theta}}{2i}.$$ \n\nQueste identità sono del tutto inaspettate a partire dalla definizione geometrica di seno e coseno. Si notino l’analogia e le differenze con le relazioni corrispondenti per le funzioni iperboliche\n\n$$\\cosh x = \\frac{e^x + e^{-x}}{2},$$\n\n$$\\sinh x = \\frac{e^x - e^{-x}}{2}.$$ \n\nLe identità di Eulero possono essere utilizzate per definire le funzioni trigonometriche nel campo complesso.\n\n**Definizione.** Dato un numero complesso $z$, definiamo\n\n$$\\cos z = \\frac{e^{iz} + e^{-iz}}{2},$$\n\n$$\\sin z = \\frac{e^{iz} - e^{-iz}}{2i}.$$\nProposizione 6.1. Se $z = x + iy$, $x, y \\in \\mathbb{R}$, si ha\n\n\\begin{align*}\n\\cos z &= \\cos x \\cosh y - i \\sin x \\sinh y \\\\\n\\sin z &= \\sin x \\cosh y + i \\cos x \\sinh y.\n\\end{align*}\n\nLasciamo al lettore la cura di dimostrare la Proposizione 6.1 utilizzando la definizione di $\\cos z$, $\\sin z$ e dell’esponenziale complessa.\n\nLe altre funzioni trigonometriche (tangente, secante, cosecante, ...) si definiscono mediante il seno e il coseno. Per concludere osserviamo che valgono le seguenti identità\n\n\\begin{align*}\n\\cos ix &= \\cosh x \\\\\n\\sin ix &= i \\sinh x.\n\\end{align*}", "id": "./materials/11.pdf" }, { "contents": "The objective of this question is to calculate the volume of solid generated by revolution of a planar region. Before proceeding into the solution, it is advised to check the theoretical part behind it.\n\n\\[ y = x^2 \\] is a upward facing parabola with vertex (0,0).\n\n\\[ x = 2 \\] is a straight line.\n\nThe straight line \\( x = 2 \\) intersects the curve \\( y = x^2 \\) on (2, 4).\n\nAccording to the question, we are supposed to revolve the region around the x-axis. On Revolving around the \\( x- \\) axis, a solid of revolution is obtained.\nRemember that, the volume of the solid of revolution formed by revolving region around the $x$-axis is given by,\n\n$$V = \\pi \\int_a^b f^2(x) - g^2(x) \\, dx,$$\n\nwhere $f(x)$ is the upper curve and $g(x)$ is the lower curve and $x \\in [a, b]$.\n\nIn this case, the upper function is $f(x) = x^2$ and lower function is $g(x) = 0$ and $x \\in [0, 2]$.\n\n$$V = \\pi \\int_0^2 x^4 \\, dx$$\n\n$$= \\pi \\left[ \\frac{x^5}{5} \\right]_0^2$$\n\n$$= \\pi \\cdot \\left( \\frac{2^5}{5} \\right)$$\n\n$$= \\frac{32\\pi}{5} \\text{ cubic units}$$", "id": "./materials/110.pdf" }, { "contents": "The objective of this question is to calculate the volume of solid generated by revolution of a planar region. As a hint, the region bounded by the curves and the solid of revolution is given below:\n\nAccording to the question, we are supposed to revolve the region around the x-axis. On Revolving around the x-axis, a solid of revolution is obtained.", "id": "./materials/111.pdf" }, { "contents": "The objective of this question is to calculate the volume of solid generated by revolution of a planar region. As a hint, the region bounded by the curves and the solid of revolution is given below:\n\nAccording to the question, we are supposed to revolve the region around the x-axis. On Revolving around the x-axis, a solid of revolution is obtained.", "id": "./materials/112.pdf" }, { "contents": "The objective of this question is to calculate the volume of solid generated by revolution of a planar region. Before proceeding into the solution, it is advised to check the theoretical part behind it.\n\n\\[ x = \\sqrt{4 - y} \\] is a downward facing parabola on positive x-axis with vertex (0,4).\n\n\\[ y = 4 \\] and \\[ y = 1 \\] are a straight line.\n\nAccording to the question, we are supposed to revolve the region around the y-axis. On Revolving around the y-axis, a solid of revolution is obtained.\nRemember that, the volume of the solid of revolution formed by revolving region around the $y$-axis is given by,\n\n$$V = \\pi \\int_{a}^{b} f^2(y) - g^2(y) \\, dy,$$\n\nwhere $f(y)$ is the curve on the right side and $g(y)$ is the curve on the left side and $y \\in [a, b]$.\n\nIn this case, the function on the right side is $f(y) = \\sqrt{4 - y}$ and function on the left side is $g(y) = 0$ and $y \\in [1, 4]$.\n\n$$V = \\pi \\int_{1}^{4} (\\sqrt{4 - y})^2 \\, dy$$\n\n$$= \\pi \\int_{1}^{4} (4 - y) \\, dy$$\n\n$$= \\pi \\left[ 4y - \\frac{y^2}{2} \\right]_{1}^{4}$$\n\n$$= \\frac{9\\pi}{2} \\text{ cubic units}$$", "id": "./materials/113.pdf" }, { "contents": "The objective of this question is to calculate the volume of solid generated by revolution of a planar region. Before proceeding into the solution, it is advised to check the theoretical part behind it.\n\n\\[ x = \\sqrt{4 - y} \\] is a downward facing parabola on positive x-axis with vertex (0,4).\n\n\\[ y = 4 \\] and \\[ y = 1 \\] are a straight line.\n\nAccording to the question, we are supposed to revolve the region around the y-axis. On Revolving around the y-axis, a solid of revolution is obtained.\nRemember that, the volume of the solid of revolution formed by revolving region around the $y$-axis is given by,\n\n$$V = \\pi \\int_{a}^{b} f^2(y) - g^2(y) \\, dy,$$\n\nwhere $f(y)$ is the curve on the right side and $g(y)$ is the curve on the left side and $y \\in [a, b]$.\n\nIn this case, the function on the right side is $f(y) = \\sqrt{4 - y}$ and function on the left side is $g(y) = 0$ and $y \\in [1, 4]$.\n\n$$V = \\pi \\int_{1}^{4} (\\sqrt{4 - y})^2 \\, dy$$\n\n$$= \\pi \\int_{1}^{4} (4 - y) \\, dy$$\n\n$$= \\pi \\left[ 4y - \\frac{y^2}{2} \\right]_{1}^{4}$$\n\n$$= \\frac{9\\pi}{2} \\text{ cubic units}$$", "id": "./materials/114.pdf" }, { "contents": "The objective of this question is to calculate the volume of solid generated by revolution of a planar region. As a hint, the region bounded by the curves and the solid of revolution is given below:\n\nAccording to the question, we are supposed to revolve the region around the x-axis. On Revolving around the x-axis, a solid of revolution is obtained.", "id": "./materials/115.pdf" }, { "contents": "Now, \\( f(x) = 8 - x \\)\n\n\\[ g(x) = 3 \\]\n\nNow,\n\n\\[\n\\text{Volume of revolution (V)} = \\pi \\int_{0}^{3} \\left( f(x)^2 - g(x)^2 \\right) \\, dx\n\\]\n\n\\[\n= \\pi \\int_{0}^{3} (8-x)^2 - (3)^2 \\, dx\n\\]\n\n\\[\n= \\pi \\int_{0}^{3} (8-x)^2 \\, dx - \\pi \\int_{0}^{3} 9 \\, dx\n\\]\n\n\\[\n= \\pi \\left[ \\frac{(8-x)^3}{3} \\right]_{0}^{3} - \\pi \\left[ 9x \\right]_{0}^{3}\n\\]\n\n\\[\n= -\\pi \\left[ \\frac{(8-x)^3}{3} \\right]_{0}^{3} - \\pi \\left[ 9x \\right]_{0}^{3}\n\\]\n\n\\[\n= -\\pi \\left( \\frac{125}{3} - \\frac{512}{3} \\right)\n\\]\n\n\\[\n= -\\pi \\times 27\n\\]\n\n\\[\n= 129\\pi - 27\\pi\n\\]\n\n\\[\n= 102\\pi \\text{ cubic units}\n\\]\n* Remember that, when we the revolution of axis is through a line, the distance is measured from that line.", "id": "./materials/116.pdf" }, { "contents": "The objective of this question is to calculate the volume of solid generated by revolution of a planar region. As a hint, the region bounded by the curves and the solid of revolution is given below:\n\nAccording to the question, we are supposed to revolve the region around the $y$-axis. On Revolving around the $y$— axis, a solid of revolution is obtained.", "id": "./materials/117.pdf" }, { "contents": "The objective of this question is to calculate the volume of solid generated by revolution of a planar region. As a hint, the region bounded by the curves and the solid of revolution is given below:\n\nAccording to the question, we are supposed to revolve the region around the x-axis. On Revolving around the x-axis, a solid of revolution is obtained.\nRemember that, the volume of the solid of revolution formed by revolving the region around the $x$-axis is given by,\n\n$$V = \\pi \\int_{a}^{b} f^2(x) - g^2(x) \\, dx,$$\n\nwhere $f(x)$ is the upper curve and $g(x)$ is the lower curve and $x \\in [a, b]$.\n\nIn this case, the upper function is $f(x) = 2 - \\frac{x}{2}$ and the lower function is $g(x) = 0$ and $x \\in [1, 2]$. ", "id": "./materials/118.pdf" }, { "contents": "The objective of this question is to calculate the volume of solid generated by revolution of a planar region. As a hint, the region bounded by the curves and the solid of revolution is given below:\n\nAccording to the question, we are supposed to revolve the region around the $y$-axis. On Revolving around the $y$– axis, a solid of revolution is obtained.\nRemember that, the volume of the solid of revolution formed by revolving the region around the $y$-axis is given by,\n\n$$V = \\pi \\int_a^b f^2(y) - g^2(y) \\, dy,$$\n\nwhere $f(y)$ is the curve on the right side and $g(y)$ is the curve on the left side and $y \\in [a, b]$.\n\nIn this case, the function on the right side is $f(y) = \\sqrt{y}$ and the function on the left side is $g(y) = \\frac{y}{4}$ and $y \\in [0, 16]$. ", "id": "./materials/119.pdf" }, { "contents": "**Trigonometric Functions**\n\n- **Cosine**\n - Definition: \\( \\cos(x) = \\frac{\\sin(x)}{\\sin(x)} \\)\n - \\( \\cos(x) = \\frac{\\cos(x)}{\\sin(x)} \\)\n\n- **Sine**\n - \\( \\sin(x) \\)\n\n- **Circle**\n - Center: \\((0,0)\\)\n - Radius: \\(1\\)\nBasic properties\n\n1) \\( \\cos^2(\\alpha) + \\sin^2(\\alpha) = 1 \\) (Pythagorean's thm)\n\n2) \\( \\cos(\\alpha + \\beta) = \\cos(\\alpha) \\cos(\\beta) - \\sin(\\alpha) \\sin(\\beta) \\)\n\n2') \\( \\sin(\\alpha + \\beta) = \\sin(\\alpha) \\cos(\\beta) + \\sin(\\beta) \\cos(\\alpha) \\)\n\n3) \\( \\sin(\\alpha + 2\\pi) = \\sin(\\alpha) \\), \\( \\cos(\\alpha + 2\\pi) = \\cos(\\alpha) \\)\n\n4) (Analytic property) if \\( 0 < \\alpha < \\frac{\\pi}{2} \\)\n\n\\[ 0 < \\sin(\\alpha) < \\alpha < \\tan(\\alpha) \\]\n\\[ \\alpha \\text{ is the length of the blue arc. (property)} \\]\n\n\\[ 2 \\cdot PH = PP' < \\text{length of the arc } \\overline{PAp} \\]\n\n\\[ 2 \\alpha \\]\n\n\\[ \\sin(\\alpha) = PH = \\frac{PP'}{2} < \\frac{2\\alpha}{2} = \\alpha \\]\nPOA is similar to TOA. In particular\n\n\\[ \\frac{PH}{OH} = \\frac{TA}{OA} \\]\n\n\\[ TA = \\frac{PH \\cdot OA}{OH} = \\frac{\\sin(x)}{\\cos(x)} \\cdot 1 = \\frac{f(x)}{g(x)} \\]\n\nArea of the whole circle = \\( \\pi r^2 = \\pi \\)\n\nArea of \\( \\text{POPA} \\) = Area of the circle \\( \\frac{2a}{2\\pi} = \\frac{2a}{2\\pi} \\)\n\nArea of \\( \\text{TOT} \\) = \\( \\frac{1}{2} \\cdot TT \\cdot AO = \\frac{1}{2} \\cdot 2 \\cdot f(x) \\cdot 1 = f(x) \\)\n\nBut Area of \\( \\text{TOT} \\) = \\( \\frac{1}{2} \\cdot TT \\cdot AO = \\frac{1}{2} \\cdot 2 \\cdot f(x) \\cdot 1 = f(x) \\)\n\\[ y(x) > x. \\]\n\n**Exercises**\n\n\\[\n\\begin{align*}\n\\text{Given} & \\quad \\cos\\left(\\frac{\\pi}{2}\\right) = 0 \\quad \\sin\\left(\\frac{\\pi}{2}\\right) = 1 \\\\\n& \\quad \\cos(\\pi) = -1 \\quad \\sin(\\pi) = 0\n\\end{align*}\n\\]\n\nProve:\n\n\\[\n\\begin{align*}\n\\cos(\\pi + x) &= -\\cos(x) \\\\\n\\sin(\\pi + x) &= -\\sin(x) \\\\\n\\cos\\left(\\frac{\\pi}{2} - x\\right) &= \\sin(x) \\\\\n\\cos(2x) &= \\cos^2(x) - \\sin^2(x) \\\\\n\\sin(2x) &= 2\\sin(x)\\cos(x)\n\\end{align*}\n\\]\n\n\\[\n\\begin{align*}\n\\sin(x) + \\sin(y) &= 2\\sin\\left(\\frac{x+y}{2}\\right)\\cos\\left(\\frac{x-y}{2}\\right) \\\\\n\\cos(x) + \\cos(y) &= 2\\cos\\left(\\frac{x+y}{2}\\right)\\cos\\left(\\frac{x-y}{2}\\right) \\\\\n\\end{align*}\n\\]\n\n\\[ (x) + y \\quad \\alpha = \\frac{x+y}{2}, \\quad \\beta = \\pm \\left(\\frac{x-y}{2}\\right) \\]\n\\[ \\cos(3x) = 4 \\cos^3(x) - 3 \\cos(x) \\]\n\n\\[ \\tan(2x) = \\frac{2 \\tan(x)}{1 - \\tan^2(x)} \\]\n\n\\[ \\cos(\\pi + x) = ? \\]\n\n\\[ \\cos(\\alpha + \\beta) = \\cos \\alpha \\cos \\beta - \\sin \\alpha \\sin \\beta \\]\n\n\\[ \\alpha = \\pi \\]\n\n\\[ \\beta = x \\]\n\n\\[ \\cos(\\pi + x) = \\cos \\pi \\cdot \\cos x - \\sin \\pi \\cdot \\sin x \\]\n\n\\[ = (-1) \\cdot \\cos x - 0 \\cdot \\sin x = -\\cos(x) \\]\n\\[ \\cos \\left( \\frac{\\pi}{2} - x \\right) \\]\n\n\\[ \\cos (\\alpha + \\beta) = \\cos \\alpha \\cos \\beta - \\sin \\alpha \\sin \\beta \\]\n\n\\[ \\alpha = \\frac{\\pi}{2} \\]\n\n\\[ \\beta = -x \\]\n\n\\[ \\cos \\left( \\frac{\\pi}{2} - x \\right) = \\cos \\left( \\frac{\\pi}{2} + (-x) \\right) = \\]\n\n\\[ = \\cos \\left( \\frac{\\pi}{2} \\right) \\cdot \\cos (-x) - \\sin \\left( \\frac{\\pi}{2} \\right) \\sin (-x) \\]\n\n\\[ = 0 \\cdot \\cos (-x) - 1 \\cdot \\sin (-x) \\]\n\n\\[ = - \\sin (-x) = - \\left( - \\sin (x) \\right) = \\sin (x) \\]\n\n[Another route: use formula for difference of \\( \\cos \\)]\n\\[ p = (\\cos \\alpha, \\sin \\alpha) \\]\n\nsince \\( p \\) is the symm. w.r.t. x-axis\n\n\\[ p' = (\\cos \\alpha, -\\sin \\alpha) = (\\cos(-\\alpha), \\sin(-\\alpha)) \\]\n\ndef. of \\( \\cos, \\sin \\)\n\n\\[ \\cos(-\\alpha) = \\cos(\\alpha), \\quad \\sin(-\\alpha) = -\\sin(\\alpha) \\]\n\n**Remark.**\n\n- If a function \\( f \\) has the property that \\( f(x) = f(-x) \\) \\( \\forall x \\in \\mathbb{R} \\) \\( \\Rightarrow \\) \\( f \\) is **EVEN**\n\n- If \\( f(-x) = -f(x) \\) \\( \\forall x \\in \\mathbb{R} \\) \\( \\Rightarrow \\) \\( f \\) is **ODD**\n\\[ \\cos(3x) = \\cos(2x + x) = \\]\n\\[ = \\cos(2x) \\cos(x) - \\sin(2x) \\sin(x) \\]\n\\[ = \\left[ \\cos^2(x) - \\sin^2(x) \\right] \\cos(x) - \\left[ 2 \\cdot \\cos(x) \\cdot \\sin(x) \\right] \\sin(x) \\]\n\\[ = \\cos^3(x) - \\sin^2(x) \\cdot \\cos(x) - 2 \\cdot \\sin^2(x) \\cdot \\cos(x) \\]\n\\[ = \\cos^3(x) - 3 \\sin^2(x) \\cdot \\cos(x) \\]\n\\[ = \\cos^3(x) - 3 \\cdot (1 - \\cos^2(x)) \\cdot \\cos(x) \\]\n\\[ = 4 \\cdot (\\cos(x))^3 - 3 \\cdot \\cos(x) \\]\n\n\\[ \\tan(\\alpha + \\beta) = \\frac{\\sin(\\alpha + \\beta)}{\\cos(\\alpha + \\beta)} = \\frac{\\sin(\\alpha) \\cdot \\cos(\\beta) + \\sin(\\beta) \\cdot \\cos(\\alpha)}{\\cos(\\alpha) \\cdot \\cos(\\beta) - \\sin(\\alpha) \\cdot \\sin(\\beta)} \\]\n\n\\[ = \\frac{\\tan(\\alpha) \\cdot \\tan(\\beta)}{1 - \\tan(\\alpha) \\cdot \\tan(\\beta)} \\]\n\n\\[ = \\tan(\\alpha) + \\tan(\\beta) \\]\n\n\\[ = \\frac{\\tan(\\alpha) + \\tan(\\beta)}{1 - \\tan(\\alpha) \\cdot \\tan(\\beta)} \\]\n\\[ a = \\beta = x \\]\n\n\\[ \\frac{1}{g(2x)} = \\frac{2 + g(x)}{1 - \\frac{1}{2}g(x) \\cdot \\frac{1}{2}g(x)} = \\frac{2^{-1}g(x)}{1 - (g(x))^2} \\]\n**Sum to Product Formula**\n\n\\[\n\\cos(x) + \\cos(y) = 2 \\cdot \\cos\\left(\\frac{x+y}{2}\\right) \\cos\\left(\\frac{x-y}{2}\\right)\n\\]\n\n\\[\n\\alpha = \\frac{x+y}{2} \\quad \\beta = \\frac{x-y}{2}\n\\]\n\n\\[\n\\alpha + \\beta = \\frac{x+y}{2} + \\frac{x-y}{2} = x\n\\]\n\n\\[\n\\alpha - \\beta = \\frac{x+y}{2} - \\frac{x-y}{2} = y\n\\]\n\n\\[\n\\cos(x) = \\cos(\\alpha + \\beta) = \\cos(\\alpha) \\cos(\\beta) - \\sin(\\alpha) \\sin(\\beta)\n\\]\n\n\\[\n\\cos(y) = \\cos(\\alpha - \\beta) = \\cos(\\alpha) \\cos(\\beta) + \\sin(\\alpha) \\sin(\\beta)\n\\]\n\n\\[\n\\cos(x) + \\cos(y) = 2 \\cos(\\alpha) \\cos(\\beta) = 2 \\cdot \\cos\\left(\\frac{x+y}{2}\\right) \\cos\\left(\\frac{x-y}{2}\\right)\n\\]\nProduct-to-sum formulas\n\n\\[ \\cos(\\alpha) \\cos(\\beta) = \\frac{\\cos(\\alpha + \\beta) + \\cos(\\alpha - \\beta)}{2} \\]\n\n\\[ \\sin(\\alpha) \\cos(\\beta) = \\frac{\\sin(\\alpha + \\beta) + \\sin(\\alpha - \\beta)}{2} \\]", "id": "./materials/12.pdf" }, { "contents": "The objective of this question is to calculate the volume of solid generated by revolution of a planar region.\n\nIn this question, we are supposed to revolve the region around \\( y = -3 \\). Therefore, the equations have to be redefined by assuming that \\( y = -3 \\) is our new origin.\n\nSo, \\( f(x) = 8 - x \\) is a straight line.\n\n\\( g(x) = 3 \\) is a straight line.\n\nAccording to the question, we are supposed to revolve the region around the line \\( y = -3 \\). On Revolving around the \\( y = -3 \\), a solid of revolution is obtained.\nRemember that, the volume of the solid of revolution formed by revolving region around the $x$-axis is given by,\n\n$$V = \\pi \\int_a^b f^2(x) - g^2(x) \\, dx,$$\n\nwhere $f(x)$ is the upper curve and $g(x)$ is the lower curve and $x \\in [a, b]$.\n\nIn this case, the upper function is $f(x) = 8 - x$ and lower function is $g(x) = 3$ and $x \\in [0, 3]$.\n\n$$V = \\pi \\int_0^3 (8 - x)^2 - 3^2 \\, dx$$\n\n$$= \\pi \\int_0^3 (8 - x)^2 - \\pi \\int_0^3 3^2 \\, dx$$\n\n$$= \\pi \\left[ \\frac{(8 - x)^3}{3} \\right]_0^3 - \\pi \\left[ 9x \\right]_0^3$$\n\n$$= \\pi \\left( \\frac{125}{3} - \\frac{512}{3} \\right) - 27\\pi$$\n\n$$= 129\\pi - 27\\pi$$\n\n$$= 102\\pi \\text{ cubic units}$$", "id": "./materials/120.pdf" }, { "contents": "Evaluate \\( \\int_{1}^{4} \\ln \\left( \\frac{x}{2} \\right) \\, dx \\)\n\n**HINT:**\n\nUse integration by parts:\n\n\\[\n\\int f(x)g(x) \\, dx = F(x)g(x) - \\int F(x)g'(x) \\, dx + C, \\text{ such that } F(x) = \\int f(x) \\, dx\n\\]\n\nThis integral is not a common integral so we cannot find it in the integral table. Therefore, it is necessary to introduce something in the integrand in order to solve it.\n\n\\[\n\\int_{1}^{4} \\ln \\left( \\frac{x}{2} \\right) \\, dx = \\int_{1}^{4} 1 \\cdot \\ln \\left( \\frac{x}{2} \\right) \\, dx\n\\]\n\nNow, take \\( f(x) = 1 \\) and \\( g(x) = \\ln \\left( \\frac{x}{2} \\right) \\).\n\nWe chose \\( g(x) = \\ln \\left( \\frac{x}{2} \\right) \\) following the LIATE rule.", "id": "./materials/121.pdf" }, { "contents": "Every definite integration problem begins with checking if the conditions for the Fundamental Theorem of Calculus are met or not.\n\nEvaluate \\( \\int_{\\sqrt{3}}^{2} \\frac{\\sqrt{4-x^2}}{x} \\, dx \\)\n\nConverting to form \\( R(x, \\sqrt{1-x^2}) \\)\n\nSo,\n\n\\[\nI(x) = \\int \\frac{\\sqrt{4-x^2}}{x} \\, dx = \\int \\frac{\\sqrt{(4-x^2)x^2}}{x} \\, dx\n\\]\n\n\\[\n= 2 \\int \\frac{\\sqrt{1-(\\frac{x}{2})^2}}{x} \\, dx\n\\]\n\nPerforming trigonometric substitution,\n\n\\[\n\\frac{x}{2} = \\sin(t)\n\\]\n\n\\[\n(\\Rightarrow) x = 2 \\sin(t)\n\\]\n\n\\[\n(\\Rightarrow) dx = 2 \\cos(t) \\, dt\n\\]\n\n\\[\n= 2 \\int \\frac{\\sqrt{\\cos^2(t)}}{\\sin(t)} \\cos(t) \\, dt\n\\]\n\n\\[\n= 2 \\int \\frac{\\cos(t) \\cdot \\cos(t)}{\\sin(t)} \\, dt\n\\]\n\n\\[\n= 2 \\int \\frac{\\cos^2(t)}{\\sin(t)} \\, dt\n\\]\n\n\\[\n= 2 \\int \\frac{1-\\sin^2(t)}{\\sin(t)} \\, dt\n\\]\n\\[ = 2 \\int \\csc(t) \\cdot \\sin(t) \\, dt \\]\n\\[ = 2 \\int \\csc(t) \\, dt - 2 \\int \\sin(t) \\, dt \\]\n\\[ = 2 \\ln |\\csc(t) - \\cot(t)| + 2 \\sin \\cos(t) + C \\]\n\nConverting to function dependent on \\( x \\):\n\n\\[ \\sin(t) = \\frac{x}{2} \\]\n\\[ \\csc(t) = \\frac{2}{x} \\]\n\\[ \\cos(t) = \\sqrt{1 - \\sin^2(t)} \\]\n\\[ = \\sqrt{1 - \\frac{x^2}{4}} \\]\n\\[ = \\frac{\\sqrt{4-x^2}}{2} \\]\n\\[ \\cot(t) = \\frac{\\cos(t)}{\\sin(t)} \\]\n\\[ = \\frac{\\sqrt{4-x^2}}{x} \\]\n\n\\[ = 2 \\ln \\left| \\frac{2}{x} - \\frac{\\sqrt{4-x^2}}{x} \\right| + \\frac{x}{2} \\sqrt{4-x^2} + C \\]\n\\[ = 2 \\ln \\left| \\frac{2 - \\sqrt{4-x^2}}{x} \\right| + \\sqrt{4-x^2} + C \\]\nNow,\n\\[\n\\int_{\\sqrt{3}}^{2} \\frac{\\sqrt{4-x^2}}{x} \\, dx = \\left[ I(x) \\right]_{\\sqrt{3}}^{2}\n\\]\n\\[\n= \\left[ 2 \\ln \\left( \\frac{2 - \\sqrt{4-x^2}}{x} + \\sqrt{4-x^2} \\right) \\right]_{\\sqrt{3}}^{2}\n\\]\n\\[\n= \\left( 2 \\ln \\left( \\frac{2 - \\sqrt{4-4}}{2} + \\sqrt{4-4} \\right) \\right) - \\left( 2 \\ln \\left( \\frac{2 - \\sqrt{4-3}}{\\sqrt{3}} + \\sqrt{4-3} \\right) \\right)\n\\]\n\\[\n= - \\left( 2 \\ln \\left( \\frac{1}{\\sqrt{3}} \\right) + 1 \\right)\n\\]\n\\[\n= -2 \\ln \\left( \\frac{1}{\\sqrt{3}} \\right) - 1\n\\]\n\\[\n= -2 \\ln(3) - 1\n\\]\n\\[\n= -2 \\times \\left( -\\frac{1}{2} \\right) \\ln(3) - 1\n\\]\n\\[\n= \\ln(3) - 1\n\\]", "id": "./materials/122.pdf" }, { "contents": "Evaluate \\[ \\int_0^1 \\frac{x-2}{(x+1)(x+2)} \\, dx \\]\n\n* All the conditions for Fundamental theorem of calculus are met.\n\nSince, \\( m < n \\), the partial fractions should be obtained.\n\nFor \\( I(x) = \\int \\frac{x-2}{(x+1)(x+2)} \\, dx \\), the partial fractions are,\n\n\\[\n\\frac{x-2}{(x+1)(x+2)} = \\frac{A}{x+1} + \\frac{B}{x+2}\n\\]\n\n\\( \\Rightarrow \\)\n\n\\[\nx - 2 = A(x+2) + B(x+1)\n\\]\n\n\\( \\Rightarrow \\)\n\n\\[\nx - 2 = Ax + 2A + Bx + B\n\\]\n\n\\( \\Rightarrow \\)\n\n\\[\nx - 2 = x(A + B) + 2A + B\n\\]\n\nComparing coefficients of left and right hand side.\n\n\\[\nA + B = 1 \\quad \\ldots \\quad (i)\n\\]\n\n\\[\n2A + B = -2 \\quad \\ldots \\quad (ii)\n\\]\n\nSolving eqn (i) and (ii), we get,\n\n\\[\nA = -3\n\\]\n\n\\[\nB = 4\n\\]\n\\[ I(x) = \\int \\frac{-3}{x+1} + \\frac{4}{x+2} \\, dx \\]\n\n\\[ = -3 \\ln |x+1| + 4 \\ln |x+2| + C \\]\n\nNow,\n\n\\[ \\int_{0}^{1} \\frac{x-2}{(x+1)(x+2)} \\, dx = \\left[ I(x) \\right]_{0}^{1} \\]\n\n\\[ = \\left[ -3 \\ln |x+1| + 4 \\ln |x+2| \\right]_{0}^{1} \\]\n\n\\[ = -3 \\ln 2 + 4 \\ln 3 - (-3 \\ln 1 + 4 \\ln 2) \\]\n\n\\[ = -3 \\ln 2 + 4 \\ln 3 - 4 \\ln 2 \\]\n\n\\[ = 4 \\ln 3 - 7 \\ln 2 \\]", "id": "./materials/123.pdf" }, { "contents": "$y = 2x$ is a straight line which passes through the origin.\n\n$y = \\frac{8}{x}$ is a curve with both vertical and horizontal asymptote.\n\n$x = 4$ is a straight line.\n\nThe straight line $y = 2x$ intersects the curve $y = \\frac{8}{x}$ on $x = 2$ and $x = -2$\n\n**Remember that**, Area bounded by the curves is given by,\n\n$$\\text{Area} = \\int_{a}^{b} f(x) - g(x) \\, dx,$$\n\nwhere $f(x)$ is the upper curve and $g(x)$ is the lower curve and $x \\in [a, b]$. \nIn this case, the upper function is \\( f(x) = 2x \\) and lower function is \\( g(x) = \\frac{8}{x} \\) and \\( x \\in [2, 4] \\).\n\n\\[\n\\text{Area} = \\int_{a}^{b} f(x) - g(x) \\, dx \\\\\n= \\int_{2}^{4} 2x - \\frac{8}{x} \\, dx \\\\\n= \\left[ x^2 - 8 \\ln(x) \\right]_{2}^{4} \\\\\n= \\left[ x^2 \\right]_{2}^{4} - \\left[ 8 \\ln(x) \\right]_{2}^{4} \\\\\n= (16 - 4) - 8(\\ln(4) - \\ln(2)) \\\\\n= 12 - 8 \\ln(2) \\text{ square units}\n\\]", "id": "./materials/124.pdf" }, { "contents": "The objective of this question is to calculate the volume of solid generated by revolution of a planar region. As a hint, the region bounded by the curves and the solid of revolution are given below:\n\nAccording to the question, we are supposed to revolve the region around the x-axis. On Revolving around the x-axis, a solid of revolution is obtained.\nRemember that, the volume of the solid of revolution formed by revolving the region around the $x$-axis is given by,\n\n$$V = \\pi \\int_{a}^{b} f^2(x) - g^2(x) \\, dx,$$\n\nwhere $f(x)$ is the upper curve and $g(x)$ is the lower curve and $x \\in [a, b]$.\n\nIn this case, the upper function is $f(x) = 2 - \\frac{x}{2}$ and the lower function is $g(x) = 0$ and $x \\in [1, 2]$. ", "id": "./materials/125.pdf" }, { "contents": "Hint:\n\nThe region bounded by the curves is highlighted in blue.\n\nContinued on the next page\nRemember that, Area bounded by the curves is given by,\n\n\\[ \\text{Area} = \\int_{a}^{b} f(x) - g(x) \\, dx, \\]\n\nwhere \\( f(x) \\) is the upper curve and \\( g(x) \\) is the lower curve and \\( x \\in [a, b] \\).\n\nIn this case, the upper function is \\( f(x) = x^2 \\) and the lower function is \\( g(x) = x^3 \\) and \\( x \\in [0, 1] \\).", "id": "./materials/126.pdf" }, { "contents": "As a hint, solution to an indirect definite integral is provided.\n\n\\[ \\int e^x \\sin(x) \\, dx \\]\n\n**Solution** Take \\( e^x \\) as the first function and \\( \\sin x \\) as the second function. Then, integrating by parts, we have\n\n\\[\nI = \\int e^x \\sin(x) \\, dx = e^x(-\\cos(x)) + \\int e^x \\cos(x) \\, dx\n\\]\n\n\\[\n= -e^x \\cos(x) + I_1 \\text{(say)}\n\\]\n\nTake \\( e^x \\) and \\( \\cos(x) \\) as the first and second functions, respectively, in \\( I_1 \\). Then, solving \\( I_1 \\), we get\n\n\\[\nI = e^x \\sin(x) - \\int e^x \\sin(x) \\, dx\n\\]\n\nSubstituting the value of \\( I_1 \\) in \\( I \\), we get\n\n\\[\nI = -e^x \\cos(x) + e^x \\sin(x) - I\n\\]\n\nwhich can be written as,\n\n\\[\n2I = e^x(\\sin x - \\cos x)\n\\]\n\nHence,\n\n\\[\nI = \\int e^x \\sin(x) \\, dx = \\frac{e^x}{2}(\\sin(x) - \\cos(x)) + C\n\\]\n\nAlternatively, above integral can also be determined by taking \\( \\sin(x) \\) as the first function and \\( e^x \\) as the second function.", "id": "./materials/127.pdf" }, { "contents": "As a hint, solution to an indefinite integral is provided.\n\n\\[ \\int e^x \\sin(x) \\, dx \\]\n\n**Solution** Take \\( e^x \\) as the first function and \\( \\sin x \\) as the second function. Then, integrating by parts, we have\n\n\\[\nI = \\int e^x \\sin(x) \\, dx = e^x(-\\cos(x)) + \\int e^x \\cos(x) \\, dx\n\\]\n\n\\[\n= -e^x \\cos(x) + I_1 \\text{ (say)}\n\\]\n\nTake \\( e^x \\) and \\( \\cos(x) \\) as the first and second functions, respectively, in \\( I_1 \\). Then, solving \\( I_1 \\), we get\n\n\\[\nI = e^x \\sin(x) - \\int e^x \\sin(x) \\, dx\n\\]\n\nSubstituting the value of \\( I_1 \\) in \\( I \\), we get\n\n\\[\nI = -e^x \\cos(x) + e^x \\sin(x) - I\n\\]\n\nwhich can be written as,\n\n\\[\n2I = e^x(\\sin x - \\cos x)\n\\]\n\nHence,\n\n\\[\nI = \\int e^x \\sin(x) \\, dx = \\frac{e^x}{2}(\\sin(x) - \\cos(x)) + C\n\\]\n\nAlternatively, above integral can also be determined by taking \\( \\sin(x) \\) as the first function and \\( e^x \\) as the second function.", "id": "./materials/128.pdf" }, { "contents": "Integration by Parts\n\nLet consider the product rule for derivatives:\n\n\\[(F \\cdot g)' = F' \\cdot g + F \\cdot g'\\]\n\n\\[\\Leftrightarrow \\int (F \\cdot g)' \\, dx = \\int F' \\cdot g + F \\cdot g' \\, dx\\]\n\n\\[\\Leftrightarrow F \\cdot g = \\int F' \\cdot g \\, dx + \\int F \\cdot g' \\, dx\\]\n\n\\[\\Leftrightarrow F \\cdot g = \\int f \\cdot g \\, dx + \\int F \\cdot g' \\, dx, \\quad \\text{if } F(x) = \\int f \\, dx\\]\n\n\\[\\Leftrightarrow \\int f(x)g(x) \\, dx = F(x)g(x) - \\int F(x)g'(x) \\, dx \\quad \\text{if } F(x) = \\int f \\, dx\\]\n\nThen the formula to integrate by parts is\nIntegration by Parts\n\n\\[ \\int f(x)g(x) \\, dx = F(x)g(x) - \\int F(x)g'(x) \\, dx, \\]\n\nsuch that, \\( F(x) = \\int f(x) \\, dx \\)\n\nRemarks:\n\n- Useful to integrate products which involve:\n - polynomial and exponential functions;\n - trigonometric and exponential functions or trigonometric and polynomial functions;\n - inverse trigonometric functions;\n - logarithmic functions.\n\n- Sometimes it is necessary to apply the method of integration by parts multiple times before a result is obtained.\nIntegration by Parts\n\nExamples\n\n\\[ \\int x (x - 1)^9 \\, dx \\]\n\n\\[ \\int e^x (x + 1) \\, dx \\]\n\n\\[ \\int \\ln(x) \\, dx = \\int \\frac{1}{f} \\cdot \\ln(x) \\, dx \\]\n\n\\[ \\int \\arctg(x) \\, dx = \\int \\frac{1}{f} \\cdot \\arctg(x) \\, dx \\]\n\n\\[ \\int e^x \\cdot \\cos(x) \\, dx \\]", "id": "./materials/129.pdf" }, { "contents": "**Functions**\n\nA function is 3 things:\n\n- A starting set\n- An arrival set\n- A rule that assigns to each element of the starting set a unique element in the arrival set\n\n\\[\n\\begin{array}{c}\n\\text{DOMAIN} \\\\\n\\downarrow \\\\\n\\text{f} \\\\\n\\downarrow \\\\\n\\text{CO-DOMAIN} \\\\\n\\end{array}\n\\]\n\n\\[f : A \\rightarrow B\\]\nProperties of functions\n\n\\( f \\) is **INJECTIVE** - if distinct elements in the domain (one-to-one) go to distinct element in the co-domain\n\n- if \\( \\forall a_1 \\neq a_2, a_1, a_2 \\in A \\)\n \\[ \\Rightarrow f(a_1) \\neq f(a_2) \\]\n\n\\( f \\) is **SURJECTIVE** - if every element in the codomain (onto) is reached by some element in the domain\n\n- \\( \\forall b \\in B, \\exists a \\in A \\text{ s.t. } f(a) = b \\)\nA function $f$ is bijective if it is injective and surjective.\n\nA function $g : B \\to A$ is the \"inverse\" of $f$ if\n\n$$g(f(a)) = a$$\n\nand\n\n$$f \\circ g = \\text{id}_B$$\n\n$$g \\circ f = \\text{id}_A$$\nBeware! Injectivity, surjectivity and bijectivity depend on the domain so, in particular\n\n\\[ \\cos : \\mathbb{R} \\to \\mathbb{R} \\text{ is not inj.} \\]\n\\[ \\text{not surj.} \\]\n\nbut \\[ \\cos : [0, \\pi] \\to \\mathbb{R} \\text{ is inj. and surj.} \\]\nand thus \\[ \\text{in-ct.} \\]\n\n\\[ \\arccos : [-1, 1] \\to [0, \\pi] \\text{ is its inverse} \\]\n\\[ f(x) = x^2 \\]\n\n\\[ f : A \\rightarrow B \\]\n\n\\[ \\forall x_1 \\neq x_2 \\Rightarrow f(x_1) \\neq f(x_2) \\quad (x_1, x_2 \\in A) \\]\n\n| A | B | INJ.? | SURJ.? | BIS.? |\n|-----|-----|-------|--------|-------|\n| IR | IR | NO! | NO! | NO |\n| IR | IR | YES | NO! | NO |\n| IR | IR | NO | YES | NO |\n| IR | IR | NO | YES | NO |\n| IR | IR | YES | YES | BIS |\n| Z | Z | YES | NO | NO |\n| Q | Q | YES | NO | NO |\nDef. \\( \\text{Range}(f) = \\{ f(a) : a \\in A \\} \\)\n\n\\( f \\) surjective if \\( \\text{Range}(f) = B \\)\n\n\\( f(x) = x^2 \\quad f : \\mathbb{N} \\rightarrow \\mathbb{N} \\quad \\text{Range}(f) = \\{ a^2 : a \\in \\mathbb{N} \\} = \\{ 0, 1, 4, 9, 16, \\ldots \\} = \\{ \\text{perfect squares} \\} \\neq \\mathbb{N} \\)\nImage and counter-image of a set\n\n\\[ f: A \\rightarrow B \\]\n\nfor every \\( C \\subseteq A \\)\n\n\\[ f(C) = \\{ b \\in B : \\exists c \\in C \\text{ s.t. } b = f(c) \\} \\]\n\n\\[ = \\{ f(c) : c \\in C \\} \\]\n\\[ A \\setminus D \\subseteq B \\]\n\n\\[ f^{-1}(D) = \\{ a \\in A \\text{ s.t. } f(a) \\in D \\} \\]", "id": "./materials/13.pdf" }, { "contents": "Properties of the Integral Operator\n\nLet $C$ and $k$ be constants.\n\nProperties:\n\n1. $\\int F'(x) \\, dx = F(x) + C$.\n2. $\\frac{d}{dx} \\left( \\int f(x) \\, dx \\right) = f(x)$.\n3. $\\int k \\cdot f(x) \\, dx = k \\int f(x) \\, dx \\quad k \\in \\mathbb{R}$.\n4. $\\int f(x) + g(x) \\, dx = \\int f(x) \\, dx + \\int g(x) \\, dx$. \n# Table of Integrals\n\nLet $f$ and $g$ be integrable functions; $C$ and $k$ real constants and $a > 0$, $a \\neq 1 \\in \\mathbb{R}$.\n\n| | | |\n|---|---|---|\n| 1. | $\\int k \\, dx = kx + C$ | 10. | $\\int f' \\csc^2(f) \\, dx = -\\cotg(f) + C$ |\n| 2. | $\\int f' f^n \\, dx = \\frac{f^{n+1}}{n+1} + C$, $n \\neq -1$ | 11. | $\\int f' \\sec(f) \\, dx = \\ln |\\sec(f) + \\tg(f)| + C$, $\\sec(f) + \\tg(f) \\neq 0$ |\n| 3. | $\\int \\frac{f'}{f} \\, dx = \\ln |f| + C$ | 12. | $\\int f' \\csc(f) \\, dx = \\ln |\\csc(f) - \\cotg(f)| + C$, $\\csc(f) - \\cotg(f) \\neq 0$ |\n| 4. | $\\int f' a^f \\, dx = \\frac{a^f}{\\ln(a)} + C$ | 13. | $\\int f' \\sec(f) \\tg(f) \\, dx = \\sec(f) + C$ |\n| 5. | $\\int f' \\cos(f) \\, dx = \\sin(f) + C$ | 14. | $\\int f' \\csc(f) \\cotg(f) \\, dx = -\\csc(f) + C$ |\n| 6. | $\\int f' \\sin(f) \\, dx = -\\cos(f) + C$ | 15. | $\\int \\frac{f'}{a^2 + f^2} \\, dx = \\frac{1}{a} \\arctg \\left( \\frac{f}{a} \\right) + C$ |\n| 7. | $\\int f' \\tg(f) \\, dx = \\ln |\\sec(f)| + C$ | | |\n| 8. | $\\int f' \\cotg(f) \\, dx = \\ln |\\csc(f)| + C$ | 16. | $\\int \\frac{f'}{\\sqrt{1 - f^2}} \\, dx = \\arcsin(f) + C$ |\n| 9. | $\\int f' \\sec^2(f) \\, dx = \\tg(f) + C$ | | |", "id": "./materials/130.pdf" }, { "contents": "Method of Partial Fractions\n\nIt is used to integrate rational functions $f(x) = \\frac{P(x)}{Q(x)}$ where $P(x)$ and $Q(x)$ are polynomial functions and the power of $P(x)$ is less than the power of $Q(x)$.\n\nExamples\n\n1. $\\int \\frac{x}{x^3 - x} \\, dx$\n2. $\\int \\frac{x}{x^2 - x - 6} \\, dx$\n3. $\\int \\frac{1}{x^2 - 1} \\, dx$\nMethod of Partial Fractions\n\nAlert\n\nIf a rational function \\( \\frac{R(x)}{Q(x)} \\) is such that the power of \\( R(x) \\) is greater than the power of \\( Q(x) \\), then one must use long division and write the rational function in the form\n\n\\[\n\\frac{R(x)}{Q(x)} = a_0x^n + a_1x^{n-1} + \\cdots + a_{n-1}x + a_n + \\frac{P(x)}{Q(x)}\n\\]\n\nwhere now \\( P(x) \\) is a remainder term with the power of \\( P(x) \\) less than the power of \\( Q(x) \\) and our objective is to integrate each term of the above representation.\nCase 1: power of $P(x) \\geq Q(x)$\n\nConsider $\\int \\frac{P(x)}{Q(x)} \\, dx$\n\n1. If power of $P(x) \\geq$ power of $Q(x)$\n\n Apply a long division.\n\n $P(x)$ \\hspace{1cm} $Q(x)$\n \\hline\n $R(x)$ \\hspace{1cm} $q(x)$\n\n And, then\n\n $\\int \\frac{P(x)}{Q(x)} \\, dx = \\int q(x) \\, dx + \\int \\frac{R(x)}{Q(x)} \\, dx$\n\n formula table\nCase 2: power of $P(x) < Q(x)$\n\nConsider $\\int \\frac{P(x)}{Q(x)} \\, dx$\n\n1. The denominator $Q(x)$ has only first power factors, none of which are repeated. Then, $Q(x)$ has the form\n\n$$Q(x) = (x - x_0)(x - x_1)(x - x_2) \\cdots (x - x_n)$$\n\nwhere $x_0 \\neq x_1 \\neq x_2 \\neq \\cdots \\neq x_n$. One can then write\n\n$$\\frac{P(x)}{Q(x)} = \\frac{A}{x - x_0} + \\frac{B}{x - x_1} + \\cdots + \\frac{C}{x - x_n}$$\n\nwhere $A, B, \\cdots, C$ are constants to be determined.\nCase 2: power of $P(x) < Q(x)$\n\nConsider $\\int \\frac{P(x)}{Q(x)} \\, dx$\n\n2. The denominator $Q(x)$ has only first power factors, but some of these factors may be repeated factors. For example, the denominator $Q(x)$ might have a form such as\n\n$$Q(x) = (x - x_0)^k (x - x_1)^p \\cdots (x - x_n)^m$$\n\nwhere $k, p, \\cdots, m$ are integers.\n\nHere the denominator has repeated factors of orders $k, p, \\cdots, m$. In this case one can write the rational function in the form:\nCase 2: power of $P(x) < Q(x)$\n\n2. (continue)\n\n\\[\n\\frac{P(x)}{Q(x)} = \\frac{A_1}{x-x_0} + \\frac{A_2}{(x-x_0)^2} + \\cdots + \\frac{A_k}{(x-x_0)^k} + \\\\\n+ \\frac{B_1}{x-x_1} + \\frac{B_2}{(x-x_1)^2} + \\cdots + \\frac{A_p}{(x-x_1)^p} + \\\\\n+ \\cdots + \\\\\n+ \\frac{C_1}{x-x_n} + \\frac{C_2}{(x-x_n)^2} + \\cdots + \\frac{C_m}{(x-x_n)^m}\n\\]\n\nwhere $A_1, \\cdots, A_k, B_1, \\cdots, B_p, \\cdots, C_1, \\cdots, C_m$ are constants to be determined.\nCase 2: power of $P(x) < Q(x)$\n\n3. The denominator $Q(x)$ has one or more quadratic factors of the form $ax^2 + bx + c$ none of which are repeated. In this case, for each quadratic factor there corresponds a partial fraction of the form\n\n$$\\frac{P(x)}{Q(x)} = \\frac{A_0x + B_0}{ax^2 + bx + c}$$\n\nwhere $A_0$ and $B_0$ are constants to be determined.\nCase 2: power of $P(x) < Q(x)$\n\n4. The denominator $Q(x)$ has one or more quadratic factors, some of which are repeated quadratic factors. In this case, for each repeated quadratic factor $(ax^2 + bx + c)^k$ there corresponds a sum of partial fractions of the form\n\n$$\\frac{P(x)}{Q(x)} = \\frac{A_1x + B_1}{ax^2 + bx + c} + \\frac{A_2x + B_2}{(ax^2 + bx + c)^2} + \\cdots + \\frac{A_kx + B_k}{(ax^2 + bx + c)^k}$$\n\nwhere $A_1, B_1, \\cdots, A_k, B_k$ are constants to be determined.\nIntegral of a Rational function — Summary\n\nLet’s consider the case 2 (power of denominator bigger than the power of numerator) and the following general form of partial fractions:\n\n\\[\n\\frac{P(x)}{Q(x)} = \\frac{A_p}{(x-\\alpha)^p} + \\frac{A_{p-1}}{(x-\\alpha)^{p-1}} + \\cdots + \\frac{A_2}{(x-\\alpha)^2} + \\frac{A_1}{x-\\alpha} + \\\\\n+ \\frac{B_q}{(x-\\beta)^q} + \\frac{B_{q-1}}{(x-\\beta)^{q-1}} + \\cdots + \\frac{B_2}{(x-\\beta)^2} + \\frac{B_1}{x-\\beta} + \\cdots + \\\\\n+ \\frac{C_r+D_r x}{((x-a)^2+b^2)^r} + \\frac{C_{r-1}+D_{r-1} x}{((x-a)^2+b^2)^{r-1}} + \\cdots + \\frac{C_2+D_2 x}{((x-a)^2+b^2)^2} + \\frac{C_1+D_1 x}{(x-a)^2+b^2}\n\\]\n\n(1)\n\nImportant:\n\nAll the constants must be determined before applying the integral calculus.\nTechniques to find the constants\n\nConsider Case 2.\n\nRemark:\nThe 3 below sub-cases can appear in the integration. For each factor we should write the partial fractions as described above.\n\nTechniques to find the constants\n\n- Undetermined coefficients method\n- The cover-up method\n- Differentiation Rule\nCover-up method\n\n- If $Q(x) = (x - \\alpha)^p Q_1(x)$ and $\\alpha \\in \\mathbb{R}$\n \n $$A_p = \\left[ \\frac{P(x)}{Q(x)} \\right]_{x=\\alpha}$$\n\n- If $Q(x) = ((x - a)^2 + b^2)^r Q_1(x)$ and $x = a + bi \\in \\mathbb{C}$\n \n $$\\left[ C_r + D_r x = \\frac{P(x)}{Q_1(x)} \\right]_{x=a+bi}$$\nCover-up method\n\nExample\n\n\\[\n\\frac{6x - 1}{x^2 - 4x + 3} = \\frac{6x - 1}{(x - 3)(x - 1)} = \\frac{A}{x - 3} + \\frac{B}{x - 1}\n\\]\n\nThe constants \\(A\\) and \\(B\\) can be determined by cover-up method:\n\n\\[\nA = \\left[ \\frac{6x - 1}{x - 1} \\right]_{x=3} = \\frac{17}{2}\n\\]\n\n\\[\nB = \\left[ \\frac{6x - 1}{x - 3} \\right]_{x=1} = \\frac{5}{-2} = -\\frac{5}{2}\n\\]\nCover-up method\n\nExample\n\n\\[\n\\frac{6x - 1}{(x - 3)^2(x - 1)} = \\frac{A_1}{x - 3} + \\frac{A_2}{(x - 3)^2} + \\frac{B_1}{x - 1}\n\\]\n\nThe constants \\(A_2\\) e \\(B_1\\) can be determined by cover-up method:\n\n\\[\nA_2 = \\left[ \\frac{6x - 1}{x - 1} \\right]_{x=3} = \\frac{26}{2} = 13\n\\]\n\n\\[\nB_1 = \\left[ \\frac{6x - 1}{(x - 3)^2} \\right]_{x=1} = \\frac{6}{4} = -\\frac{3}{2}\n\\]\n\nBe careful:\n\nThe constant \\(A_1\\) can not be determined using cover-up method\nDifferentiation method\n\nUseful when the denominator has roots (real or complex) with multiplicity greater than one.\n\n1. Put the same denominator in the equation 1 (page 46).\n2. Assign to $x$ the roots values and find some coefficients.\n3. Differentiate both sides of the equation and repeat step 2.\n4. Differentiate again both sides of the equation until all the coefficients are found.\nDifferentiation method\n\nFor real roots\n\nIf \\( Q(x) = (x - \\alpha)^p Q_1(x) \\) and \\( \\alpha \\in \\mathbb{R} \\)\n\n\\[\n\\left[ \\frac{1}{r!} \\cdot \\frac{d^r}{dx^r} \\left( \\frac{P(x)}{Q_1(x)} \\right) \\right]_{x=\\alpha} = A_{p-r}, \\quad 0 \\leq r \\leq p - 1\n\\]\n\nRemark: The derivative of order 0 is \\( \\frac{d^0}{dx^0} (f(x)) = f(x) \\)\nDifferentiation method\n\nExamples\n\n1. Consider \\( \\frac{x^2 + 1}{(x - 1)^3(x - 2)} = \\frac{A_3}{(x - 1)^3} + \\frac{A_2}{(x - 1)^2} + \\frac{A_1}{x - 1} + \\frac{B_1}{x - 2} \\).\n Determine all the constants using the differentiation rule.\n\n2. Consider \\( \\frac{2}{(x^2 + 1)^2x} = \\frac{C_2 + D_2x}{(x^2 + 1)^2} + \\frac{C_1 + D_1x}{x^2 + 1} + \\frac{B_1}{x} \\).\n Determine all the constants using the undetermined coefficients method.\nSolution of First Example 1\n\nIt is intended to find \\[ \\int \\frac{x^2 + 1}{(x - 1)^3(x - 2)} \\, dx \\]\n\nFirst, write the rational function as sums of partial fractions:\n\\[\n\\frac{x^2 + 1}{(x - 1)^3(x - 2)} = \\frac{A_3}{(x - 1)^3} + \\frac{A_2}{(x - 1)^2} + \\frac{A_1}{x - 1} + \\frac{B_1}{x - 2}.\n\\]\n\nBy comparison we obtain the following:\n\n- \\( P(x) = x^2 + 1 \\).\n- \\( (x - \\alpha)^p = (x - 1)^3 \\), then \\( \\alpha = 1 \\land p = 3 \\).\n- \\( Q_1(x) = x - 2 \\).\nSolution of First Example II\n\nSince \\( p = 3 \\land 0 \\leq r \\leq p - 1 \\implies 0 \\leq r \\leq 2. \\)\n\n\\[\nA_{p-r} = \\left[ \\frac{1}{r!} \\cdot \\frac{d^r}{dx^r} \\left( \\frac{P(x)}{Q_1(x)} \\right) \\right]_{x=\\alpha}, \\quad 0 \\leq r \\leq 2.\n\\]\n\nThen, for:\n\n- \\( r = 0, \\ A_3 = \\left[ \\frac{1}{0!} \\cdot \\frac{d^0}{dx^0} \\left( \\frac{x^2+1}{x-2} \\right) \\right]_{x=1} = \\left[ \\left( \\frac{x^2+1}{x-2} \\right) \\right]_{x=1} = -2. \\)\n- \\( r = 1, \\ A_2 = \\left[ \\frac{1}{1!} \\cdot \\left( \\frac{x^2+1}{x-2} \\right)' \\right]_{x=1} = \\left[ \\left( \\frac{x^2-4x-1}{(x-2)^2} \\right) \\right]_{x=1} = -4 \\)\n- \\( r = 2, \\ A_1 = \\left[ \\frac{1}{2!} \\cdot \\left( \\frac{x^2+1}{x-2} \\right)'' \\right]_{x=1} = \\left[ \\frac{1}{2!} \\left( \\frac{x^2-4x-1}{(x-2)^2} \\right)' \\right]_{x=1} = \\ldots = -5 \\)\nSolution of First Example III\n\n$B_1$ can be determined by hidden rule:\n\n$$B_1 = \\left[ \\frac{x^2 + 1}{(x - 1)^3} \\right]_{x=2} = \\frac{5}{1} = 5$$\n\nSecond, replace the constants and find the related integrals:\n\n$$\\int \\frac{x^2 + 1}{(x - 1)^3(x - 2)} \\, dx = \\int \\frac{A_3}{(x - 1)^3} \\, dx + \\int \\frac{A_2}{(x - 1)^2} \\, dx + \\int \\frac{A_1}{x - 1} \\, dx + \\int \\frac{B_1}{x - 2} \\, dx.$$", "id": "./materials/131.pdf" }, { "contents": "Integration by substitution\n\nIf \\( u = g(x) \\) is a continuous and differentiable function in the interval \\( I \\) then\n\n\\[\n\\int f(g(x))g'(x) \\, dx = \\int f(t) \\, dt\n\\]\n\nUseful\n\n1. to simplify the integrand.\n2. when the integrand involve irrational fractions.\n3. when a particular expression is repeated.\nIntegration by substitution\n\nExample\n\n\\[ \\int e^{\\sqrt{x}} \\, dx = \\int e^t \\cdot 2t \\, dt = \\cdots \\]\n\nDo the substitution:\n\n\\[ t = \\sqrt{x} \\]\n\n\\[ \\Rightarrow \\quad dt = \\frac{1}{2\\sqrt{x}} \\, dx \\]\n\n\\[ \\Rightarrow \\quad 2t \\, dt = dx \\]", "id": "./materials/132.pdf" }, { "contents": "Integration by Trigonometric Substitution\n\nExample\n\n\\[ \\int \\sqrt{1 - x^2} \\, dx = \\ldots = \\int \\sqrt{1 - \\sin^2(t) \\cos(t)} \\, dt = \\ldots \\]\n\nDo the following substitution:\n\n\\[ \\sin(t) = x \\]\n\\[ \\Rightarrow \\cos(t) \\, dt = dx \\]\n\n- This substitution leads to a simplification in the radicand.\n- At the end it is necessary to apply an inverse substitution (to obtain the first variable).\nIntegration by Trigonometric Substitution\n\nTable of Trigonometric Substitution\n\n| Integrand | Substitution | Trigonometric Identity |\n|-----------------|--------------|------------------------|\n| $\\sqrt{1 - f^2}$ | $f = \\sin(t)$ | $1 - \\sin^2(t) = \\cos^2(t)$ |\n| | $-\\frac{\\pi}{2} \\leq t \\leq \\frac{\\pi}{2}$ | |\n| $\\sqrt{1 + f^2}$ | $f = \\tan(t)$ | $1 + \\tan^2(t) = \\sec^2(t)$ |\n| | $-\\frac{\\pi}{2} < t < \\frac{\\pi}{2}$ | |\n| $\\sqrt{f^2 - 1}$ | $f = \\sec(t)$ | $\\sec^2(t) - 1 = \\tan^2(t)$ |\n| | $0 \\leq t < \\frac{\\pi}{2}$ ou $\\pi \\leq t < \\frac{3\\pi}{2}$ | |", "id": "./materials/133.pdf" }, { "contents": "Integration of Trigonometric (power) Functions — Rules\n\nP4. \\[ \\int \\sin^2(mx) \\, dx = \\frac{1}{2m} (mx - \\sin(mx) \\cos(mx)) + C \\]\n\nP5. \\[ \\int \\cos^2(mx) \\, dx = \\frac{1}{2m} (mx + \\sin(mx) \\cos(mx)) + C \\]\n\nP6. \\[ \\int \\sin^n(x) \\, dx = -\\frac{\\sin^{n-1}(x) \\cos(x)}{n} + \\frac{n-1}{n} \\int \\sin^{n-2}(x) \\, dx \\]\n\nP7. \\[ \\int \\cos^n(x) \\, dx = \\frac{\\cos^{n-1}(x) \\sin(x)}{n} + \\frac{n-1}{n} \\int \\cos^{n-2}(x) \\, dx \\]\n\nP8. \\[ \\int \\sin^m(x) \\cos^n(x) \\, dx = \\frac{\\sin^{m+1}(x) \\cos^{n-1}(x)}{n+m} + \\frac{n-1}{n+m} \\int \\sin^m(x) \\cos^{n-2}(x) \\, dx, \\quad \\text{for } m \\neq 1 \\text{ or } n \\neq 1 \\]\n\nP9. \\[ \\int \\tan^n(x) \\, dx = \\frac{\\tan^{n-1}(x)}{n-1} - \\int \\tan^{n-2}(x) \\, dx, \\quad n \\neq 1 \\]\n\nP10. \\[ \\int \\cot^n(x) \\, dx = -\\frac{\\cot^{n-1}(x)}{n-1} - \\int \\cot^{n-2}(x) \\, dx, \\quad n \\neq 1 \\]\n\nP11. \\[ \\int \\sec^n(x) \\, dx = \\frac{\\tan(x) \\sec^{n-2}(x)}{n-1} + \\frac{n-2}{n-1} \\int \\sec^{n-2}(x) \\, dx, \\quad n \\neq 1 \\]\n\nP12. \\[ \\int \\csc^n(x) \\, dx = -\\frac{\\cot(x) \\csc^{n-2}(x)}{n-1} + \\frac{n-2}{n-1} \\int \\csc^{n-2}(x) \\, dx, \\quad n \\neq 1 \\]", "id": "./materials/134.pdf" }, { "contents": "Direct integration (or direct inspection)\n\nExamples:\n\n- \\( \\int 3 \\, dx = 3x + C \\)\n- \\( \\int \\frac{1}{2} \\, dx = \\frac{x}{2} + C \\)\n- \\( \\int -5 \\, dx = -5x + C \\)\n- \\( \\int 10^3 \\, dx = 10^3 x + C \\)\n\nFormula:\n\n\\[ \\int k \\, dx = kx + C \\]\n\n\\( k \\) a constant\n\n\\[ (kx + C)' = k \\]\n\\[ \\int x \\, dx = \\frac{x^2}{2} + C \\]\n\n\\[ \\int f(x) \\, dx = \\frac{f^{m+1}}{m+1} + C \\]\n\n\\[ f(x) = x \\]\n\\[ m = 1 \\Rightarrow m+1 = 2 \\]\n\\[ f'(x) = 1 \\]\n\n\\[ \\int x^3 \\, dx = \\frac{x^4}{4} + C \\]\n\n\\[ \\int x^{1/2} \\, dx = \\frac{x^{3/2}}{3/2} + C = \\frac{2}{3} x^{3/2} + C \\]\n\n\\[ m = \\frac{1}{2} \\Rightarrow m+1 = \\frac{3}{2} \\]\n\\[ \\int 1 \\cdot (x + 1)^3 \\, dx = \\frac{(x + 1)^4}{4} + C \\]\n\n\\[ \\int 2x \\cdot (x^2 + 1)^{10} \\, dx = \\frac{(x^2 + 1)^{11}}{11} + C \\]\n\n\\[ m + 1 = 10 + 1 = 11 \\]\n\n\\[ \\int x^2 \\cdot (x^3 + 2)^5 \\, dx = \\frac{1}{3} \\int 3x^2 \\cdot (x^3 + 2)^5 \\, dx = \\frac{1}{3} \\cdot \\frac{(x^3 + 2)^6}{6} + C \\]\n\n\\[ \\Rightarrow f'(x) = 3x^2 \\]\n\n\\[ = \\frac{(x^3 + 2)^6}{18} + C \\]", "id": "./materials/135.pdf" }, { "contents": "Find\n\\[ \\int \\ln(\\sqrt{x}) \\, dx = \\]\n\\[ = \\int \\ln(t) \\cdot \\frac{2}{t} \\, dt \\]\nby \\[ \\ln(t^2) = 2 \\ln(t) \\]\n\\[ = \\frac{1}{2} \\int \\ln(t) \\, dt \\]\n\\[ = \\frac{1}{2} \\ln(t) - \\frac{1}{2} t + C \\]\n\\[ = x \\cdot \\ln(\\sqrt{x}) - \\frac{x}{2} + C \\]\n\nSubstitution:\n\\[ x^{1/2} = t \\]\n\\[ \\Rightarrow \\frac{1}{2} x^{-1/2} \\, dx = dt \\]\n\\[ \\Rightarrow \\frac{1}{2} \\int dx = dt \\]\n\\[ \\Rightarrow dx = 2 \\, dt \\]\n\\[ f(t) = 2t \\Rightarrow \\int 2t \\, dt = t^2 + C \\]\n\\[ g(1) = \\ln(4) \\Rightarrow g'(1) = \\frac{1}{4} \\]", "id": "./materials/136.pdf" }, { "contents": "Find \\( \\int 2x + 2x^2 - \\frac{1}{x} \\, dx \\) = \\( \\star \\)\n\nwe know that:\n\\[\n\\int f \\pm g \\, dx = \\int f \\, dx \\pm \\int g \\, dx\n\\]\n\nThen,\n\\[\n\\star = \\int 2x \\, dx + \\int 2x^2 \\, dx - \\int \\frac{1}{x} \\, dx =\n\\]\n\nNow, \\( \\int k \\cdot f \\, dx = k \\cdot \\int f \\, dx \\), \\( k \\) a constant\n\n\\[\n= 2 \\int x \\, dx + 2 \\int x^2 \\, dx - \\int \\frac{1}{x} \\, dx\n\\]\n\n\\[\n= 2 \\cdot \\frac{x^2}{2} + 2 \\cdot \\frac{x^3}{3} - \\ln |x| + C\n\\]\n\n\\[\n= x^2 + \\frac{2}{3} x^3 - \\ln |x| + C\n\\]\nFind\n\\[ \\int \\ln(\\sqrt{x}) \\, dx = \\]\n\\[ = \\int \\ln(t) \\cdot \\frac{2}{t} \\, dt \\]\nby \\[ \\ln(t^2) = 2 \\ln(t) \\]\n\\[ = \\frac{1}{2} \\int \\ln(t) \\, dt \\]\n\\[ = \\frac{1}{2} \\ln(t) - \\frac{1}{2} t + C \\]\n\\[ = x \\cdot \\ln(\\sqrt{x}) - \\frac{x}{2} + C \\]\n\nSubstitution:\n\\[ x^{1/2} = t \\]\n\\[ \\Rightarrow \\frac{1}{2} x^{-1/2} \\, dx = dt \\]\n\\[ \\Rightarrow \\frac{1}{2} \\int dx = dt \\]\n\\[ \\Rightarrow dx = 2t \\, dt \\]\n\\[ f(t) = 2t \\Rightarrow \\int 2t \\, dt = t^2 + C \\]\n\\[ g(1) = \\ln(4) \\Rightarrow g'(1) = \\frac{1}{t} \\]", "id": "./materials/137.pdf" }, { "contents": "Find \\( \\int \\sqrt{x^2 + 2x} \\, dx = \\int \\frac{\\sqrt{x^2 + 2x + 1} - 1}{(x + 1)^2} \\, dx \\)\n\n\\[\n= \\int \\sqrt{(x+1)^2 - 1} \\, dx\n\\]\n\n\\[\n= \\int \\sqrt{\\sec^2(t) - 1} \\cdot \\sec(t) \\cdot \\tan(t) \\, dt\n\\]\n\n\\[\n= \\int \\tan(t) \\cdot \\sec(t) \\cdot \\tan(t) \\, dt\n\\]\n\n\\[\n= \\int \\tan^2(t) \\cdot \\sec(t) \\, dt\n\\]\n\n\\[\n= \\int (\\sec^2(t) - 1) \\cdot \\sec(t) \\, dt\n\\]\n\n\\[\n= \\int \\sec^3(t) - \\sec(t) \\, dt\n\\]\n\\[\n\\int \\sec^3(t) \\, dt = \\int \\sec(t) \\, dt \\\\\n\\text{direct}\n\\]\n\n\\[\n\\begin{align*}\n\\text{we know:} \\\\\n\\int \\sec^m(t) \\, dt &= \\frac{\\tan(t) \\sec^{m-2}(t)}{m-1} + \\frac{m-2}{m-1} \\int \\sec^{m-2}(t) \\, dt \\\\\n&\\quad \\text{for } m \\neq 1.\n\\end{align*}\n\\]\n\n\\[\n= \\frac{\\tan(t) \\cdot \\sec(t)}{2} + \\frac{1}{2} \\int \\sec(t) \\, dt - \\int \\sec(t) \\, dt\n\\]\n\n\\[\n= \\frac{\\tan(t) \\cdot \\sec(t)}{2} - \\frac{1}{2} \\ln |\\sec(t) + \\tan(t)| + C = \\Theta\n\\]\nBut, \\( x + 1 = \\sec(\\theta) \\)\n\nand we need to find \\( \\tan(\\theta) \\)!\n\nWe know,\n\n\\[\n1 + \\tan^2(\\theta) = \\sec^2(\\theta)\n\\]\n\n\\[\n\\Rightarrow \\tan^2(\\theta) = \\sec^2(\\theta) - 1\n\\]\n\n\\[\n\\Rightarrow \\tan(\\theta) = \\sqrt{\\sec^2(\\theta) - 1}\n\\]\n\n\\[\n\\Rightarrow \\tan(\\theta) = \\sqrt{(x+1)^2 - 1}\n\\]\n\n\\[\n\\Rightarrow \\tan(\\theta) = \\sqrt{x^2 + 2x}\n\\]\n\nThen,\n\n\\[\n\\Theta = \\frac{\\sqrt{x^2 + 2x} \\cdot (x+1)}{2} - \\frac{1}{2} \\ln |x+1 + \\sqrt{x^2 + 2x}| + C\n\\]", "id": "./materials/138.pdf" }, { "contents": "Can a binary relation be both symmetric and anti-symmetric?\n\nRoshan Poudel\n\nInstituto Politécnico de Bragança, Bragança, Portugal\nBinary Relations\n\nCartesian Product\n\nFor any two sets $X$ and $Y$, the Cartesian product of $X$ by $Y$ is defined as:\n\n$$X \\times Y = \\{(a, b) : a \\in X \\land b \\in Y\\}$$\n\nBinary Relations\n\n- If $X$ and $Y$ are two sets, then a binary relation from $X$ to $Y$ is a subset of $X \\times Y$.\n- A subset of $X \\times X$ is called a binary relation in $X$.\n- The empty set ($\\emptyset$) and the cartesian product $X \\times Y$ are binary relations from $X$ to $Y$.\n- If $R$ is a binary relation from $X$ to $Y$ and $a \\in X$, $b \\in Y$, we write $(a, b) \\in R$ or $aRb$. \nReflexive Relations\n\n**Definition**\n\nA binary relation $R$ defined in set $X$ is reflexive if it relates every element of $X$ to itself.\n\n$$R \\text{ is reflexive iff } \\forall a \\in X \\implies aRa$$\n\n**Example**\n\nFor a set $A = \\{1, 2, 3\\}$, the relation $R$ on $A$ defined as $R = \\{(1, 1), (1, 2), (1, 3), (2, 2), (3, 3)\\}$, is reflexive because $(1, 1), (2, 2), (3, 3)$ are in the relation.\nTransitive Relations\n\n**Definition**\n\nA binary relation $R$ defined in a set $X$ such that for all $a$, $b$ and $c$ in $X$, if $aRb$ and $bRc$ then $aRc$, is said to be transitive.\n\n$$R \\text{ is transitive iff } \\forall a, b, c \\in X, aRb \\land bRc \\implies aRc$$\n\n**Example**\n\nFor a set $A = \\{1, 2, 3\\}$, $R = \\{(1, 1), (1, 2), (2, 3), (1, 3), (3, 3)\\}$ is transitive because:\n\n- For every $a$, $b$, $c$, $aRb$ and $bRc$ implies $aRc$. Actually, $(1, 2)$ and $(2, 3)$ are in $R$ and so is $(1, 3)$, $(1, 1)$ and $(1, 2)$ are in $R$ and so is $(1, 2)$, $(1, 1)$ and $(1, 3)$ are in $R$ and so is $(1, 3)$, $(2, 3)$ and $(3, 3)$ are in $R$ and so is $(2, 3)$, $(1, 3)$ and $(3, 3)$ are in $R$ and so does $(1, 3)$.\n\n**Note:** If only $aRb$ exists without $bRc$ then it is not necessary\nSymmetric Relations\n\n**Definition**\n\nA binary relation $R$ defined in a set $X$ is said to be symmetric in $X$ if and only if for any $a$ and $b$ in $X$, $aRb$ implies $bRa$.\n\n$$R \\text{ is Symmetric iff } \\forall a, b \\in X, aRb \\implies bRa$$\n\n**Example**\n\nFor a set $A = \\{1, 2, 3\\}$, relation $R = \\{(1, 2), (2, 1), (2, 3), (3, 2), (3, 3)\\}$ is symmetric because:\n\n- For every $aRb$ there exists $bRa$. Actually, $(1, 2)$ and $(2, 1)$ both exist in $R$, $(2, 3)$ and $(3, 2)$ both exist in $R$.\n- For $(3, 3)$ the symmetric is also $(3, 3) \\in R$. \nA binary relation $R$ defined in a set $X$ is said to be anti-symmetric in $X$ if and only if for any $a$ and $b$ in $X$, $aRb$, $bRa$ implies $a = b$.\n\n$$R \\text{ is anti-symmetric iff } \\forall a, b \\in X, aRb \\land bRa \\implies a = b$$\n\nIf only $aRb$ exist and $bRa$ does not, then it is not necessary for $a = b$ for the relation $R$ to be anti-symmetric.\n\nNote: anti-symmetric doesn’t mean not symmetric.\nExample\n\nFor a set $A = \\{1, 2, 3\\}$, relation $R = \\{(1, 1), (2, 1), (1, 3), (3, 3)\\}$ is anti-symmetric because:\n\n- $(1, 1)$ and $(3, 3)$ both fit in the condition if $aRb$ and $bRa$ then $a = b$.\n- Furthermore, $(2, 1)$ and $(1, 3)$, their symmetric ones doesn’t exist in $R$ so they do not need to be equal for $R$ to be symmetric.\nA binary relation that is reflexive, symmetric and transitive is called an equivalence relation.\n\nThe equivalence class of an element $a$ of $X$ is the set of the elements of $X$ that relate to $a$:\n\n$$[a]_R = \\{ x \\in A : xRa \\}$$\n\nElement $a$ is said to represent such class.\nExample\n\nLet us consider a set $A = \\{a, b, c\\}$. Is\n\n$\\{R = (a, a), (b, b), (c, c), (a, c), (c, a)\\}$ an equivalence relation in $A$?\n\n- Since $(a, a), (b, b)$ and $(c, c)$ are all in $R$, $R$ is reflexive.\n- For all the pairs in $R$, the symmetric pair is also in $R$. For example $(a, c)$ has $(c, a)$, and the same happens for the other pairs of $R$. So, $R$ is also symmetric.\n- If $aRb$ and $bRc$ there is also $aRc$. For example, there is $aRa$ and $aRc$ and there is also $aRc$. This applies for all other possible combinations of pairs so, $R$ is also transitive.\n\nAs $R$ is reflexive, symmetric and transitive, then $R$ is an equivalence Relation.\nExample - Equivalent Classes\n\nIn the relation $R$ above, what are the equivalent classes of $[a]$, $[b]$ and $[c]$?\n\n1. In $R$, $a$ is related with $a$ and $c$, so, $[a] = \\{a, c\\}$\n2. In $R$, $b$ is related with $b$ only, so $[b] = \\{b\\}$\n3. In $R$, $c$ is related with $a$ and $c$, so, $[c] = \\{a, c\\}$.\n\nTherefore, the set of all equivalence classes for the equivalence relation $R$ is $\\{\\{a, c\\}, \\{b\\}\\}$. \nPartial Order\n\nDefinition\nA binary relation that is reflexive, anti-symmetric and transitive is called a partial order.\n\nExample\nIs a relation $R = \\{(1, 2), (1, 1), (2, 2), (2, 3), (1, 3), (3, 3)\\}$ in $X = \\{1, 2, 3\\}$ a partial order?\n\n- $R$ is reflexive as $(1, 1), (2, 2)$ and $(3, 3)$ all belong in $R$.\n- The only pairs whose symmetric also exists in $R$ are $(1, 1), (2, 2), (3, 3)$. so, here for all $aRb$ and $bRa$ then $a = b$. so, $R$ is anti-symmetric\n- If for all $a, b, c \\in X$, $aRb$ and $bRc$, there is also $aRc$. Like there is $aRa$ and $aRc$ then there is also $aRc$, so $R$ is also transitive.\n\n$R$ is reflexive, anti-symmetric and transitive. so, $R$ is a partial order.\nYes, a relation can be both symmetric and anti-symmetric at the same time. Or it can be neither as well.\n\n**Explanation**\n\nLet us consider a set $A = \\{1, 2, 3\\}$ and relation $R = \\{(1, 1), (2, 2), (3, 3)\\}$ in $A$. Let’s see if $R$ can be both symmetric and anti-symmetric:\n\n- For $(1, 1)$, the symmetric pair is also $(1, 1)$. The same happens for all the pairs $(x, x)$ in $R$, so the relation $R$ is symmetric.\n\n- Since the elements of $R$ are pairs of the type $(x, x)$, they satisfy the requirement ‘if $(a, b) \\in R$ and $(b, a) \\in R$ then $a = b$’ which is the condition required for anti-symmetry, so $R$ is also anti-symmetric.", "id": "./materials/139.pdf" }, { "contents": "LIMITS OF FUNCTIONS\n\n\\[ f : \\mathbb{R} \\rightarrow \\mathbb{R} \\]\n\n\\[ \\lim_{x \\to x_0} f(x) \\quad \\lim_{x \\to +\\infty} f(x) \\quad \\lim_{x \\to -\\infty} f(x) \\]\n\n\\[ \\lim_{x \\to x_0^+} f(x) \\quad \\lim_{x \\to x_0^-} f(x) \\]\nFor each of them there are 4 possibilities\n\n\\[\n\\lim_{x \\to a} f(x) = \\begin{cases} \n l & \\text{1. } l \\in \\mathbb{R} \\text{ finite real number} \\\\\n +\\infty & \\text{2.} \\\\\n -\\infty & \\text{3.} \\\\\n \\text{IT DOESN'T EXIST} & \\text{4.}\n\\end{cases}\n\\]\n\n2. \\( \\forall M \\in \\mathbb{R} \\exists k \\in \\mathbb{R} \\) such that \\( \\forall x \\geq k \\) we have \\( f(x) \\geq M \\)\n3. \\( \\forall M \\in \\mathbb{R} \\exists k \\in \\mathbb{R} \\) such that \\( \\forall x \\geq k \\)\n\n\\[ f(x) \\leq -M \\]\n\n1. \\( \\forall \\varepsilon > 0 \\exists k \\in \\mathbb{R} \\) s.t.\n\n\\[ \\forall x \\geq k \\quad |f(x) - l| < \\varepsilon \\]\n\n\\[ l - \\varepsilon \\leq f(x) \\leq l + \\varepsilon \\quad (\\varepsilon) \\]\n\\[ \\lim_{x \\to \\infty} f(x) = l^+ \\] it means that we are converging \"from above\"\n\nDef. is the same as before but the condition \\((*)\\) becomes \\(l \\leq f(x) \\leq l + \\varepsilon\\)\n\n\\[ \\lim_{x \\to \\infty} f(x) = l^- \\]\n\n\\[ \\begin{align*}\n(\\&) & \\text{ because } \\\\\n& l - \\varepsilon \\leq f(x) \\leq l\n\\end{align*} \\]\n\\[\n\\lim_{x \\to -\\infty} f(x) = \\begin{cases} \n\\infty & \\text{if } x \\to +\\infty \\\\\n-\\infty & \\text{if } x \\to -\\infty \\\\\n\\text{Doesn't exist} & \\text{otherwise}\n\\end{cases}\n\\]\n\n(Similar to before)\n\n(3) \\(\\forall M \\in \\mathbb{R} \\exists k \\in \\mathbb{R}\\)\n\ns.t. \\(\\forall x \\in \\mathbb{R} \\ x \\leq k\\)\n\nWe have \\(f(x) \\leq -M\\)\n\nand so on...\n\\[\n\\lim_{x \\to x_0} f(x) = \\begin{cases} \n\\infty & \\text{if } x \\in \\mathbb{R} \\\\\n-\\infty & \\text{if } x \\not\\in \\mathbb{R} \\\\\n\\text{Doesn't exist} & \\text{if } x = x_0\n\\end{cases}\n\\]\n\n(2) \\quad \\forall M \\in \\mathbb{R} \\quad \\exists \\delta > 0 \\quad \\forall x \\in (x_0 - \\delta, x_0 + \\delta) \\quad \\text{we have } f(x) > M\n\\( \\forall M \\in \\mathbb{R} \\quad \\exists \\varepsilon > 0 \\quad \\text{s.t.} \\quad \\forall x \\in (-3 + x_0, x_0 + 3) \\quad \\text{we have} \\quad f(x) \\leq -M \\)\n\n\\( \\lim_{x \\to x_0^+} f(x) = +\\infty \\)\n\nVery similar to (3), the only thing that changes is \\( x \\in (x_0, x_0 + \\varepsilon) \\)\n\\[ A \\geq 0 \\quad \\text{if} \\quad 0 \\geq 0 \\]\n\ns.t. \\[ A \\times 6 \\left( x_0 - 5, x_0 + 5 \\right) \\]\n\nwe have\n\n\\[ |f(x) - l| \\leq \\varepsilon \\]\n\n(\\text{that is,} \\quad l - \\varepsilon \\leq f(x) \\leq l + \\varepsilon)\nWhy do we care about limits?\n\n1) See next lecture — in order to do derivatives\n\n2) It can happen that we are faced with the problem of finding \"the asymptotic behavior of a function.\"\n\nClassical example: \\( f(x) = \\frac{\\sin(x)}{x} \\) vs. \\( f(x) = x^2 \\)\n\nQ: What is this point?", "id": "./materials/14.pdf" }, { "contents": "Let $U = \\{1, 2, 3, 4, \\ldots, 20\\}$,\n$A = \\{x: 5 < x \\leq 10\\}$,\n$B = \\{x: 8 \\leq x \\leq 15\\}$\n$C = \\{x: 1 \\leq x \\leq 5\\}$.\nFind $(A \\cap B \\cap C) \\cup C$.\n\n- $A = \\{6, 7, 8, 9, 10\\}$\n- $B = \\{8, 9, 10, 11, 12, 13, 14, 15\\}$\n- $C = \\{1, 2, 3, 4, 5\\}$\n\nNow finding $(A \\cap B \\cap C)$\n\n- $(A \\cap B \\cap C) = \\emptyset$ as there are no common elements in $A$, $B$ and $C$\n\nSo, calculating $(A \\cap B \\cap C) \\cup C$.\n\n- $(A \\cap B \\cap C) \\cup C = \\emptyset \\cup C = C$\n\n- so, $(A \\cap B \\cap C) \\cup C = \\{1, 2, 3, 4, 5\\}$", "id": "./materials/140.pdf" }, { "contents": "If $|A| = 25$, $|B| = 20$, $|A \\cap B| = 10$ and $|\\cup| = 40$. Then, Find $|A - B|$ and $|B - A|$.\n\n- $|A| = 25$, $|B| = 20$, $|\\cup| = 40$ and $|A \\cap B| = 10$\n\n- We know, $|\\cup| = |A| + |B| + |A \\cap B| - |A \\cup B|$\n\nSo, filling in all known values\n\n- $40 = 20 + 25 + 10 - |A \\cup B|$\n\n- so, $|A \\cup B| = 55 - 25 = 15$\n\nNow finding $|A - B|$\n\n- $|A| = |A - B| + |A \\cup B|$\n\n- $|A - B| = |A| - |A \\cup B|$\n\n- $|A - B| = 25 - 15 = 10$\n\nsimilarly,\n\n- $|B| = |B - A| + |A \\cup B|$\n\n- $|B - A| = 20 - 15 = 5$", "id": "./materials/141.pdf" }, { "contents": "2.2 Graphical Displays of Sample Data\n\nDotplots, Stem-and-Leaf Diagrams (Stemplots), Histograms, Boxplots, Bar Charts, Pie Charts, Pareto Diagrams, …\n\nExample: Random variable $X =$ “Age (years) of individuals at Memorial Union.”\n\nConsider the following sorted random sample of $n = 20$ ages:\n\n$$\\{18, 19, 19, 19, 20, 21, 21, 23, 24, 24, 26, 27, 31, 35, 35, 37, 38, 42, 46, 59\\}$$\n\n- **Dotplot**\n\n ![Dotplot Diagram]\n\n **Comment:** Uses all of the values. Simple, but crude; does not summarize the data.\n\n- **Stemplot**\n\n | Stem | Leaves |\n |------|--------|\n | Tens | Ones |\n | 1 | 8 9 9 9 |\n | 2 | 0 1 1 3 4 4 6 7 |\n | 3 | 1 5 5 7 8 |\n | 4 | 2 6 |\n | 5 | 9 |\n\n **Comment:** Uses all of the values more effectively. Grouping summarizes the data better.\nHistograms\n\n| Class Interval | Frequency (#occurrences) |\n|----------------|-------------------------|\n| [10, 20) | 4 |\n| [20, 30) | 8 |\n| [30, 40) | 5 |\n| [40, 50) | 2 |\n| [50, 60) | 1 |\n\nn = 20\n\nFrequency Histogram\n| Class Interval | Absolute Frequency (#occurrences) | Relative Frequency (Frequency ÷ n) |\n|----------------|----------------------------------|-----------------------------------|\n| [10, 20) | 4 | \\( \\frac{4}{20} = 0.20 \\) |\n| [20, 30) | 8 | \\( \\frac{8}{20} = 0.40 \\) |\n| [30, 40) | 5 | \\( \\frac{5}{20} = 0.25 \\) |\n| [40, 50) | 2 | \\( \\frac{2}{20} = 0.10 \\) |\n| [50, 60) | 1 | \\( \\frac{1}{20} = 0.05 \\) |\n\n\\( n = 20 \\)\n\n\\( \\frac{20}{20} = 1.00 \\)\n\nRelative frequencies are always between 0 and 1, and their sum is always = 1!\n\n**Relative Frequency Histogram**\n\n![Relative Frequency Histogram](image)\nOften, it is of interest to determine the total relative frequency, up to a certain value. For example, we see here that 0.60 of the age data are under 30 years, 0.85 are under 40 years, etc. The resulting cumulative distribution, which always increases monotonically from 0 to 1, can be represented by the discontinuous “step function” or “staircase function” in the first graph below. By connecting the right endpoints of the steps, we obtain a continuous polygonal graph called the ogive (pronounced “o-jive”), shown in the second graph. This has the advantage of approximating the rate at which the cumulative distribution increases within the intervals. For example, suppose we wish to know the median age, i.e., the age that divides the values into equal halves, above and below. It is clear from the original data that 25 does this job, but if data are unavailable, we can still estimate it from the ogive. Imagine drawing a flat line from 0.5 on the vertical axis until it hits the graph, then straight down to the horizontal “Age” axis somewhere in the interval [20, 30); it is this value we seek. But the cumulative distribution up to 20 years is 0.2, and up to 30 years is 0.6... a rise of 0.4 in 10 years, or 0.04 per year, on average. To reach 0.5 from 0.2 – an increase of 0.3 – would thus require a ratio of 0.3 / 0.04 = 7.5 years from 20 years, or 27.5 years. Medians and other percentiles will be addressed in the next section.\n\n| Class Interval | Absolute Frequency (#occurrences) | Relative Frequency (Frequency ÷ n) | Cumulative Relative Frequency |\n|----------------|----------------------------------|------------------------------------|------------------------------|\n| [0, 10) | 0 | 0.00 | 0.00 |\n| [10, 20) | 4 | 0.20 | 0.20 = 0.00 + 0.20 |\n| [20, 30) | 8 | 0.40 | 0.60 = 0.20 + 0.40 |\n| [30, 40) | 5 | 0.25 | 0.85 = 0.60 + 0.25 |\n| [40, 50) | 2 | 0.10 | 0.95 = 0.85 + 0.10 |\n| [50, 60) | 1 | 0.05 | 1.00 = 0.95 + 0.05 |\n\n\\[ n = 20 \\] 1.00\nProblem! Suppose that all ages 30 and older are “lumped” into a single class interval:\n\n\\{18, 19, 19, 19, 20, 21, 21, 23, 24, 24, 26, 27, 31, 35, 35, 37, 38, 42, 46, 59\\}\n\n| Class Interval | Absolute Frequency (#occurrences) | Relative Frequency (Frequency ÷ n) |\n|----------------|----------------------------------|-----------------------------------|\n| [10, 20) | 4 | \\(\\frac{4}{20} = 0.20\\) |\n| [20, 30) | 8 | \\(\\frac{8}{20} = 0.40\\) |\n| [30, 60) | 8 | \\(\\frac{8}{20} = 0.40\\) |\n\n\\(n = 20\\)\n\nRelative Frequency Histogram\n\nIf this outlier (59) were larger, the histogram would be even more distorted!\n**Remedy:** Let...\n\n\\[\n\\text{Area of each class rectangle} = \\text{Relative Frequency} \\\\\n\\text{Height of rectangle} \\times \\text{Class Width}\n\\]\n\nTherefore...\n\n\\[\n\\text{Density} = \\frac{\\text{Relative Frequency}}{\\text{Class Width}}\n\\]\n\n| Class Interval | Absolute Frequency (#occurrences) | Relative Frequency (Frequency ÷ n) | Density (Rel Freq ÷ Class Width) |\n|----------------|----------------------------------|------------------------------------|----------------------------------|\n| [10, 20); width = 10 | 4 | \\(\\frac{4}{20} = 0.20\\) | \\(\\frac{0.20}{10} = 0.02\\) |\n| [20, 30); width = 10 | 8 | \\(\\frac{8}{20} = 0.40\\) | \\(\\frac{0.40}{10} = 0.04\\) |\n| [30, 60); width = 30 | 8 | \\(\\frac{8}{20} = 0.40\\) | \\(\\frac{0.40}{30} = 0.01333\\ldots\\) |\n\n\\(n = 20\\)\n\n\\(\\frac{20}{20} = 1.00\\)\n\n**Density Histogram**\n\nTotal Area = 1!", "id": "./materials/144.pdf" }, { "contents": "2.3 Summary Statistics – Measures of Center and Spread\n\n**POPULATION**\n\nRandom Variable $X$, numerical\n- True “center” = ???\n- True “spread” = ???\n\n**SAMPLE, size $n$**\n\n- Measures of center\n - median, mode, mean\n- Measures of spread\n - range, variance, standard deviation\n\n**Distribution of $X$**\n\n- $X$ discrete\n- $X$ continuous\n\n**parameters**\n\n(“population characteristics”)\n\n- unknown fixed numerical values\n- usually denoted by Greek letters, e.g., $\\theta$ (“theta”)\n\n**Statistical Inference**\n\n**statistics**\n\n(“sample characteristics”)\n\n- known (or computable) numerical values obtained from sample data\n- estimators of parameters, e.g., $\\hat{\\theta}$\n- usually denoted by corresponding Roman letters\nMeasures of Center\n\nFor a given numerical random variable $X$, assume that a random sample $\\{x_1, x_2, \\ldots, x_n\\}$ has been selected, and sorted from lowest to highest values, i.e.,\n\n$$x_1 \\leq x_2 \\leq \\ldots \\leq x_{n-1} \\leq x_n$$\n\n- **sample median** = the numerical “middle” value, in the sense that half the data values are smaller, half are larger.\n\n If $n$ is odd, take the value in position $\\# \\frac{n+1}{2}$.\n\n If $n$ is even, take the average of the two closest neighboring data values, left (position $\\# \\frac{n}{2}$) and right (position $\\# \\frac{n}{2} + 1$).\n\nComments:\n\n- The sample median is robust (insensitive) with respect to the presence of outliers.\n\n- More generally, can also define **quartiles** ($Q_1 = 25\\%$ cutoff, $Q_2 = 50\\%$ cutoff = median, $Q_3 = 75\\%$ cutoff), or **percentiles** (a.k.a. **quantiles**), which divide the data values into any given $p\\%$ vs. $(100 - p)\\%$ split. Example: SAT scores\n\n- **sample mode** = the data value with the largest frequency ($f_{\\text{max}}$)\n\n Comment: The sample mode is robust to outliers.\n\nIf present, repeated sample data values can be neatly consolidated in a **frequency table**, vis-à-vis the corresponding dotplot. (If a value $x_i$ is not repeated, then its $f_i = 1$.)\n\n| $x_i$ | $f_i$ | $p(x_i) = f_i / n$ |\n|-------|-------|-------------------|\n| $x_1$ | $f_1$ | $f(x_1)$ |\n| $x_2$ | $f_2$ | $f(x_2)$ |\n| $\\vdots$ | $\\vdots$ | $\\vdots$ |\n| $x_k$ | $f_k$ | $f(x_k)$ |\n\nIf $n = 1$, then $f_1 = 1$. If $n > 1$, then $f_i < 1$ for $i \\neq 1$.\nExample: \\( n = 12 \\) random sample values of \\( X = \\) “Body Temperature (°F)”: \n\n\\[\n\\{98.5, 98.6, 98.6, 98.6, 98.6, 98.6, 98.9, 98.9, 98.9, 99.1, 99.1, 99.2\\}\n\\]\n\n| \\( x_i \\) | \\( f_i \\) | \\( p(x_i) \\) |\n|---|---|---|\n| 98.5 | 1 | 1/12 |\n| 98.6 | 5 | 5/12 |\n| 98.9 | 3 | 3/12 |\n| 99.1 | 2 | 2/12 |\n| 99.2 | 1 | 1/12 |\n\n\\[ n = 12 \\]\n\n- **Sample median** = \\( \\frac{98.6 + 98.9}{2} \\) = 98.75°F (six data values on either side)\n- **Sample mode** = 98.6°F\n- **Sample mean** = \\( \\frac{1}{12} \\left[ (98.5)(1) + (98.6)(5) + (98.9)(3) + (99.1)(2) + (99.2)(1) \\right] \\)\n\nor, \\( = (98.5) \\frac{1}{12} + (98.6) \\frac{5}{12} + (98.9) \\frac{3}{12} + (99.1) \\frac{2}{12} + (99.2) \\frac{1}{12} = 98.8°F \\)\n\n- **Sample mean** = the “weighted average” of all the data values\n\n\\[\n\\bar{x} = \\frac{1}{n} \\sum_{i=1}^{k} x_i f_i , \\quad \\text{where } f_i \\text{ is the absolute frequency of } x_i\n\\]\n\n\\[\n= \\sum_{i=1}^{k} x_i p(x_i) , \\quad \\text{where } p(x_i) = \\frac{f_i}{n} \\text{ is the relative frequency of } x_i\n\\]\n\nComments:\n\n- The sample mean is the **center of mass**, or “balance point,” of the data values.\n- The sample mean is sensitive to outliers. One common remedy for this…\n\n**Trimmed mean**: Compute the sample mean after deleting a predetermined number or percentage of outliers from each end of the data set, e.g., “10% trimmed mean.” Robust to outliers by construction.\n**Grouped Data** – Suppose the original values had been “lumped” into categories.\n\n**Example:** Recall the grouped “Memorial Union age” data set…\n\n| $x_i$ | Class Interval | Frequency $f_i$ | Relative Frequency $f_i/n$ | Density $(\\text{Rel Freq} \\div \\text{Class Width})$ |\n|-------|----------------|-----------------|-----------------------------|-----------------------------------------------|\n| 15 | [10, 20) | 4 | 0.20 | 0.02 |\n| 25 | [20, 30) | 8 | 0.40 | 0.04 |\n| 45 | [30, 60) | 8 | 0.40 | 0.013 |\n\n$n = 20$ \n\n- **group mean:** Same formula as above, with $x_i = \\text{midpoint of } i^{\\text{th}} \\text{ class interval}$. \n \\[\n \\bar{x}_{\\text{group}} = \\frac{1}{20} \\left[ (15)(4) + (25)(8) + (45)(8) \\right] = 31.0 \\text{ years}\n \\]\n\n**Exercise:** Compare this value with the ungrouped sample mean $\\bar{x} = 29.2 \\text{ years}$.\n\n- **group median** (& other quantiles):\n\n **Density Histogram**\n\n By definition, the median $Q$ divides the data set into equal halves, i.e., 0.50 above and below. In this example, it must therefore lie in the class interval [20, 30), and divide the 0.40 area of the corresponding class rectangle as shown. Since the 0.10 “strip” is $\\frac{1}{4}$ of that area, it proportionally follows that $Q$ must lie at $\\frac{1}{4}$ of the class width $30 - 20 = 10$, or 2.5, from the right endpoint of 30. That is, $Q = 30 - 2.5$, or $Q = 27.5 \\text{ years}$. \n (Check that the ungrouped median = 25 years.)\nFormal approach ~\n\nFirst, identify which class interval \\([a, b]\\) contains the desired quantile \\(Q\\) (e.g., median, quartile, etc.), and determine the respective left and right areas \\(A\\) and \\(B\\) into which it divides the corresponding class rectangle. Equating proportions for \\(\\text{Density} = \\frac{A+B}{b-a}\\), we obtain\n\n\\[\n\\text{Density} = \\frac{A}{Q-a} = \\frac{B}{b-Q},\n\\]\n\nfrom which it follows that\n\n\\[\nQ = a + \\frac{A}{\\text{Density}} \\quad \\text{or} \\quad Q = b - \\frac{B}{\\text{Density}} \\quad \\text{or} \\quad Q = \\frac{Ab+Ba}{A+B}.\n\\]\n\nFor example, in the grouped “Memorial Union age” data, we have \\(a = 20\\), \\(b = 30\\), and \\(A = 0.30\\), \\(B = 0.10\\). Substituting these values into any of the equivalent formulas above yields the median \\(Q_2 = 27.5\\).\n\n**Exercise:** Now that \\(Q_2\\) is found, use the formula again to find the first and third quartiles \\(Q_1\\) and \\(Q_3\\), respectively.\n\nNote also from above, we obtain the useful formulas\n\n\\[\nA = (Q-a) \\times \\text{Density}\n\\]\n\n\\[\nB = (b-Q) \\times \\text{Density}\n\\]\n\nfor calculating the areas \\(A\\) and \\(B\\), when a value of \\(Q\\) is given! This can be used when finding the area between two quantiles \\(Q_1\\) and \\(Q_2\\). (See next page for another way.)\nAlternative approach → First, form this column:\n\n| Class Interval | Frequency $f_i$ | Relative Frequency $f_i / n$ | Cumulative Relative Frequency $F_i = \\frac{f_1}{n} + \\frac{f_2}{n} + \\ldots + \\frac{f_i}{n}$ |\n|----------------|-----------------|-----------------------------|----------------------------------------------------------------------------------|\n| $I_0$ | 0 | 0 | 0 |\n| $I_1$ | $f_1$ | $f_1 / n$ | $F_1$ |\n| $I_2$ | $f_2$ | $f_2 / n$ | $F_2$ |\n| $\\vdots$ | $\\vdots$ | $\\vdots$ | $\\vdots$ |\n| $I_i$ | $f_i$ | $f_i / n$ | $F_{low} < 0.5$ |\n| $Q = ?$ | | | 0.5 |\n| $[a, b)$ | $f_{i+1}$ | $f_{i+1} / n$ | $F_{high} > 0.5$ |\n| $\\vdots$ | $\\vdots$ | $\\vdots$ | $\\vdots$ |\n| $I_k$ | $f_k$ | $f_k / n$ | 1 |\n\nThen\n\n$$Q = a + \\left( \\frac{0.5 - F_{low}}{F_{high} - F_{low}} \\right) (b - a)$$\n\nor\n\n$$Q = b - \\left( \\frac{F_{high} - 0.5}{F_{high} - F_{low}} \\right) (b - a).$$\n\nAgain, in the grouped “Memorial Union age” data, we have $a = 20$, $b = 30$, $F_{low} = 0.2$, and $F_{high} = 0.6$ (why?). Substituting these values into either formula yields the median $Q_2 = 27.5$. ✓\n\nTo find $Q_1$, replace the 0.5 in the formula by 0.25; to find $Q_3$, replace the 0.5 in the formula by 0.75, etc.\n\nConversely, if a quantile $Q$ in an interval $[a, b)$ is given, then we can solve for the cumulative relative frequency $F(Q)$ up to that quantile value:\n\n$$F(Q) = F(a) + \\left( \\frac{F(b) - F(a)}{b - a} \\right) (Q - a).$$\n\nIt follows that the relative frequency (i.e., area) between two quantiles $Q_1$ and $Q_2$ is equal to the difference between their cumulative relative frequencies: $F(Q_2) - F(Q_1)$. \n**Shapes of Distributions**\n\n**Symmetric distributions** correspond to values that are spread equally about a “center.”\n\n\\[\n\\text{mean} = \\text{median}\n\\]\n\n**Examples:** (Drawn for “smoothed histograms” of a random variable \\(X\\).)\n\n- **uniform**\n- **triangular**\n- **bell-shaped**\n\n**Note:** An important special case of the “bell-shaped” curve is the **normal distribution**, a.k.a. **Gaussian distribution**. **Example:** \\(X = \\text{IQ score}\\)\n\nOtherwise, if more outliers of \\(X\\) occur on one side of the median than the other, the corresponding distribution will be **skewed** in that direction, forming a **tail**.\n\n- **skewed to the left** (negatively skewed)\n- **skewed to the right** (positively skewed)\n\n**Examples:** \\(X = \\text{“calcium level (mg)”}\\) \\(X = \\text{“serum cholesterol level (mg/dL)”}\\)\n\nFurthermore, distributions can also be classified according to the number of “peaks”:\n\n- **unimodal**\n- **bimodal**\n- **multimodal**\nMeasures of Spread\n\nAgain assume that a numerical random sample \\( \\{x_1, x_2, \\ldots, x_n\\} \\) has been selected, and sorted from lowest to highest values, i.e.,\n\n\\[\nx_1 \\leq x_2 \\leq \\ldots \\leq x_{n-1} \\leq x_n\n\\]\n\n- **sample range** = \\( x_n - x_1 \\) (highest value – lowest value)\n\nComments:\n\n- Uses only the two most extreme values. Very crude estimator of spread.\n- The sample range is extremely sensitive to outliers. One common remedy …\n\n**Interquartile range (IQR)** = \\( Q_3 - Q_1 \\). Robust to outliers by construction.\n\n- If the original data are grouped into \\( k \\) class intervals \\([a_1, a_2), [a_2, a_3), \\ldots, [a_k, a_{k+1})\\), then the **group range** = \\( a_{k+1} - a_1 \\). A similar calculation holds for **group IQR**.\n\nExample: The “Body Temperature” data set has a **sample range** = 99.2 – 98.5 = 0.7°F.\n\n\\[\n\\{98.5, 98.6, 98.6, 98.6, 98.6, 98.6, 98.9, 98.9, 98.9, 99.1, 99.1, 99.2\\}\n\\]\n\n| \\( x_i \\) | \\( f_i \\) |\n|---|---|\n| 98.5 | 1 |\n| 98.6 | 5 |\n| 98.9 | 3 |\n| 99.1 | 2 |\n| 99.2 | 1 |\n\n\\( n = 12 \\)\nFor a much less crude measure of spread that uses all the data, first consider the following…\n\n**Definition:** \\( x_i - \\bar{x} \\) = **individual deviation** of the \\( i^{th} \\) sample data value from the sample mean\n\n| \\( x_i \\) | \\( x_i - \\bar{x} \\) | \\( f_i \\) |\n|---|---|---|\n| 98.5 | -0.3 | 1 |\n| 98.6 | -0.2 | 5 |\n| 98.9 | +0.1 | 3 |\n| 99.1 | +0.3 | 2 |\n| 99.2 | +0.4 | 1 |\n\n\\( n = 12 \\)\n\nNaively, an estimate of the spread of the data values might be calculated as the average of these \\( n = 12 \\) individual deviations from the mean. However, this will **always** yield zero!\n\n**FACT:**\n\n\\[\n\\sum_{i=1}^{k} (x_i - \\bar{x}) f_i = 0,\n\\]\n\ni.e., the sum of the deviations is always zero.\n\n**Check:** In this example, the sum = \\((-0.3)(1) + (-0.2)(5) + (0.1)(3) + (0.3)(2) + (0.4)(1) = 0. \\checkmark\\)\n\n**Exercise:** Prove this general fact algebraically.\n\n**Interpretation:** The sample mean is the **center of mass**, or “balance point,” of the data values.\nBest remedy: To make them non-negative, square the deviations before summing.\n\n- **sample variance**\n\n\\[\ns^2 = \\frac{1}{n-1} \\sum_{i=1}^{k} (x_i - \\bar{x})^2 f_i\n\\]\n\n\\(s^2\\) is not on the same scale as the data values!\n\n- **sample standard deviation**\n\n\\[\ns = +\\sqrt{s^2}\n\\]\n\n\\(s\\) is on the same scale as the data values.\n\nExample:\n\n| \\(x_i\\) | \\(x_i - \\bar{x}\\) | \\((x_i - \\bar{x})^2\\) | \\(f_i\\) |\n|--------|-----------------|-----------------|------|\n| 98.5 | -0.3 | +0.09 | 1 |\n| 98.6 | -0.2 | +0.04 | 5 |\n| 98.9 | +0.1 | +0.01 | 3 |\n| 99.1 | +0.3 | +0.09 | 2 |\n| 99.2 | +0.4 | +0.16 | 1 |\n\n\\(n = 12\\)\n\nThen…\n\n\\[\ns^2 = \\frac{1}{11} \\left[ (0.09)(1) + (0.04)(5) + (0.01)(3) + (0.09)(2) + (0.16)(1) \\right] = 0.06 \\text{ (°F)}^2,\n\\]\n\nso that…\n\n\\[\ns = \\sqrt{0.06} = 0.245\\text{°F}.\n\\]\n\nBody Temp has a small amount of variance.\n\nComments:\n\n- \\(s^2 = \\frac{\\sum (x_i - \\bar{x})^2 f_i}{n-1}\\) has the important frequently-recurring form \\(\\frac{SS}{df}\\), where \\(SS = \\text{“Sum of Squares”}\\) (sometimes also denoted \\(S_{xx}\\)) and \\(df = \\text{“degrees of freedom”} = n - 1\\), since the \\(n\\) individual deviations have a single constraint. (Namely, their sum must equal zero.)\n\n- Same formulas are used for grouped data, with \\(\\bar{x}_{\\text{group}}\\), and \\(x_i = \\text{class interval midpoint}\\).\n\n**Exercise:** Compute \\(s\\) for the grouped and ungrouped Memorial Union age data.\n\n- A related measure of spread is the **absolute deviation**, defined as \\(\\frac{1}{n} \\sum |x_i - \\bar{x}| f_i\\), but its statistical properties are not as well-behaved as the **standard deviation**. Also, see **Appendix > Geometric Viewpoint > Mean and Variance**, for a way to understand the “sum of squares” formula via the Pythagorean Theorem (!), as well as a useful alternate computational formula for the **sample variance**.\nTypical “Grouped Data” Exam Problem\n\n| Age Intervals | Frequencies |\n|---------------|-------------|\n| [0, 18) | - |\n| [18, 24) | 208 |\n| [24, 30) | 156 |\n| [30, 40) | 104 |\n| [40, 60) | 52 |\n| | 520 |\n\nGiven the sample frequency table of age intervals shown above; answer the following.\n\n1. Sketch the density histogram. (See Lecture Notes, page 2.2-6)\n2. Sketch the graph of the cumulative distribution. (page 2.2-4)\n3. What proportion of the sample is under 36 yrs old? (pages 2.3-5 bottom, 2.3-6 bottom)\n4. What proportion of the sample is under 45 yrs old? (same)\n5. What proportion of the sample is between 36 and 45 yrs old? (same)\n6. Calculate the values of the following grouped summary statistics.\n\n Quartiles $Q_1$, $Q_2$, $Q_3$ and IQR (pages 2.3-4 to 2.3-6)\n Mean (page 2.3-4)\n Variance (page 2.3-10, second comment on bottom)\n Standard deviation (same)\n\nSolutions at [http://www.stat.wisc.edu/~ifischer/Grouped_Data_Sols.pdf](http://www.stat.wisc.edu/~ifischer/Grouped_Data_Sols.pdf)", "id": "./materials/145.pdf" }, { "contents": "6. Statistical Inference and Hypothesis Testing\n\n6.1 One Sample\n\n§ 6.1.1 Mean\n\n**STUDY POPULATION** = Cancer patients on new drug treatment\n\nRandom Variable: \\( X = \\) “Survival time” (months)\nAssume \\( X \\sim N(\\mu, \\sigma) \\), with unknown mean \\( \\mu \\), but known \\( \\sigma = 6 \\) months.\n\nPopulation Distribution of \\( X \\)\n\n\\[ \\sigma = 6 \\]\n\n\\[ \\mu \\]\n\nWhat can be said about the mean \\( \\mu \\) of this study population?\n\n**RANDOM SAMPLE**, \\( n = 64 \\)\n\n\\( \\{X_1, X_2, X_3, X_4, X_5, \\ldots, X_{64}\\} \\)\n\nSampling Distribution of \\( \\bar{X} \\)\n\n\\[ \\frac{\\sigma}{\\sqrt{n}} = \\frac{6}{\\sqrt{64}} = 0.75 \\]\n\n\\( \\bar{X} \\) is called a “point estimate” of \\( \\mu \\)\n**Objective 1:** **Parameter Estimation** ~ Calculate an interval estimate of $\\mu$, centered at the point estimate $\\bar{x}$, that contains $\\mu$ with a high probability, say 95%. (Hence, $1 - \\alpha = 0.95$, so that $\\alpha = 0.05$.)\n\nThat is, for any random sample, solve for $d$:\n\n$$P(\\bar{X} - d \\leq \\mu \\leq \\bar{X} + d) = 0.95$$\n\ni.e., via some algebra,\n\n$$P(\\mu - d \\leq \\bar{X} \\leq \\mu + d) = 0.95.$$ \n\nBut recall that $Z = \\frac{\\bar{X} - \\mu}{\\sigma/\\sqrt{n}} \\sim N(0, 1)$. Therefore,\n\n$$P\\left(\\frac{-d}{\\sigma/\\sqrt{n}} \\leq Z \\leq \\frac{+d}{\\sigma/\\sqrt{n}}\\right) = 0.95$$\n\nHence, $\\frac{+d}{\\sigma/\\sqrt{n}} = z_{0.025} \\Rightarrow d = z_{0.025} \\times \\frac{\\sigma}{\\sqrt{n}} = (1.96)(0.75 \\text{ months}) = 1.47 \\text{ months}.$\n\n95% margin of error\n95% Confidence Interval for $\\mu$\n\n$$\\left( x - z_{0.025} \\frac{\\sigma}{\\sqrt{n}}, \\ x + z_{0.025} \\frac{\\sigma}{\\sqrt{n}} \\right)$$\n\n95% Confidence Limits\n\nwhere the critical value $z_{0.025} = 1.96$.\n\nTherefore, the margin of error (and thus, the size of the confidence interval) remains the same, from sample to sample.\n\nExample:\n\n| Sample | Mean $\\bar{x}$ | 95% CI |\n|--------|----------------|-----------------|\n| 1 | 26.0 mos | (26 - 1.47, 26 + 1.47) = 24.53 26 27.47 |\n| 2 | 27.0 mos | (27 - 1.47, 27 + 1.47) = 25.53 27 28.47 |\n\netc.\n\n**Interpretation:** Based on Sample 1, the true mean $\\mu$ of the “new treatment” population is between 24.53 and 27.47 months, with 95% “confidence.” Based on Sample 2, the true mean $\\mu$ is between 25.53 and 28.47 months, with 95% “confidence,” etc. The ratio of #CI’s that contain $\\mu$ to Total #CI’s $\\rightarrow 0.95$, as more and more samples are chosen, i.e., “The probability that a random CI contains the population mean $\\mu$ is equal to 0.95.” In practice however, the common (but technically incorrect) interpretation is that “the probability that a fixed CI (such as the ones found above) contains $\\mu$ is 95%.” In reality, the parameter $\\mu$ is constant; once calculated, a single fixed confidence interval either contains it or not.\nFor any significance level $\\alpha$ (and hence confidence level $1 - \\alpha$), we similarly define the...\n\n\\[\n(1 - \\alpha) \\times 100\\% \\text{ Confidence Interval for } \\mu\n\\]\n\n\\[\n\\left( \\bar{x} - Z_{\\alpha/2} \\frac{\\sigma}{\\sqrt{n}}, \\bar{x} + Z_{\\alpha/2} \\frac{\\sigma}{\\sqrt{n}} \\right)\n\\]\n\nwhere $Z_{\\alpha/2}$ is the critical value that divides the area under the standard normal distribution $N(0, 1)$ as shown. Recall that for $\\alpha = 0.10, 0.05, 0.01$ (i.e., $1 - \\alpha = 0.90, 0.95, 0.99$), the corresponding critical values are $z_{0.05} = 1.645$, $z_{0.025} = 1.960$, and $z_{0.005} = 2.576$, respectively. The quantity $Z_{\\alpha/2} \\frac{\\sigma}{\\sqrt{n}}$ is the two-sided margin of error.\n\nTherefore, as the significance level $\\alpha$ decreases (i.e., as the confidence level $1 - \\alpha$ increases), it follows that the margin of error increases, and thus the corresponding confidence interval widens. Likewise, as the significance level $\\alpha$ increases (i.e., as the confidence level $1 - \\alpha$ decreases), it follows that the margin of error decreases, and thus the corresponding confidence interval narrows.\n\nExercise: Why is it not realistic to ask for a 100% confidence interval (i.e., “certainty”)?\n\nExercise: Calculate the 90% and 99% confidence intervals for Samples 1 and 2 in the preceding example, and compare with the 95% confidence intervals.\nWe are now in a position to be able to conduct **Statistical Inference** on the population, via a formal process known as\n\n**Objective 2a: Hypothesis Testing** ~ “How does this new treatment compare with a ‘control’ treatment?” In particular, how can we use a **confidence interval** to decide this?\n\n**STANDARD POPULATION** = Cancer patients on standard drug treatment\n\nRandom Variable: \\( X = \\) “Survival time” (months)\n\nSuppose \\( X \\) is known to have **mean = 25** months.\n\nPopulation Distribution of \\( X \\)\n\n\\[ \\sigma = 6 \\]\n\n\\[ 25 \\]\n\nHow does this compare with the **mean \\( \\mu \\)** of the study population?\n\n**Technical Notes:** Although this is drawn as a bell curve, we don’t really care how the variable \\( X \\) is distributed in this population, as long as it is normally distributed in the study population of interest, an assumption we will learn how to check later, from the data. Likewise, we don’t really care about the value of the standard deviation \\( \\sigma \\) of this population, only of the study population. However, in the absence of other information, it is sometimes assumed (not altogether unreasonably) that the two are at least comparable in value. And if this is indeed a standard treatment, it has presumably been around for a while and given to many patients, during which time much data has been collected, and thus very accurate parameter estimates have been calculated. Nevertheless, for the vast majority of studies, it is still relatively uncommon that this is the case; in practice, very little if any information is known about any **population** standard deviation \\( \\sigma \\). In lieu of this value then, \\( \\sigma \\) is usually well-estimated by the **sample** standard deviation \\( s \\) with little change, if the sample is sufficiently “large,” but small samples present special problems. These issues will be dealt with later; for now, we will simply assume that the value of \\( \\sigma \\) is known.\nHence, let us consider the situation where, before any sampling is done, it is actually the experimenter’s intention to see if there is a statistically significant difference between the unknown mean survival time $\\mu$ of the “new treatment” population, and the known mean survival time of 25 months of the “standard treatment” population. (See page 1-1!) That is, the sample data will be used to determine whether or not to reject the formal...\n\nNull Hypothesis $H_0$: $\\mu = 25$\n\nversus the\n\nAlternative Hypothesis $H_A$: $\\mu \\neq 25$\n\nat the $\\alpha = 0.05$ significance level (i.e., the 95% confidence level).\n\nSample 1: 95% CI does contain $\\mu = 25$. Therefore, the data support $H_0$, and we cannot reject it at the $\\alpha = .05$ level. Based on this sample, the new drug does not result in a mean survival time that is significantly different from 25 months. Further study?\n\nSample 2: 95% CI does not contain $\\mu = 25$. Therefore, the data do not support $H_0$, and we can reject it at the $\\alpha = .05$ level. Based on this sample, the new drug does result in a mean survival time that is significantly different from 25 months. A genuine treatment effect.\n\nIn general...\n\nNull Hypothesis $H_0$: $\\mu = \\mu_0$\n\nversus the\n\nAlternative Hypothesis $H_A$: $\\mu \\neq \\mu_0$\n\nDecision Rule: If the $(1 - \\alpha) \\times 100\\%$ confidence interval contains the value $\\mu_0$, then the difference is not statistically significant; “accept” the null hypothesis at the $\\alpha$ level of significance. If it does not contain the value $\\mu_0$, then the difference is statistically significant; reject the null hypothesis in favor of the alternative at the $\\alpha$ significance level.\n**Objective 2b**: Calculate which sample mean values $\\bar{x}$ will lead to rejecting or not rejecting (i.e., “accepting” or “retaining”) the null hypothesis.\n\nFrom equation $\\star$ above, and the calculated margin of error $= 1.47$, we have...\n\n$$P(\\mu - 1.47 \\leq \\bar{x} \\leq \\mu + 1.47) = 0.95.$$ \n\nNow, IF the null hypothesis : $\\mu = 25$ is indeed true, then substituting this value gives...\n\n$$P(23.53 \\leq \\bar{x} \\leq 26.47) = 0.95.$$ \n\n**Interpretation**: If the mean survival time $\\bar{x}$ of a random sample of $n = 64$ patients is between 23.53 and 26.47, then the difference from 25 is “not statistically significant” (at the $\\alpha = .05$ significance level), and we retain the null hypothesis. However, if $\\bar{x}$ is either less than 23.53, or greater than 26.47, then the difference from 25 will be “statistically significant” (at $\\alpha = .05$), and we reject the null hypothesis in favor of the alternative. More specifically, if the former, then the result is significantly lower than the standard treatment average (i.e., new treatment is detrimental); if the latter, then the result is significantly higher than the standard treatment average (i.e., new treatment is beneficial).\n\nIn general...\n\n$$(1 - \\alpha) \\times 100\\% \\text{ Acceptance Region for } H_0: \\mu = \\mu_0$$\n\n$$\\left(\\mu_0 - Z_{\\alpha/2} \\frac{\\sigma}{\\sqrt{n}}, \\mu_0 + Z_{\\alpha/2} \\frac{\\sigma}{\\sqrt{n}}\\right)$$\n\n**Decision Rule**: If the $(1 - \\alpha) \\times 100\\%$ acceptance region contains the value $\\bar{x}$, then the difference is not statistically significant; “accept” the null hypothesis at the $\\alpha$ significance level. If it does not contain the value $\\bar{x}$, then the difference is statistically significant; reject the null hypothesis in favor of the alternative at the $\\alpha$ significance level.\nError Rates Associated with Accepting / Rejecting a Null Hypothesis\n\n(vis-à-vis Neyman-Pearson)\n\n- Confidence Level -\n\n\\[ P(\\text{Accept } H_0 \\mid H_0 \\text{ true}) = 1 - \\alpha \\]\n\n- Significance Level -\n\n\\[ P(\\text{Reject } H_0 \\mid H_0 \\text{ true}) = \\alpha \\]\n\nType I Error\n\nLikewise,\n\n\\[ P(\\text{Accept } H_0 \\mid H_0 \\text{ false}) = \\beta \\]\n\nType II Error\n\n- Power -\n\n\\[ P(\\text{Reject } H_0 \\mid H_\\alpha: \\mu = \\mu_1) = 1 - \\beta \\]\n**Objective 2c:** “How probable is my experimental result, if the null hypothesis is true?” Consider a sample mean value \\( \\bar{x} \\). Again assuming that the null hypothesis: \\( \\mu = \\mu_0 \\) is indeed true, calculate the **p-value** of the sample = the probability that any random sample mean is this far away or farther, in the direction of the alternative hypothesis. That is, how significant is the decision about \\( H_0 \\), at level \\( \\alpha \\)?\n\n\\[\nZ = \\frac{\\bar{X} - \\mu_0}{\\sigma / \\sqrt{n}} \\sim N(0, 1)\n\\]\n\n**Sample 1:** p-value = \\( P(\\bar{X} \\leq 24 \\text{ or } \\bar{X} \\geq 26) \\)\n\\[\n= P(\\bar{X} \\leq 24) + P(\\bar{X} \\geq 26)\n= 2 \\times P(\\bar{X} \\geq 26)\n= 2 \\times P\\left(Z \\geq \\frac{26 - 25}{0.75}\\right)\n= 2 \\times P(Z \\geq 1.333)\n= 2 \\times 0.0912\n= 0.1824 > 0.05 = \\alpha\n\\]\n\n**Sample 2:** p-value = \\( P(\\bar{X} \\leq 23 \\text{ or } \\bar{X} \\geq 27) \\)\n\\[\n= P(\\bar{X} \\leq 23) + P(\\bar{X} \\geq 27)\n= 2 \\times P(\\bar{X} \\geq 27)\n= 2 \\times P\\left(Z \\geq \\frac{27 - 25}{0.75}\\right)\n= 2 \\times P(Z \\geq 2.667)\n= 2 \\times 0.0038\n= 0.0076 < 0.05 = \\alpha\n\\]\n\n**Decision Rule:** If the p-value of the sample is greater than the significance level \\( \\alpha \\), then the difference is not statistically significant; “accept” the null hypothesis at this level. If the p-value is less than \\( \\alpha \\), then the difference is statistically significant; reject the null hypothesis in favor of the alternative at this level.\n\n**Guide to statistical significance of p-values for \\( \\alpha = .05 \\):**\n\n| p-value | Significance |\n|---------|--------------|\n| \\( 0 \\leq p \\leq .001 \\) | extremely strong |\n| \\( .005 \\leq p \\leq .01 \\) | strong |\n| \\( .01 \\leq p \\leq .05 \\) | moderate |\n| \\( .05 \\leq p \\leq .10 \\) | borderline |\n| \\( .10 \\leq p \\leq 1 \\) | not significant |\n\nRecall that \\( Z = 1.96 \\) is the \\( \\alpha = .05 \\) cutoff z-score!\nSummary of findings: Even though the data from both samples suggest a generally longer “mean survival time” among the “new treatment” population over the “standard treatment” population, the formal conclusions and interpretations are different. Based on Sample 1 patients ($\\bar{x} = 26$), the difference between the mean survival time $\\mu$ of the study population, and the mean survival time of 25 months of the standard population, is not statistically significant, and may in fact simply be due to random chance. Based on Sample 2 patients ($\\bar{x} = 27$) however, the difference between the mean age $\\mu$ of the study population, and the mean age of 25 months of the standard population, is indeed statistically significant, on the longer side. Here, the increased survival times serve as empirical evidence of a genuine, beneficial “treatment effect” of the new drug.\n\nComment: For the sake of argument, suppose that a third sample of patients is selected, and to the experimenter’s surprise, the sample mean survival time is calculated to be only $\\bar{x} = 23$ months. Note that the p-value of this sample is the same as Sample 2, with $\\bar{x} = 27$ months, namely, $0.0076 < 0.05 = \\alpha$. Therefore, as far as inference is concerned, the formal conclusion is the same, namely, reject $H_0: \\mu = 25$ months. However, the practical interpretation is very different! While we do have statistical significance as before, these patients survived considerably shorter than the standard average, i.e., the treatment had an unexpected effect of decreasing survival times, rather than increasing them. (This kind of unanticipated result is more common than you might think, especially with investigational drugs, which is one reason for formal hypothesis testing, before drawing a conclusion.)\nModification: Consider now the (unlikely?) situation where the experimenter knows that the new drug will not result in a “mean survival time” \\( \\mu \\) that is significantly less than 25 months, and would specifically like to determine if there is a statistically significant increase. That is, he/she formulates the following one-sided null hypothesis to be rejected, and complementary alternative:\n\nNull Hypothesis \\( H_0: \\mu \\leq 25 \\)\n\nversus the\n\nAlternative Hypothesis \\( H_A: \\mu > 25 \\)\n\nat the \\( \\alpha = 0.05 \\) significance level (i.e., the 95% confidence level).\n\nIn this case, the acceptance region for \\( H_0 \\) consists of sample mean values \\( \\bar{x} \\) that are less than the null-value of \\( \\mu_0 = 25 \\), plus the one-sided margin of error \\( = z_\\alpha \\frac{\\sigma}{\\sqrt{n}} = z_{0.05} \\frac{6}{\\sqrt{64}} = (1.645)(0.75) = 1.234 \\), hence 26.234. Note that \\( \\alpha \\) replaces \\( \\alpha/2 \\) here!\n\nSample 1: \\( p \\)-value = \\( P(\\bar{X} \\geq 26) \\)\n\n\\[ = P(Z \\geq 1.333) \\]\n\n\\[ = 0.0912 > 0.05 = \\alpha \\]\n\n(accept)\n\nSample 2: \\( p \\)-value = \\( P(\\bar{X} \\geq 27) \\)\n\n\\[ = P(Z \\geq 2.667) \\]\n\n\\[ = 0.0038 < 0.05 = \\alpha \\]\n\n(fairly strong rejection)\n\nNote that these one-sided \\( p \\)-values are exactly half of their corresponding two-sided \\( p \\)-values found above, potentially making the null hypothesis easier to reject. However, there are subtleties that arise in one-sided tests that do not arise in two-sided tests...\nConsider again the third sample of patients, whose sample mean is unexpectedly calculated to be only $\\bar{x} = 23$ months. Unlike the previous two samples, this evidence is in strong agreement with the null hypothesis $H_0: \\mu \\leq 25$ that the “mean survival time” is 25 months or less. This is confirmed by the p-value of the sample, whose definition (recall above) is “the probability that any random sample mean is this far away or farther, in the direction of the alternative hypothesis” which, in this case, is the right-sided $H_A: \\mu > 25$. Hence, \n\n$$p\\text{-value} = P(\\bar{X} \\geq 23) = P(Z \\geq -2.667) = 1 - 0.0038 = 0.9962 \\gg 0.05 = \\alpha$$\n\nwhich, as just observed informally, indicates a strong “acceptance” of the null hypothesis.\n\n**Exercise:** What is the one-sided p-value if the sample mean $\\bar{x} = 24$ mos? Conclusions?\n\n**A word of caution:** One-sided tests are less conservative than two-sided tests, and should be used sparingly, especially when it is a priori unknown if the mean response $\\mu$ is likely to be significantly larger or smaller than the null-value $\\mu_0$, e.g., testing the effect of a new drug. More appropriate to use when it can be clearly assumed from the circumstances that the conclusion would only be of practical significance if $\\mu$ is either higher or lower (but not both) than some tolerance or threshold level $\\mu_0$, e.g., toxicity testing, where only higher levels are of concern.\n\n**SUMMARY:** To test any null hypothesis for one mean $\\mu$, via the p-value of a sample...\n\n- **Step I:** Draw a picture of a bell curve, centered at the “null value” $\\mu_0$.\n- **Step II:** Calculate your sample mean $\\bar{x}$, and plot it on the horizontal $X$ axis.\n- **Step III:** From $\\bar{x}$, find the area(s) in the direction(s) of $H_A$ ($<, >$, or both tails), by first transforming $\\bar{x}$ to a z-score, and using the z-table. This is your p-value. SEE NEXT PAGE!\n- **Step IV:** Compare $p$ with the significance level $\\alpha$. If $<$, reject $H_0$. If $>$, retain $H_0$.\n- **Step V:** Interpret your conclusion in the context of the given situation!\nP-VALUES MADE EASY\n\n**Def:** Suppose a null hypothesis $H_0$ about a population mean $\\mu$ is to be tested, at a significance level $\\alpha$ (= .05, usually), using a known sample mean $x$ from an experiment. The **p-value** of the sample is the probability that a general random sample yields a mean $\\bar{X}$ that differs from the hypothesized “null value” $\\mu_0$, by an amount which is as large as – or larger than – the difference between our known $\\bar{X}$ value and $\\mu_0$. Thus, a small p-value (i.e., $< \\alpha$) indicates that our sample provides evidence against the null hypothesis, and we may reject it; the smaller the p-value, the stronger the rejection, and the more “statistically significant” the finding. A p-value $> \\alpha$ indicates that our sample does not provide evidence against the null hypothesis, and so we may not reject it. Moreover, a large p-value (i.e., $\\approx 1$) indicates empirical evidence in support of the null hypothesis, and we may retain, or even “accept” it. Follow these simple steps:\n\n**STEP 1.** From your sample mean $\\bar{X}$, calculate the standardized $z$-score $z = \\frac{\\bar{X} - \\mu_0}{\\sigma / \\sqrt{n}}$.\n\n**STEP 2.** What form is your alternative hypothesis?\n\n- $H_A: \\mu < \\mu_0$ (1-sided, left)........ $p$-value = tabulated entry corresponding to $z$-score\n = left shaded area, whether $z < 0$ or $z > 0$\n (illustrated)\n\n- $H_A: \\mu > \\mu_0$ (1-sided, right)...... $p$-value = 1 – tabulated entry corresponding to $z$-score\n = right shaded area, whether $z < 0$ or $z > 0$\n (illustrated)\n\n**Example:** Toxic levels of arsenic in drinking water? Test $H_0: \\mu < 10$ ppb (safe) vs. $H_A: \\mu \\geq 10$ ppb (unsafe), at $\\alpha = .05$. Assume $N(\\mu, \\sigma)$, with $\\sigma = 1.6$ ppb. A sample of $n = 64$ readings that average to $\\bar{x} = 10.1$ ppb would have a $z$-score $= 0.1/0.2 = 0.5$, which corresponds to a $p$-value $= 1 - 0.69146 = 0.30854 > .05$, hence not significant; toxicity has not been formally shown. (Unsafe levels are $x \\geq 10.33$ ppb. Why?)\n\n- $H_A: \\mu \\neq \\mu_0$ (2-sided)\n - If $z$-score is negative........ $p$-value $= 2 \\times$ tabulated entry corresponding to $z$-score\n $= 2 \\times$ left-tailed shaded area\n - If $z$-score is positive........ $p$-value $= 2 \\times (1 -$ tabulated entry corresponding to $z$-score)\n $= 2 \\times$ right-tailed shaded area\n\n**STEP 3.**\n\n- If the p-value is less than $\\alpha$ (= .05, usually), then **REJECT NULL HYPOTHESIS** - EXPERIMENTAL RESULT IS STATISTICALLY SIGNIFICANT AT THIS LEVEL!\n\n- If the p-value is greater than $\\alpha$ (= .05, usually), then **RETAIN NULL HYPOTHESIS** - EXPERIMENTAL RESULT IS NOT STATISTICALLY SIGNIFICANT AT THIS LEVEL!\n\n**STEP 4.** **IMPORTANT** - Interpret results in context. (Note: For many, this is the hardest step of all!)\nP-VALUES MADE SUPER EASY\n\nSTATBOT 301, MODEL Z\n\nSUBJECT: BASIC CALCULATION OF P-VALUES FOR Z-TEST\n\nCALCULATE... from $H_0$\n\nTest Statistic\n\n\"z-score\" = \\( \\frac{\\bar{X} - \\mu_0}{\\sigma/\\sqrt{n}} \\)\n\nRemember that the Z-table corresponds to the \"cumulative\" area to the left of any z-score.\n\nCheck the direction of the alternative hypothesis!\n\n$H_A$: $\\mu < \\mu_0$\n\n$H_A$: $\\mu \\neq \\mu_0$?\n\n$H_A$: $\\mu > \\mu_0$\n\ntable entry\n\nsign of z-score?\n\n- 2 × (table entry)\n\n+ 2 × (1 – table entry)\n\n1 – table entry\nPower and Sample Size Calculations\n\nRecall: $X =$ survival time (mos) $\\sim N(\\mu, \\sigma)$, with $\\sigma = 6$ (given). Testing null hypothesis $H_0: \\mu = 25$ (versus the 2-sided alternative $H_A: \\mu \\neq 25$), at the $\\alpha = .05$ significance level. Also recall that, by definition, power $= 1 - \\beta = P(\\text{Reject } H_0 \\mid H_0 \\text{ is false, i.e., } \\mu \\neq 25)$. Indeed, suppose that the mean survival time of “new treatment” patients is actually suspected to be $H_A: \\mu = 28$ mos. In this case, what is the resulting power to distinguish the difference and reject $H_0$, using a sample of $n = 64$ patients (as in the previous examples)?\n\nThese diagrams compare the null distribution for $\\mu = 25$, with the alternative distribution corresponding to $\\mu = 28$ in the rejection region of the null hypothesis. By definition, $\\beta = P(\\text{Accept } H_0 \\mid H_A: \\mu = 28)$, and its complement – the power to distinguish these two distributions from one another – is $1 - \\beta = P(\\text{Reject } H_0 \\mid H_A: \\mu = 28)$, as shown by the gold-shaded areas below. However, the “left-tail” component of this area is negligible, leaving the remaining “right-tail” area equal to $1 - \\beta$ by itself, approximately. Hence, this corresponds to the critical value $-z_\\beta$ in the standard normal $Z$-distribution (see inset), which transforms back to $28 - 0.75 z_\\beta$ in the $\\bar{X}$-distribution. Comparing this boundary value in both diagrams, we see that:\n\n$$28 - 0.75 z_\\beta = 26.47$$\n\nand solving yields $-z_\\beta = -2.04$. Thus, $\\beta = 0.0207$, and the associated power $= 1 - \\beta = 0.9793$, or 98%. Hence, we would expect to be able to detect significance 98% of the time, using 64 patients.\n**General Formulation:**\n\nProcurement of drug samples for testing purposes, or patient recruitment for clinical trials, can be extremely time-consuming and expensive. How to determine the minimum sample size \\( n \\) required to reject the null hypothesis \\( H_0: \\mu = \\mu_0 \\), in favor of an alternative value \\( H_A: \\mu = \\mu_1 \\), with a desired power \\( 1 - \\beta \\), at a specified significance level \\( \\alpha \\)? (And conversely, how to determine the power \\( 1 - \\beta \\) for a given sample size \\( n \\), as above?)\n\n| \\( H_0 \\) true | \\( H_0 \\) false |\n|-----------------|-----------------|\n| Reject \\( H_0 \\) | \\( \\times \\) Type I error, probability = \\( \\alpha \\) (significance level) | \\( \\checkmark \\) probability = \\( 1 - \\beta \\) (power) |\n| Accept \\( H_0 \\) | \\( \\checkmark \\) probability = \\( 1 - \\alpha \\) (confidence level) | \\( \\times \\) Type II error, probability = \\( \\beta \\) (1 – power) |\n\nThat is, \\( \\text{confidence level} = 1 - \\alpha = P(\\text{Accept } H_0: \\mu = \\mu_0 \\mid H_0 \\text{ is true}) \\), and \\( \\text{power} = 1 - \\beta = P(\\text{Reject } H_0: \\mu = \\mu_0 \\mid H_A: \\mu = \\mu_1) \\).\n\n**Null Distribution**\n\n\\[ X \\sim N\\left(\\mu_0, \\frac{\\sigma}{\\sqrt{n}}\\right) \\]\n\n**Alternative Distribution**\n\n\\[ X \\sim N\\left(\\mu_1, \\frac{\\sigma}{\\sqrt{n}}\\right) \\]\nHence (compare with (⋆) above),\n\n\\[ \\mu_1 - Z_\\beta \\frac{\\sigma}{\\sqrt{n}} = \\mu_0 + Z_{\\alpha/2} \\frac{\\sigma}{\\sqrt{n}}. \\]\n\nSolving for \\( n \\) yields the following.\n\nIn order to be able to detect a statistically significant difference (at level \\( \\alpha \\)) between the null population distribution having mean \\( \\mu_0 \\), and an alternative population distribution having mean \\( \\mu_1 \\), with a power of \\( 1 - \\beta \\), we require a minimum sample size of\n\n\\[ n = \\left( \\frac{Z_{\\alpha/2} + Z_\\beta}{\\Delta} \\right)^2, \\]\n\nwhere \\( \\Delta = \\frac{|\\mu_1 - \\mu_0|}{\\sigma} \\) is the “scaled difference” between \\( \\mu_0 \\) and \\( \\mu_1 \\).\n\nComments:\n\n- This formula corresponds to a two-sided hypothesis test. For a one-sided test, simply replace \\( \\alpha/2 \\) by \\( \\alpha \\). Recall that if \\( \\alpha = .05 \\), then \\( z_{.025} = 1.960 \\) and \\( z_{.05} = 1.645 \\).\n- If \\( \\sigma \\) is not known, then it can be replaced above by \\( s \\), the sample standard deviation, provided the resulting sample size turns out to be \\( n \\geq 30 \\), to be consistent with CLT. However, if the result is \\( n < 30 \\), then add 2 to compensate. [Modified from: Lachin, J. M. (1981), Introduction to sample size determination and power analysis for clinical trials. Controlled Clinical Trials, 2(2), 93-113.]\n\nWhat affects sample size, and how? With all other values being equal...\n\n- As power \\( 1 - \\beta \\) increases, \\( n \\) increases; as \\( 1 - \\beta \\) decreases, \\( n \\) decreases.\n- As the difference \\( \\Delta \\) decreases, \\( n \\) increases; as \\( \\Delta \\) increases, \\( n \\) decreases.\n\nExercise: Also show that \\( n \\) increases...\n\n- as \\( \\sigma \\) increases. [Hint: It may be useful to draw a picture, similar to above.]\n- as \\( \\alpha \\) decreases. [Hint: It may be useful to recall that \\( \\alpha \\) is the Type I Error rate, or equivalently, that \\( 1 - \\alpha \\) is the confidence level.]\nExamples: Recall that in our study, $\\mu_0 = 25$ months, $\\sigma = 6$ months.\n\nSuppose we wish to detect a statistically significant difference (at level $\\alpha = .05 \\Rightarrow z_{.025} = 1.960$) between this null distribution, and an alternative distribution having...\n\n- $\\mu_1 = 28$ months, with 90% power ($1 - \\beta = .90 \\Rightarrow \\beta = .10 \\Rightarrow z_{.10} = 1.282$). Then the scaled difference $\\Delta = \\frac{|28 - 25|}{6} = 0.5$, and\n \\[ n = \\left( \\frac{1.960 + 1.282}{0.5} \\right)^2 = 42.04, \\quad \\text{so} \\quad n \\geq 43 \\text{ patients}. \\]\n\n- $\\mu_1 = 28$ months, with 95% power ($1 - \\beta = .95 \\Rightarrow \\beta = .05 \\Rightarrow z_{.05} = 1.645$). Then,\n \\[ n = \\left( \\frac{1.960 + 1.645}{0.5} \\right)^2 = 51.98, \\quad \\text{so} \\quad n \\geq 52 \\text{ patients}. \\]\n\n- $\\mu_1 = 27$ months, with 95% power (so again, $z_{.05} = 1.645$). Then $\\Delta = \\frac{|27 - 25|}{6} = 0.333$,\n \\[ n = \\left( \\frac{1.960 + 1.645}{0.333} \\right)^2 = 116.96, \\quad \\text{so} \\quad n \\geq 117 \\text{ patients}. \\]\n\n---\n\n**Table of Sample Sizes* for Two-Sided Tests ($\\alpha = .05$)**\n\n| $\\Delta$ | 80% | 85% | 90% | 95% | 99% |\n|----------|------|------|------|------|------|\n| 0.1 | 785 | 898 | 1051 | 1300 | 1838 |\n| 0.125 | 503 | 575 | 673 | 832 | 1176 |\n| 0.15 | 349 | 400 | 467 | 578 | 817 |\n| 0.175 | 257 | 294 | 344 | 425 | 600 |\n| 0.2 | 197 | 225 | 263 | 325 | 460 |\n| 0.25 | 126 | 144 | 169 | 208 | 294 |\n| 0.3 | 88 | 100 | 117 | 145 | 205 |\n| 0.35 | 65 | 74 | 86 | 107 | 150 |\n| 0.4 | 50 | 57 | 66 | 82 | 115 |\n| 0.45 | 39 | 45 | 52 | 65 | 91 |\n| 0.5 | 32 | 36 | 43 | 52 | 74 |\n| 0.6 | 24 | 27 | 30 | 37 | 52 |\n| 0.7 | 19 | 21 | 24 | 29 | 38 |\n| 0.8 | 15 | 17 | 19 | 23 | 31 |\n| 0.9 | 12 | 14 | 15 | 19 | 25 |\n| 1.0 | 10 | 11 | 13 | 15 | 21 |\n\n* Shaded cells indicate that 2 was added to compensate for small $n$. \n**Power Curves** – A visual way to relate power and sample size.\n\n**Question:** Why is power not equal to 0 if $\\Delta = 0$?\nComments:\n\n- Due to time and/or budget constraints for example, a study may end before optimal sample size is reached. Given the current value of \\( n \\), the corresponding power can then be determined by the graph above, or computed exactly via the following formula.\n\n\\[\n\\text{Power} = 1 - \\beta = P(Z \\leq -z_{\\alpha/2} + \\Delta \\sqrt{n})\n\\]\n\nThe \\( z \\)-score can be +, -, or 0.\n\nExample: As in the original study, let \\( \\alpha = .05 \\), \\( \\Delta = \\frac{|28 - 25|}{6} = 0.5 \\), and \\( n = 64 \\). Then the \\( z \\)-score \\( = -1.96 + 0.5 \\sqrt{64} = 2.04 \\), so power \\( = 1 - \\beta = P(Z \\leq 2.04) = 0.9793 \\), or 98%. The probability of committing a Type 2 error \\( = \\beta = 0.0207 \\), or 2%. See page 6.1-15.\n\nExercise: How much power exists if the sample size is \\( n = 25? \\ 16? \\ 9? \\ 4? \\ 1? \\)\n\n- Generally, a minimum of 80% power is acceptable for reporting purposes.\n\n- Note: Larger sample size \\( \\Rightarrow \\) longer study time \\( \\Rightarrow \\) longer wait for results. In clinical trials and other medical studies, formal protocols exist for early study termination.\n\n- Also, to achieve a target sample size, practical issues must be considered (e.g., parking, meals, bed space,...). Moreover, may have to recruit many more individuals due to eventual censoring (e.g., move-aways, noncompliance,...) or death. $$$$$$$ issues...\n\n- Research proposals must have power and sample size calculations in their “methods” section, in order to receive institutional approval, support, and eventual journal publication.\nSmall Samples: Student’s t-distribution\n\nRecall that, vis-à-vis the Central Limit Theorem: \\( X \\sim N(\\mu, \\sigma) \\Rightarrow \\bar{X} \\sim N\\left(\\mu, \\frac{\\sigma}{\\sqrt{n}}\\right) \\), for any \\( n \\).\n\nTest statistic...\n\n- \\( \\sigma \\) known: \\( Z = \\frac{\\bar{X} - \\mu}{\\sigma/\\sqrt{n}} \\sim N(0, 1) \\).\n\n- \\( \\sigma \\) unknown, \\( n \\geq 30 \\): \\( Z = \\frac{\\bar{X} - \\mu}{s/\\sqrt{n}} \\sim N(0, 1) \\) approximately\n\n- \\( \\sigma \\) unknown, \\( n < 30 \\): \\( T = \\frac{\\bar{X} - \\mu}{s/\\sqrt{n}} \\sim t_{n-1} \\) ← Note: Can use for \\( n \\geq 30 \\) as well.\n\nStudent’s t-distribution, with \\( \\nu = n - 1 \\) degrees of freedom \\( df = 1, 2, 3, \\ldots \\)\n\n(Due to William S. Gossett (1876 - 1937), Guinness Brewery, Ireland, anonymously publishing under the pseudonym “Student” in 1908.)\n\n\\[ N(0, 1): \\varphi(z) = \\frac{1}{\\sqrt{2\\pi}} e^{-z^2/2} \\]\n\n\\[ t_{n-1}: \\quad f_{t}(t) = \\frac{1}{\\sqrt{(n-1)\\pi}} \\frac{\\Gamma\\left(\\frac{n}{2}\\right)}{\\Gamma\\left(\\frac{n-1}{2}\\right)} \\left(1 + \\frac{t^2}{n-1}\\right)^{-n/2} \\]\n\n\\( df = 1 \\) is also known as the Cauchy distribution.\n\nAs \\( df \\to \\infty \\), it follows that \\( T \\sim t_{df} \\to Z \\sim N(0, 1) \\).\nExample: Again recall that in our study, the variable $X =$ “survival time” was assumed to be normally distributed among cancer patients, with $\\sigma = 6$ months. The null hypothesis $H_0: \\mu = 25$ months was tested with a random sample of $n = 64$ patients; a sample mean of $\\bar{x} = 27.0$ months was shown to be statistically significant ($p = .0076$), i.e., sufficient evidence to reject the null hypothesis, suggesting a genuine difference, at the $\\alpha = .05$ level.\n\nNow suppose that $\\sigma$ is unknown and, like $\\mu$, must also be estimated from sample data. Further suppose that the sample size is small, say $n = 25$ patients, with which to test the same null hypothesis $H_0: \\mu = 25$, versus the two-sided alternative $H_A: \\mu \\neq 25$, at the $\\alpha = .05$ significance level. Imagine that a sample mean $\\bar{x} = 27.4$ months, and a sample standard deviation $s = 6.25$ months, are obtained. The greater mean survival time appears promising. However...\n\n\\[ \\text{s.e.} = \\frac{s}{\\sqrt{n}} = \\frac{6.25 \\text{ mos}}{\\sqrt{25}} = 1.25 \\text{ months} \\]\n\n\\[ (> \\text{s.e.} = 0.75 \\text{ months}) \\]\n\n\\[ \\text{critical value} = t_{24,.025} = 2.064 \\]\n\n\\[ \\text{Margin of Error} = (2.064)(1.25 \\text{ mos}) = 2.58 \\text{ months} \\]\n\n\\[ (> 1.47 \\text{ months, previously}) \\]\n\nSo...\n\n- **95% Confidence Interval for $\\mu = (27.4 - 2.58, 27.4 + 2.58) = (24.82, 29.98) \\text{ months}$, which does contain the null value $\\mu = 25 \\Rightarrow \\text{Accept } H_0$... No significance shown!**\n\n- **95% Acceptance Region for $H_0 = (25 - 2.58, 25 + 2.58) = (22.42, 27.58) \\text{ months}$, which does contain the sample mean $\\bar{x} = 27.4 \\Rightarrow \\text{Accept } H_0$... No significance shown!**\n\n\\[ \\text{p-value} = 2 \\text{ P}(\\bar{X} \\geq 27.4) = 2 \\text{ P}(T_{24} \\geq \\frac{27.4 - 25}{1.25}) \\]\n\n\\[ = 2 \\text{ P}(T_{24} \\geq 1.92) = 2(0.0334) = 0.0668, \\text{ which is greater than } \\alpha = .05 \\Rightarrow \\text{Accept } H_0... \\text{ No significance shown!} \\]\n\nWhy? The inability to reject is a typical consequence of small sample size, thus low power!\n\nAlso see **Appendix > Statistical Inference > Mean, One Sample** for more info and many more examples on this material.\nExample: A very simplified explanation of how fMRI works\n\nFunctional Magnetic Resonance Imaging (fMRI) is one technique of visually mapping areas of the human cerebral cortex in real time. First, a three-dimensional computer-generated image of the brain is divided into cube-shaped voxels (i.e., “volume elements” – analogous to square “picture elements,” or pixels, in a two-dimensional image), about 2-4 mm on a side, each voxel containing thousands of neurons. While the patient is asked to concentrate on a specific mental task, increased cerebral blood flow releases oxygen to activated neurons at a greater rate than to inactive ones (the so-called “hemodynamic response”), and the resulting magnetic resonance signal can be detected. In one version, each voxel signal is compared with the mean of its neighboring voxels; if there is a statistically significant difference in the measurements, then the original voxel is assigned one of several colors, depending on the intensity of the signal (e.g., as determined by the p-value); see figures.\n\nSuppose the variable $X =$ “Cerebral Blood Flow (CBF)” typically follows a normal distribution with mean $\\mu = 0.5$ ml/g/min at baseline. Further, suppose that the $n = 6$ neighbors surrounding a particular voxel (i.e., front and back, left and right, top and bottom) yields a sample mean of $\\bar{x} = 0.767$ ml/g/min, and sample standard deviation of $s = 0.082$ ml/g/min. Calculate the two-sided p-value of this sample (using baseline as the null hypothesis for simplicity), and determine what color should be assigned to the central voxel, using the scale shown.\n\nSolution: $X =$ “Cerebral Blood Flow (CBF)” is normally distributed, $H_0: \\mu = 0.5$ ml/g/min\n\n$n = 6 \\quad \\bar{x} = 0.767$ ml/g/min \\quad s = 0.082$ ml/g/min\n\nAs the population standard deviation $\\sigma$ is unknown, and the sample size $n$ is small, the t-test on $df = 6 - 1 = 5$ degrees of freedom is appropriate.\n\nUsing standard error estimate $\\hat{s.e.} = \\frac{s}{\\sqrt{n}} = \\frac{0.082 \\text{ ml/g/min}}{\\sqrt{6}} = 0.03348$ ml/g/min yields\n\n$p$-value $= 2 \\cdot P(\\bar{X} \\geq 0.767) = 2 \\cdot P\\left(T_5 \\geq \\frac{0.767 - 0.5}{0.03348}\\right) = 2 \\cdot P(T_5 \\geq 7.976) = 2 \\cdot (0.00025) = 0.0005$\n\nThis is strongly significant at any reasonable level $\\alpha$. According to the scale, the voxel should be assigned the color RED.\n**STATBOT 301, MODEL T**\n\n**Subject: Basic Calculation of p-values for T-test**\n\n**Calculate...**\n\nTest Statistic\n\n“t-score” = \\( \\frac{\\bar{X} - \\mu_0}{s/\\sqrt{n}} \\)\n\n**Alternative Hypothesis**\n\n- \\( H_A: \\mu < \\mu_0 \\)\n- \\( H_A: \\mu \\neq \\mu_0 \\)\n- \\( H_A: \\mu > \\mu_0 \\)\n\nRemember that the T-table corresponds to the area to the right of a positive t-score.\n\n\\[ t\\text{-score} = \\frac{\\bar{X} - \\mu_0}{s/\\sqrt{n}} \\]\n\n\\[ 1 - \\text{table entry} \\]\n\n\\[ 2 \\times \\text{table entry} \\]\n\n\\[ \\text{table entry} \\]\n\n\\[ \\text{table entry for } |t\\text{-score}| \\]\n\n\\[ 2 \\times \\text{table entry for } |t\\text{-score}| \\]\n\n\\[ 1 - \\text{table entry for } |t\\text{-score}| \\]\nChecks for normality ~ Is the ongoing assumption that the sample data come from a normally-distributed population reasonable?\n\n- **Quantiles:** As we have already seen, ≈ 68% within ±1 s.d. of mean, ≈ 95% within ±2 s.d. of mean, ≈ 99.7% within ±3 s.d. of mean, etc. Other percentiles can also be checked informally, or more formally via...\n\n- **Normal Scores Plot:** The graph of the quantiles of the n ordered (low-to-high) observations, versus the n known z-scores that divide the total area under N(0, 1) equally (representing an ideal sample from the standard normal distribution), should resemble a straight line. Highly skewed data would generate a curved plot. Also known as a probability plot or Q-Q plot (for “Quantile-Quantile”), this is a popular method.\n\nExample: Suppose n = 24 ages (years). Calculate the .04 quantiles of the sample, and plot them against the 24 known (i.e., “theoretical”) .04 quantiles of the standard normal distribution (below).\nSample 1:\n{6, 8, 11, 12, 15, 17, 20, 20, 21, 23, 24, 24, 26, 28, 29, 30, 31, 32, 34, 37, 40, 41, 42, 45}\n\nThe Q-Q plot of this sample (see first graph, below) reveals a more or less linear trend between the quantiles, which indicates that it is not unreasonable to assume that these data are derived from a population whose ages are indeed normally distributed.\n\nSample 2:\n{6, 6, 8, 8, 9, 10, 10, 10, 11, 11, 13, 16, 20, 21, 23, 28, 31, 32, 36, 38, 40, 44, 47, 50}\n\nThe Q-Q plot of this sample (see second graph, below) reveals an obvious deviation from normality. Moreover, the general “concave up” nonlinearity seems to suggest that the data are positively skewed (i.e., skewed to the right), and in fact, this is the case. Applying statistical tests that rely on the normality assumption to data sets that are not so distributed could very well yield erroneous results!\n\nFormal tests for normality include:\n- Anderson-Darling\n- Shapiro-Wilk\n- Lilliefors (a special case of Kolmogorov-Smirnov)\nRemedies for non-normality ~ What can be done if the normality assumption is violated, or difficult to verify (as in a very small sample)?\n\n- **Transformations:** Functions such as $Y = \\sqrt{X}$ or $Y = \\ln(X)$, can transform a positively-skewed variable $X$ into a normally distributed variable $Y$. (These functions “spread out” small values, and “squeeze together” large values. In the latter case, the original variable $X$ is said to be log-normal.)\n\n**Exercise:** Sketch separately the dotplot of $X$, and the dotplot of $Y = \\ln(X)$ (to two decimal places), and compare.\n\n| $X$ | $Y = \\ln(X)$ | Frequency |\n|-----|--------------|-----------|\n| 1 | 1 | 1 |\n| 2 | 2 | 2 |\n| 3 | 3 | 3 |\n| 4 | 4 | 4 |\n| 5 | 5 | 5 |\n| 6 | 5 | 5 |\n| 7 | 4 | 4 |\n| 8 | 4 | 4 |\n| 9 | 3 | 3 |\n| 10 | 3 | 3 |\n| 11 | 3 | 3 |\n| 12 | 2 | 2 |\n| 13 | 2 | 2 |\n| 14 | 2 | 2 |\n| 15 | 2 | 2 |\n| 16 | 1 | 1 |\n| 17 | 1 | 1 |\n| 18 | 1 | 1 |\n| 19 | 1 | 1 |\n| 20 | 1 | 1 |\n\n- **Nonparametric Tests:** Statistical tests (on the median, rather than the mean) that are free of any assumptions on the underlying distribution of the population random variable. Slightly less powerful than the corresponding parametric tests, tedious to carry out by hand, but their generality makes them very useful, especially for small samples where normality can be difficult to verify.\n\n- **Sign Test** (crude), **Wilcoxon Signed Rank Test** (preferred)\nGENERAL SUMMARY...\n\nStep-by-Step Hypothesis Testing\nOne Sample Mean \\( H_0: \\mu \\text{ vs. } \\mu_0 \\)\n\nIs random variable approximately normally distributed (or mildly skewed)?\n\nYes\n\nIs \\( \\sigma \\) known?\n\nYes\n\nUse \\( Z \\)-test (with \\( \\sigma \\))\n\n\\[ Z = \\frac{\\bar{X} - \\mu_0}{\\sigma/\\sqrt{n}} \\sim N(0,1) \\]\n\nNo\n\nUse \\( Z \\)-test or \\( t \\)-test (with \\( \\hat{\\sigma} = s \\))\n\n\\[ Z = \\frac{\\bar{X} - \\mu_0}{s/\\sqrt{n}} \\sim N(0,1) \\]\n\nNo, or don’t know\n\nYes\n\nIs \\( n \\geq 30 \\)?\n\nYes\n\nUse \\( t \\)-test (with \\( \\hat{\\sigma} = s \\))\n\n\\[ T = \\frac{\\bar{X} - \\mu_0}{s/\\sqrt{n}} \\sim t_{n-1} \\]\n\n(used most often in practice)\n\nNo\n\nUse a transformation, or a nonparametric test, e.g., Wilcoxon Signed Rank Test\n\nCONTINUE...\n**p-value:** “How do I know in which direction to move, to find the p-value?”\n\nSee STATBOT, page 6.1-14 (Z) and page 6.1-24 (T), or...\n\n**Alternative Hypothesis**\n\n| 1-sided, left | 2-sided | 1-sided, right |\n|---------------|---------|---------------|\n| $H_A: <$ | $H_A: \\neq$ | $H_A: >$ |\n\n- The **p-value** of an experiment is the probability (hence always between 0 and 1) of obtaining a random sample with an outcome that is as, or more, extreme than the one actually obtained, if the null hypothesis is true.\n\n- Starting from the value of the test statistic (i.e., z-score or t-score), the p-value is computed in the direction of the alternative hypothesis (either $<$, $>$, or both), which usually reflects the investigator’s belief or suspicion, if any.\n\n- If the p-value is “small,” then the sample data provides evidence that tends to refute the null hypothesis; in particular, if the p-value is less than the significance level $\\alpha$, then the null hypothesis can be rejected, and the result is **statistically significant** at that level. However, if the p-value is greater than $\\alpha$, then the null hypothesis is retained; the result is not statistically significant at that level. Furthermore, if the p-value is “large” (i.e., close to 1), then the sample data actually provides evidence that tends to support the null hypothesis.\n§ 6.1.2 Variance\n\nGiven: Null Hypothesis \\( H_0: \\sigma^2 = \\sigma_0^2 \\) (constant value)\n\nversus Alternative Hypothesis \\( H_A: \\sigma^2 \\neq \\sigma_0^2 \\)\n\nTwo-sided Alternative\nEither \\( \\sigma^2 < \\sigma_0^2 \\) or \\( \\sigma^2 > \\sigma_0^2 \\)\n\nTest statistic:\n\\[\nX^2 = \\frac{(n-1)s^2}{\\sigma_0^2} \\sim \\chi^2_{n-1}\n\\]\n\nSampling Distribution of \\( X^2 \\):\nChi-Squared Distribution, with \\( \\nu = n - 1 \\) degrees of freedom \\( df = 1, 2, 3, \\ldots \\)\n\nNote that the chi-squared distribution is not symmetric, but skewed to the right. We will not pursue the details for finding an acceptance region and confidence intervals for \\( \\sigma^2 \\) here. But this distribution will appear again, in the context of hypothesis testing for equal proportions.\n§ 6.1.3 Proportion\n\n**POPULATION**\n\n**Binary random variable**\n\n\\[\nY = \\begin{cases} \n1, & \\text{Success with probability } \\pi \\\\\n0, & \\text{Failure with probability } 1 - \\pi \n\\end{cases}\n\\]\n\n**Experiment:** \\( n \\) independent trials\n\n**SAMPLE**\n\n**Random Variable:** \\( X = \\# \\text{Successes} \\sim \\text{Bin}(n, \\pi) \\)\n\nRecall: Assuming \\( n \\geq 30, n\\pi \\geq 15, \\) and \\( n(1 - \\pi) \\geq 15, \\)\n\n\\[\nX \\sim N\\left( n\\pi, \\sqrt{n\\pi(1 - \\pi)} \\right), \\text{ approximately. (see §4.2)}\n\\]\n\nTherefore, dividing by \\( n \\)...\n\n\\[\n\\hat{\\pi} = \\frac{X}{n} \\sim N\\left( \\pi, \\frac{\\pi(1 - \\pi)}{n} \\right), \\text{ approximately.}\n\\]\n\n**Problem!** The expression for the standard error involves the very parameter \\( \\pi \\) upon which we are performing statistical inference. (This did not happen with inference on the mean \\( \\mu \\), where the standard error is s.e. = \\( \\sigma / \\sqrt{n} \\), which does not depend on \\( \\mu \\).)\n\n\\[\n\\text{Illustration of the bell curves } N\\left( \\pi, \\sqrt{\\frac{\\pi(1 - \\pi)}{n}} \\right)\n\\]\n\nfor \\( n = 100 \\), as proportion \\( \\pi \\) ranges from 0 to 1. Note how, rather than being fixed at a constant value, the “spread” s.e. is smallest when \\( \\pi \\) is close to 0 or 1 (i.e., when success in the population is either very rare or very common), and is maximum when \\( \\pi = 0.5 \\) (i.e., when both success and failure are equally likely).\n\n**Also see Problem 4.4/10.** This property of nonconstant variance has further implications; see “Logistic Regression” in section 7.3.\nExample: Refer back to the coin toss example of section 1.1, where a random sample of \\( n = 100 \\) independent trials is performed in order to acquire information about the probability \\( P(\\text{Heads}) = \\pi \\). Suppose that \\( X = 64 \\) Heads are obtained. Then the sample-based point estimate of \\( \\pi \\) is calculated as \\( \\hat{\\pi} = X / n = 64/100 = 0.64 \\). To improve this to an interval estimate, we can compute the...\n\n\\[\n(1 - \\alpha) \\times 100\\% \\text{ Confidence Interval for } \\pi\n\\]\n\n\\[\n\\left( \\hat{\\pi} - z_{\\alpha/2} \\sqrt{\\frac{\\hat{\\pi} (1 - \\hat{\\pi})}{n}}, \\quad \\hat{\\pi} + z_{\\alpha/2} \\sqrt{\\frac{\\hat{\\pi} (1 - \\hat{\\pi})}{n}} \\right)\n\\]\n\n\\[\\text{95\\% Confidence Interval for } \\pi\\]\n\n95\\% limits = \\( 0.64 \\pm z_{0.025} \\sqrt{\\frac{(0.64)(0.36)}{100}} = 0.64 \\pm 1.96 \\times 0.048 \\)\n\n\\[\\therefore \\text{95\\% CI} = (0.546, 0.734) \\text{ contains the true value of } \\pi, \\text{ with 95\\% confidence.}\\]\n\nIs the coin fair at the \\( \\alpha = .05 \\) level?\n\nNull Hypothesis \\( H_0: \\pi = 0.5 \\)\n\nvs. Alternative Hypothesis \\( H_A: \\pi \\neq 0.5 \\)\n\nAs the 95\\% CI does not contain the null-value \\( \\pi = 0.5 \\), \\( H_0 \\) can be rejected at the \\( \\alpha = .05 \\) level, i.e., the coin is not fair.\n\n\\[\\text{95\\% Acceptance Region for } H_0: \\pi = \\pi_0\\]\n\n\\[\n\\left( \\pi_0 - z_{\\alpha/2} \\sqrt{\\frac{\\pi_0 (1 - \\pi_0)}{n}}, \\quad \\pi_0 + z_{\\alpha/2} \\sqrt{\\frac{\\pi_0 (1 - \\pi_0)}{n}} \\right)\n\\]\n\n\\[\\text{95\\% Acceptance Region for } H_0: \\pi = 0.50\\]\n\n95\\% limits = \\( 0.50 \\pm z_{0.025} \\sqrt{\\frac{(0.50)(0.50)}{100}} = 0.50 \\pm 1.96 \\times 0.050 \\)\n\n\\[\\therefore \\text{95\\% AR} = (0.402, 0.598)\\]\n\nAs the 95\\% AR does not contain the sample proportion \\( \\hat{\\pi} = 0.64 \\), \\( H_0 \\) can be rejected at the \\( \\alpha = .05 \\) level, i.e., the coin is not fair.\n\\[ Z = \\frac{\\hat{\\pi} - \\pi_0}{\\sqrt{\\frac{\\pi_0(1 - \\pi_0)}{n}}} \\sim N(0, 1) \\]\n\n\\[ \\text{p-value} = 2 \\cdot P(\\hat{\\pi} \\geq 0.64) = 2 \\cdot P\\left(Z \\geq \\frac{0.64 - 0.50}{0.05}\\right) = 2 \\cdot P(Z \\geq 2.8) = 2 \\cdot 0.0026 = 0.0052 \\]\n\nAs \\( p \\ll \\alpha = 0.05 \\), \\( H_0 \\) can be strongly rejected at this level, i.e., the coin is not fair.\nComments:\n\n- A continuity correction factor of $\\pm \\frac{0.5}{n}$ may be added to the numerator of the Z test statistic above, in accordance with the “normal approximation to the binomial distribution” – see 4.2 of these Lecture Notes. (The “n” in the denominator is there because we are here dealing with proportion of success $\\hat{\\pi} = X / n$, rather than just number of successes $X$.)\n\n- Power and sample size calculations are similar to those of inference for the mean, and will not be pursued here.\n\nSee Appendix > Statistical Inference > General Parameters and FORMULA TABLES, and Appendix > Statistical Inference > Means and Proportions, One and Two Samples.", "id": "./materials/146.pdf" }, { "contents": "6.2 Two Samples\n\n§ 6.2.1 Means\n\nFirst assume that the samples are randomly selected from two populations that are independent, i.e., no relation exists between individuals of one population and the other, relative to the random variable, or any lurking or confounding variables that might have an effect on this variable.\n\nModel: Phase III Randomized Clinical Trial (RCT)\n\nMeasuring the effect of treatment (e.g., drug) versus control (e.g., placebo) on a response variable $X$, to determine if there is any significant difference between them.\n\nControl Arm | Treatment Arm\n---|---\nAssume $X_1 \\sim N(\\mu_1, \\sigma_1)$ | Assume $X_2 \\sim N(\\mu_2, \\sigma_2)$\n\nThen... ↓ CLT ↓\n\nSample, size $n_1$ | Sample, size $n_2$\n\n$\\bar{X}_1 \\sim N\\left(\\mu_1, \\frac{\\sigma_1}{\\sqrt{n_1}}\\right)$ | $\\bar{X}_2 \\sim N\\left(\\mu_2, \\frac{\\sigma_2}{\\sqrt{n_2}}\\right)$\n\nSo... $\\bar{X}_1 - \\bar{X}_2 \\sim N\\left(\\mu_1 - \\mu_2, \\sqrt{\\frac{\\sigma_1^2}{n_1} + \\frac{\\sigma_2^2}{n_2}}\\right)$\n\n$H_0$: $\\mu_1 - \\mu_2 = 0$\n(There is no difference in mean response between the two populations.)\n\nNull Distribution\n\nComments:\n\n- Recall from 4.1: If $Y_1$ and $Y_2$ are independent, then $\\text{Var}(Y_1 - Y_2) = \\text{Var}(Y_1) + \\text{Var}(Y_2)$.\n- If $n_1 = n_2$, the samples are said to be (numerically) balanced.\n- The null hypothesis $H_0$: $\\mu_1 - \\mu_2 = 0$ can be replaced by $H_0$: $\\mu_1 - \\mu_2 = \\mu_0$ if necessary, in order to compare against a specific constant difference $\\mu_0$ (e.g., 10 cholesterol points), with the corresponding modifications below.\n- $\\text{s.e.} = \\sqrt{\\frac{\\sigma_1^2}{n_1} + \\frac{\\sigma_2^2}{n_2}}$ can be replaced by $\\text{s.e.} = \\sqrt{\\frac{s_1^2}{n_1} + \\frac{s_2^2}{n_2}}$, provided $n_1 \\geq 30, n_2 \\geq 30$. \nExample: \\( X = \\) “cholesterol level (mg/dL)”\n\nTest \\( H_0: \\mu_1 - \\mu_2 = 0 \\) vs. \\( H_A: \\mu_1 - \\mu_2 \\neq 0 \\) for significance at the \\( \\alpha = .05 \\) level.\n\n| Placebo | Drug |\n|---------|------|\n| \\( n_1 = 80 \\) | \\( n_2 = 60 \\) |\n| \\( \\bar{x}_1 = 240 \\) | \\( \\bar{x}_2 = 229 \\) |\n| \\( s_1^2 = 1200 \\) | \\( s_2^2 = 600 \\) |\n\n\\[\n\\frac{s_1^2}{n_1} = \\frac{1200}{80} = 15, \\quad \\frac{s_2^2}{n_2} = \\frac{600}{60} = 10 \\quad \\Rightarrow \\quad \\text{s.e.} = \\sqrt{\\frac{s_1^2}{n_1} + \\frac{s_2^2}{n_2}} = \\sqrt{25} = 5\n\\]\n\n\\[\n(1 - \\alpha) \\times 100\\% \\text{ Confidence Interval for } \\mu_1 - \\mu_2\n\\]\n\n\\[\n(\\bar{x}_1 - \\bar{x}_2) - z_{\\alpha/2} \\sqrt{\\frac{s_1^2}{n_1} + \\frac{s_2^2}{n_2}}, \\quad (\\bar{x}_1 - \\bar{x}_2) + z_{\\alpha/2} \\sqrt{\\frac{s_1^2}{n_1} + \\frac{s_2^2}{n_2}}\n\\]\n\n\\[\n95\\% \\text{ Confidence Interval for } \\mu_1 - \\mu_2\n\\]\n\n95\\% limits = \\( 11 \\pm (1.96)(5) = 11 \\pm 9.8 \\) ← margin of error\n\n\\[\n\\Rightarrow 95\\% \\text{ CI} = (1.2, 20.8), \\text{ which does not contain } 0 \\Rightarrow \\text{ Reject } H_0. \\text{ Drug works!}\n\\]\n\n\\[\n(1 - \\alpha) \\times 100\\% \\text{ Acceptance Region for } H_0: \\mu_1 - \\mu_2 = \\mu_0\n\\]\n\n\\[\n(\\mu_0 - z_{\\alpha/2} \\sqrt{\\frac{s_1^2}{n_1} + \\frac{s_2^2}{n_2}}, \\mu_0 + z_{\\alpha/2} \\sqrt{\\frac{s_1^2}{n_1} + \\frac{s_2^2}{n_2}})\n\\]\n\n\\[\n95\\% \\text{ Acceptance Region for } H_0: \\mu_1 - \\mu_2 = 0\n\\]\n\n95\\% limits = \\( 0 \\pm (1.96)(5) = \\pm 9.8 \\) ← margin of error\n\n\\[\n\\Rightarrow 95\\% \\text{ AR} = (-9.8, +9.8), \\text{ which does not contain } 11 \\Rightarrow \\text{ Reject } H_0. \\text{ Drug works!}\n\\]\n\n\\[\n\\text{Test Statistic}\n\\]\n\n\\[\nZ = \\frac{(\\bar{X}_1 - \\bar{X}_2) - \\mu_0}{\\sqrt{\\frac{s_1^2}{n_1} + \\frac{s_2^2}{n_2}}} \\sim N(0, 1)\n\\]\n\n\\[\n\\text{p-value} = 2 P(\\bar{X}_1 - \\bar{X}_2 \\geq 11)\n\\]\n\n\\[\n= 2 P(Z \\geq \\frac{11 - 0}{5})\n\\]\n\n\\[\n= 2 P(Z \\geq 2.2)\n\\]\n\n\\[\n= 2(.0139)\n\\]\n\n\\[\n= .0278 < .05 = \\alpha\n\\]\n\n\\[\n\\Rightarrow \\text{ Reject } H_0. \\text{ Drug works!}\n\\]\nNull Distribution\n\n\\[ \\bar{X}_1 - \\bar{X}_2 \\sim N(0, 5) \\]\n\n\\[ \\mu_1 - \\mu_2 = 0 \\]\n\n0 is not in the 95% Confidence Interval = (1.2, 20.8)\n\n11 is not in the 95% Acceptance Region = (-9.8, 9.8)\nSmall samples: What if \\( n_1 < 30 \\) and/or \\( n_2 < 30 \\)? Then use the \\( t \\)-distribution, provided…\n\n\\[ H_0: \\sigma_1^2 = \\sigma_2^2 \\] \n(equivariance, homoscedasticity)\n\nTechnically, this requires a formal test using the \\( F \\)-distribution; see next section (§ 6.2.2). However, an informal criterion is often used:\n\n\\[ \\frac{1}{4} < F = \\frac{s_1^2}{s_2^2} < 4. \\]\n\nIf equivariance is accepted, then the common value of \\( \\sigma_1^2 \\) and \\( \\sigma_2^2 \\) can be estimated by the weighted mean of \\( s_1^2 \\) and \\( s_2^2 \\), the pooled sample variance:\n\n\\[ s_{\\text{pooled}}^2 = \\frac{df_1 s_1^2 + df_2 s_2^2}{df_1 + df_2}, \\text{ where } df_1 = n_1 - 1 \\text{ and } df_2 = n_2 - 1, \\]\n\ni.e.,\n\n\\[ s_{\\text{pooled}}^2 = \\frac{(n_1 - 1) s_1^2 + (n_2 - 1) s_2^2}{n_1 + n_2 - 2} = \\frac{SS}{df}. \\]\n\nTherefore, in this case, we have \\( \\text{s.e.} = \\sqrt{\\frac{\\sigma_1^2}{n_1} + \\frac{\\sigma_2^2}{n_2}} \\) estimated by\n\n\\[ \\text{s.e.} = \\sqrt{s_{\\text{pooled}}^2 \\left( \\frac{1}{n_1} + \\frac{1}{n_2} \\right)} \\]\n\ni.e.,\n\n\\[ \\text{s.e.} = s_{\\text{pooled}} \\sqrt{\\frac{1}{n_1} + \\frac{1}{n_2}}. \\]\n\nIf equivariance (but not normality) is rejected, then an approximate \\( t \\)-test can be used, with the approximate degrees of freedom \\( df \\) given by\n\n\\[ \\left( \\frac{s_1^2}{n_1} + \\frac{s_2^2}{n_2} \\right)^2 \\]\n\n\\[ \\frac{(s_1^2/n_1)^2}{n_1 - 1} + \\frac{(s_2^2/n_2)^2}{n_2 - 1}. \\]\n\nThis is known as the Smith-Satterwaite Test. (Also used is the Welch Test.)\nExample: \\( X = \\text{\"cholesterol level (mg/dL)\"} \\)\n\nTest: \\( H_0: \\mu_1 - \\mu_2 = 0 \\) vs. \\( H_A: \\mu_1 - \\mu_2 \\neq 0 \\) for significance at the \\( \\alpha = .05 \\) level.\n\n| Placebo | Drug |\n|---------|------|\n| \\( n_1 = 8 \\) | \\( n_2 = 10 \\) |\n| \\( \\bar{x}_1 = 230 \\) | \\( \\bar{x}_2 = 200 \\) |\n| \\( s_1^2 = 775 \\) | \\( s_2^2 = 1175 \\) |\n\n\\[ \\bar{x}_1 - \\bar{x}_2 = 30 \\]\n\n\\[ F = \\frac{s_1^2}{s_2^2} = 0.66, \\]\n\nwhich is between 0.25 and 4.\n\nEquivariance accepted \\( \\Rightarrow t\\text{-test} \\)\n\n\\[ \\text{Pooled Variance} \\]\n\n\\[ s_{\\text{pooled}}^2 = \\frac{(8 - 1)(775) + (10 - 1)(1175)}{8 + 10 - 2} = \\frac{16000}{16} = 1000 \\]\n\n\\[ \\uparrow \\]\n\n\\[ \\text{df} \\]\n\nNote that \\( s_{\\text{pooled}}^2 = 1000 \\) is indeed between the variances \\( s_1^2 = 775 \\) and \\( s_2^2 = 1175 \\).\n\n\\[ \\text{Standard Error} \\]\n\n\\[ \\text{s.e.} = \\sqrt{1000 \\left( \\frac{1}{8} + \\frac{1}{10} \\right)} = 15 \\]\n\n\\[ \\text{Margin of Error} = (2.120)(15) = 31.8 \\]\n\n\\[ \\text{Critical Value} \\]\n\n\\[ t_{16,.025} = 2.120 \\]\n\\[ (1 - \\alpha) \\times 100\\% \\text{ Confidence Interval for } \\mu_1 - \\mu_2 \\]\n\\[\n(\\bar{x}_1 - \\bar{x}_2) - t_{df, \\alpha/2} \\sqrt{\\frac{s_{pooled}^2}{n_1} + \\frac{s_{pooled}^2}{n_2}}, \\quad (\\bar{x}_1 - \\bar{x}_2) + t_{df, \\alpha/2} \\sqrt{\\frac{s_{pooled}^2}{n_1} + \\frac{s_{pooled}^2}{n_2}}\n\\]\nwhere \\( df = n_1 + n_2 - 2 \\)\n\n\\[ 95\\% \\text{ Confidence Interval for } \\mu_1 - \\mu_2 \\]\n95\\% limits = \\( 30 \\pm 31.8 \\) \\( \\leftarrow \\) margin of error\n\\[ \\therefore 95\\% \\text{ CI} = (-1.8, 61.8), \\text{ which contains } 0 \\Rightarrow \\text{Accept } H_0. \\]\n\n\\[ (1 - \\alpha) \\times 100\\% \\text{ Acceptance Region for } H_0: \\mu_1 - \\mu_2 = \\mu_0 \\]\n\\[\n(\\mu_0 - t_{df, \\alpha/2} \\sqrt{\\frac{s_{pooled}^2}{n_1} + \\frac{s_{pooled}^2}{n_2}}, \\mu_0 + t_{df, \\alpha/2} \\sqrt{\\frac{s_{pooled}^2}{n_1} + \\frac{s_{pooled}^2}{n_2}})\n\\]\nwhere \\( df = n_1 + n_2 - 2 \\)\n\n\\[ 95\\% \\text{ Acceptance Region for } H_0: \\mu_1 - \\mu_2 = 0 \\]\n95\\% limits = \\( 0 \\pm 31.8 \\) \\( \\leftarrow \\) margin of error\n\\[ \\therefore 95\\% \\text{ AR} = (-31.8, +31.8), \\text{ which contains } 30 \\Rightarrow \\text{Accept } H_0. \\]\n\n\\[ \\text{Test Statistic} \\]\n\\[\nT = \\frac{(\\bar{x}_1 - \\bar{x}_2) - \\mu_0}{\\sqrt{\\frac{s_{pooled}^2}{n_1} + \\frac{s_{pooled}^2}{n_2}}} \\sim t_{df}\n\\]\nwhere \\( df = n_1 + n_2 - 2 \\)\n\n\\[ \\text{p-value} = 2 P(\\bar{X}_1 - \\bar{X}_2 \\geq 30) \\]\n\\[ = 2 P\\left(T_{16} \\geq \\frac{30 - 0}{15}\\right) \\]\n\\[ = 2 P(T_{16} \\geq 2.0) \\]\n\\[ = 2(.0314) \\]\n\\[ = .0628 > .05 = \\alpha \\]\n\\[ \\Rightarrow \\text{Accept } H_0. \\]\n\nOnce again, low sample size implies low power to reject the null hypothesis. The tests do not show significance, and we cannot conclude that the drug works, based on the data from these small samples. Perhaps a larger study is indicated...\nNow consider the case where the two samples are **dependent**. That is, each observation in the first sample is **paired**, or **matched**, in a natural way on a corresponding observation in the second sample.\n\n**Examples:**\n\n- Individuals may be matched on characteristics such as age, sex, race, and/or other variables that might confound the intended response.\n- Individuals may be matched on personal relations such as siblings (similar genetics, e.g., twin studies), spouses (similar environment), etc.\n- Observations may be connected physically (e.g., left arm vs. right arm), or connected in time (e.g., before treatment vs. after treatment).\n\n\\[\nH_0: \\mu_1 - \\mu_2 = 0\n\\]\n\nAssume\n\\[\nX \\sim N(\\mu_1, \\sigma_1)\n\\]\n\nAssume\n\\[\nY \\sim N(\\mu_2, \\sigma_2)\n\\]\n\nSubtract...\n\\[\nD = X - Y \\sim N(\\mu, \\sigma)\n\\]\n\nwhere \\(\\mu_D = \\mu_1 - \\mu_2\\)\n\nSample, size \\(n\\)\n\n| # | \\(x_1\\) | \\(y_1\\) |\n|----|--------|--------|\n| 1 | | |\n| 2 | | |\n| 3 | | |\n| ...| | |\n| \\(n\\) | \\(x_n\\) | \\(y_n\\) |\n\nCalculate the difference \\(d_i = x_i - y_i\\) of each matched pair of observations, thereby forming a single collapsed sample \\(\\{d_1, d_2, d_3, \\ldots, d_n\\}\\), and apply the appropriate one-sample Z- or t- test to the equivalent null hypothesis \\(H_0: \\mu_D = 0\\).\nChecks for normality include normal scores plot (probability plot, Q-Q plot), etc., just as with one sample.\n\nRemedies for non-normality include transformations (e.g., logarithmic or square root), or nonparametric tests.\n\n- Independent Samples: Wilcoxon Rank Sum Test (=Mann-Whitney U Test)\n- Dependent Samples: Sign Test, Wilcoxon Signed Rank Test (just as with one sample)\nStep-by-Step Hypothesis Testing\nTwo Sample Means \\( H_0: \\mu_1 - \\mu_2 \\) vs. 0\n\nIndependent or Paired?\n\nYes\n\nAre \\( X_1 \\) and \\( X_2 \\) approximately normally distributed (or mildly skewed)?\n\nYes\n\nAre \\( \\sigma_1, \\sigma_2 \\) known?\n\nYes\n\nUse \\( Z \\)-test (with \\( \\sigma_1, \\sigma_2 \\))\n\n\\[\nZ = \\frac{(\\bar{X}_1 - \\bar{X}_2) - \\mu_0}{\\sqrt{\\frac{\\sigma_1^2}{n_1} + \\frac{\\sigma_2^2}{n_2}}}\n\\]\n\nNo\n\nNo, or don’t know\n\nCompute \\( D = X_1 - X_2 \\) for each \\( i = 1, 2, \\ldots, n \\).\n\nThen calculate...\n\n- sample mean \\( \\bar{d} = \\frac{1}{n} \\sum d_i \\)\n- sample variance \\( s_d^2 = \\frac{1}{n-1} \\sum (d_i - \\bar{d})^2 \\)\n\n... and GO TO “One Sample Mean” testing of \\( H_0: \\mu_D = 0 \\), section 6.1.1.\n\nNo\n\nUse a transformation, or a nonparametric test, e.g., Wilcoxon Rank Sum Test\n\nIndependent\n\nYes\n\nAre \\( n_1 \\geq 30 \\) and \\( n_2 \\geq 30 \\)?\n\nYes\n\nUse \\( Z \\)-test or \\( t \\)-test (with \\( \\sigma_1^2 = s_1^2, \\sigma_2^2 = s_2^2 \\))\n\n\\[\nZ = \\frac{(\\bar{X}_1 - \\bar{X}_2) - \\mu_0}{\\sqrt{\\frac{s_1^2}{n_1} + \\frac{s_2^2}{n_2}}}\n\\]\n\nNo\n\nPaired\n\nNo\n\nEquivariance: \\( \\sigma_1^2 = \\sigma_2^2 \\)?\n\nCompute \\( F = \\frac{s_1^2}{s_2^2} \\).\n\nIs \\( 1/4 < F < 4 \\)?\n\nYes\n\nUse \\( t \\)-test (with \\( \\sigma_1^2 = \\sigma_2^2 = s_{\\text{pooled}}^2 \\))\n\n\\[\nT_{n_1+n_2-2} = \\frac{(\\bar{X}_1 - \\bar{X}_2) - \\mu_0}{\\sqrt{s_{\\text{pooled}}^2 (1/n_1 + 1/n_2)}}\n\\]\n\n\\[\ns_{\\text{pooled}}^2 = \\frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}\n\\]\n\nNo\n\nUse an approximate \\( t \\)-test, e.g., Satterwaith Test\n\n... GO TO PAGE 6.1-28\n§ 6.2.2 Variances\n\nSuppose $X_1 \\sim N(\\mu_1, \\sigma_1)$ and $X_2 \\sim N(\\mu_2, \\sigma_2)$.\n\nNull Hypothesis $H_0$: $\\sigma_1^2 = \\sigma_2^2$\n\nversus\n\nAlternative Hypothesis $H_A$: $\\sigma_1^2 \\neq \\sigma_2^2$\n\nTest Statistic\n\n$$F = \\frac{s_1^2}{s_2^2} \\sim F_{\\nu_1, \\nu_2}$$\n\nwhere $\\nu_1 = n_1 - 1$ and $\\nu_2 = n_2 - 1$ are the corresponding numerator and denominator degrees of freedom, respectively.\n\nFormal test: Reject $H_0$ if the $F$-statistic is significantly different from 1.\n\nInformal criterion: Accept $H_0$ if the $F$-statistic is between 0.25 and 4.\n\nComment: Another test, more robust to departures from the normality assumption than the $F$-test, is Levene’s Test, a $t$-test of the absolute deviations of each sample. It can be generalized to more than two samples (see section 6.3.2).\n§ 6.2.3 Proportions\n\nPOPULATION\n\n**Binary random variable**\n\n\\[ I_1 = 1 \\text{ or } 0, \\text{ with} \\]\n\n\\[ P(I_1 = 1) = \\pi_1, \\quad P(I_1 = 0) = 1 - \\pi_1 \\]\n\n\\[ I_2 = 1 \\text{ or } 0, \\text{ with} \\]\n\n\\[ P(I_2 = 1) = \\pi_2, \\quad P(I_2 = 0) = 1 - \\pi_2 \\]\n\n\\[ \\hat{\\pi}_1 = \\frac{X_1}{n_1} \\sim N\\left(\\pi_1, \\sqrt{\\frac{\\pi_1 (1 - \\pi_1)}{n_1}}\\right), \\text{ approx.} \\]\n\n\\[ \\hat{\\pi}_2 = \\frac{X_2}{n_2} \\sim N\\left(\\pi_2, \\sqrt{\\frac{\\pi_2 (1 - \\pi_2)}{n_2}}\\right), \\text{ approx.} \\]\n\nTherefore, approximately...\n\n\\[ \\hat{\\pi}_1 - \\hat{\\pi}_2 \\sim N\\left(\\pi_1 - \\pi_2, \\sqrt{\\frac{\\pi_1 (1 - \\pi_1)}{n_1} + \\frac{\\pi_2 (1 - \\pi_2)}{n_2}}\\right). \\]\n\n\\[ \\uparrow \\text{ standard error s.e.} \\]\n\nConfidence intervals are computed in the usual way, using the estimate\n\n\\[ \\text{s.e.} = \\sqrt{\\frac{\\hat{\\pi}_1 (1 - \\hat{\\pi}_1)}{n_1} + \\frac{\\hat{\\pi}_2 (1 - \\hat{\\pi}_2)}{n_2}}, \\]\n\nas follows:\n(1 − α) × 100% Confidence Interval for \\( \\pi_1 - \\pi_2 \\)\n\n\\[\n\\left( \\hat{\\pi}_1 - \\hat{\\pi}_2 \\right) - Z_{\\alpha/2} \\sqrt{\\frac{\\hat{\\pi}_1(1-\\hat{\\pi}_1)}{n_1} + \\frac{\\hat{\\pi}_2(1-\\hat{\\pi}_2)}{n_2}}, \\quad \\left( \\hat{\\pi}_1 - \\hat{\\pi}_2 \\right) + Z_{\\alpha/2} \\sqrt{\\frac{\\hat{\\pi}_1(1-\\hat{\\pi}_1)}{n_1} + \\frac{\\hat{\\pi}_2(1-\\hat{\\pi}_2)}{n_2}}\n\\]\n\nUnlike the one-sample case, the same estimate for the standard error can also be used in computing the acceptance region for the null hypothesis \\( H_0: \\pi_1 - \\pi_2 = \\pi_0 \\), as well as the test statistic for the p-value, provided the null value \\( \\pi_0 \\neq 0 \\). HOWEVER, if testing for equality between two proportions via the null hypothesis \\( H_0: \\pi_1 - \\pi_2 = 0 \\), then their common value should be estimated by the more stable weighted mean of \\( \\hat{\\pi}_1 \\) and \\( \\hat{\\pi}_2 \\), the pooled sample proportion:\n\n\\[\n\\hat{\\pi}_{\\text{pooled}} = \\frac{X_1 + X_2}{n_1 + n_2} = \\frac{n_1 \\hat{\\pi}_1 + n_2 \\hat{\\pi}_2}{n_1 + n_2}.\n\\]\n\nSubstituting yields...\n\n\\[\n\\text{s.e.}_0 = \\sqrt{\\frac{\\hat{\\pi}_{\\text{pooled}}(1-\\hat{\\pi}_{\\text{pooled}})}{n_1} + \\frac{\\hat{\\pi}_{\\text{pooled}}(1-\\hat{\\pi}_{\\text{pooled}})}{n_2}}\n\\]\n\ni.e.,\n\n\\[\n\\text{s.e.}_0 = \\sqrt{\\hat{\\pi}_{\\text{pooled}}(1-\\hat{\\pi}_{\\text{pooled}})} \\sqrt{\\frac{1}{n_1} + \\frac{1}{n_2}}.\n\\]\n\nHence...\n\n(1 − α) × 100% Acceptance Region for \\( H_0: \\pi_1 - \\pi_2 = 0 \\)\n\n\\[\n\\left( 0 - Z_{\\alpha/2} \\sqrt{\\hat{\\pi}_{\\text{pooled}}(1-\\hat{\\pi}_{\\text{pooled}})} \\sqrt{\\frac{1}{n_1} + \\frac{1}{n_2}}, \\quad 0 + Z_{\\alpha/2} \\sqrt{\\hat{\\pi}_{\\text{pooled}}(1-\\hat{\\pi}_{\\text{pooled}})} \\sqrt{\\frac{1}{n_1} + \\frac{1}{n_2}} \\right)\n\\]\n\nTest Statistic for \\( H_0: \\pi_1 - \\pi_2 = 0 \\)\n\n\\[\nZ = \\frac{(\\hat{\\pi}_1 - \\hat{\\pi}_2) - 0}{\\sqrt{\\hat{\\pi}_{\\text{pooled}}(1-\\hat{\\pi}_{\\text{pooled}})} \\sqrt{\\frac{1}{n_1} + \\frac{1}{n_2}}} \\sim \\text{N}(0, 1)\n\\]\nExample: Consider a group of 720 patients who undergo physical therapy for arthritis. A daily supplement of glucosamine and chondroitin is given to \\( n_1 = 400 \\) of them in addition to the physical therapy; after four weeks of treatment, \\( X_1 = 332 \\) show measurable signs of improvement (increased ROM, etc.). The remaining \\( n_2 = 320 \\) patients receive physical therapy only; after four weeks, \\( X_2 = 244 \\) show improvement. Does this difference represent a statistically significant treatment effect? Calculate the p-value, and form a conclusion at the \\( \\alpha = .05 \\) significance level.\n\n\\[\n\\begin{array}{c|c}\n\\text{PT + Supplement} & \\text{PT only} \\\\\n\\hline\nn_1 = 400 & n_2 = 320 \\\\\nX_1 = 332 & X_2 = 244 \\\\\n\\end{array}\n\\]\n\n\\[\n\\hat{\\pi}_1 = \\frac{332}{400} = 0.83, \\quad \\hat{\\pi}_2 = \\frac{244}{320} = 0.7625\n\\]\n\n\\[\n\\hat{\\pi}_{\\text{pooled}} = \\frac{332 + 244}{400 + 320} = \\frac{576}{720} = 0.8\n\\]\n\nand thus \\( 1 - \\hat{\\pi}_{\\text{pooled}} = \\frac{144}{720} = 0.2 \\)\n\nTherefore, \\( \\text{p-value} = 2 \\cdot P(\\hat{\\pi}_1 - \\hat{\\pi}_2 \\geq 0.0675) = 2 \\cdot P\\left(Z \\geq \\frac{0.0675 - 0}{0.03}\\right) = 2 \\cdot P(Z \\geq 2.25) = 2(0.0122) = 0.0244 \\).\n\nConclusion: As this value is smaller than \\( \\alpha = .05 \\), we can reject the null hypothesis that the two proportions are equal. There does indeed seem to be a moderately significant treatment difference between the two groups.\nExercise: Instead of $H_0: \\pi_1 - \\pi_2 = 0$ vs. $H_A: \\pi_1 - \\pi_2 \\neq 0$, test the null hypothesis for a 5% difference, i.e., $H_0: \\pi_1 - \\pi_2 = .05$ vs. $H_A: \\pi_1 - \\pi_2 \\neq .05$, at $\\alpha = .05$. [Note that the pooled proportion $\\hat{\\pi}_{pooled}$ is no longer appropriate to use in the expression for the standard error under the null hypothesis, since $H_0$ is not claiming that the two proportions $\\pi_1$ and $\\pi_2$ are equal (to a common value); see notes above.] Conclusion?\n\nExercise: Instead of $H_0: \\pi_1 - \\pi_2 = 0$ vs. $H_A: \\pi_1 - \\pi_2 \\neq 0$, test the one-sided null hypothesis $H_0: \\pi_1 - \\pi_2 \\leq 0$ vs. $H_A: \\pi_1 - \\pi_2 > 0$ at $\\alpha = .05$. Conclusion?\n\nExercise: Suppose that in a second experiment, $n_1 = 400$ patients receive a new drug that targets B-lymphocytes, while the remaining $n_2 = 320$ receive a placebo, both in addition to physical therapy. After four weeks, $X_1 = 376$ and $X_2 = 272$ show improvement, respectively. Formally test the null hypothesis of equal proportions at the $\\alpha = .05$ level. Conclusion?\n\nExercise: Finally suppose that in a third experiment, $n_1 = 400$ patients receive “magnet therapy,” while the remaining $n_2 = 320$ do not, both in addition to physical therapy. After four weeks, $X_1 = 300$ and $X_2 = 240$ show improvement, respectively. Formally test the null hypothesis of equal proportions at the $\\alpha = .05$ level. Conclusion?\n\nSee...\n\nAppendix > Statistical Inference > General Parameters and FORMULA TABLES.\nAlternate Method: Chi-Squared ($\\chi^2$) Test\n\nAs before, let the binary variable $I = 1$ for improvement, $I = 0$ for no improvement, with probability $\\pi$ and $1 - \\pi$, respectively. Now define a second binary variable $J = 1$ for the “PT + Drug” group, and $J = 0$ for the “PT only” group. Thus, there are four possible disjoint events: “$I = 0$ and $J = 0$,” “$I = 0$ and $J = 1$,” “$I = 1$ and $J = 0$,” and “$I = 1$ and $J = 1$.” The number of times these events occur in the random sample can be arranged in a $2 \\times 2$ contingency table that consists of four cells (NW, NE, SW, and SE) as demonstrated below, and compared with their corresponding expected values based on the null hypothesis.\n\n**Observed Values**\n\n| Status (I) | PT + Drug | PT only | Row marginal totals |\n|------------|-----------|---------|---------------------|\n| Improvement | 332 | 244 | 576 |\n| No Improvement | 68 | 76 | 144 |\n| **Total** | **400** | **320** | **720** |\n\nversus...\n\n**Expected Values**\n\nunder $H_0$: $\\pi_1 = \\pi_2$\n\n$\\hat{\\pi}_{pooled} = \\frac{576}{720} = 0.8$\n\n**Group (J)**\n\n| Status (I) | PT + Drug | PT only | Column marginal totals |\n|------------|-----------|---------|-----------------------|\n| Improvement | $\\frac{400 \\times 576}{720} = 320.0$ | $\\frac{320 \\times 576}{720} = 256.0$ | 576 |\n| No Improvement | $\\frac{400 \\times 144}{720} = 80.0$ | $\\frac{320 \\times 144}{720} = 64.0$ | 144 |\n| **Total** | **400.0** | **320.0** | **720** |\n\nInformal reasoning: Consider the first cell, improvement in the 400 patients of the “PT + Drug” group. The null hypothesis conjectures that the probability of improvement is equal in both groups, and this common value is estimated by the pooled proportion $\\frac{576}{720}$. Hence, the expected number (under $H_0$) of improved patients in the “PT + Drug” group is $400 \\times \\frac{576}{720}$, etc.\n\nNote that, by construction, $H_0: \\frac{320}{400} = \\frac{256}{320}$, the pooled proportion.\nIdeally, if all the observed values = all the expected values, then this statistic would = 0, and the corresponding p-value = 1. As it is,\n\n\\[ X^2 = \\sum \\frac{(\\text{Obs} - \\text{Exp})^2}{\\text{Exp}} \\sim \\chi^2_1 \\]\n\nTherefore, the p-value = \\( P(\\chi^2_1 \\geq 5.0625) = .0244 \\), as before. Reject \\( H_0 \\).\n\nComments:\n\n- Chi-squared Test is valid, provided Expected Values \\( \\geq 5 \\). (Otherwise, the score is inflated.) For small expected values in a \\( 2 \\times 2 \\) table, defer to Fisher’s Exact Test.\n\n- Chi-squared statistic with Yates continuity correction to reduce spurious significance:\n\n\\[ X^2 = \\sum \\frac{(|\\text{Obs} - \\text{Exp}| - 0.5)^2}{\\text{Exp}} \\]\n\n- Chi-squared Test is strictly for the two-sided \\( H_0: \\pi_1 - \\pi_2 = 0 \\) vs. \\( H_A: \\pi_1 - \\pi_2 \\neq 0 \\). It cannot be modified to a one-sided test, or to \\( H_0: \\pi_1 - \\pi_2 = \\pi_0 \\) vs. \\( H_A: \\pi_1 - \\pi_2 \\neq \\pi_0 \\).\nHow could we solve this problem using R? The code (which can be shortened a bit):\n\n```r\n# Lines preceded by the pound sign are read as comments, # and ignored by R.\n# The following set of commands builds the 2-by-2 contingency table, # column by column (with optional headings), and displays it as # output (my boldface).\n\nTx.vs.Control = matrix(c(332, 68, 244, 76), ncol = 2, nrow = 2, dimnames = list(\"Status\" = c(\"Improvement\", \"No Improvement\"), \"Group\" = c(\"PT + Drug\", \"PT\")))\n\nTx.vs.Control\n\n| Group | PT + Drug | PT |\n|-------------|-----------|----|\n| Improvement | 332 | 244|\n| No Improvement | 68 | 76 |\n\n# A shorter alternative that outputs a simpler table:\n\nImprovement = c(332, 244)\nNo_Improvement = c(68, 76)\nTx.vs.Control = rbind(Improvement, No_Improvement)\n\nTx.vs.Control\n\n[,1] [,2]\nImprovement 332 244\nNo_Improvement 68 76\n\n# The actual Chi-squared Test itself. Since using a correction # factor is the default, the F option specifies that no such # factor is to be used in this example.\n\nchisq.test(Tx.vs.Control, correct = F)\n\nPearson's Chi-squared test\n\ndata: Tx.vs.Control\nX-squared = 5.0625, df = 1, p-value = 0.02445\n\nNote how the output includes the Chi-squared test statistic, degrees of freedom, and p-value, all of which agree with our previous manual calculations.\n**Application: Case-Control Study Design**\n\nDetermines if an association exists between disease D and risk factor exposure E.\n\n**Chi-Squared Test**\n\n\\[ H_0: \\pi_{E+|D+} = \\pi_{E+|D-} \\]\n\nRandomly select a sample of cases and controls, and categorize each member according to whether or not he/she was exposed to the risk factor.\n\n**McNemar’s Test**\n\n\\[ H_0: \\pi_{E+|D+} = \\pi_{E+|D-} \\]\n\nMatch each case with a corresponding control on age, sex, race, and any other confounding variables that may affect the outcome. Note that this requires a balanced sample: \\( n_1 = n_2 \\).\n\nSee Appendix > Statistical Inference > Means and Proportions, One and Two Samples.\nTo quantify the strength of association between D and E, we turn to the notion of…\n\n**Odds Ratios - Revisited**\n\nRecall:\n\n**POPULATION**\n\n**Case-Control Studies:**\n\n\\[\n\\text{OR} = \\frac{\\text{odds(Exposure | Disease)}}{\\text{odds(Exposure | No Disease)}} = \\frac{P(E+ | D+) / P(E- | D+)}{P(E+ | D-) / P(E- | D-)}\n\\]\n\n**Cohort Studies:**\n\n\\[\n\\text{OR} = \\frac{\\text{odds(Disease | Exposure)}}{\\text{odds(Disease | No Exposure)}} = \\frac{P(D+ | E+) / P(D- | E+)}{P(D+ | E-) / P(D- | E-)}\n\\]\n\n**H₀:** OR = 1 ⇔ No association exists between D, E.\n\nversus…\n\n**Hₐ:** OR ≠ 1 ⇔ An association exists between D, E.\n\nAlas, the probability distribution of the odds ratio OR is distinctly skewed to the right. However, its natural logarithm, ln(OR), is approximately normally distributed, which makes it more useful for conducting the Test of Association above. Namely…\n\n\\[\n(1 - \\alpha) \\times 100\\% \\text{ Confidence Limits for } \\ln(\\text{OR})\n\\]\n\n\\[\n\\ln(\\text{OR}) \\pm (z_{\\alpha/2}) \\text{ s.e.}\n\\]\n\n\\[\n\\text{E} \\quad \\text{where } \\text{s.e.} = \\sqrt{\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d}}\n\\]\n\n\\[\n(1 - \\alpha) \\times 100\\% \\text{ Confidence Limits for OR}\n\\]\nExamples: Test $H_0$: $OR = 1$ versus $H_A$: $OR \\neq 1$ at the $\\alpha = .05$ significance level.\n\n\\[\n\\begin{array}{c|cc}\n & D+ & D- \\\\\nE+ & 8 & 10 \\\\\nE- & 10 & 32 \\\\\n\\end{array}\n\\]\n\n\\[\n\\hat{OR} = \\frac{(8)(32)}{(10)(10)} = 2.56\n\\]\n\n\\[\\ln(2.56) = 0.94\\]\n\n\\[\\text{s.e.} = \\sqrt{\\frac{1}{8} + \\frac{1}{10} + \\frac{1}{10} + \\frac{1}{32}} = 0.6 \\Rightarrow 95\\% \\text{ Margin of Error} = (1.96)(0.6) = 1.176\\]\n\n95\\% Confidence Interval for $\\ln(OR) = (0.94 - 1.176, 0.94 + 1.176) = (-0.236, 2.116)$\n\nand so...\n\n95\\% Confidence Interval for $OR = (e^{-0.236}, e^{2.116}) = (0.79, 8.30)$\n\nConclusion: As this interval does contain the null value $OR = 1$, we cannot reject the hypothesis of non-association at the 5\\% significance level.\n\n\\[\n\\begin{array}{c|cc}\n & D+ & D- \\\\\nE+ & 40 & 50 \\\\\nE- & 50 & 160 \\\\\n\\end{array}\n\\]\n\n\\[\n\\hat{OR} = \\frac{(40)(160)}{(50)(50)} = 2.56\n\\]\n\n\\[\\ln(2.56) = 0.94\\]\n\n\\[\\text{s.e.} = \\sqrt{\\frac{1}{40} + \\frac{1}{50} + \\frac{1}{50} + \\frac{1}{160}} = 0.267 \\Rightarrow 95\\% \\text{ Margin of Error} = (1.96)(0.267) = 0.523\\]\n\n95\\% Confidence Interval for $\\ln(OR) = (0.94 - 0.523, 0.94 + 0.523) = (0.417, 1.463)$\n\nand so...\n\n95\\% Confidence Interval for $OR = (e^{0.417}, e^{1.463}) = (1.52, 4.32)$\n\nConclusion: As this interval does not contain the null value $OR = 1$, we can reject the hypothesis of non-association at the 5\\% level. With 95\\% confidence, the odds of disease are between 1.52 and 4.32 times higher among the exposed than the unexposed.\n\nComments:\n\n- If any of $a$, $b$, $c$, or $d = 0$, then use $\\text{s.e.} = \\sqrt{\\frac{1}{a+0.5} + \\frac{1}{b+0.5} + \\frac{1}{c+0.5} + \\frac{1}{d+0.5}}$.\n\n- If $OR < 1$, this suggests that exposure might have a protective effect, e.g., daily calcium supplements (yes/no) and osteoporosis (yes/no).\n**Summary Odds Ratio**\n\nCombining $2 \\times 2$ tables corresponding to distinct strata.\n\n**Examples:**\n\n| | Males | | Females | | All |\n|-------|-------|-------|---------|-------|-----|\n| | D+ | D− | D+ | D− | |\n| E+ | 10 | 50 | 10 | 10 | 20 |\n| E− | 10 | 150 | 60 | 60 | 70 |\n\n$\\text{OR}_1 = 3$\n\n$\\text{OR}_2 = 1$\n\n$\\text{OR} = 1$\n\n| | Males | | Females | | All |\n|-------|-------|-------|---------|-------|-----|\n| | D+ | D− | D+ | D− | |\n| E+ | 80 | 20 | 10 | 20 | 90 |\n| E− | 20 | 10 | 20 | 80 | 40 |\n\n$\\text{OR}_1 = 2$\n\n$\\text{OR}_2 = 2$\n\n$\\text{OR} = 5.0625$\n\n| | Males | | Females | | All |\n|-------|-------|-------|---------|-------|-----|\n| | D+ | D− | D+ | D− | |\n| E+ | 60 | 100 | 50 | 10 | 110 |\n| E− | 10 | 50 | 100 | 60 | 110 |\n\n$\\text{OR}_1 = 3$\n\n$\\text{OR}_2 = 3$\n\n$\\text{OR} = 1$\n\nThese examples illustrate the phenomenon known as **Simpson’s Paradox**.\n\nIgnoring a confounding variable (e.g., gender) may obscure an association that exists within each stratum, but not observed in the pooled data, and thus must be adjusted for. When is it acceptable to combine data from two or more such strata? How is the summary odds ratio $\\text{OR}_{\\text{summary}}$ estimated? And how is it tested for association?\nIn general...\n\n| Stratum 1 | Stratum 2 |\n|-----------|-----------|\n| D+ | D+ |\n| E+ | a₁ | a₂ |\n| E− | b₁ | b₂ |\n| | c₁ | c₂ |\n| | d₁ | d₂ |\n\n\\[ \\hat{OR}_1 = \\frac{a_1 d_1}{b_1 c_1} \\]\n\n\\[ \\hat{OR}_2 = \\frac{a_2 d_2}{b_2 c_2} \\]\n\nI. Calculate the estimates of \\( \\hat{OR}_1 \\) and \\( \\hat{OR}_2 \\) for each stratum, as shown.\n\nII. Can the strata be combined? Conduct a “Breslow-Day” (Chi-squared) Test of Homogeneity for\n\n\\[ H_0: \\hat{OR}_1 = \\hat{OR}_2. \\]\n\nIII. If accepted, calculate the Mantel-Haenszel Estimate of \\( \\hat{OR}_{\\text{summary}} \\):\n\n\\[ \\hat{OR}_{\\text{MH}} = \\frac{\\frac{a_1 d_1}{n_1} + \\frac{a_2 d_2}{n_2}}{\\frac{b_1 c_1}{n_1} + \\frac{b_2 c_2}{n_2}}. \\]\n\nIV. Finally, conduct a Test of Association for the combined strata\n\n\\[ H_0: \\hat{OR}_{\\text{summary}} = 1 \\]\n\neither via confidence interval, or special \\( \\chi^2 \\)-test (shown below).\n\nExample:\n\n| Males | Females |\n|-------|---------|\n| D+ | D+ |\n| E+ | 10 | 40 |\n| E− | 30 | 60 |\n| D− | 20 | 50 |\n| | 90 | 90 |\n\n\\[ \\hat{OR}_1 = 1.5 \\]\n\n\\[ \\hat{OR}_2 = 1.2 \\]\n\nAssuming that the Test of Homogeneity \\( H_0: \\hat{OR}_1 = \\hat{OR}_2 \\) is conducted and accepted,\n\n\\[ \\hat{OR}_{\\text{MH}} = \\frac{(10)(90)}{150} + \\frac{(40)(90)}{240} = \\frac{6 + 15}{4 + 12.5} = \\frac{21}{16.5} = 1.273. \\]\n\nExercise: Show algebraically that \\( \\hat{OR}_{\\text{MH}} \\) is a weighted average of \\( \\hat{OR}_1 \\) and \\( \\hat{OR}_2 \\).\nTo conduct a formal Chi-squared Test of Association $H_0: \\text{OR}_{\\text{summary}} = 1$, we calculate, for the $2 \\times 2$ contingency table in each stratum $i = 1, 2, \\ldots, s$.\n\n| Observed # diseased | Expected # diseased | Variance |\n|---------------------|---------------------|----------|\n| D+ | $a_i$ | $R_{1i} \\rightarrow E_{1i} = \\frac{R_{1i} C_{1i}}{n_i}$ |\n| D− | $b_i$ | $E_{2i} = \\frac{R_{2i} C_{1i}}{n_i}$ |\n| E+ | $c_i$ | $R_{2i} \\rightarrow E_{1i} = \\frac{R_{1i} C_{2i}}{n_i}$ |\n| E− | $d_i$ | $E_{2i} = \\frac{R_{2i} C_{2i}}{n_i}$ |\n\nTherefore, summing over all strata $i = 1, 2, \\ldots, s$, we obtain the following:\n\n| Observed total, Diseased | Expected total, Diseased | Total Variance |\n|--------------------------|--------------------------|----------------|\n| Exposed: $O_1 = \\sum a_i$ | Exposed: $E_1 = \\sum E_{1i}$ | $V = \\sum V_i$ |\n| Not Exposed: $O_2 = \\sum c_i$ | Not Exposed: $E_2 = \\sum E_{2i}$ | |\n\nand the formal test statistic for significance is given by\n\n$$X^2 = \\frac{(O_1 - E_1)^2}{V} \\sim \\chi^2_1.$$ \n\nThis formulation will appear again in the context of the Log-Rank Test in the area of Survival Analysis (section 8.3).\n\n**Example (cont’d):**\n\nFor stratum 1 (males), $E_{11} = \\frac{(30)(40)}{150} = 8$ and $V_1 = \\frac{(30)(120)(40)(110)}{150^2 (149)} = 4.725$.\n\nFor stratum 2 (females), $E_{12} = \\frac{(90)(100)}{240} = 37.5$ and $V_2 = \\frac{(90)(150)(100)(140)}{240^2 (239)} = 13.729$.\n\nTherefore, $O_1 = 50$, $E_1 = 45.5$, and $V = 18.454$, so that $X^2 = \\frac{(4.5)^2}{18.454} = 1.097$ on 1 degree of freedom, from which it follows that the null hypothesis $H_0: \\text{OR}_{\\text{summary}} = 1$ cannot be rejected at the $\\alpha = .05$ significance level, i.e., there is not enough empirical evidence to conclude that an association exists between disease $D$ and exposure $E$.\n\n**Comment:** This entire discussion on Odds Ratios OR can be modified to Relative Risk RR (defined only for a cohort study), with the following changes: $\\text{s.e.} = \\sqrt{\\frac{1}{a} - \\frac{1}{R_1} + \\frac{1}{c} - \\frac{1}{R_2}}$, as well as $b$ replaced with row marginal $R_1$, and $d$ replaced with row marginal $R_2$, in all other formulas. [Recall, for instance, that $\\text{OR} = \\frac{ad}{bc}$, whereas $\\text{RR} = \\frac{aR_2}{R_1c}$, etc.]", "id": "./materials/147.pdf" }, { "contents": "6.4 Problems\n\nNOTE: Before starting these problems, it might be useful to review pages 1.3-1 and 2.1-1.\n\n1. Suppose that a random sample of \\( n = 102 \\) children is selected from the population of newborn infants in Mexico. The probability that a child in this population weighs at most 2500 grams is presumed to be \\( \\pi = 0.15 \\). Calculate the probability that thirteen or fewer of the infants weigh at most 2500 grams, using...\n\n(a) the exact binomial distribution (Tip: Use the function \\texttt{pbinom} in R),\n\n(b) the normal approximation to the binomial distribution (with continuity correction).\n\nSuppose we wish to test the null hypothesis \\( H_0: \\pi = 0.15 \\) versus the alternative \\( H_A: \\pi \\neq 0.15 \\), and that in this random sample of \\( n = 102 \\) children, we find thirteen whose weights are under 2500 grams. Use this information to decide whether or not to reject \\( H_0 \\) at the \\( \\alpha = .05 \\) significance level, and interpret your conclusion in context.\n\n(c) Calculate the \\textit{p-value}, using the “normal approximation to the binomial” with continuity correction. (Hint: See (b).) Also compute the 95% confidence interval.\n\n(d) Calculate the exact \\textit{p-value}, via the function \\texttt{binom.test} in R.\n\n2. A new “smart pill” is tested on \\( n = 36 \\) individuals randomly sampled from a certain population whose IQ scores are known to be normally distributed, with mean \\( \\mu = 100 \\) and standard deviation \\( \\sigma = 27 \\). After treatment, the sample mean IQ score is calculated to be \\( \\bar{x} = 109.9 \\), and a two-sided test of the null hypothesis \\( H_0: \\mu = 100 \\) versus the alternative hypothesis \\( H_A: \\mu \\neq 100 \\) is performed, to see if there is any statistically significant difference from the mean IQ score of the original population. Using this information, answer the following.\n\n(a) Calculate the \\textit{p-value} of the sample.\n\n(b) Fill in the following table, concluding with the decision either to reject or not reject the null hypothesis \\( H_0 \\) at the given significance level \\( \\alpha \\).\n\n| Significance Level \\( \\alpha \\) | Confidence Level \\( 1 - \\alpha \\) | Confidence Interval | Decision about \\( H_0 \\) |\n|-------------------------------|---------------------------------|---------------------|--------------------------|\n| .10 | | | |\n| .05 | | | |\n| .01 | | | |\n\n(c) Extend these observations to more general circumstances. Namely, as the significance level decreases, what happens to the ability to reject a null hypothesis? Explain why this is so, in terms of the \\textit{p-value} and generated confidence intervals.\n3. Consider the distribution of serum cholesterol levels for all 20- to 74-year-old males living in the United States. The mean of this population is 211 mg/dL, and the standard deviation is 46.0 mg/dL. In a study of a subpopulation of such males who smoke and are hypertensive, it is assumed (not unreasonably) that the distribution of serum cholesterol levels is normally distributed, with unknown mean \\( \\mu \\), but with the same standard deviation \\( \\sigma \\) as the original population.\n\n(a) Formulate the null hypothesis and complementary alternative hypothesis, for testing whether the unknown mean serum cholesterol level \\( \\mu \\) of the subpopulation of hypertensive male smokers is equal to the known mean serum cholesterol level of 211 mg/dL of the general population of 20- to 74-year-old males.\n\n(b) In the study, a random sample of size \\( n = 12 \\) hypertensive smokers was selected, and found to have a sample mean cholesterol level of \\( \\bar{x} = 217 \\) mg/dL. Construct a 95% confidence interval for the true mean cholesterol level of this subpopulation.\n\n(c) Calculate the p-value of this sample, at the \\( \\alpha = .05 \\) significance level.\n\n(d) Based on your answers in parts (b) and (c), is the null hypothesis rejected in favor of the alternative hypothesis, at the \\( \\alpha = .05 \\) significance level? Interpret your conclusion: What exactly has been demonstrated, based on the empirical evidence?\n\n(e) Determine the 95% acceptance region and complementary rejection region for the null hypothesis. Is this consistent with your findings in part (d)? Why?\n\n4. Consider a random sample of ten children selected from a population of infants receiving antacids that contain aluminum, in order to treat peptic or digestive disorders. The distribution of plasma aluminum levels is known to be approximately normal; however its mean \\( \\mu \\) and standard deviation \\( \\sigma \\) are not known. The mean aluminum level for the sample of \\( n = 10 \\) infants is found to be \\( \\bar{x} = 37.20 \\) \\( \\mu g/l \\) and the sample standard deviation is \\( s = 7.13 \\) \\( \\mu g/l \\). Furthermore, the mean plasma aluminum level for the population of infants not receiving antacids is known to be only 4.13 \\( \\mu g/l \\).\n\n(a) Formulate the null hypothesis and complementary alternative hypothesis, for a two-sided test of whether the mean plasma aluminum level of the population of infants receiving antacids is equal to the mean plasma aluminum level of the population of infants not receiving antacids.\n\n(b) Construct a 95% confidence interval for the true mean plasma aluminum level of the population of infants receiving antacids.\n\n(c) Calculate the p-value of this sample (as best as possible), at the \\( \\alpha = .05 \\) significance level.\n\n(d) Based on your answers in parts (b) and (c), is the null hypothesis rejected in favor of the alternative hypothesis, at the \\( \\alpha = .05 \\) significance level? Interpret your conclusion: What exactly has been demonstrated, based on the empirical evidence?\n\n(e) With the knowledge that significantly elevated plasma aluminum levels are toxic to human beings, reformulate the null hypothesis and complementary alternative hypothesis, for the appropriate one-sided test of the mean plasma aluminum levels. With the same sample data as above, how does the new p-value compare with that found in part (c), and what is the resulting conclusion and interpretation?\n5. Refer to Problem 4.4/2.\n\n(a) Suppose we wish to formally test the null hypothesis $H_0: \\mu = 25$ against the alternative $H_A: \\mu \\neq 25$, at the $\\alpha = .05$ significance level, by using the random sample of $n = 80$ given.\n\n- Calculate the p-value, and verify that in fact, this sample leads to an incorrect conclusion.\n \n [[Hint: Use the Central Limit Theorem to approximate the sampling distribution of $\\bar{X}$ with the normal distribution $N(\\mu, \\sigma/\\sqrt{n})$.]] Which type of error (Type I or Type II) is committed here, and why?\n\n(b) Now suppose we wish to formally test the null hypothesis $H_0: \\mu = 27$ against the specific alternative $H_A: \\mu = 25$, at the $\\alpha = .05$ significance level, using the same random sample of $n = 80$ trials.\n\n- How much power exists (i.e., what is the probability) of inferring the correct conclusion?\n- Calculate the p-value, and verify that, once again, this sample in fact leads to an incorrect conclusion. [[Use the same hint as in part (a).]] Which type of error (Type I or Type II) is committed here, and why?\n\n6. Two physicians are having a disagreement about the effectiveness of chicken soup in relieving common cold symptoms. While both agree that the number of symptomatic days generally follows a normal distribution, physician A claims that most colds last about a week; chicken soup makes no difference, whereas physician B argues that it does. They decide to settle the matter by performing a formal two-sided test of the null hypothesis $H_0: \\mu = 7$ days, versus the alternative $H_A: \\mu \\neq 7$ days.\n\n(a) After treating a random sample of $n = 16$ cold patients with chicken soup, they calculate a mean number of symptomatic days $\\bar{x} = 5.5$, and standard deviation $s = 3.0$ days. Using either the 95% confidence interval or the p-value (or both), verify that the null hypothesis cannot be rejected at the $\\alpha = .05$ significance level.\n\n(b) Physician A is delighted, but can predict physician B’s rebuttal: “The sample size was too small! There wasn’t enough power to detect a statistically significant difference between $\\mu = 7$ days, and say $\\mu = 5$ days, even if there was one present!” Calculate the minimum sample size required in order to achieve at least 99% power of detecting such a genuine difference, if indeed one actually exists. (Note: Use $s$ to estimate $\\sigma$.)\n\n(c) Suppose that, after treating a random sample of $n = 49$ patients, they calculate the mean number of symptomatic days $\\bar{x} = 5.5$ (as before), and standard deviation $s = 2.8$ days. Using either the 95% confidence interval or the p-value (or both), verify that the null hypothesis can now be rejected at the $\\alpha = .05$ significance level.\n\nFYI: The long-claimed ability of chicken soup – sometimes referred to as “Jewish penicillin” – to combat colds has actually been the subject of several well-known published studies, starting with a 1978 seminal paper written by researchers at Mount Sinai Hospital in NYC. The heat does serve to break up chest congestion, but it turns out that there are many other surprising cold-fighting benefits, far beyond just that. “Who knew?” Evidently… Mama. See http://well.blogs.nytimes.com/2007/10/12/the-science-of-chicken-soup/.\n7. **Toxicity Testing.** [Tip: See page 6.1-28] According to the EPA (Environmental Protection Agency), drinking water can contain no more than 10 ppb (parts per billion) of arsenic, in order to be considered safe for human consumption.* Suppose that the concentration $X$ of arsenic in a typical water source is known to be normally distributed, with an unknown mean $\\mu$ and standard deviation $\\sigma$. A random sample of $n = 121$ independent measurements is to be taken, from which the sample mean $\\bar{x}$ and sample standard deviation $s$ are calculated, and used in formal hypothesis testing. The following sample data for four water sources are obtained:\n\n- Source 1: $\\bar{x} = 11.43$ ppb, $s = 5.5$ ppb\n- Source 2: $\\bar{x} = 8.57$ ppb, $s = 5.5$ ppb\n- Source 3: $\\bar{x} = 9.10$ ppb, $s = 5.5$ ppb\n- Source 4: $\\bar{x} = 10.90$ ppb, $s = 5.5$ ppb\n\n(a) For each water source, answer the following questions to test the null hypothesis $H_0: \\mu = 10$ ppb, vs. the two-sided alternative hypothesis $H_A: \\mu \\neq 10$ ppb, at the $\\alpha = .05$ significance level.\n\n(i) **Just by intuitive inspection**, i.e., without first conducting any formal calculations, does this sample mean suggest that the water might be safe, or unsafe, to drink? Why??\n\n(ii) Calculate the $p$-value of this sample (to the closest entries of the appropriate table), and use it to draw a formal conclusion about whether or not the null hypothesis can be rejected in favor of the alternative, at the $\\alpha = .05$ significance level.\n\n(iii) **Interpret**: According to your findings, is the result statistically significant? That is... **Is the water unsafe to drink?** Does this agree with your informal reasoning in (i)?\n\n(b) For the hypothesis test in (a), what is the two-sided 5% rejection region for this $H_0$? Is it consistent with your findings?\n\n(c) One-sided hypothesis tests can be justifiably used in some contexts, such as situations where one direction (either $\\leq$ or $\\geq$) is impossible (for example, a human knee cannot flex backwards), or irrelevant, as in “toxicity testing” here. We are really not concerned if the mean is significantly below 10 ppb, only above. With this in mind, repeat the instructions in (a) above, to test the left-sided null hypothesis $H_0: \\mu \\leq 10$ ppb (i.e., safe) versus the right-sided alternative $H_A: \\mu > 10$ ppb (i.e., unsafe) at the $\\alpha = .05$ significance level.\n\n(d) Suppose a fifth water source yields $\\bar{x} = 10.6445$ ppb and $s = 5.5$ ppb. Repeat part (c).\n\n(e) For the hypothesis test in (c), what is the exact cutoff ppb level for $\\bar{x}$, above which we can conclude that the water is unsafe? (Compare Sources 4 and 5, for example.) That is, what is the one-sided 5% rejection region for this $H_0$? **Is it consistent with your findings?**\n\n(f) Summarize these results, and make some general conclusions regarding advantages and disadvantages of using a one-sided test, versus a two-sided test, in this context. [Hint: Compare the practical results in (a) and (c) for Source 4, for example.]\n\n* This is known as the Maximum Contaminant Level (MCL).\n8. Do the Exercise on page 6.1-20.\n\n9. (a) In R, type the following command to generate a data set called “x” of 1000 random values.\n\n\\[ x = \\text{rf}(1000, 5, 20) \\]\n\nObtain a graph of its frequency histogram by typing \\texttt{hist(x)}. Include this graph as part of your submitted homework assignment. (Do not include the 1000 data values!)\n\nNext construct a “normal q-q plot” by typing \\texttt{qqnorm(x, pch = 19)}. Include this plot as part of your submitted homework assignment.\n\n(b) Now define a new data set called “y” by taking the (natural) logarithm of x.\n\n\\[ y = \\log(x) \\]\n\nObtain a graph of its frequency histogram by typing \\texttt{hist(y)}. Include this graph as part of your submitted homework assignment. (Do not include the 1000 data values!)\n\nThen construct a “normal q-q plot” by typing \\texttt{qqnorm(y, pch = 19)}. Include this plot as part of your submitted homework assignment.\n\n(c) Summarize the results in (a) and (b). In particular, from their respective histograms and q-q plots, what general observation can be made regarding the distributions of x and y = \\log(x)? (Hint: See pages 6.1-25 through 6.1-27.)\n10. In this problem, assume that population cholesterol level is normally distributed.\n\n(a) Consider a small clinical trial, designed to measure the efficacy of a new cholesterol-lowering drug against a placebo. A group of six high-cholesterol patients is randomized to either a treatment arm or a control arm, resulting in two numerically balanced samples of \\( n_1 = n_2 = 3 \\) patients each, in order to test the null hypothesis \\( H_0: \\mu_1 = \\mu_2 \\) vs. the alternative \\( H_A: \\mu_1 \\neq \\mu_2 \\). Suppose that the data below are obtained.\n\n| Placebo | Drug |\n|---------|------|\n| 220 | 180 |\n| 240 | 200 |\n| 290 | 220 |\n\nObtain the 95% confidence interval for \\( \\mu_1 - \\mu_2 \\), and the p-value of the data, and use each to decide whether or not to reject \\( H_0 \\) at the \\( \\alpha = .05 \\) significance level. Conclusion?\n\n(b) Now imagine that the same drug is tested using another pilot study, with a different design. Serum cholesterol levels of \\( n = 3 \\) patients are measured at the beginning of the study, then re-measured after a six month treatment period on the drug, in order to test the null hypothesis \\( H_0: \\mu_1 = \\mu_2 \\) versus the alternative \\( H_A: \\mu_1 \\neq \\mu_2 \\). Suppose that the data below are obtained.\n\n| Baseline | End of Study |\n|----------|--------------|\n| 220 | 180 |\n| 240 | 200 |\n| 290 | 220 |\n\nObtain the 95% confidence interval for \\( \\mu_1 - \\mu_2 \\), and the p-value of the data, and use each to decide whether or not to reject \\( H_0 \\) at the \\( \\alpha = .05 \\) significance level. Conclusion?\n\n(c) Compare and contrast these two study designs and their results.\n\n(d) Redo (a) and (b) using R (see hint). Show agreement between your answers and the output.\n11. In order to determine whether children with cystic fibrosis have a normal level of iron in their blood on average, a study is performed to detect any significant difference in mean serum iron levels between this population and the population of healthy children, both of which are approximately normally distributed with unknown standard deviations. A random sample of \\( n_1 = 9 \\) healthy children has mean serum iron level \\( \\bar{x}_1 = 18.9 \\, \\mu\\text{mol/l} \\) and standard deviation \\( s_1 = 5.9 \\, \\mu\\text{mol/l} \\); a sample of \\( n_2 = 13 \\) children with cystic fibrosis has mean serum iron level \\( \\bar{x}_2 = 11.9 \\, \\mu\\text{mol/l} \\) and standard deviation \\( s_2 = 6.3 \\, \\mu\\text{mol/l} \\).\n\n(a) Formulate the null hypothesis and complementary alternative hypothesis, for testing whether the mean serum iron level \\( \\mu_1 \\) of the population of healthy children is equal to the mean serum iron level \\( \\mu_2 \\) of children with cystic fibrosis.\n\n(b) Construct the 95% confidence interval for the mean serum iron level difference \\( \\mu_1 - \\mu_2 \\).\n\n(c) Calculate the p-value for this experiment, under the null hypothesis.\n\n(d) Based on your answers in parts (b) and (c), is the null hypothesis rejected in favor of the alternative hypothesis, at the \\( \\alpha = .05 \\) significance level? Interpret your conclusion: What exactly has been demonstrated, based on the sample evidence?\n\n12. Methylphenidate is a drug that is widely used in the treatment of attention deficit disorder (ADD). As part of a crossover study, ten children between the ages of 7 and 12 who suffered from this disorder were assigned to receive the drug and ten were given a placebo. After a fixed period of time, treatment was withdrawn from all 20 children and, after a “washout period” of no treatment for either group, subsequently resumed after switching the treatments between the two groups. Measures of each child’s attention and behavioral status, both on the drug and on the placebo, were obtained using an instrument called the Parent Rating Scale. Distributions of these scores are approximately normal with unknown means and standard deviations. In general, lower scores indicate an increase in attention. It is found that the random sample of \\( n = 20 \\) children enrolled in the study has a sample mean attention rating score of \\( \\bar{x}_{\\text{methyl}} = 10.8 \\) and standard deviation \\( s_{\\text{methyl}} = 2.9 \\) when taking methylphenidate, and mean rating score \\( \\bar{x}_{\\text{placebo}} = 14.0 \\) and standard deviation \\( s_{\\text{placebo}} = 4.8 \\) when taking the placebo.\n\n(a) Calculate the 95% confidence interval for \\( \\mu_{\\text{placebo}} \\), the mean attention rating score of the population of children taking the placebo.\n\n(b) Calculate the 95% confidence interval for \\( \\mu_{\\text{methyl}} \\), the mean attention rating score of the population of children taking the drug.\n\n(c) Comparing these two confidence intervals side-by-side, develop an informal conclusion about the efficacy of methylphenidate, based on this experiment. Why can this not be used as a formal test of the hypothesis \\( H_0: \\mu_{\\text{placebo}} = \\mu_{\\text{methyl}} \\) vs. the alternative \\( H_A: \\mu_{\\text{placebo}} \\neq \\mu_{\\text{methyl}} \\) at the \\( \\alpha = .05 \\) significance level? (Hint: See next problem.)\n13. A formal hypothesis test for two-sample means using the confidence interval for $\\mu_1 - \\mu_2$ is generally NOT equivalent to an informal side-by-side comparison of the individual confidence intervals for $\\mu_1$ and $\\mu_2$ for detecting overlap between them.\n\n(a) Suppose that two population random variables $X_1$ and $X_2$ are normally distributed, each with standard deviation $\\sigma = 50$. We wish to test the null hypothesis $H_0: \\mu_1 = \\mu_2$ versus the alternative $H_0: \\mu_1 \\neq \\mu_2$, at the $\\alpha = .05$ significance level. Two independent, random samples are selected, each of size $n = 100$, and it is found that the corresponding means are $\\bar{x}_1 = 215$ and $\\bar{x}_2 = 200$, respectively. Show that even though the two individual 95% confidence intervals for $\\mu_1$ and $\\mu_2$ overlap, the formal 95% confidence interval for the mean difference $\\mu_1 - \\mu_2$ does not contain the value 0, and hence the null hypothesis can be rejected. (See middle figure below.)\n\n(b) In general, suppose that $X_1 \\sim N(\\mu_1, \\sigma)$ and $X_2 \\sim N(\\mu_2, \\sigma)$, with equal $\\sigma$ (for simplicity). In order to test the null hypothesis $H_0: \\mu_1 = \\mu_2$ versus the two-sided alternative $H_0: \\mu_1 \\neq \\mu_2$ at the $\\alpha$ significance level, two random samples are selected, each of the same size $n$ (for simplicity), resulting in corresponding means $\\bar{x}_1$ and $\\bar{x}_2$, respectively. Let $CI_{\\mu_1}$ and $CI_{\\mu_2}$ be the respective 100(1 - $\\alpha$)% confidence intervals, and let $d = \\frac{|\\bar{x}_1 - \\bar{x}_2|}{z_{\\alpha/2}(\\sigma/\\sqrt{n})}$. (Note that the denominator is simply the margin of error for the confidence intervals.) Also let $CI_{\\mu_1 - \\mu_2}$ be the 100(1 - $\\alpha$)% confidence interval for the true mean difference $\\mu_1 - \\mu_2$. Prove:\n\n- If $d < \\sqrt{2}$, then $0 \\in CI_{\\mu_1 - \\mu_2}$ (i.e., “accept” $H_0$), and $CI_{\\mu_1} \\cap CI_{\\mu_2} \\neq \\emptyset$ (i.e., overlap).\n\n- If $\\sqrt{2} < d < 2$, then $0 \\notin CI_{\\mu_1 - \\mu_2}$ (i.e., reject $H_0$), but $CI_{\\mu_1} \\cap CI_{\\mu_2} \\neq \\emptyset$ (i.e., overlap)!\n\n- If $d > 2$, then $0 \\notin CI_{\\mu_1 - \\mu_2}$ (i.e., reject $H_0$), and $CI_{\\mu_1} \\cap CI_{\\mu_2} = \\emptyset$ (i.e., no overlap).\n14. Z-tests and Chi-squared Tests\n\n(a) Test of Independence (1 population, 2 random responses). Imagine that a marketing research study surveys a random sample of \\( n = 2000 \\) consumers about their responses regarding two brands (A and B) of a certain product, with the following observed results.\n\n| Do You Like Brand B? | Yes | No |\n|----------------------|-----|----|\n| Do You Like Brand A? | | |\n| Yes | 335 | 915| 1250|\n| No | 165 | 585| 750 |\n| | 500 | 1500| 2000|\n\nFirst consider the null hypothesis \\( H_0: \\pi_{AB} = \\pi_{A|B} \\), that is, in this consumer population, “The probability of liking A, given that B is liked, is equal to probability of liking A, given that B is not liked.”\n\n\\[ \\iff \\text{There is no association between liking A and liking B.} \\]\n\n\\[ \\iff \\text{Liking A and liking B are independent of each other.} \\]\n\n[Why? See Problem 3.5/22(a.)]\n\nCalculate the point estimate \\( \\hat{\\pi}_{AB} - \\hat{\\pi}_{A|B} \\). Determine the Z-score of this sample (and thus whether or not \\( H_0 \\) is rejected at \\( \\alpha = .05 \\)). Conclusion?\n\nNow consider the null hypothesis \\( H_0: \\pi_{BA} = \\pi_{B|A} \\), that is, in this consumer population, “The probability of liking B, given that A is liked, is equal to probability of liking B, given that A is not liked.”\n\n\\[ \\iff \\text{There is no association between liking B and liking A.} \\]\n\n\\[ \\iff \\text{Liking B and liking A are independent of each other.} \\]\n\nCalculate the point estimate \\( \\hat{\\pi}_{BA} - \\hat{\\pi}_{B|A} \\). Determine the Z-score of this sample (and thus whether or not \\( H_0 \\) is rejected at \\( \\alpha = .05 \\)). How does it compare with the previous Z-score? Conclusion?\n\nCompute the Chi-squared score. How does it compare with the preceding Z-scores? Conclusion?\n(b) **Test of Homogeneity (2 populations, 1 random response).** Suppose that, for the sake of simplicity, the same data are obtained in a survey that compares the probability $\\pi$ of liking Brand A between two populations.\n\n| Do You Like Brand A? | City 1 | City 2 |\n|----------------------|--------|--------|\n| Yes | 335 | 915 | 1250 |\n| No | 165 | 585 | 750 |\n| | 500 | 1500 | 2000 |\n\nHere, the null hypothesis is $H_0: \\pi_{A|\\text{City 1}} = \\pi_{A|\\text{City 2}}$, that is,\n\n\"The probability of liking A in the City 1 population is equal to probability of liking A in the City 2 population.\"\n\n$\\Leftrightarrow$ \"City 1 and City 2 populations are **homogeneous** with respect to liking A.\"\n\n$\\Leftrightarrow$ \"There is no association between city and liking A.\"\n\nHow do these corresponding Z and Chi-squared test statistics compare with those in (a)?\n\nConclusion?\n15. Consider the following $2 \\times 2$ contingency table taken from a retrospective case-control study that investigates the proportion of diabetes sufferers among acute myocardial infarction (heart attack) victims in the Navajo population residing in the United States.\n\n| MI | Diabetes | No Diabetes | Total |\n|----------|----------|-------------|-------|\n| Yes | 46 | 25 | 71 |\n| No | 98 | 119 | 217 |\n| Total | 144 | 144 | 288 |\n\n(a) Conduct a Chi-squared Test for the null hypothesis $H_0: \\pi_{\\text{Diabetes | MI}} = \\pi_{\\text{Diabetes | No MI}}$ versus the alternative $H_a: \\pi_{\\text{Diabetes | MI}} \\neq \\pi_{\\text{Diabetes | No MI}}$. Determine whether or not we can reject the null hypothesis at the $\\alpha = .01$ significance level. Interpret your conclusion: At the $\\alpha = .01$ significance level, what exactly has been demonstrated about the proportion of diabetics among the two categories of heart disease in this population?\n\n(b) In the study design above, the 144 victims of myocardial infarction (cases) and the 144 individuals free of heart disease (controls) were actually age- and gender-matched. The members of each case-control pair were then asked whether they had ever been diagnosed with diabetes. Of the 46 individuals who had experienced MI and who were diabetic, it turned out that 9 were paired with diabetics and 37 with non-diabetics. Of the 98 individuals who had experienced MI but who were not diabetic, it turned out that 16 were paired with diabetics and 82 with non-diabetics. Therefore, each cell in the resulting $2 \\times 2$ contingency table below corresponds to the combination of responses for age- and gender-matched case-control pairs, rather than individuals.\n\n| MI | Diabetes | No Diabetes | Totals |\n|----------|----------|-------------|--------|\n| No MI | 9 | 16 | 25 |\n| Diabetes | 37 | 82 | 119 |\n| Totals | 46 | 98 | 144 |\n\nConduct a McNemar Test for the null hypothesis $H_0$: “The number of ‘diabetic, MI case’ - ‘non-diabetic, non-MI control’ pairs, is equal to the number of ‘non-diabetic, MI case’ - ‘diabetic, non-MI control’ pairs, who have been matched on age and gender;” or more succinctly, $H_0$: “There is no association between diabetes and myocardial infarction in the Navajo population, adjusting for age and gender.” Determine whether or not we can reject the null hypothesis at the $\\alpha = .01$ significance level. Interpret your conclusion: At the $\\alpha = .01$ significance level, what exactly has been demonstrated about the association between diabetes and myocardial infarction in this population?\n\n(c) Why does the McNemar Test only consider discordant case-control pairs? Hint: What, if anything, would a concordant pair (i.e., either both individuals in a ‘MI case - No MI control’ pair are diabetic, or both are non-diabetic) reveal about a diabetes-MI association, and why?\n\n(d) Redo this problem with R, using `chisq.test` and `mcnemar.test`. \n16. The following data are taken from a study that attempts to determine whether the use of electronic fetal monitoring (\"exposure\") during labor affects the frequency of caesarian section deliveries (\"disease\"). Of the 5824 infants included in the study, 2850 were electronically monitored during labor and 2974 were not. Results are displayed in the $2 \\times 2$ contingency table below.\n\n| Caesarian Delivery | Yes | No | Totals |\n|--------------------|-----|-----|--------|\n| EFM Exposure | | | |\n| Yes | 358 | 2492| 2850 |\n| No | 229 | 2745| 2974 |\n| Totals | 587 | 5237| 5824 |\n\n(a) Calculate a **point estimate** for the population odds ratio $OR$, and interpret.\n\n(b) Compute a **95% confidence interval** for the population odds ratio $OR$.\n\n(c) Based on your answer in part (b), show that the null hypothesis $H_0$: $OR = 1$ can be rejected in favor of the alternative $H_A$: $OR \\neq 1$, at the $\\alpha = .05$ significance level. **Interpret this conclusion:** What exactly has been demonstrated about the association between electronic fetal monitoring and caesarian section delivery? Be precise.\n\n(d) Does this imply that electronic monitoring somehow causes a caesarian delivery? Can the association possibly be explained any other way? If so, how?\n17. The following data come from two separate studies, both conducted in San Francisco, that investigate various risk factors for epithelial ovarian cancer:\n\n| Study 1 | Disease Status | Study 2 | Disease Status |\n|---------|----------------|---------|----------------|\n| | Cancer | No Cancer | Total | Cancer | No Cancer | Total |\n| Term Pregnancies | | | | | | |\n| None | 31 | 93 | 124 | 39 | 74 | 113 |\n| One or More | 80 | 379 | 459 | 149 | 465 | 614 |\n| Total | 111 | 472 | 583 | 188 | 539 | 727 |\n\n(a) Compute point estimates $\\hat{OR}_1$ and $\\hat{OR}_2$ of the respective odds ratios $OR_1$ and $OR_2$ of the two studies, and interpret.\n\n(b) In order to determine whether or not we may combine information from the two tables, it is first necessary to conduct a Test of Homogeneity on the null hypothesis $H_0$: $OR_1 = OR_2$, vs. the alternative $H_A$: $OR_1 \\neq OR_2$, by performing the following steps.\n\nStep 1: First, calculate $l_1 = \\ln(\\hat{OR}_1)$ and $l_2 = \\ln(\\hat{OR}_2)$, in the usual way.\n\nStep 2: Next, using the definition of $\\hat{s.e.}$ given in the notes, calculate the weights\n\n$$w_1 = \\frac{1}{\\hat{s.e.}_1^2} \\quad \\text{and} \\quad w_2 = \\frac{1}{\\hat{s.e.}_2^2}.$$ \n\nStep 3: Compute the weighted mean of $l_1$ and $l_2$:\n\n$$L = \\frac{w_1 l_1 + w_2 l_2}{w_1 + w_2}.$$ \n\nStep 4: Finally, calculate the test statistic\n\n$$\\chi^2 = w_1 (l_1 - L)^2 + w_2 (l_2 - L)^2,$$\n\nwhich follows an approximate $\\chi^2$ distribution, with 1 degree of freedom.\n\nStep 5: Use this information to show that the null hypothesis cannot be rejected at the $\\alpha = .05$ significance level, and that the information from the two tables may therefore be combined.\n\n(c) Hence, calculate the Mantel-Haenszel estimate of the summary odds ratio:\n\n$$\\hat{OR}_{\\text{summary}} = \\frac{(a_1 d_1 / n_1) + (a_2 d_2 / n_2)}{(b_1 c_1 / n_1) + (b_2 c_2 / n_2)}.$$\n(d) To compute a 95% confidence interval for the summary odds ratio $OR_{\\text{summary}}$, we must first verify that the sample sizes in the two studies are large enough to ensure that the method used is valid.\n\nStep 1: Verify that the expected number of observations of the $(i, j)^{th}$ cell in the first table, plus the expected number of observations of the corresponding $(i, j)^{th}$ cell in the second table, is greater than or equal to 5, for $i = 1, 2$ and $j = 1, 2$. Recall that the expected number of the $(i, j)^{th}$ cell is given by $E_{ij} = R_i C_j / n$.\n\nStep 2: By its definition, the quantity $L$ computed in part (b) is a weighted mean of log-odds ratios, and already represents a point estimate of $\\ln(OR_{\\text{summary}})$. The estimated standard error of $L$ is given by\n\n$$\\text{s.e.}(L) = \\frac{1}{\\sqrt{w_1 + w_2}}.$$ \n\nStep 3: From these two values in Step 2, construct a 95% confidence interval for $\\ln(OR_{\\text{summary}})$, and exponentiate it to derive a 95% confidence interval for $OR_{\\text{summary}}$ itself.\n\n(e) Also compute the value of the Chi-squared test statistic for $OR_{\\text{summary}}$ given at the end of § 6.2.3.\n\n(f) Use the confidence interval in (d), and/or the $\\chi^2$ statistic in (e), to perform a Test of Association of the null hypothesis $H_0$: $OR_{\\text{summary}} = 1$, versus the alternative $H_A$: $OR_{\\text{summary}} \\neq 1$, at the $\\alpha = .05$ significance level. Interpret your conclusion: What exactly has been demonstrated about the association between the number of term pregnancies and the odds of developing epithelial ovarian cancer? Be precise.\n\n(g) Redo this problem in R, using the code found in the link below, and compare results.\n\nhttp://www.stat.wisc.edu/~ifischer/Intro_Stat/Lecture_Notes/Rcode/\n18. Suppose a survey determines the political orientation of 60 men in a certain community:\n\n| | Left | Middle | Right |\n|-------|------|--------|-------|\n| Men | 12 | 18 | 30 |\n| | | | 60 |\n\nAmong these men, calculate the proportion belonging to each political category. Then show that a Chi-squared Test of the null hypothesis of equal proportions\n\n\\[ H_0: \\pi_{\\text{Left | Men}} = \\pi_{\\text{Mid | Men}} = \\pi_{\\text{Right | Men}} \\]\n\nleads to its rejection at the \\( \\alpha = .05 \\) significance level. Conclusion?\n\n(b) Suppose the survey also determines the political orientation of 540 women in the same community:\n\n| | Left | Middle | Right |\n|-------|------|--------|-------|\n| Women | 108 | 162 | 270 |\n| | | | 540 |\n\nAmong these women, calculate the proportion belonging to each political category. How do these proportions compare with those in (a)? Show that a Chi-squared Test of the null hypothesis of equal proportions\n\n\\[ H_0: \\pi_{\\text{Left | Women}} = \\pi_{\\text{Mid | Women}} = \\pi_{\\text{Right | Women}} \\]\n\nleads to its rejection at the \\( \\alpha = .05 \\) significance level. Conclusion?\n\n(c) Suppose the two survey results are combined:\n\n| | Left | Middle | Right |\n|-------|------|--------|-------|\n| Men | 12 | 18 | 30 |\n| | | | 60 |\n| Women | 108 | 162 | 270 |\n| | 120 | 180 | 300 |\n| | 600 | | |\n\nAmong the individuals in each gender (i.e., row), the proportion belonging to each political category (i.e., column) of course match those found in (a) and (b), respectively. Therefore, show that a Chi-squared Test of the null hypothesis of equal proportions\n\n\\[ H_0: \\pi_{\\text{Left | Men}} = \\pi_{\\text{Left | Women}} \\text{ AND } \\pi_{\\text{Mid | Men}} = \\pi_{\\text{Mid | Women}} \\text{ AND } \\pi_{\\text{Right | Men}} = \\pi_{\\text{Right | Women}} \\]\n\nleads to a 100% acceptance at the \\( \\alpha = .05 \\) significance level. Conclusion?\n\n**NOTE:** The closely-resembling null hypothesis\n\n\\[ H_0: \\pi_{\\text{Men | Left}} = \\pi_{\\text{Women | Left}} \\text{ AND } \\pi_{\\text{Men | Mid}} = \\pi_{\\text{Women | Mid}} \\text{ AND } \\pi_{\\text{Men | Right}} = \\pi_{\\text{Women | Right}} \\]\n\ntests for equal proportions of men and women within each political category, which is very different from the above. Based on sample proportions (0.1 vs. 0.9), it is likely to be rejected, but each column would need to be formally tested by a separate Goodness-of-Fit.\n(d) Among the individuals in each political category (i.e., column), calculate the proportion of men, and show that they are all equal to each other.\n\nAmong the individuals in each political category (i.e., column), calculate the proportion of women, and show that they are all equal to each other.\n\nTherefore, show that a Chi-squared Test of the null hypothesis of equal proportions\n\n\\[ H_0: \\pi_{\\text{Men} | \\text{Left}} = \\pi_{\\text{Men} | \\text{Mid}} = \\pi_{\\text{Men} | \\text{Right}} \\quad \\text{AND} \\quad \\pi_{\\text{Women} | \\text{Left}} = \\pi_{\\text{Women} | \\text{Mid}} = \\pi_{\\text{Women} | \\text{Right}} \\]\n\nalso leads to a 100% acceptance at the \\( \\alpha = .05 \\) significance level. Conclusion?\n\n**MORAL:** There is more than one type of null hypothesis on proportions to which the Chi-squared Test can be applied.\n\n19. In a random sample of \\( n = 1200 \\) consumers who are surveyed about their ice cream flavor preferences, 416 indicate that they prefer vanilla, 419 prefer chocolate, and 365 prefer strawberry.\n\n(a) Conduct a Chi-squared “Goodness-of-Fit” Test of the null hypothesis of equal proportions\n\n\\[ H_0: \\pi_{\\text{Vanilla}} = \\pi_{\\text{Chocolate}} = \\pi_{\\text{Strawberry}} \\]\n\nof flavor preferences, at the \\( \\alpha = .05 \\) significance level.\n\n| Vanilla | Chocolate | Strawberry | Totals |\n|---------|-----------|------------|--------|\n| 416 | 419 | 365 | 1200 |\n\n(b) Suppose that the sample of \\( n = 1200 \\) consumers is equally divided between males and females, yielding the results shown below. Conduct a Chi-squared Test of the null hypothesis that flavor preference is not associated with gender, at the \\( \\alpha = .05 \\) level.\n\n| Males | Vanilla | Chocolate | Strawberry | Totals |\n|-------|---------|-----------|------------|--------|\n| | 200 | 190 | 210 | 600 |\n| Females | 216 | 229 | 155 | 600 |\n| Totals | 416 | 419 | 365 | 1200 |\n\n(c) Redo (a) and (b) with R, using `chisq.test`. Show agreement with your calculations!\nIn the late 1980s, the pharmaceutical company Upjohn received approval from the Food and Drug Administration to market Rogaine™, a 2% minoxidil solution, for the treatment of androgenetic alopecia (male pattern hair loss). Upjohn’s advertising campaign for Rogaine included the results of a double-blind randomized clinical trial, conducted with 1431 patients in 27 centers across the United States. The results of this study at the end of four months are summarized in the $2 \\times 5$ contingency table below, where the two row categories represent the treatment arm and control arm respectively, and each column represents a response category, the degree of hair growth reported. [Source: Ronald L. Iman, A Data-Based Approach to Statistics, Duxbury Press]\n\n| Degree of Hair Growth | Rogaine | Placebo | Total |\n|-----------------------|---------|---------|-------|\n| No Growth | 301 | 423 | 714 |\n| New Vellus | 172 | 150 | 322 |\n| Minimal Growth | 178 | 114 | 292 |\n| Moderate Growth | 58 | 29 | 87 |\n| Dense Growth | 5 | 1 | 6 |\n| Total | 724 | 717 | 1431 |\n\n(a) Conduct a Chi-squared Test of the null hypothesis $H_0: \\pi_{\\text{Rogaine}} = \\pi_{\\text{Placebo}}$ versus the alternative hypothesis $H_A: \\pi_{\\text{Rogaine}} \\neq \\pi_{\\text{Placebo}}$ across the five hair growth categories. (That is, $H_0: \\pi_{\\text{No Growth | Rogaine}} = \\pi_{\\text{No Growth | Placebo}}$ and $\\pi_{\\text{New Vellus | Rogaine}} = \\pi_{\\text{New Vellus | Placebo}}$ and ... and $\\pi_{\\text{Dense Growth | Rogaine}} = \\pi_{\\text{Dense Growth | Placebo}}$.) Infer whether or not we can reject the null hypothesis at the $\\alpha = .01$ significance level. Interpret in context: At the $\\alpha = .01$ significance level, what exactly has been demonstrated about the efficacy of Rogaine versus placebo?\n\n(b) Form a $2 \\times 2$ contingency table by combining the last four columns into a single column labeled Growth. Conduct a Chi-squared Test for the null hypothesis $H_0: \\pi_{\\text{Rogaine}} = \\pi_{\\text{Placebo}}$ versus the alternative $H_A: \\pi_{\\text{Rogaine}} \\neq \\pi_{\\text{Placebo}}$ between the resulting No Growth versus Growth binary response categories. (That is, $H_0: \\pi_{\\text{Growth | Rogaine}} = \\pi_{\\text{Growth | Placebo}}$.) Infer whether or not we can reject the null hypothesis at the $\\alpha = .01$ significance level. Interpret in context: At the $\\alpha = .01$ significance level, what exactly has been demonstrated about the efficacy of Rogaine versus placebo?\n\n(c) Calculate the p-value using a two-sample Z-test of the null hypothesis in part (b), and show that the square of the corresponding z-score is equal to the Chi-squared test statistic found in (b). Verify that the same conclusion about $H_0$ is reached, at the $\\alpha = .01$ significance level.\n\n(d) Redo this problem with R, using chisq.test. Show agreement with your calculations!\n21. Male patients with coronary artery disease were recruited from three different medical centers – the Johns Hopkins University School of Medicine, The Rancho Los Amigos Medical Center, and the St. Louis University School of Medicine – to investigate the effects of carbon monoxide exposure. One of the baseline characteristics considered in the study was pulmonary lung function, as measured by $X = \\text{\"Forced Expiratory Volume in one second,\"}$ or FEV$_1$. The data are summarized below.\n\n| Johns Hopkins | Rancho Los Amigos | St. Louis |\n|---------------|------------------|-----------|\n| $n_1 = 21$ | $n_2 = 16$ | $n_2 = 23$|\n| $\\bar{x}_1 = 2.63$ liters | $\\bar{x}_2 = 3.03$ liters | $\\bar{x}_3 = 2.88$ liters |\n| $s_1^2 = 0.246$ liters$^2$ | $s_2^2 = 0.274$ liters$^2$ | $s_3^2 = 0.248$ liters$^2$ |\n\nBased on histograms of the raw data (not shown), it is reasonable to assume that the FEV$_1$ measurements of the three populations from which these samples were obtained are each approximately normally distributed, i.e., $X_1 \\sim N(\\mu_1, \\sigma_1)$, $X_2 \\sim N(\\mu_2, \\sigma_2)$, and $X_3 \\sim N(\\mu_3, \\sigma_3)$. Furthermore, because the three sample variances are so close in value, it is reasonable to assume equivariance of the three populations, that is, $\\sigma_1^2 = \\sigma_2^2 = \\sigma_3^2$. With these assumptions, answer the following.\n\n(a) Compute the pooled estimate of the common variance $\\sigma^2$ “within groups” via the formula\n\n$$s_{\\text{within}}^2 = \\frac{\\text{MS}_{\\text{Error}}}{\\text{df}_{\\text{Error}}} = \\frac{(n_1 - 1) s_1^2 + (n_2 - 1) s_2^2 + \\ldots + (n_k - 1) s_k^2}{n - k}.$$ \n\n(b) Compute the grand mean of the $k = 3$ groups via the formula\n\n$$\\bar{x} = \\frac{n_1 \\bar{x}_1 + n_2 \\bar{x}_2 + \\ldots + n_k \\bar{x}_k}{n},$$\n\nwhere the combined sample size $n = n_1 + n_2 + \\ldots + n_k$.\n\nFrom this, calculate the estimate of the variance “between groups” via the formula\n\n$$s_{\\text{between}}^2 = \\frac{\\text{MS}_{\\text{Treatment}}}{\\text{df}_{\\text{Treatment}}} = \\frac{n_1 (\\bar{x}_1 - \\bar{x})^2 + n_2 (\\bar{x}_2 - \\bar{x})^2 + \\ldots + n_k (\\bar{x}_k - \\bar{x})^2}{k - 1}.$$ \n\n(c) Using this information, construct a complete ANOVA table, including the F-statistic, and corresponding p-value, relative to .05 (i.e., < .05, > .05, or = .05). Infer whether or not we can reject $H_0: \\mu_1 = \\mu_2 = \\mu_3$ at the $\\alpha = .05$ level of significance. Interpret in context: Exactly what has been demonstrated about the baseline FEV$_1$ levels of the three groups?\n22. Generalization of Problem 2.5/8\n\n(a) Suppose a random sample of size \\( n_1 \\) has a mean \\( \\bar{x}_1 \\) and variance \\( s_1^2 \\), and a second random sample of size \\( n_2 \\) has a mean \\( \\bar{x}_2 \\) and variance \\( s_2^2 \\). If the two samples are combined into a single sample, then algebraically express its mean \\( \\bar{x}_{\\text{total}} \\) and variance \\( s_{\\text{total}}^2 \\) in terms of the preceding variables. (Hint: If you think of this in the right way, it’s easier than it looks.)\n\n(b) In a study of the medical expenses at a particular hospital, it is determined from a sample of 4000 patients that a certain laboratory procedure incurs a mean cost of $30, with a standard deviation of $10. It is realized however, that these values inadvertently excluded 1000 patients for whom the cost was $0. When these patients are included in the study, what is the adjusted cost of the mean and standard deviation?\n\n23.\n\n(a) For a generic \\( 2 \\times 2 \\) contingency table such as the one shown, prove that the Chi-squared test statistic reduces to\n\n\\[\n\\chi^2 = \\frac{n(ad - bc)^2}{R_1R_2C_1C_2}.\n\\]\n\n(b) Suppose that a \\( z \\)-test of two equal proportions results in the generic sample values shown in this table. Prove that the square of the \\( z \\)-score is equal to the Chi-squared score in (a).\n\n24. Problem 5.3/1 illustrates one way that the normal and \\( t \\) distributions differ, as similar as their graphs may appear (drawn to scale, below). Essentially, any \\( t \\)-curve has heavier tails than the bell curve, indicating a higher density of outliers in the distribution. (So much higher in fact, that the mean does not exist!) Another way is to see this is to check the \\( t \\)-distribution for normality, via a Q-Q plot. The posted \\( R \\) code for this problem graphs such a plot for a standard normal distribution (with predictable results), and for a \\( t \\)-distribution with 1 degree of freedom (a.k.a. the Cauchy distribution). Run this code five times each, and comment on the results!\n\n```r\ncurve(dnorm(x, -3, 3, lwd = 2, col = \"darkgreen\"))\ncurve(dt(x, 1), -3, 3, ylim = range(0, .4), lwd = 2, col = \"darkgreen\")\n```\n25. (a) In R, type the following command to generate a data set called “x” of 1000 random values.\n\n\\[ x = rf(1000, 5, 20) \\]\n\nObtain a graph of its frequency histogram by typing `hist(x)`. Include this graph as part of your submitted homework assignment. (Do not include the 1000 data values!)\n\n(b) Next construct a “normal q-q plot” by typing the following.\n\n\\[ qqnorm(x, pch = 19) \\]\n\\[ qqline(x) \\]\n\nInclude this plot as part of your submitted homework assignment.\n\nNow define a new data set called “y” by taking the (natural) logarithm of x.\n\n\\[ y = \\log(x) \\]\n\nObtain a graph of its frequency histogram by typing `hist(y)`. Include this graph as part of your submitted homework assignment. (Do not include the 1000 data values!)\n\nThen construct a “normal q-q plot” by typing the following.\n\n\\[ qqnorm(y, pch = 19) \\]\n\\[ qqline(y) \\]\n\nInclude this plot as part of your submitted homework assignment.\n\n(c) Summarize the results in (a) and (b). In particular, from their respective histograms and q-q plots, what general observation can be made regarding the distributions of x and y = \\log(x)? (Hint: See pages 6.1-25 through 6.1-27.)\n\n26. Refer to the posted Rcode folder for this problem. Please answer all questions.\n\n27. Refer to the posted Rcode folder for this problem. Please answer all questions.", "id": "./materials/148.pdf" }, { "contents": "7.2 Linear Correlation and Regression\n\n**POPULATION**\n\nRandom Variables $X, Y$: **numerical**\n\n**Definition:** Population Linear Correlation Coefficient of $X, Y$\n\n$$\\rho = \\frac{\\sigma_{XY}}{\\sigma_X \\sigma_Y}$$\n\n**FACT:** $-1 \\leq \\rho \\leq +1$\n\n**SAMPLE, size $n$**\n\n**Definition:** Sample Linear Correlation Coefficient of $X, Y$\n\n$$\\hat{\\rho} = r = \\frac{s_{XY}}{s_X s_Y}$$\n\nExample: $r = \\frac{600}{\\sqrt{250} \\sqrt{1750}} = 0.907$ strong, positive linear correlation\n\n**FACT:** $-1 \\leq r \\leq +1$\n\nAny set of data points $(x_i, y_i), \\ i = 1, 2, \\ldots, n,$ having $r > 0$ (likewise, $r < 0$) is said to have a **positive linear correlation** (likewise, **negative linear correlation**). The linear correlation can be **strong**, **moderate**, or **weak**, depending on the magnitude. The closer $r$ is to $+1$ (likewise, $-1$), the more strongly the points follow a straight line having some positive (likewise, negative) slope. The closer $r$ is to 0, the weaker the linear correlation; if $r = 0$, then EITHER the points are uncorrelated (see 7.1), OR they are correlated, but nonlinearly (e.g., $Y = X^2$).\n\n**Exercise:** Draw a scatterplot of the following $n = 7$ data points, and compute $r$.\n\n$(-3, 9), (-2, 4), (-1, 1), (0, 0), (1, 1), (2, 4), (3, 9)$\n(Pearson’s) Sample Linear Correlation Coefficient \\[ r = \\frac{s_{xy}}{s_x s_y} \\]\n\nSome important exceptions to the “typical” cases above:\n\n- \\( r = 0 \\), but \\( X \\) and \\( Y \\) are correlated, nonlinearly\n- \\( r > 0 \\) in each of the two individual subgroups, but \\( r < 0 \\) when combined\n- \\( r > 0 \\), only due to the effect of one influential outlier; if removed, then data are uncorrelated (\\( r = 0 \\))\nStatistical Inference for $\\rho$\n\nSuppose we now wish to conduct a formal test of...\n\n| Hypothesis | $H_0$: $\\rho = 0$ | $\\iff$ | “There is no linear correlation between $X$ and $Y$.” |\n|------------|-------------------|--------|--------------------------------------------------|\n| vs. | $H_A$: $\\rho \\neq 0$ | $\\iff$ | “There is a linear correlation between $X$ and $Y$.” |\n\nTest Statistic\n\n$$T = \\frac{r \\sqrt{n - 2}}{\\sqrt{1 - r^2}} \\sim t_{n-2}$$\n\nExample: $p$-value $= 2 \\cdot P(T_3 \\geq \\frac{.907 \\sqrt{3}}{\\sqrt{1 - (.907)^2}}) = 2 \\cdot P(T_3 \\geq 3.733) = 2(.017) = .034$\n\nAs $p < \\alpha = .05$, the null hypothesis of no linear correlation can be rejected at this level.\n\nComments:\n\n- Defining the numerator “sums of squares” $S_{xx} = (n - 1) s_x^2$, $S_{yy} = (n - 1) s_y^2$, and $S_{xy} = (n - 1) s_{xy}$, the correlation coefficient can also be written as $r = \\frac{S_{xy}}{\\sqrt{S_{xx} S_{yy}}}$.\n\n- The general null hypothesis $H_0$: $\\rho = \\rho_0$ requires a more complicated $Z$-test, which first applies the so-called Fisher transformation, and will not be presented here.\n\n- The assumption on $X$ and $Y$ is that their joint distribution is bivariate normal, which is difficult to check fully in practice. However, a consequence of this assumption is that $X$ and $Y$ are linearly uncorrelated (i.e., $\\rho = 0$) if and only if $X$ and $Y$ are independent. That is, it overlooks the possibility that $X$ and $Y$ might have a nonlinear correlation. The moral: $\\rho$ - and therefore the Pearson sample linear correlation coefficient $r$ calculated above - only captures the strength of linear correlation. A more sophisticated measure, the multiple correlation coefficient, can detect nonlinear correlation, or correlation in several variables. Also, the nonparametric Spearman rank-correlation coefficient can be used as a substitute.\n\n- Correlation does not imply causation! (E.g., $X =$ “children’s foot size” is indeed positively correlated with $Y =$ “IQ score,” but is this really cause-and-effect????) The ideal way to establish causality is via a well-designed randomized clinical trial, but this is not always possible, or even desirable. (E.g., $X =$ smoking vs. $Y =$ lung cancer)\nSimple Linear Regression and the Method of Least Squares\n\n\\[ k = 2 \\text{ parameters, } \\text{\"regression coefficients\"} \\]\n\nPredictor Variable, Explanatory Variable\n\n\\[ Y = \\beta_0 + \\beta_1 X + \\varepsilon \\]\n\n\"Response = (Linear) Model + Error\"\n\nIf a linear association exists between variables \\( X \\) and \\( Y \\), then it can be written as\n\n\\[ \\hat{Y} = \\hat{\\beta}_0 + \\hat{\\beta}_1 X \\]\n\nintercept = \\( b_0 \\) \\hspace{1cm} \\( b_1 \\) = slope\n\nSample-based estimator of response\n\nThat is, given the \"response vector\" \\( Y \\), we wish to find the linear estimate \\( \\hat{Y} \\) that makes the magnitude of the difference \\( \\hat{\\varepsilon} = Y - \\hat{Y} \\) as small as possible.\n\\[ Y = \\beta_0 + \\beta_1 X + \\varepsilon \\quad \\Rightarrow \\quad \\hat{Y} = \\hat{\\beta}_0 + \\hat{\\beta}_1 X \\]\n\nHow should we define the line that “best” fits the data, and obtain its coefficients \\( \\hat{\\beta}_0 \\) and \\( \\hat{\\beta}_1 \\)?\n\nFor any line, errors \\( \\varepsilon_i, \\ i = 1, 2, \\ldots, n \\), can be estimated by the residuals \\( \\hat{\\varepsilon}_i = \\varepsilon_i = y_i - \\hat{y}_i \\).\n\nThe least squares regression line is the unique line that minimizes the Error (or Residual) Sum of Squares \\( \\text{SS}_{\\text{Error}} = \\sum_{i=1}^{n} \\varepsilon_i^2 = \\sum_{i=1}^{n} (y_i - \\hat{y}_i)^2 \\).\n\n\\[\n\\begin{align*}\n\\text{Slope:} & \\quad \\hat{\\beta}_1 = b_1 = \\frac{s_{xy}}{s_x^2} \\\\\n\\text{Intercept:} & \\quad \\hat{\\beta}_0 = b_0 = \\bar{y} - b_1 \\bar{x}\n\\end{align*}\n\\]\n\n\\[ \\hat{Y} = b_0 + b_1 X \\]\n\nExample (cont’d): Slope \\( b_1 = \\frac{600}{250} = 2.4 \\) \\quad Intercept \\( b_0 = 240 - (2.4)(80) = 48 \\)\n\nTherefore, the least squares regression line is given by the equation \\( \\hat{Y} = 48 + 2.4 X \\).\nNote that the sum of the residuals is equal to zero. But the sum of their squares,\n\n$$\\|\\varepsilon\\|^2 = \\text{SS}_{\\text{Error}} = (+18)^2 + (-16)^2 + (-20)^2 + (+16)^2 + (+2)^2 = 1240$$\n\nis, by construction, the smallest such value of all possible regression lines that could have been used to estimate the data. Note also that the center of mass \\((80, 240)\\) lies on the least squares regression line.\n\n**Example:** The population cholesterol level corresponding to \\(x^* = 75\\) fat grams is estimated by \\(\\hat{y} = 48 + 2.4(75) = 228\\) mg/dL. But how precise is this value? (Later...)\nStatistical Inference for $\\beta_0$ and $\\beta_1$\n\nIt is possible to test for significance of the intercept parameter $\\beta_0$ and slope parameter $\\beta_1$ of the least squares regression line, using the following:\n\n\\[\n(1 - \\alpha) \\times 100\\% \\text{ Confidence Limits}\n\\]\n\nFor $\\beta_0$: \\[b_0 \\pm t_{n-2, \\alpha/2} \\cdot s_e \\sqrt{\\frac{1}{n} + \\frac{(\\bar{x})^2}{S_{xx}}}\\]\n\nFor $\\beta_1$: \\[b_1 \\pm t_{n-2, \\alpha/2} \\cdot s_e \\frac{1}{\\sqrt{S_{xx}}}\\]\n\n\\[\n\\text{Test Statistic}\n\\]\n\nFor $\\beta_0$: \\[T = \\left(\\frac{b_0 - \\beta_0}{s_e}\\right) \\sqrt{\\frac{n S_{xx}}{S_{xx} + n (\\bar{x})^2}} \\sim t_{n-2}\\]\n\nFor $\\beta_1$: \\[T = \\left(\\frac{b_1 - \\beta_1}{s_e}\\right) \\sqrt{S_{xx}} \\sim t_{n-2}\\]\n\nwhere $s_e^2 = \\frac{SS_{\\text{error}}}{n-2}$ is the so-called standard error of estimate, and $S_{xx} = (n - 1) s_x^2$.\n\n(Note: $s_e^2$ is also written as MSE or $MS_{\\text{Error}}$, the “mean square error” of the regression; see ANOVA below.)\n\nExample: Calculate the p-value of the slope parameter $\\beta_1$, under...\n\nNull Hypothesis $H_0$: $\\beta_1 = 0$ $\\iff$ “There is no linear association between X and Y.”\n\nvs.\n\nAlternative Hyp. $H_A$: $\\beta_1 \\neq 0$ $\\iff$ “There is a linear association between X and Y.”\n\nFirst, $s_e^2 = \\frac{1240}{3} = 413.333$, so $s_e = 20.331$. And $S_{xx} = (4)(250) = 1000$. So...\n\n\\[\np\\text{-value} = 2 \\cdot P(T_3 \\geq \\frac{2.4 - 0}{20.331} \\sqrt{1000}) = 2 \\cdot P(T_3 \\geq 3.733) = 2 (.017) = .034\n\\]\n\nAs $p < \\alpha = .05$, the null hypothesis of no linear association can be rejected at this level.\n\nNote that the T-statistic (3.733), and hence the resulting p-value (.034), is identical to the test of significance of the linear correlation coefficient $H_0$: $\\rho = 0$ conducted above!\n\nExercise: Calculate the 95% confidence interval for $\\beta_1$, and use it to test $H_0$: $\\beta_1 = 0$. \nConfidence and Prediction Intervals\n\nRecall that, from the discussion in the previous section, a regression problem such as this may be viewed in the formal context of starting with \\( n \\) normally-distributed populations, each having a conditional mean \\( \\mu_{Y|X=x} \\), \\( i = 1, 2, \\ldots, n \\). From this, we then obtain a linear model that allows us to derive an estimate of the response variable via \\( \\hat{Y} = b_0 + b_1 X \\), for any value \\( X = x^* \\) (with certain restrictions to be discussed later), i.e., \\( \\hat{Y} = b_0 + b_1 x^* \\). There are two standard possible interpretations for this fitted value. First, \\( \\hat{Y} \\) can be regarded simply as a “predicted value” of the response variable \\( Y \\), for a randomly selected individual from the specific normally-distributed population corresponding to \\( X = x^* \\), and can be improved via a so-called prediction interval.\n\n**Exercise:** Confirm that the 95% prediction interval for \\( \\hat{Y} = 228 \\) (when \\( x^* = 75 \\)) is \\((156.3977, 299.6023)\\).\n\n**Example (\\( \\alpha = .05 \\)):**\n\n| \\( X \\) | fit | Lower | Upper |\n|--------|-----|-------|-------|\n| 60 | 192 | 110.1589 | 273.8411 |\n| 70 | 216 | 142.2294 | 289.7706 |\n| 80 | 240 | 169.1235 | 310.8765 |\n| 90 | 264 | 190.2294 | 337.7706 |\n| 100 | 288 | 206.1589 | 369.8411 |\nThe second interpretation is that \\( \\hat{y} \\) can be regarded as a point estimate of the conditional mean \\( \\mu_{Y|X=x^*} \\) of this population, and can be improved via a confidence interval.\n\n\\[\n(1 - \\alpha) \\times 100\\% \\text{ Confidence Limits for } \\mu_{Y|X=x^*} \\\\\n(b_0 + b_1 x^*) \\pm t_{n-2, \\alpha/2} \\cdot s_e \\sqrt{\\frac{1}{n} + \\frac{(x^* - \\bar{x})^2}{S_{xx}}}\n\\]\n\nThis diagram illustrates the associated 95% confidence interval around \\( \\hat{y} = b_0 + b_1 x^* \\), which contains the true conditional mean \\( \\mu_{Y|X=x^*} \\) with 95% probability. Note that it is narrower than the corresponding prediction interval above.\n\n**Exercise:** Confirm that the 95% confidence interval for \\( \\hat{y} = 228 \\) (when \\( x^* = 75 \\)) is (197.2133, 258.6867).\n\n**Note:** Both approaches are based on the fact that there is, in principle, variability in the coefficients \\( b_0 \\) and \\( b_1 \\) themselves, from one sample of \\( n \\) data points to another. Thus, for fixed \\( x^* \\), the object \\( \\hat{y} = b_0 + b_1 x^* \\) can actually be treated as a random variable in its own right, with a computable sampling distribution.\n\nAlso, we define the general conditional mean \\( \\mu_{Y|X} \\) - i.e., conditional expectation \\( E[Y|X] \\) - as \\( \\mu_{Y|X=x^*} \\) - i.e., \\( E[Y|X=x^*] \\) - for all appropriate \\( x^* \\), rather than a specific one.\nExample ($\\alpha = .05$):\n\n| $X$ | fit | Lower | Upper |\n|-----|-----|-----------|-----------|\n| 60 | 192 | 141.8827 | 242.1173 |\n| 70 | 216 | 180.5617 | 251.4383 |\n| 80 | 240 | 211.0648 | 268.9352 |\n| 90 | 264 | 228.5617 | 299.4383 |\n| 100 | 288 | 237.8827 | 338.1173 |\n\nComments:\n\n- Note that, because individual responses have greater variability than mean responses (recall the Central Limit Theorem, for example), we expect prediction intervals to be wider than the corresponding confidence intervals, and indeed, this is the case. The two formulas differ by a term of “1 +” in the standard error of the former, resulting in a larger margin of error.\n\n- Note also from the formulas that both types of interval are narrowest when $x^* = \\bar{x}$, and grow steadily wider as $x^*$ moves farther away from $\\bar{x}$. (This is evident in the graph of the 95% confidence intervals above.) Great care should be taken if $x^*$ is outside the domain of sample values! For example, when fat grams $x = 0$, the linear model predicts an unrealistic cholesterol level of $\\hat{y} = 48$, and the margin of error is uselessly large. The linear model is not a good predictor there.\nANOVA Formulation\n\nAs with comparison of multiple treatment means (§6.3.3), regression can also be interpreted in the general context of analysis of variance. That is, because\n\n\\[\n\\text{Response} = \\text{Model} + \\text{Error},\n\\]\n\nit follows that the total variation in the original response data can be partitioned into a source of variation due to the model, plus a source of variation for whatever remains. We now calculate the three “Sums of Squares (SS)” that measure the variation of the system and its two component sources, and their associated degrees of freedom (df).\n\n1. **Total Sum of Squares** = sum of the squared deviations of each observed response value \\(y_i\\) from the mean response value \\(\\bar{y}\\).\n\n\\[\n\\begin{align*}\n\\text{SS}_{\\text{Total}} &= (210 - 240)^2 + (200 - 240)^2 + (220 - 240)^2 + (280 - 240)^2 + (290 - 240)^2 = 7000 \\\\\n\\text{df}_{\\text{Total}} &= 5 - 1 = 4 \\quad \\text{Reason: } n \\text{ data values} - 1\n\\end{align*}\n\\]\n\nNote that, by definition, \\(s_y^2 = \\frac{\\text{SS}_{\\text{Total}}}{\\text{df}_{\\text{Total}}} = \\frac{7000}{4} = 1750\\), as given in the beginning of this example in 7.1.\n\n2. **Regression Sum of Squares** = sum of the squared deviations of each fitted response value \\(\\hat{y}_i\\) from the mean response value \\(\\bar{y}\\).\n\n\\[\n\\begin{align*}\n\\text{SS}_{\\text{Reg}} &= (192 - 240)^2 + (216 - 240)^2 + (240 - 240)^2 + (264 - 240)^2 + (288 - 240)^2 = 5760 \\\\\n\\text{df}_{\\text{Reg}} &= 1 \\quad \\text{Reason: As the regression model is linear, its degrees of freedom = one less than the } k = 2 \\text{ parameters we are trying to estimate (} \\beta_0 \\text{ and } \\beta_1).}\n\\end{align*}\n\\]\n\n3. **Error Sum of Squares** = sum of the squared deviations of each observed response \\(y_i\\) from its corresponding fitted response \\(\\hat{y}_i\\) (i.e., the sum of the squared residuals).\n\n\\[\n\\begin{align*}\n\\text{SS}_{\\text{Error}} &= (210 - 192)^2 + (200 - 216)^2 + (220 - 240)^2 + (280 - 264)^2 + (290 - 288)^2 = 1240 \\\\\n\\text{df}_{\\text{Error}} &= 5 - 2 = 3 \\quad \\text{Reason: } n \\text{ data values} - k \\text{ regression parameters in model}\n\\end{align*}\n\\]\n\n\\[\n\\begin{align*}\n\\text{SS}_{\\text{Total}} &= \\text{SS}_{\\text{Reg}} + \\text{SS}_{\\text{Error}} \\\\\n\\text{df}_{\\text{Total}} &= \\text{df}_{\\text{Reg}} + \\text{df}_{\\text{Error}}\n\\end{align*}\n\\]\nNull Hypothesis \\( H_0: \\beta_1 = 0 \\) ⇔ \"There is no linear association between X and Y.\"\n\nvs.\n\nAlternative Hyp. \\( H_A: \\beta_1 \\neq 0 \\) ⇔ \"There is a linear association between X and Y.\"\n\nat the \\( \\alpha = .05 \\) significance level, which is consistent with our earlier findings.\n\nComment: Again, note that \\( 13.94 = (\\pm 3.733)^2 \\), i.e., \\( F_{1,3} = t_3^2 \\Rightarrow \\) equivalent tests.\nHow well does the model fit? Out of a total response variation of 7000, the linear regression model accounts for 5760, with the remaining 1240 unaccounted for (perhaps explainable by a better model, or simply due to random chance). We can therefore assess how well the model fits the data by calculating the ratio \\( \\frac{SS_{\\text{Reg}}}{SS_{\\text{Total}}} = \\frac{5760}{7000} = 0.823 \\). That is, 82.3% of the total response variation is due to the linear association between the variables, as determined by the least squares regression line, with the remaining 17.7% unaccounted for. (Note: This does NOT mean that 82.3% of the original data points lie on the line. This is clearly false; from the scatterplot, it is clear that none of the points lies on the regression line!)\n\nMoreover, note that \\( 0.823 = (0.907)^2 = r^2 \\), the square of the correlation coefficient calculated before! This relation is true in general...\n\n**Coefficient of Determination**\n\n\\[\nr^2 = \\frac{SS_{\\text{Reg}}}{SS_{\\text{Total}}} = 1 - \\frac{SS_{\\text{Err}}}{SS_{\\text{Total}}}\n\\]\n\nThis value (always between 0 and 1) indicates the proportion of total response variation that is accounted for by the least squares regression model.\n\nComment: In practice, it is tempting to over-rely on the coefficient of determination as the sole indicator of linear fit to a data set. As with the correlation coefficient \\( r \\) itself, a reasonably high \\( r^2 \\) value is suggestive of a linear trend, or a strong linear component, but should not be used as the definitive measure.\n\n**Exercise:** Sketch the \\( n = 5 \\) data points \\((X, Y)\\)\n\n\\[(0, 0), (1, 1), (2, 4), (3, 9), (4, 16)\\]\n\nin a scatterplot, and calculate the coefficient of determination \\( r^2 \\) in two ways:\n\n1. By squaring the linear correlation coefficient \\( r \\).\n2. By explicitly calculating the ratio \\( \\frac{SS_{\\text{Reg}}}{SS_{\\text{Total}}} \\) from the regression line.\n\nShow agreement of your answers, and that, despite a value of \\( r^2 \\) very close to 1, the exact association between \\( X \\) and \\( Y \\) is actually a nonlinear one. Compare the linear estimate of \\( Y \\) when \\( X = 5 \\), with its exact value.\n\nAlso see **Appendix > Geometric Viewpoint > Least Squares Approximation**.\nRegression Diagnostics - Checking the Assumptions\n\nResponse = Model + Error\n\nTrue Responses: \\( Y = \\beta_0 + \\beta_1 X + \\varepsilon \\) \\( \\iff \\) \\( Y_i = \\beta_0 + \\beta_1 x_i + \\varepsilon_i \\), \\( i = 1, 2, \\ldots, n \\)\n\nFitted Responses: \\( \\hat{Y} = \\hat{b}_0 + \\hat{b}_1 X \\) \\( \\iff \\) \\( \\hat{y}_i = \\hat{b}_0 + \\hat{b}_1 x_i \\), \\( i = 1, 2, \\ldots, n \\)\n\nResiduals: \\( \\hat{\\varepsilon} = Y - \\hat{Y} \\) \\( \\iff \\) \\( \\hat{\\varepsilon}_i = e_i = y_i - \\hat{y}_i \\), \\( i = 1, 2, \\ldots, n \\)\n\n1. **The model is “correct.”**\n\n Perhaps a better word is “useful,” since correctness is difficult to establish without a theoretical justification, based on known mathematical and scientific principles.\n\n **Check:** Scatterplot(s) for general behavior; \\( r^2 \\approx 1 \\), overall balance of simplicity vs. complexity of model, and robustness of response variable explanation.\n\n2. **Errors \\( \\varepsilon_i \\) are independent of each other, \\( i = 1, 2, \\ldots, n \\).**\n\n This condition is equivalent to the assumption that the responses \\( y_i \\) are independent of one other. Alas, it is somewhat problematic to check in practice; formal statistical tests are limited. Often, but not always, it is implicit in the design of the experiment. Other times, errors (and hence, responses) may be autocorrelated with each other. Example: \\( Y = \\) “systolic blood pressure (mm Hg)” at times \\( t = 0 \\) and \\( t = 1 \\) minute later. Specialized time-series techniques exist for these cases, but are not pursued here.\n\n3. **Errors \\( \\varepsilon_i \\) are normally distributed with mean 0, and equal variances \\( \\sigma_1^2 = \\sigma_2^2 = \\ldots = \\sigma_n^2 \\) (\\( = \\sigma^2 \\)), i.e., \\( \\varepsilon_i \\sim N(0, \\sigma) \\), \\( i = 1, 2, \\ldots, n \\).**\n\n This condition is equivalent to the original normality assumption on the responses \\( y_i \\). Informally, if for each fixed \\( x_i \\), the true response \\( y_i \\) is normally distributed with mean \\( \\mu_{Y|X=x} \\) and variance \\( \\sigma^2 \\) – i.e, \\( y_i \\sim N(\\mu_{Y|X=x}, \\sigma) \\) – then the error \\( \\varepsilon_i \\) that remains upon “subtracting out” the true model value \\( \\beta_0 + \\beta_1 x_i \\) (see boxed equation above) turns out also to be normally distributed, with mean 0 and the same variance \\( \\sigma^2 \\) – i.e., \\( \\varepsilon_i \\sim N(0, \\sigma) \\). Formal details are left to the mathematically brave to complete.\nCheck: Residual plot (residuals $e_i$ vs. fitted values $\\hat{y}_i$) for a general random appearance, evenly distributed about zero. (Can also check the normal probability plot.)\n\nTypical residual plots that violate Assumptions 1-3:\n\n- **Nonlinearity**: Nonlinear trend can often be described with a polynomial regression model, e.g., $Y = \\beta_0 + \\beta_1 X + \\beta_2 X^2 + \\epsilon$. If a residual plot resembles the last figure, this is a possible indication that more than one predictor variable may be necessary to explain the response, e.g., $Y = \\beta_0 + \\beta_1 X_1 + \\beta_2 X_2 + \\epsilon$, multiple linear regression.\n- **Dependent errors**: Nonconstant variance can be handled by Weighted Least Squares (WLS) – versus Ordinary Least Squares (OLS) above – or by using a transformation of the data, which can also alleviate nonlinearity, as well as violations of the third assumption that the errors are normally distributed.\nExample: Regress $Y =$ “human age (years)” on $X =$ “dog age (years),” based on the following $n = 20$ data points, for adult dogs 23-34 lbs.: \n\n| X | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n|----|----|----|----|----|----|----|----|----|----|----|\n| Y | 15 | 21 | 27 | 32 | 37 | 42 | 46 | 51 | 55 | 59 |\n| | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 |\n| | 63 | 67 | 71 | 76 | 80 | 85 | 91 | 97 | 103| 111|\n\nSadie\n\n\\[ \\hat{Y} = 12.1 + 4.7X \\]\n\nResiduals:\n\n| Min | 1Q | Median | 3Q | Max |\n|----------|--------|--------|--------|---------|\n| -2.61353 | -1.57124 | 0.08947 | 1.16654 | 4.87143 |\n\nCoefficients:\n\n| Estimate | Std. Error | t value | Pr(>|t|) |\n|-----------|------------|---------|----------|\n| (Intercept) | 12.06842 | 0.87794 | 13.75 | 5.5e-11 *** |\n| X | 4.70301 | 0.07329 | 64.17 | < 2e-16 *** |\n\nMultiple R-Squared: 0.9956, Adjusted R-squared: 0.9954\nF-statistic: 4118 on 1 and 18 degrees of freedom, p-value: 0\nThe residual plot exhibits a clear nonlinear trend, despite the excellent fit of the linear model. It is possible to take this into account using, say, a cubic (i.e., third-degree) polynomial, but this then begs the question: How complicated should we make the regression model?\n\nMy assistant and I, thinking hard about regression models.", "id": "./materials/149.pdf" }, { "contents": "End digression.\n\n**Numerical sets**\n\n\\[ \\mathbb{N} = \\{0, 1, 2, 3, \\ldots \\} \\] natural numbers\n\n\\[ \\mathbb{Z} = \\{0, 1, -1, 2, -2, 3, -3, \\ldots \\} \\] integer numbers\n\n\\[ \\mathbb{Q} = \\left\\{ \\frac{m}{n} : m \\in \\mathbb{Z}, n \\in \\mathbb{N}, n \\neq 0 \\right\\} \\] rational numbers\n\n\\[ \\mathbb{R} = \\text{\"every number\"} \\]\n\\[ \\mathbb{N} \\subseteq \\mathbb{Z} \\subseteq \\mathbb{Q} \\subseteq \\mathbb{R} \\]\n\n**Algebraic properties of \\( \\mathbb{R} \\)**\n\nIn \\( \\mathbb{R} \\), we have defined two operations:\n\n- \\( + \\) and \\( \\cdot \\) (sum and product), with the following properties:\n\n 1. **Commutativity**\n \\[ a + b = b + a \\quad \\forall a, b \\in \\mathbb{R} \\]\n\n 2. **Associativity**\n \\[ a + (b + c) = (a + b) + c \\quad \\forall a, b, c \\in \\mathbb{R} \\]\n\n 3. **Identity element** (neutral element)\n \\[ a + 0 = a \\quad \\forall a \\in \\mathbb{R} \\]\n\\[ 0 + a = a + 0 = a \\quad \\text{also in } \\mathbb{N}, \\mathbb{Z}, \\mathbb{Q} \\]\n\n\\[(S4) \\quad \\forall a \\in \\mathbb{R} \\quad \\text{there exists } (-a) \\in \\mathbb{R} \\quad \\text{inverse} \\]\n\n\\[ e + (-a) = (-e) + a = 0 \\quad \\text{with respect to } + \\]\n\n\\[(M1) \\quad a \\cdot b = b \\cdot a \\quad \\forall a, b \\in \\mathbb{R} \\quad \\text{(also in } \\mathbb{N}, \\mathbb{Z}, \\mathbb{Q}) \\]\n\n\\[(M2) \\quad a \\cdot (b \\cdot c) = (a \\cdot b) \\cdot c \\quad \\forall a, b, c \\in \\mathbb{R} \\quad \\text{(also in } \\mathbb{N}, \\mathbb{Z}, \\mathbb{Q}) \\]\n\n\\[(M3) \\quad \\exists \\text{ \"neutral element\" } 1 \\in \\mathbb{R} \\quad \\text{(also in } \\mathbb{N}, \\mathbb{Z}, \\mathbb{Q}) \\]\n\n\\[ a \\cdot 1 = 1 \\cdot a = a \\]\n\n\\[(M4) \\quad \\forall a \\in \\mathbb{R} \\quad a \\neq 0 \\quad \\text{there exist } \\frac{1}{a} \\in \\mathbb{R} \\quad \\text{not in } \\mathbb{N} \\]\n\n\\[ a \\cdot \\left( \\frac{1}{a} \\right) = \\left( \\frac{1}{a} \\right) \\cdot a = 1 \\quad \\text{ok in } \\mathbb{R} \\]\n(D) \\( a \\cdot (b + c) = a \\cdot b + a \\cdot c \\quad \\forall a, b, c \\in \\mathbb{R} \\) \n(distributivity) \n(Also in \\( \\mathbb{R}^2, \\mathbb{R}^3 \\))\n\n**Ordering on \\( \\mathbb{R} \\)** \n\\( \\forall a, b \\in \\mathbb{R} \\)\n\nEither \\( a \\leq b \\) or \\( b \\leq a \\)\n\n(OA1) \\( a \\leq b \\quad c \\in \\mathbb{R} \\)\n\n\\( a + c \\leq b + c \\)\n\n(I can add the same quantity to both sides on both sides doesn't change)\n(0A2) \\[ a \\leq b \\]\n\n\\[ a \\cdot c \\leq b \\cdot c \\quad \\text{if} \\quad c > 0 \\]\n\n\\[ a \\cdot c \\geq b \\cdot c \\quad \\text{if} \\quad c < 0 \\]\n\n(0.0) \\[ \\forall a, b \\in \\mathbb{R} \\quad a \\leq b \\text{ or } b \\leq a \\]\n\n(01) \\[ a \\leq b \\text{ and } b \\leq c \\implies a \\leq c \\quad \\text{(transitivity)} \\]\n\n(02) \\[ a \\leq b \\text{ and } b \\leq a \\implies a = b \\]\n\nSo, up to now \\( \\mathbb{R} \\) and \\( \\mathbb{Q} \\) aren't different.\n\\[ \\mathbb{Q} = \\left\\{ \\frac{m}{n} : m \\in \\mathbb{Z}, n \\in \\mathbb{N}, n \\neq 0 \\right\\} \\]\n\n\\[ \\frac{5}{3} \\in \\mathbb{Q} \\quad 5 \\in \\mathbb{Z} \\quad 3 \\in \\mathbb{N} \\quad 3 \\neq 0 \\]\n\n\\[ \\mathbb{R} \\setminus \\mathbb{Q} = \\text{\"irrational numbers\"} \\ni \\sqrt{2} \\]\n\n\\[ \\exists q \\in \\mathbb{Q} \\text{ s.t. } q^2 = 2 \\quad \\text{No!} \\]\nCompleteness (Continuity Axiom)\n\n\\[ A \\subseteq \\mathbb{R} \\quad B \\subseteq \\mathbb{R} \\]\n\nLet us suppose that \\( \\forall a \\in A, \\forall b \\in B \\) we have \\( a \\leq b \\), the set \\( A \\) is \"on the left\" of \\( B \\).\n\nIs \\( c \\in \\mathbb{R} \\) \"in the middle\"?\n\n\\[ c \\text{ is a separator for } A \\text{ and } B \\text{ if} \\]\n\n\\[ \\forall a \\in A \\quad a \\leq c \\]\n\n\\[ \\forall b \\in B \\quad c \\leq b \\]\nKey property of \\( \\mathbb{R} \\): for every \\( A, B \\) such that \\( A \\) is \"on the left\" of \\( B \\), there exists a separator for \\( A \\) and \\( B \\). (Continuity axiom)\n\nObs. \\( \\mathbb{Q} \\) does NOT satisfy the cont. axiom.\n\n\\[\nA = \\{ x \\in \\mathbb{Q} : x > 0, \\ x^2 < 2 \\}\n\\]\n\n\\[\nB = \\{ x \\in \\mathbb{Q} : x > 0, \\ x^2 > 2 \\}\n\\]\n\n\\( A \\) is \"on the left\" of \\( B \\). But it doesn't exist a separator \\( c \\in \\mathbb{Q} \\), because\nIf it would exist, we should have $c^2 = 2$. ", "id": "./materials/15.pdf" }, { "contents": "LECTURE NOTES\nON\nPARAMETRIC & NON-PARAMETRIC TESTS\nFOR\nSOCIAL SCIENTISTS/ PARTICIPANTS\nOF\nRESEARCH METHODOLOGY WORKSHOP\nBBAU, LUCKNOW\n\nBy\n\nDr. Rajeev Pandey\nProfessor in Statistics\nUniversity of Lucknow\nLucknow – 226 007\nWhat is Statistics?\nStatistics is neither really a science nor a branch of Mathematics. It is perhaps best considered as a meta-science (or meta-language) for dealing with data collection, analysis, and interpretation. As such its scope is enormous and it provides much guiding insight in many branches of science, business.\n\n- **Statistics** is the science of collecting, organizing, analyzing, interpreting, and presenting data. (Old Definition).\n- A **statistic** is a single measure (number) used to summarize a sample data set. For example, the average height of students in this class.\n- A **statistician** is an expert with at least a master’s degree in mathematics or statistics or a trained professional in a related field.\n\nStatistics is a tool for creating new understanding from a set of numbers.\n\nStatistics is a science of getting informed decisions.\n\nA Taxonomy of Statistics\n**Statistical Inference:** is to draw conclusions about the Population on the basis of information available in the sample which has been drawn from the population by a random sampling technique/ procedure. There are two branches Statistical Inference namely ESTIMATION & TESTING OF HYPOTHESIS.\n\nIn ESTIMATION, we try to find an estimate of any population characteristic while in TESTING OF HYPOTHESIS, we try to test the statement about any population characteristic. Here, our main concern is with “TESTING OF HYPOTHESIS”.\n\n**Some Basic Definitions:**\n\n**Population:** Any collection of individuals under study is said to be Population (Universe). The individuals are often called the members or the units of the population may be may be physical objects or measurements expressed numerically or otherwise.\n\n**Sample:** A part or small section selected from the population is called a sample and process of such selection is called sampling.\n\n(The fundamental object of sampling is to get as much information as possible of the whole universe by examining only a part of it. An attempt is thus made through sampling to give the maximum information about parent universe with the minimum effort).\n\n**Parameters:** Statistical measurements such as Mean, Variance etc. of the population are called parameters.\n\n**Statistic:** It a statistical measure computed from sample observations alone. The theoretical distribution of a statistics is called its sampling distribution. Standard deviation of the sampling distribution of a statistic is called Standard Error.\n\n**Hypothesis:** is a statement given by an individual. Usually it is required to make decisions about populations on the basis of sample information. Such decisions are called Statistical Decisions. In attempting to reach decisions it is often necessary to make assumption about population involved. Such assumptions, which are not necessarily true, are called statistical hypothesis.\n\n**Parametric Hypothesis:** A statistical hypothesis which refers only to values of unknown parameters of population is usually called a parametric hypothesis.\n\n**Null Hypothesis and Alternative Hypothesis:** A hypothesis which is tested under the assumption that it is true is called a null hypothesis and is denoted by $H_0$. Thus a hypothesis which is tested for possible rejection under the assumption that it is true is known as Null Hypothesis. The hypothesis which differs from the given Null Hypothesis $H_0$ and is accepted when $H_0$ is rejected is called an alternative hypothesis and is denoted by $H_1$ (The hypothesis against which we test the null hypothesis, is an alternative hypothesis).\n**Simple and Composite Hypothesis:** A parametric hypothesis which describes a distribution completely is called a simple hypothesis otherwise it is called composite hypothesis. For example; In case of Normal Distribution $N(\\mu, \\sigma^2)$, $\\mu = 5$, $\\sigma = 3$ is simple hypothesis whereas $\\mu = 5$ is a composite hypothesis as nothing have been said about $\\sigma$.\n\nSimilarly, $\\mu < 5$, $\\sigma = 3$ is a composite hypothesis.\n\nLet $H_0: \\mu = 5$ be the null hypothesis, then\n- $H_1: \\mu \\neq 5$ is two sided composite alternative hypothesis.\n- $H_1: \\mu < 5$ is one sided (Left) composite alternative hypothesis.\n- $H_1: \\mu > 5$ is one sided (Right) composite alternative hypothesis.\n\n**Test:** Test is a rule through we test the null hypothesis against the given alternative hypothesis.\n\n**Tests of Significance:** Procedure which enables us to decide, on the basis of sample information whether to accept or reject the hypothesis or to determine whether observed sampling results differ significantly from expected results are called tests of significance, rules of decisions or tests of hypothesis.\n\n**Level of Significance:** The probability level below which we reject the hypothesis is called level of significance. The levels of significance usually employed in testing of hypothesis are 5% and 1%.\n\n**P-Value:** The p-value is the level of marginal significance within a statistical hypothesis test representing the probability of the occurrence of a given event. The p-value is used as an alternative to rejection points to provide the smallest level of significance at which the null hypothesis would be rejected.\nA p value is used in hypothesis testing to help you support or reject the null hypothesis. The p value is the evidence against a null hypothesis. The smaller the p-value, the stronger the evidence that you should reject the null hypothesis.\n\nP values are expressed as decimals although it may be easier to understand what they are if you convert them to a percentage. For example, a p value of 0.0254 is 2.54%. This means there is a 2.54% chance your results could be random (i.e. happened by chance). That’s pretty tiny. On the other hand, a large p-value of .9(90%) means your results have a 90% probability of being completely random and not due to anything in your experiment. Therefore, the smaller the p-value, the more important (“significant”) your results.\n\n**Critical Region and Acceptance Region:** A region (corresponding to a statistic t) is called the sample space. The part of sample space which amounts to rejection of null hypothesis $H_0$, is called critical region or region of rejection.\n\nIf $X = (x_1, x_2, ..., x_n)$ is the random vector observed and $W_c$ is the critical region (which corresponds the rejection of the hypothesis according to a prescribed test procedure) of the sample space $W$, then $W_a = W - W_c$ of the sample space is called the acceptance region.\n\n**Two Types of Errors in Testing of a Hypothesis:** While testing a hypothesis $H_0$, the following four situations may arise:\n\n(a) The test statistic may fall in the critical region even if $H_0$ is true then we shall be led to reject $H_0$ when it is true.\n\n(b) The test statistic may fall in the acceptance region when $H_0$ is true we shall be led to accept $H_0$.\n\n(c) The test statistic may fall in the critical region when $H_0$ is not true i.e. $H_1$ is true, then we shall be led to reject $H_0$ when $H_1$ is true.\n\n(d) The test statistic may fall in the acceptance region even if $H_0$ is not true, then we shall be led to accept $H_0$ when it is not true.\n\nIt is quite obvious that the decisions taken in (b) and (c) are correct while the decisions taken in (a) and (d) are incorrect.\n\nThe wrong decision of rejecting a null hypothesis $H_0$, when it is true is called the **Type I Error** i.e. we reject $H_0$ when it is true. Similarly, the wrong decision of accepting the null hypothesis $H_0$ when it is not true is called the **Type II Error** i.e. we accept $H_0$ when $H_1$ is true.\n\n**Probability Forms:**\n\n$$P(\\text{reject } H_0 \\text{ when it is true}) = P(\\text{reject } H_0/H_0) = \\alpha$$\n\nand\n\n$$P(\\text{accept } H_0 \\text{ when it is wrong}) = P(\\text{reject } H_0/H_1) = \\beta$$\n\nThe $\\alpha$ and $\\beta$ are called the size of Type I Error and size of Type II Error respectively.\nRules or Procedure for Testing of Hypothesis:\nA test is a statistical procedure or a rule for deciding whether to accept or reject the hypothesis on the basis of sample values obtained.\nFollowing is the Procedure for testing of Hypothesis:-\n(a) Mention the null hypothesis $H_0$ to be tested along with an alternative hypothesis $H_1$.\n(b) Make some assumptions such as the sample is random, the population is normal, the variance of two different populations are equal or unknown etc.\n(c) Then find the most appropriate test statistic together with its sampling distribution. A statistic whose primary role is that of providing a test of some hypothesis is called a test statistic.\n(d) On the basis of the sampling distribution make a decision to either accept or reject the null hypothesis $H_0$.\n(e) Take a random sample and compute the test statistic. If the calculated value of the test statistic falls in the acceptance region, then accept the null hypothesis $H_0$. If it falls in the region of rejection (or Critical Region), reject the null hypothesis $H_0$.\n\nPower Function of a Test: The Power Function of a test of a statistical hypothesis $H_0: \\theta = \\theta_0$, say, against alternative hypothesis $H_1: \\theta > \\theta_0$, $\\theta < \\theta_0$, $\\theta \\neq \\theta_0$ is a function of the parameter, under consideration, which gives the probability that the test statistic will fall in the critical region when $\\theta$ is the true value of the parameter.\n\nDeduction: $P(\\theta) = P(\\text{rejecting } H_0 \\text{ when } H_1 \\text{ is true})$\n$= P(W \\text{ belong to } W_c / H_1) = 1 - P(\\text{accept } H_0 / H_1)$\n$= 1 - \\beta(\\theta)$\nThe value of the power function at a particular value of the parameter is called the power of the test.\n\nBest Critical Region: In testing the hypothesis $H_0: \\theta = \\theta_0$ against the given alternative $H_1: \\theta = \\theta_1$, the critical region is best if the type II error is minimum or the power is maximum when compared to every other possible critical region of size $\\alpha$. A test defined by this critical region is called most powerful test.\n\nPARAMETRIC and NON-PARAMETRIC TESTS\n\nIn the literal meaning of the terms, a parametric statistical test is one that makes assumptions about the parameters (defining properties) of the population distribution(s) from which one's data are drawn, while a non-parametric test is one that makes no such assumptions.\n\nPARAMETRIC TESTS: Most of the statistical tests we perform are based on a set of assumptions. When these assumptions are violated the results of the analysis can be misleading or completely erroneous.\nTypical assumptions are:\n\n- **Normality**: Data have a normal distribution (or at least is symmetric)\n- **Homogeneity of variances**: Data from multiple groups have the same variance\n- **Linearity**: Data have a linear relationship\n- **Independence**: Data are independent\n\nWe explore in detail what it means for data to be normally distributed in Normal Distribution, but in general it means that the graph of the data has the shape of a bell curve. Such data is symmetric around its mean and has kurtosis equal to zero. In Testing for Normality and Symmetry we provide tests to determine whether data meet this assumption.\n\nSome tests (e.g. ANOVA) require that the groups of data being studied have the same variance. In Homogeneity of Variances we provide some tests for determining whether groups of data have the same variance.\n\nSome tests (e.g. Regression) require that there be a linear correlation between the dependent and independent variables. Generally linearity can be tested graphically using scatter diagrams or via other techniques explored in Correlation, Regression and Multiple Regression.\n• Many statistical methods require that the numeric variables we are working with have an approximate **normal distribution**.\n\n• Standardized normal distribution with empirical rule percentages.\n\n• For example, t-tests, F-tests, and regression analyses all require in some sense that the numeric variables are approximately normally distributed.\n**t-test**\n\n**Introduction:**\n\n- The t-test is a basic test that is limited to two groups. For multiple groups, you would have to compare each pair of groups. For example with three groups there would be three tests (AB, AC, BC) whilst with seven groups there would be need of 21 tests.\n- The basic principle is to test the null hypothesis that means of the two groups are equal.\n\n**The t-test assumes:**\n\n- A normal distribution (parametric data)\n- Underlying variances are equal (if not, use welch’s test)\n- It is used when there is random assignment and only two sets of measurement to compare.\n\n**There are two main types of t-test:**\n\n- **Independent – measures – t-test:** when samples are not matched.\n- **Match – pair – t-test:** when samples appear in pairs (eg. before and after)\n\n- **A single – sample t-test** compares a sample against a known figure. For example when measures of a manufactured item are compared against the required standard.\n\n**APPLICATIONS:**\n\n- To compare the mean of a sample with population mean. (Simple t-test)\n- To compare the mean of one sample with the independent sample. (Independent Sample t-test)\n- To compare between the values (readings) of one sample but in two occasions. (Paired sample t-test)\n\n**Independent Samples t-Test (or 2-Sample t-Test)**\n\nThe independent samples t-test is probably the single most widely used test in statistics. It is used to compare differences between separate groups. In Psychology, these groups are often composed by randomly assigning research participants to conditions. However, this test can also be used to explore differences in naturally occurring groups. For example, we may be interested in differences of emotional intelligence between males and females.\n\nAny differences between groups can be explored with the independent t-test, as long as the tested members of each group are reasonably representative of the population.\n\n**There are some technical requirements as well. PRINCIPALLY, EACH VARIABLE MUST COME FROM A NORMAL (OR NEARLY NORMAL) DISTRIBUTION.**\n**Example:** Suppose we put people on 2 diets: the pizza diet and the beer diet.\n\nParticipants are randomly assigned to either 1-week of eating exclusively pizza or 1-week of exclusively drinking beer. Of course, this would be unethical, because pizza and beer should always be consumed together, but this is just an example.\n\nAt the end of the week, we measure weight gain by each participant. Which diet causes more weight gain?\n\nIn other words, the null hypothesis is: Ho: wt. gain pizza diet = wt. gain beer diet.\n\n(The null hypothesis is the opposite of what we hope to find. In this case, our research hypothesis is that there ARE differences between the 2 diets. Therefore, our null hypothesis is that there are NO differences between these 2 diets.)\n\n| X₁: Pizza | X₂: Beer | (X₁ - X)² | (X₂ - X)² |\n|-----------|----------|-----------|-----------|\n| 1 | 3 | 1 | 1 |\n| 2 | 4 | 0 | 0 |\n| 2 | 4 | 0 | 0 |\n| 2 | 4 | 0 | 0 |\n| 3 | 5 | 1 | 1 |\n| 2 | 4 | 0.4 | 0.4 |\n\n\\[ s_x^2 = \\frac{\\sum (X - \\bar{X})^2}{n} = 0.4 \\]\n\nThe formula for the independent samples t-test is:\n\n\\[ t = \\frac{\\bar{X}_1 - \\bar{X}_2}{\\sqrt{\\frac{S^2_{x_1}}{n_1 - 1} + \\frac{S^2_{x_2}}{n_2 - 1}}} \\]\n\\[ t = \\frac{2 - 4}{\\sqrt{\\frac{4}{4} + \\frac{4}{4}}} \\approx -4.47 \\]\n\n\\[ df = (n_1 - 1) + (n_2 - 1) = (5 - 1) + (5 - 1) = 8 \\]\n\nAfter calculating the “t” value, we need to know if it is large enough to reject the null hypothesis. The “t” is calculated under the assumption, called the null hypothesis, that there are no differences between the pizza and beer diet. If this were true, when we repeatedly sample 10 people from the population and put them in our 2 diets, most often we would calculate a “t” of “0.”\n\nThe calculated t-value is 4.47 (notice, I’ve eliminated the unnecessary “-“ sign), and the degrees of freedom are 8. In the research question we did not specify which diet should cause more weight gain, therefore this t-test is a so-called “2-tailed t.”\n\nIn the last step, we need to find the critical value for a 2-tailed “t” with 8 degrees of freedom. (This is available from tables that are in the back of any Statistics textbook).\n\nLook in the back for “Critical Values of the t-distribution,” or something similar. The value you should find is: C.V. \\( t(8), \\text{2-tailed} = 2.31 \\).\n\nThe calculated t-value of 4.47 is larger in magnitude than the C.V. of 2.31, therefore we can reject the null hypothesis. Even for a results section of journal article, this language is a bit too formal and general. It is more important to state the research result, namely:\n\nParticipants on the Beer diet (\\( M = 4.00 \\)) gained significantly more weight than those on the Pizza diet (\\( M = 2.00 \\)), \\( t(8) = 4.47, p < .05 \\) (two-tailed).\n\n**Chi-Square as a Statistical Test**\n\n- **Chi-square test**: an inferential statistics technique designed to test for significant relationships between two variables organized in a bivariate table.\n- **Chi-square requires no assumptions** about the shape of the population distribution from which a sample is drawn.\n- A statistical method used to determine goodness of fit\n - Goodness of fit refers to how close the observed data are to those predicted from a hypothesis\n- **Note**:\n - The chi square test does not prove that a hypothesis is correct\n - It evaluates to what extent the data and the hypothesis have a good fit.\n\n**Limitations of the Chi-Square Test**:\n\n- **The chi-square test** does not give us much information about the strength of the relationship or its substantive significance in the population.\n- **The chi-square test** is sensitive to sample size. The size of the calculated chi-square is directly proportional to the size of the sample, independent of the strength of the relationship between the variables.\n- **The chi-square test** is also sensitive to small expected frequencies in one or more of the cells in the table.\nStatistical Independence:\n\nIndependence (statistical): the absence of association between two cross-tabulated variables. The percentage distributions of the dependent variable within each category of the independent variable are identical.\n\nHypothesis Testing with Chi-Square:\n\nChi-square follows five steps:\n1. Making assumptions (random sampling)\n2. Stating the research and null hypotheses\n3. Selecting the sampling distribution and specifying the test statistic\n4. Computing the test statistic\n5. Making a decision and interpreting the results\n\nThe Assumptions:\n- The chi-square test requires no assumptions about the shape of the population distribution from which the sample was drawn.\n- However, like all inferential techniques it assumes random sampling.\n\nH₀: The two variables are related in the population. Gender and fear of walking alone at night are statistically dependent.\n\nH₁: The two variables are related in the population. Gender and fear of walking alone at night are statistically dependent.\n\n| Afraid | Men | Women | Total |\n|--------|---------|----------|-------|\n| No | 83.3% | 57.2% | 71.1% |\n| Yes | 16.7% | 42.8% | 28.9% |\n| Total | 100% | 100% | 100% |\n\nH₀: There is no association between the two variables. Gender and fear of walking alone at night are statistically independent.\n\nThe Concept of Expected Frequencies:\n\nExpected frequencies fₑ: the cell frequencies that would be expected in a bivariate table if the two tables were statistically independent.\n\nObserved frequencies fₒ: the cell frequencies actually observed in a bivariate table.\n\nCalculating Expected Frequencies:\n\n\\[ fₑ = \\frac{(\\text{column marginal})(\\text{row marginal})}{N} \\]\n\nTo obtain the expected frequencies for any cell in any cross-tabulation in which the two variables are assumed independent, multiply the row and column totals for that cell and divide the product by the total number of cases in the table.\n\nChi-Square (obtained):\n\nThe test statistic that summarizes the differences between the observed (fₒ) and the expected (fₑ) frequencies in a bivariate table.\nCalculating the Obtained Chi-Square:\n\n\\[ \\chi^2 = \\sum \\frac{(f_o - f_e)^2}{f_e} \\]\n\n\\( f_o \\) = expected frequencies\n\\( f_e \\) = observed frequencies\n\nThe Sampling Distribution of Chi-Square:\n\n- The distributions are **positively skewed**. The research hypothesis for the chi-square is **always** a one-tailed test.\n- Chi-square values are **always positive**. The minimum possible value is zero, with **no upper limit** to its maximum value.\n- As the number of degrees of freedom increases, the \\( c^2 \\) distribution becomes **more symmetrical**.\n\n\\[ df = (r - 1)(c - 1) \\]\n\nwhere \\( r \\) = the number of rows ; \\( c \\) = the number of columns\n\n\\( (3 - 1)(2 - 1) = 2 \\) degrees of freedom\nNON-PARAMETRIC STATISTICS\n\nThe term non-parametric was first used by Wolfowitz, 1942. To understand the idea of nonparametric statistics it is required to have a basic understanding of parametric statistics which we have already discussed. A parametric test requires a sample to be normally distributed. A nonparametric test does not rely on parametric assumptions like normality.\n\nNonparametric test create flexible demands of the data. To make standard parametric legitimate, some provisions need to fulfilled, especially for minor sample sizes. For example, the requirement of the one sample t-test is that the observation must be made from ordinarily distributed population. In case, the provision is defined, then the resultants may not be credible. However, in case of Wilcoxon Signed rank test to illustrate valid inference, normality is not required.\n\nWe can assume that the sampling distribution is normal even if we are not sure that the distribution of the variable in the population is normal, as long as our sample is large enough, (for example, 100 or more observations). However, if our selected sample is too large, then those tests can only be utilized if we are assured that the variable is disseminated normally.\n\nThe applications of tests that are based on the normality assumptions are restricted by the deficiency of accurate measurement. For example, a study measures Grade Point Average (GPA) in place of percentage Marks. This measurement scale does not measure the exact distance between the marks of two students. GPA allows us only to rank the students from “good” to “poor” students. This measurement is called the ordinal scale. Statistical techniques such as Analysis of Variance, t-test etc. assume that the data are measured either on interval or ratio scale. In such situations where data is measured on nominal or ordinal scale nonparametric tests are more useful.\n\nThus, nonparametric tests are used when either:\n\n- Sample is not normally distributed.\n- Sample size is small.\n- The variables are measured on nominal or ordinal scale.\n\nThere is at least one nonparametric equivalent for each parametric general type of test. Broadly, these tests fall into the following categories:\n\n- Test of differences between groups (independent samples)\n- Test of differences ( dependent samples)\n- Test of relationships between variables.\n\nThe concepts and procedure to undertake Run Test, Chi-Square Test, Wilcoxon Signed Rank Test, Mann-Whitney Test and Kruskal-Wallis Test are discussed here under:\n**Run Test:** Run Test is used to examine the randomness of data. Many statistical procedures require data to be randomly selected. Run Test can be explained with the example of tossing of a fair coin, where the probability of getting a head or tail is equal which is 0.5. Suppose we denote Head by “1” and tail by “0” and record the outcomes as shown below, we conduct Run Test to see whether the sample is randomly chosen or not. The null and alternative hypotheses are:\n\n- **H₀:** The sample is randomly selected.\n- **H₁:** The sample is not randomly selected.\n\nThe Run length is calculated by computing the number of 0’s or 1’s in sequence. For example the run length for 0 and 1 in the following sequence is 5 and 4 respectively and number of runs is 2.\n\n000001111\n\nWe shall conduct Run test on our data set to examine whether the students belonging to different faculties (graduation) are randomly selected or not.\n\n**Chi-Square Test:** Chi-Square Test is used to examine the association between two or more variables measured on categorical scales. Chi-Square is used most frequently to test the statistical significance of result reported in bivariate tables, and interpreting bivariate tables is integral to interpreting the results of a chi-square test.\n\nBivariate tabular (Cross tabulation) analysis is used when trying to summarize the intersections of independent and dependent variables and to examine the relationship (if any) between those variables. For example, to know if there is any association between the Gender and their Location, Chi-Square test can be applied. In this case our dependent variable is Location. We control the independent variable Gender and elicit as well as measure the dependent variable Location to test the hypothesis, whether there is some association between these two variables.\n\nThe Chi-Square test is a statistical technique to examine the association or statistical independence between the row and column variables in a two – way table. The null and alternative hypotheses for Chi-Square Test are:\n\n- **H₀:** There is no association between the row (Gender) and column (Location) variables.\n- **H₁:** There is association between the row (Gender) and column (Location) variables.\n\nMany researchers often get confused between statistical significance and strength of the relationship. People tend to think that more significant (the lower the P-value) relationship means stronger relationship. The significance level is influenced by the strength of the relationship and sample size. We require different measure to capture the strength of the relationship, or the effect size.\n\n**Mann-Whitney U Test:** Generally, the t-test for independent samples is used, if two samples are compared over their mean value for some variable interest. Nonparametric alternatives for the test are the Wald-Wolfowitz Run test, the Mann Whitney U test, and the Kolmogorov-Smirnov two sample test.\n\nMann Whitney U test compares the sums of ranks of two independent groups.\n**Wilcoxon Signed Rank Test:** Wilcoxon Signed Rank Test (also known as Wilcoxon Matched Pair Test) is the non-parametric version of dependent sample t-test or paired sample t-test. Sign test is the other nonparametric alternative to the paired sample t-test. If the variables of interest are dichotomous in nature (Male and Female or Yes and No) then McNemar’s Chi-Square test is used.\n\nWilcoxon Signed Rank Test is also a nonparametric version for one sample t-test. Wilcoxon Signed Rank Test compares the medians of the groups under two situations (paired samples) or it compares the median of the group with hypothesized median (one sample).\n\n**Kruskal–Wallis Test:** Kruskal–Wallis Test is used with multiple groups. It is the non-parametric version of one-Way ANOVA. Median test is another nonparametric alternative to one-Way ANOVA. Kruskal–Wallis Test compares medians of more than two independent groups.\n\n**Below are the complete selection of Appropriate Statistical tests for Social Scientists/ Participants of the Workshop**\nSELECTION OF APPROPRIATE STATISTICAL TESTS\n\nDescription of one Group\n\nComparison of one Group to a hypothetical value\n\nComparison of two groups\n\nComparison of three or more groups\n\nMeasure association between two groups\n\nPrediction\n\nOne sample t-test\n\nWilcoxon test\n\nChi-square or Binomial test\n\nUnpaired groups\n\npaired groups\n\nUnpaired groups\n\nMatche\n\nFrom another measured variables\n\nFrom several measured or binomial variables\n\nR, I = Ratio & Interval Data, O = Ordinal Data, N = Nominal\n\nND = Normal distribution, NND = Non Normal Distribution\n\nMean, SD\n\nMedian, Interquartile Range\n\nProportion\n\nPaired t-test\n\nWilcoxon test\n\nMcNamara’s test\n\nChi-Square test\n\nOne way ANOVA\n\nKruskal Wallis test\n\nSpearman Correlation\n\nMultiple Linear Regressions or Multiple nonlinear\n\nSimple Logistic Regression\n\nNon Parametric Regression\n\nSimple Linear Regression\n\nFriedman test\n\nRepeated measures ANOVA\n\nCochrane Q\n\nPearson Correlation\n\nContingency coefficients\nThe flow chart of commonly used statistical tests:", "id": "./materials/150.pdf" }, { "contents": "**USE OF STATISTICAL TABLES**\n\nLucy Radford, Jenny V Freeman and Stephen J Walters introduce three important statistical distributions: the standard Normal, t and Chi-squared distributions.\n\n**STANDARD NORMAL DISTRIBUTION**\n\nThe Normal distribution is widely used in statistics and has been discussed in detail previously. As the mean of a Normally distributed variable can take any value \\(-\\infty \\to \\infty\\) and the standard deviation any positive value \\(0 \\to \\infty\\), there are an infinite number of possible Normal distributions. It is therefore not feasible to print tables for each Normal distribution; however it is possible to convert any Normal distribution to the standard Normal distribution, for which tables are available. The standard Normal distribution has a mean of 0 and standard deviation of 1.\n\nAny value \\(X\\) from a Normal distribution with mean \\(\\mu\\) and standard deviation \\(\\sigma\\) can be transformed to the standard Normal distribution using the following formula:\n\n\\[\nz = \\frac{X - \\mu}{\\sigma}\n\\]\n\nThis transformed \\(X\\)-value, often called \\(z\\) or \\(z\\)-score, is also known as the standard Normal deviate, or Normal score. If an average, rather than a single value, is used the standard deviation should be divided by the square root of the sample size, \\(n\\), as shown in equation (2).\n\n\\[\nz = \\frac{\\bar{X} - \\mu}{\\sigma/\\sqrt{n}}\n\\]\n\n**TABLE 1.** Extract from two-tailed standard Normal table. Values tabulated are \\(P\\)-values corresponding to particular cut-offs and are for \\(z\\) values calculated to two decimal places.\n\n| \\(z\\) | 0.00 | 0.01 | 0.02 | 0.03 | 0.04 | 0.05 | 0.06 | 0.07 | 0.08 | 0.09 |\n|------|------|------|------|------|------|------|------|------|------|------|\n| 0.00 | 1.000 | 0.992 | 0.984 | 0.976 | 0.968 | 0.960 | 0.952 | 0.944 | 0.936 | 0.928 |\n| 0.10 | 0.920 | 0.912 | 0.904 | 0.896 | 0.888 | 0.880 | 0.872 | 0.865 | 0.857 | 0.849 |\n| 0.20 | 0.841 | 0.833 | 0.825 | 0.818 | 0.810 | 0.802 | 0.794 | 0.787 | 0.779 | 0.771 |\n| 0.30 | 0.764 | 0.756 | 0.749 | 0.741 | 0.733 | 0.726 | 0.718 | 0.711 | 0.703 | 0.696 |\n| 0.40 | 0.689 | 0.681 | 0.674 | 0.667 | 0.659 | 0.652 | 0.645 | 0.638 | 0.631 | 0.624 |\n| 0.50 | 0.617 | 0.610 | 0.603 | 0.596 | 0.589 | 0.582 | 0.575 | 0.568 | 0.561 | 0.555 |\n| 0.60 | 0.546 | 0.541 | 0.535 | 0.528 | 0.522 | 0.515 | 0.509 | 0.503 | 0.496 | 0.490 |\n| 0.70 | 0.483 | 0.477 | 0.471 | 0.465 | 0.459 | 0.453 | 0.447 | 0.441 | 0.435 | 0.429 |\n| 0.80 | 0.423 | 0.417 | 0.412 | 0.406 | 0.400 | 0.395 | 0.389 | 0.384 | 0.379 | 0.373 |\n| 0.90 | 0.368 | 0.362 | 0.357 | 0.352 | 0.347 | 0.342 | 0.337 | 0.332 | 0.327 | 0.322 |\n| 1.00 | 0.317 | 0.312 | 0.307 | 0.303 | 0.298 | 0.293 | 0.289 | 0.284 | 0.280 | 0.275 |\n| 1.10 | 0.271 | 0.267 | 0.262 | 0.258 | 0.254 | 0.250 | 0.246 | 0.242 | 0.238 | 0.234 |\n| 1.20 | 0.230 | 0.226 | 0.222 | 0.218 | 0.215 | 0.211 | 0.207 | 0.204 | 0.200 | 0.197 |\n| 1.30 | 0.193 | 0.190 | 0.186 | 0.183 | 0.180 | 0.177 | 0.173 | 0.170 | 0.167 | 0.164 |\n| 1.40 | 0.161 | 0.158 | 0.155 | 0.152 | 0.149 | 0.147 | 0.144 | 0.141 | 0.138 | 0.136 |\n| 1.50 | 0.136 | 0.131 | 0.128 | 0.126 | 0.123 | 0.121 | 0.118 | 0.116 | 0.114 | 0.111 |\n| 1.60 | 0.109 | 0.107 | 0.105 | 0.103 | 0.101 | 0.099 | 0.097 | 0.095 | 0.093 | 0.091 |\n| 1.70 | 0.089 | 0.087 | 0.085 | 0.083 | 0.081 | 0.079 | 0.078 | 0.076 | 0.075 | 0.073 |\n| 1.80 | 0.079 | 0.079 | 0.078 | 0.076 | 0.075 | 0.073 | 0.072 | 0.071 | 0.070 | 0.069 |\n| 1.90 | 0.057 | 0.056 | 0.055 | 0.054 | 0.053 | 0.052 | 0.051 | 0.050 | 0.049 | 0.048 |\n| 2.00 | 0.045 | 0.044 | 0.043 | 0.042 | 0.041 | 0.040 | 0.039 | 0.038 | 0.037 | 0.036 |\n| 2.10 | 0.035 | 0.034 | 0.034 | 0.033 | 0.032 | 0.031 | 0.030 | 0.029 | 0.028 | 0.027 |\n| 2.20 | 0.028 | 0.027 | 0.026 | 0.025 | 0.025 | 0.024 | 0.023 | 0.022 | 0.022 | 0.021 |\n| 2.30 | 0.021 | 0.021 | 0.020 | 0.019 | 0.019 | 0.018 | 0.018 | 0.017 | 0.017 | 0.016 |\n| 2.40 | 0.016 | 0.016 | 0.015 | 0.015 | 0.015 | 0.015 | 0.015 | 0.015 | 0.015 | 0.015 |\n| 2.50 | 0.012 | 0.012 | 0.011 | 0.011 | 0.011 | 0.011 | 0.011 | 0.011 | 0.011 | 0.011 |\n| 2.60 | 0.009 | 0.009 | 0.008 | 0.008 | 0.008 | 0.008 | 0.008 | 0.008 | 0.008 | 0.008 |\n| 2.70 | 0.006 | 0.006 | 0.006 | 0.006 | 0.006 | 0.006 | 0.006 | 0.006 | 0.006 | 0.006 |\n| 2.80 | 0.005 | 0.005 | 0.005 | 0.005 | 0.005 | 0.005 | 0.005 | 0.005 | 0.005 | 0.005 |\n| 2.90 | 0.003 | 0.003 | 0.003 | 0.003 | 0.003 | 0.003 | 0.003 | 0.003 | 0.003 | 0.003 |\n| 3.00 | 0.002 | 0.002 | 0.002 | 0.002 | 0.002 | 0.002 | 0.002 | 0.002 | 0.002 | 0.002 |\nFor example, the exam results for the first year of a medical degree are known to be approximately Normally distributed with mean 72 and standard deviation 8. To find the probability that a student will score 89 or more we first need to convert this value to a standard Normal deviate. In this instance, as we have a single value we use equation (1):\n\n$$z = \\frac{89 - 72}{8} = 2.13$$\n\nIf we wished to find the probability that an average of 10 scores is 75 or more we would use equation (2) to convert to the standard Normal distribution:\n\n$$z = \\frac{75 - 72}{8/\\sqrt{10}} = 1.15$$\n\nWe then use the standard Normal table to find the probabilities of observing these z-values, or values more extreme given that the population mean and standard deviation are 72 and 8 respectively.\n\nStandard Normal tables can be either one-tailed or two-tailed. In the majority of hypothesis tests the direction of the difference is not specified, leading to a two-sided (or two-tailed) test. The standard Normal table shown in table 1 is two-sided. In this two-sided table the value tabulated is the probability, $\\alpha$, that a random variable, Normally distributed with mean zero and standard deviation one, will be either greater than $z$ or less than $-z$ (as shown in the diagram at the top of the table). The total area under the curve represents the total probability space for the standard Normal distribution and sums to 1, and the shaded areas at either end are equal to $\\alpha/2$. A one-tailed probability can be calculated by halving the tabulated probabilities in table 1. As the Normal distribution is symmetrical it is not necessary for tables to include the probabilities for both positive and negative z-values.\n\n**WORKED EXAMPLES**\n\nFrom our first example above we want to know what the probability is that a student chosen at random will have a test score of 89, given a population mean of 72 and standard deviation of 8. The z-score calculated above is 2.13. In order to obtain the $P$-value that corresponds to this z-score we first look at the row in the table that corresponds to a z-score of 2.1. We then need to look down the column that is headed 0.03. The corresponding $P$-value is 0.0198. However, this is a two-sided probability and corresponds to the probability that a z-score is either $-2.13$ or $2.13$ (see figure 1). To get the probability that a student chosen at random will have a test score of at least 89 we need to halve the tabulated $P$-value. This gives a $P$-value of 0.0099.\n\nIn a previous tutorial we used the Normal approximation to the binomial to examine whether there were significant differences in the proportion of patients with healed leg ulcers at 12 weeks, between standard treatment and treatment in a specialised leg ulcer clinic. The null hypothesis was that there was no difference in healing rates between the two groups. From this test we obtained a z-score of 0.673. Looking this up in table 1 we can see that it corresponds to a two-sided $P$-value of 0.503. Thus we cannot reject the null, and we conclude that there is no reliable evidence of a difference in ulcer healing rates at 12 weeks between the two groups.\n\n**STUDENT’S $t$-DISTRIBUTION**\n\nThe $t$-test is used for continuous data to compare differences in means between two groups (either paired or unpaired). It is based on Student’s $t$-distribution (sometimes referred to as just the $t$-distribution). This distribution is particularly important when we wish to estimate the mean (or mean difference between groups) of a Normally distributed population but have only a small sample. This is because the $t$-test, based on the $t$-distribution, offers more precise estimates for small sample sizes than the tests associated with the Normal distribution. It is closely related to the Normal distribution and as the sample size tends towards infinity the probabilities of the $t$-distribution approach those of the standard Normal distribution.\n\nThe main difference between the $t$-distribution and the Normal distribution is that the $t$ depends only on one parameter, $v$, the degrees of freedom (d.f.), not on the mean or standard deviation. The degrees of freedom are based on the sample size, $n$, and are equal to $n-1$. If the $t$ statistic calculated in the test is greater than the critical value for the chosen level of statistical significance (usually $P = 0.05$) the null hypothesis for the particular test being carried out is rejected in favour of the alternative. The critical value that is compared to the $t$ statistic is taken from the table of probabilities for the $t$-distribution, an extract of which is shown in table 2.\n\nUnlike the table for the Normal distribution described above, the\n\n---\n\n**TABLE 2**\n\n| d.f. | 0.05 | 0.01 | 0.02 | 0.03 | 0.05 | 0.10 |\n|------|------|------|------|------|------|------|\n| 1 | 1.000| 6.314| 12.706| 31.821| 63.657| 636.619|\n| 2 | 0.816| 2.920| 4.303| 6.965| 9.295| 31.998|\n| 3 | 0.765| 2.353| 3.162| 4.541| 5.841| 12.941|\n| 4 | 0.741| 2.132| 2.776| 3.747| 4.604| 8.610|\n| 5 | 0.727| 2.015| 2.571| 3.365| 4.032| 6.859|\n| 6 | 0.718| 1.943| 2.447| 3.143| 3.707| 5.959|\n| 7 | 0.711| 1.895| 2.365| 2.998| 3.499| 5.040|\n| 8 | 0.706| 1.860| 2.306| 2.869| 3.355| 4.041|\n| 9 | 0.700| 1.833| 2.262| 2.821| 3.250| 3.781|\n| 10 | 0.697| 1.796| 2.201| 2.764| 3.169| 3.581|\n| 11 | 0.693| 1.761| 2.145| 2.704| 3.085| 3.440|\n| 12 | 0.689| 1.725| 2.092| 2.648| 2.998| 3.311|\n| 13 | 0.686| 1.700| 2.048| 2.602| 2.920| 3.182|\n| 14 | 0.682| 1.674| 1.995| 2.557| 2.850| 3.052|\n| 15 | 0.679| 1.650| 1.950| 2.516| 2.787| 2.920|\n| 16 | 0.676| 1.626| 1.908| 2.476| 2.724| 2.794|\n| 17 | 0.673| 1.602| 1.865| 2.437| 2.660| 2.660|\n| 18 | 0.670| 1.579| 1.821| 2.400| 2.597| 2.528|\n| 19 | 0.667| 1.556| 1.781| 2.364| 2.534| 2.396|\n| 20 | 0.664| 1.533| 1.740| 2.326| 2.472| 2.261|\n| 21 | 0.661| 1.510| 1.700| 2.288| 2.408| 2.129|\n| 22 | 0.658| 1.487| 1.660| 2.250| 2.343| 2.000|\n| 23 | 0.655| 1.464| 1.625| 2.211| 2.276| 1.871|\n| 24 | 0.652| 1.442| 1.591| 2.173| 2.208| 1.746|\n| 25 | 0.649| 1.420| 1.557| 2.136| 2.140| 1.620|\n| 30 | 0.639| 1.356| 1.475| 2.048| 1.960| 1.485|\n| 40 | 0.629| 1.296| 1.403| 1.960| 1.833| 1.363|\n| 60 | 0.620| 1.250| 1.356| 1.890| 1.746| 1.250|\n| 120 | 0.610| 1.196| 1.303| 1.833| 1.667| 1.140|\n\n**FIGURE 1.** Normal curve showing the Z-values and corresponding P-values for the data in example 1.\ntabulated values relate to particular levels of statistical significance, rather than the actual P-values. Each of the columns represents the cut-off points for declaring statistical significance for a given level of (two-sided) significance. For example, the column headed 0.05 in table 2 gives the values which a calculated t-statistic must be above in order for a result to be statistically significant at the two-sided 5 per cent level. Each row represents the cut-offs for different degrees of freedom. Any test which results in a t statistic less than the tabulated value will not be statistically significant at that level and the P-value will be greater than the value indicated in the column heading. As the t-distribution is symmetrical about the mean, it is not necessary for tables to include the probabilities for both positive and negative t statistics.\n\nConsider, for example, a t-test from which a t-value of 2.66 on 30 d.f. was obtained. Looking at the row corresponding to 30 d.f. in table 2 this value falls between the tabulated values for 0.02 (=2.457) and 0.01 (=2.75). Thus, the P-value that corresponds with this particular t-value will be less than 0.02, but greater than 0.01. In fact the actual (two-tailed) P-value is 0.012.\n\n**CHI-SQUARED DISTRIBUTION**\n\nThe final statistical table being considered in this tutorial is that of the Chi-squared distribution. There are a wide range of statistical tests that lead to use of the Chi-squared distribution, the most common of which is the Chi-squared test described in a previous tutorial. Like the t-distribution the Chi-squared distribution has only one parameter, the degrees of freedom, k. A section of the Chi-squared distribution is shown in table 3. Like the table for the t distribution described above the tabulated values are the Chi-squared values that relate to particular levels of statistical significance, rather than actual P-values. Each of the columns represents the cut-off points for declaring statistical significance for a given level of significance. For example, the column headed 0.05 in table 3 gives the values above which a calculated Chi-squared statistic must be in order for a result to be statistically significant at the two-sided 5 per cent level, for degrees of freedom ranging from 1 to 30. Any test which results in a Chi-squared statistic less than the tabulated value will not be statistically significant at that level and the P-value will be greater than the value at the top of the column. Consider, for example, a Chi-squared value of 4.2 on 1 d.f. Looking at the row corresponding to 1 d.f. in table 3 this value falls between the tabulated values for 0.05 (=3.841) and 0.02 (=5.412). Thus, the P-value that corresponds with this particular Chi-squared statistic will be less than 0.05, but greater than 0.02.\n\nAs a second example consider the results of a Chi-squared test that was used to assess whether leg ulcer healing rates differed between two different treatment groups (group 1: standard care; treatment 2: specialised leg ulcer clinic). From this significance test a Chi-squared value of 0.243 with 1 d.f. was obtained. Looking at the 1 d.f. row in table 3 it can be seen that all the values are greater than this value, including the value that corresponds with a P-value of 0.5, 0.455. Thus we can conclude that the P-value corresponding to a Chi-squared value of 0.243 is greater than 0.5; in fact the exact value is 0.62.\n\n**SUMMARY**\n\nIn this tutorial we have shown how to read statistical tables of P-values for the standard Normal, t and Chi-squared distributions, and given examples to show how the values from these tables are used to make decisions in a variety of basic statistical tests.\n\n**REFERENCES**\n\n1. Freeman JV, Julious SA. Hypothesis testing and estimation. *Scope* 2006;15(1).\n2. Freeman JV, Julious SA. Basic tests for continuous Normally distributed data. *Scope* 2006;15(3).\n3. Freeman JV, Campbell MJ. Basic test for continuous data: Mann-Whitney U and Wilcoxon signed rank sum tests. *Scope* 2006;15(4).\n4. Freeman JV, Julious SA. The analysis of categorical data. *Scope* 2007; 16(1):18–21.\n5. Freeman JV, Julious SA. The Normal distribution. *Scope* 2005; 14(4).\n6. Swinscow TDV, Campbell MJ. *Statistics at square one*. 10th ed. London: BMJ Books, 2002.\n\n**TABLE 3**\n\n| d.f. | 0.05 | 0.1 | 0.05 | 0.02 | 0.01 | 0.001 |\n|------|------|-----|------|------|------|-------|\n| 1 | 0.455| 2.706| 3.841| 5.142| 6.635| 10.827|\n| 2 | 1.386| 4.605| 5.991| 7.824| 9.210| 13.815|\n| 3 | 2.366| 6.251| 7.815| 9.837| 11.345| 16.268|\n| 4 | 3.357| 7.779| 9.488| 11.668| 13.277| 18.465|\n| 5 | 4.351| 9.236| 11.070| 13.388| 15.086| 20.517|\n| 6 | 5.346| 10.645| 12.592| 15.033| 16.812| 22.457|\n| 7 | 6.346| 12.017| 14.067| 16.622| 18.475| 24.322|\n| 8 | 7.344| 13.362| 15.507| 18.168| 20.090| 26.125|\n| 9 | 8.343| 14.684| 16.919| 19.679| 21.666| 27.877|\n| 10 | 9.342| 15.987| 18.307| 21.161| 23.209| 29.588|\n| 11 | 10.341| 17.275| 19.675| 22.618| 24.725| 31.264|\n| 12 | 11.340| 18.549| 21.026| 24.054| 26.217| 32.909|\n| 13 | 12.340| 19.812| 22.362| 25.472| 27.688| 34.528|\n| 14 | 13.339| 21.064| 23.685| 26.873| 29.141| 36.123|\n| 15 | 14.339| 22.307| 24.996| 28.259| 30.578| 37.697|\n| 16 | 15.338| 23.542| 26.296| 29.633| 32.000| 39.252|\n| 17 | 16.338| 24.769| 27.587| 30.995| 33.409| 40.790|\n| 18 | 17.338| 25.989| 28.869| 32.346| 34.805| 42.312|\n| 19 | 18.338| 27.204| 30.144| 33.667| 36.191| 43.820|\n| 20 | 19.337| 28.412| 31.410| 35.020| 37.566| 45.315|\n| 21 | 20.337| 29.615| 32.671| 36.343| 38.932| 46.797|\n| 22 | 21.337| 30.813| 33.924| 37.659| 40.289| 48.268|\n| 23 | 22.337| 32.007| 35.172| 38.968| 41.638| 49.728|\n| 24 | 23.337| 33.196| 36.415| 40.270| 42.980| 51.745|\n| 25 | 24.337| 34.382| 37.652| 41.566| 44.314| 52.820|\n| 26 | 25.336| 35.563| 38.885| 42.479| 45.642| 54.707|\n| 27 | 26.336| 36.741| 40.113| 44.140| 45.963| 55.476|\n| 28 | 27.336| 37.916| 41.337| 45.419| 48.278| 56.893|\n| 29 | 28.336| 39.087| 42.557| 46.693| 49.588| 58.302|\n| 30 | 29.336| 40.256| 43.773| 47.962| 50.892| 59.703|\n\n*†A simple trick for seeing whether a particular table is one-tailed or two-tailed is to look at the value that corresponds to a cut-off of 1%. If the tabulated P-value is 0.05 then the table is for two-tailed P-values.*", "id": "./materials/151.pdf" }, { "contents": "Today:\n\n- Inter-Quartile Range,\n- Outliers,\n- Boxplots.\n\nReading for today: Start Chapter 4.\nQuartiles and the Five Number Summary\n\n- The five numbers are the Minimum (Q0), Lower Quartile (Q1), Median (Q2), Upper Quartile (Q3), and Maximum (Q4).\n\n- Q1 means bigger than 1 Quarter of the data.\n- Q3 means bigger than 3 Quarters of the data.\n\nFor the values \\{0, 1, 2, 4, 5, 5, 7, 10, 10, 12, 13, 17, 39\\}, the five number summary is: 0 \\rightarrow 3 \\rightarrow 7 \\rightarrow 12.5 \\rightarrow 39.\nInter-Quartile Range\n\n- Even in the unimodal cases, neither the mean nor the median describes the data adequately.\n\n- The mean number of legs per Swede is 1.999, clearly there’s something more we should know.\n\n- The median of \\{30,31,32\\} is 31.\n\n- The median of \\{-10000, 31, 10000\\} is also 31.\nInter-Quartile Range\n\n- We also need measures of spread, like the Inter-Quartile Range. (Literally “range the between the quartiles”, called the IQR for short).\n- The Inter-Quartile range is calculated:\n\n\\[ \\text{IQR} = Q3 - Q1 \\]\n\n- The size of the IQR indicates how spread out the middle half of the data is.\nOutliers (1.5 x IQR Rule)\n\n- Now that we have a measure of spread, we can use it to identify values that are much farther from the center than usual.\n\n- How? Spread measures like the IQR tell us how far a typical value could be from the average, so anything much more than the typical distance can be identified.\n- We call these data points outliers.\n\nThey (figuratively) lay outside the rest of the data.\n\n- Because an outlier stands out from the rest of the data, it...\n - might not belong there, or\n - is worthy of extra attention.\n- One way to define an outlier is\n o anything below $Q1 - 1.5 \\times IQR$ or...\n o above $Q3 + 1.5 \\times IQR$.\n\nThis is called the 1.5 x IQR rule. (Important).\n- Example: \\{0, 1, 2, 4, 5, 5, 7, 10, 10, 12, 13, 17, 39\\}\n Q1 = 3, Q3 = 12.5\n IQR = 12.5 - 3 = 9.5.\n\n Q1 – 1.5xIQR = 3 – 1.5(9.5)\n = 3 -14.25 = -11.25\n\n Anything less than -11.25 is an outlier.\n\n In this case there are no outliers on the low end.\n- Example: \\{0, 1, 2, 4, 5, 5, 7, 10, 10, 12, 13, 17, 39\\}\n Q1 = 3, Q3 = 12.5\n IQR = 9.5\n\n Q3 + 1.5\\times IQR = 12.5 + 1.5\\times 9.5\n\n = 12.5 + 14.25 = 26.75\n\n Anything more than 26.75 is an outlier.\n\n 39 is the only outlier.\nMore on IQR and Outliers:\n\n- There are other ways to define outliers, but 1.5xIQR is one of the most straightforward.\n\n- If our range has a natural restriction, (like it can’t possibly be negative), it’s okay for an outlier limit to be beyond that restriction.\n\n- If a value is more than Q3 + 3*IQR or less than Q1 – 3*IQR it is sometimes called an extreme outlier.\n- The standard graph for showing the median, quartiles, and outliers of a data set is the boxplot, for \\{0, 1, 2, 4, 5, 5, 7, 10, 10, 12, 13, 17, 39\\} it looks like this:\n- The **five-number summary** is in the boxplot:\n- The box from 3 to 12.5 is the region between **Q1** and **Q3**.\n- The line going through the middle of the box at 7 is the **median**.\n- The lines going out the ends of the box are called the **whiskers**. They show the range of values that are **not outliers**.\n\n- The lower whisker goes to the lowest value, 1. The upper whisker goes to 17 because it’s the biggest value before the upper limit of 26.75 is hit.\n- The individual dot at 39 shows an outlier.\n\n- Outliers in SPSS are labelled with their row number so you can find them in data view.\n\n- In SPSS extreme outliers are shown as stars.\n\n- The farthest outliers on either side are the minimum and maximum.\n\n- If there are no outliers on a side, the end of the whisker is that minimum or maximum.\nBoxplots and Skew\n\n- Skewed distributions have more extreme values on one side, so a boxplot of a skewed distribution will have one whisker longer than the other.\n\n- There will also be more outliers on one side of the boxplot than the other.\nSide-by-side Boxplots\n\n- Boxplots can also be used to compare the distributions of two samples.\n- Example: Heights of adult men and women.\n- There is some overlap\n- In general men are taller.\n- The variance is about the same.\n- Both distributions appear to be symmetric.\nWhat exactly IS an outlier?\n\n- It’s a value far from anything else that warrants special consideration aside from the rest of the data.\n\n- Often it’s a mistake in data entry. If were recording a grade of 73%, mistyped, and recorded 3% or 730%, both of these values would be far from the rest of the data and would indicate that the data is not being represented properly.\n- If the times to finish a final exam had Q1 at 120 minutes and Q3 at 150 minutes, but someone finished in 62 minutes, that person could be a student with a stronger than recommended background for that course or someone who gave up during the exam.\n\n- In both cases, their exam wouldn’t a good representation of the exams as whole.\n\n- Sometimes outliers can tell your assumptions and expectations are wrong, like in minor hockey.\nMinor Hockey and Outliers (See Malcolm Gladwell’s Outliers) (this is for interest)\n\n- If I took samples of 25 random people over and over and got their ‘average’ birthdays, I would get a bell curve around late June or early July. (Using days since January 1 as the value of people’s birthdays) This would fit my assumption that the way I took my sample had nothing to do with when people were born, so the average birthday should be right near the middle of the year.\n- If my sample is Team Canada in the World Junior Hockey Championship, however, the average is much closer to the beginning of the year.\n- The average birthday of champion hockey players is an outlier compared to the average of other groups of 25. The data was entered correctly, so the population must be different somehow.\nIt’s possible this happened by chance, but unlikely in the context of the other samples. My assumption that birthdays and being a hockey champion are unrelated may be wrong.\nNext lecture:\n\nStandard Deviation and the Normal Curve\n(read more of chapter 4)", "id": "./materials/153.pdf" }, { "contents": "Products of vectors and orthogonal projection of one vector over another\n\nLet’s talk about two types of products of two vectors, the scalar product and the vector product.\n\nScalar product of two vectors and orthogonal projection of one vector over another\n\nDefinition: The scalar product (or dot product) $u \\cdot v$ of two vectors $u$ and $v$ is a number defined by\n\n$$||u|| ||v|| \\cos(\\theta),$$\n\nwith $\\theta = \\hat{uv} \\in [0, \\pi]$.\n\n- If $u$ and $v$ are parallel vectors, then $u \\cdot v = ||u|| ||v||$;\n- If $u$ and $v$ are antiparallel vectors, then $u \\cdot v = -||u|| ||v||$;\n- If $u$ and $v$ are two orthogonal vectors, then $u \\cdot v = 0$.\n\nNotice that $\\hat{vv} = 0$ and this implies that the dot product of a vector $a$ with itself is $v \\cdot v = ||v|| ||v||$, which gives\n\n$$||v|| = \\sqrt{v \\cdot v}.$$\n\nIn $\\mathbb{R}^n$ we have the alternative definition of scalar product:\n\nDefinition: The dot product of two vectors $v = (v_1, v_2, \\ldots, v_n), u = (u_1, u_2, \\ldots, u_n) \\in \\mathbb{R}^n$ is\n\n$$v \\cdot u = v_1u_1 + v_2u_2 + v_3u_3 + \\cdots + v_nu_n.$$\n\nExample: On the Cartesian plane, consider the vectors $i = (1, 0), j = (0, 1)$ and $v(-1, 1)$.\n\nOn the one hand, we have $i \\cdot j = (1, 0) \\cdot (0, 1) = 0$ and $j \\cdot v = (0, 1) \\cdot (-1, 1) = 1 + 0 = 1$.\n\nOn the other hand, we also have\n\n$$i \\cdot j = ||(1, 0)|| ||(0, 1)|| \\cos\\left(\\frac{\\pi}{2}\\right) = 0$$\n\nAlso\n\n$$j \\cdot v = ||(0, 1)|| ||(-1, 1)|| \\cos\\left(\\frac{\\pi}{2}\\right) = \\sqrt{2} \\times \\left(\\frac{\\sqrt{2}}{2}\\right) = 1.$$\n\n$$i \\cdot v = ||(1, 0)|| ||(-1, 1)|| \\cos\\left(\\frac{3\\pi}{2}\\right) = \\sqrt{2} \\times \\left(-\\frac{\\sqrt{2}}{2}\\right) = -1.$$\nThe dot product fulfills the following properties if \\( u, v, \\) and \\( w \\) are vectors and \\( k \\) is a real scalar:\n\n1. \\( v \\cdot u = u \\cdot v; \\)\n2. \\( v \\cdot (u + w) = (v \\cdot u) + (v \\cdot w); \\)\n3. \\( v \\cdot (ku + w) = k(v \\cdot u) + (v \\cdot w); \\)\n4. \\( k_1v \\cdot (k_2u) = k_1k_2(v \\cdot u). \\)\n\nAn inner product is a generalization of the dot product, is any operator who checks the properties above.\n\nOne important use of dot products is in projections.\n\nThe orthogonal projection of \\( u \\) onto \\( v \\) is the length of the segment \\([AD]\\) shown in the figure beside, \\( ||AD||.\\)\n\nThe vector projection of \\( u \\) onto \\( v \\) is the vector \\( \\vec{AD} \\)\n\nNote that \\( |\\text{proj}_v u| = ||v|| \\cos(\\theta) | \\) and therefore:\n\n\\[\n|\\text{proj}_v u| = ||AD|| = \\frac{|u \\cdot v|}{||v||} \\quad \\text{and} \\quad \\text{proj}_v u = \\vec{AD} = \\frac{|u \\cdot v|}{||v||^2} v.\n\\]\n\n**Vector Product (Cross Product)**\n\nThe vector product of two vectors \\( u \\) and \\( v \\) is a vector \\( u \\times v \\) that is at right angles to both and is defined by\n\n\\[\nu \\times v = ||u|| ||v|| \\sin(\\hat{u} \\hat{v}) \\hat{n}, \\quad \\text{with} \\quad ||n|| = 1 \\quad \\text{and} \\quad u, v \\perp n.\n\\]\n\nSpecifically,\n\n1. \\( u \\times v \\) is perpendicular to the vectors \\( u \\) and \\( v; \\)\n2. \\( ||u \\times v|| = ||u|| \\cdot ||v|| \\sin((u, v)); \\)\n3. \\( u \\times v \\) has sense determined by the right hand (follow with the fingers of the right hand, the rotation movement of the vector \\( u \\) to approach \\( v \\) and consider the direction of the thumb).\n\nNotice that:\n\n- \\( u \\times v \\) is orthogonal to the plane containing the vectors;\n- \\( u \\times v = 0 \\) when vectors \\( u \\) and \\( v \\) point in the same, or opposite, direction.\nIn the 3-dimensional Cartesian system, the vector product of vectors \\( u = (u_1, u_2, u_3) \\) and \\( v = (v_1, v_2, v_3) \\) is defined as\n\\[\n\\mathbf{u} \\times \\mathbf{v} = (u_2v_3 - v_2u_3, v_1u_3 - u_1v_3, u_1v_2 - v_1u_2).\n\\]\n\nIt is a vector perpendicular to the vectors \\( u \\) and \\( v \\) and can more easily be represented matrix-wise as:\n\\[\n\\mathbf{u} \\times \\mathbf{v} = \\begin{vmatrix} i & j & k \\\\ u_1 & u_2 & u_3 \\\\ v_1 & v_2 & v_3 \\end{vmatrix} = (u_2v_3 - v_2u_3)i - (u_1v_3 - v_1u_3)j + (u_1v_2 - v_1u_2)k.\n\\]\n\n**Example:** \\((1, 2, -1) \\times (2, 0, 1) = \\begin{vmatrix} i & j & k \\\\ 1 & 2 & -1 \\\\ 2 & 0 & 1 \\end{vmatrix} = 2i - 3j - 4k = (2, -3, -4)\\)\n\n**Properties:** Be the vectors \\( u, v, w \\in \\mathbb{R}^3 \\). We have\n1. \\( u \\times v \\times w = u \\times (v \\times w) \\) (associative);\n2. \\( u \\times v = -v \\times u \\) (anti-commutative);\n3. \\( u \\times v = 0 \\iff u = 0 \\lor v = 0 \\lor \\hat{(u, v)} = 0^\\circ \\lor \\hat{(u, v)} = 180^\\circ \\).\n\n**Example:** \\((1, -2, 3) \\times (-2, 4, -6) = \\begin{vmatrix} i & j & k \\\\ 1 & -2 & 3 \\\\ -2 & 4 & -6 \\end{vmatrix} = (0, 0, 0),\\)\n\nbecause the vectors \\((1, -2, 3)\\) and \\((-2, 4, -6)\\) are collinear.\n\nThe norm of the vector product \\( ||u \\times v|| = ||u|| \\cdot ||v|| \\cdot \\sin(\\angle(u, v)) \\) is the area of the parallelogram determined by \\( u \\) and \\( v \\).\n\nIn effect, according to the figure above, the area of the parallelogram is given by \\( A = ||v|| \\cdot h \\). Besides that, \\( ||u|| \\sin(\\angle(u, v)) = h \\).", "id": "./materials/154.pdf" }, { "contents": "Angle between planes or lines\n\nAngle between two lines\n\nThe angle between two lines is defined as the smallest angle between their directions.\n\nIn the figure to the side we can see that:\n\n- The angle of the straight lines \\( s \\) and \\( t \\) belonging to the \\( ABC \\) plane measures \\( 30^\\circ \\).\n- The angle of the reverse lines \\( r \\) and \\( s \\) is of \\( 90^\\circ \\) (equal to the angle between lines \\( BC \\) and \\( s \\) in the same plane).\n\nSo, the angle between two reverse lines (which do not intersect and are not parallel to each other) is the acute angle that one forms with a line parallel to the other.\n\n**Example:** Let us consider the lines\n\n\\[\n\\begin{align*}\n r : (x, y, z) &= (1, 2, 0) + k(2, 1, 3), k \\in \\mathbb{R} \\\\\n s : (x, y, z) &= (0, -1, -1) + t(3, 2, 1), t \\in \\mathbb{R}\n\\end{align*}\n\\]\n\nof \\( \\mathbb{R}^3 \\), whose directions are those of the non-collinear vectors \\( u = (2, 1, 3) \\) and \\( v = (3, 2, 1) \\), respectively. We can see that \\( r \\) and \\( s \\) do not intersect. In fact,\n\n\\[\n(1, 2, 0) + k(2, 1, 3) = (0, -1, -1) + t(3, 2, 1) \\iff \\begin{cases} 2k - 3t = -1 \\\\ k - 2t = -3 \\\\ 3k - t = -1 \\end{cases} \\iff \\begin{cases} k = \\frac{2}{7} \\\\ k = \\frac{1}{5} \\\\ t = 3k + 1 \\end{cases}\n\\]\n\nSo \\( r \\) and \\( s \\) are reverse lines.\n\nBesides that, \\( \\cos(\\hat{r}s) = |\\cos(\\hat{uv})| = \\frac{|u \\cdot v|}{|u||v|} = \\frac{6 + 2 + 3}{\\sqrt{14}\\sqrt{14}} = \\frac{11}{14} \\), that is, \\( \\hat{r}s = 23.6^\\circ \\).\n\nAngle between a line and a plane\n\nThe angle \\( \\theta \\) between a line \\( r \\) and a plane \\( p \\) is defined as the angle complementary to the acute angle between the direction vector on this line and the vector normal to the plane, according to the image on the side.\nIf \\( u \\) has the direction line \\( r \\) and \\( n \\) is a vector normal to the plane \\( p \\), then\n\n\\[\n\\sin(\\theta) = \\sin\\left(\\frac{\\pi}{2} - \\alpha\\right) = \\cos(\\alpha) = \\frac{|u \\cdot n|}{|u||n|}.\n\\]\n\n**Example:** The line \\( r : (x, y, z) = (1, -2, 0) + k(2, 2, 0), k \\in \\mathbb{R} \\) has a direction given by \\( u = (2, 2, 0) \\) and the plane \\( p : 2x + y + z - 1 = 0 \\) is orthogonal to \\( n = (2, 1, 1) \\).\n\nThen the angle formed by \\( r \\) and \\( p \\) is such that\n\n\\[\n\\sin(\\hat{r}p) = \\cos(\\hat{n}u) = \\frac{(4, 2, 2) \\cdot (2, 2, 0)}{\\sqrt{16 + 4 + 4\\sqrt{4 + 4}}} = \\frac{\\sqrt{3}}{2}.\n\\]\n\nThat is, \\( \\hat{r}p = \\frac{\\pi}{3} \\).\n\n**Angle between two planes**\n\n**Definition:** The angle between planes is equal to a angle between their normal vectors.\n\nConsider the equation plans \\( \\pi_1 : a_1x + b_1y + c_1z + d_1 = 0 \\) and \\( \\pi_2 : a_2x + b_2y + c_2z + d_2 = 0 \\).\n\nIn this case, \\( n_1 = (a_1, b_1, c_1) \\) and \\( n_2 = (a_2, b_2, c_2) \\). Then, by the scalar product, we have:\n\n\\[\n\\cos(\\alpha) = \\frac{n_1 \\cdot n_2}{|n_1||n_2|} = \\frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\\sqrt{a_1^2 + b_1^2 + c_1^2} \\cdot \\sqrt{a_2^2 + b_2^2 + c_2^2}}.\n\\]\n\n**Example:** The angle \\( \\alpha \\) formed by the \\( p_1 : 2x - y + z = 0 \\) and \\( p_2 = x + 2y - z + 1 = 0 \\) planes is such that\n\n\\[\n\\cos(\\alpha) = \\frac{(2, -1, 1) \\cdot (1, 2, -1)}{\\sqrt{6}\\sqrt{6}} = \\frac{-1}{6}.\n\\]\n\nThen \\( \\alpha = 163^\\circ \\).", "id": "./materials/155.pdf" }, { "contents": "Lines and planes in the Cartesian coordinate system\n\nLines in a Cartesian plane can be described algebraically by linear equations.\n\n**Lines in \\( \\mathbb{R}^2 \\)**\n\nIn the two-dimensional Cartesian referential, a point \\( P \\in \\mathbb{R}^2 \\) is determined by its distance each axis and by the quadrant in which it is located, that is, by its Cartesian coordinates \\((x, y)\\).\n\nGiven a vector \\( v \\) (through its length and direction) and a point \\( A \\), the sum \\( A + v \\) corresponds to a point \\( B \\), end of \\( v \\) when applied to \\( A \\).\n\nFor example, let the vector \\( v \\) in the figure on the side and be \\( A = (1, 2) \\in \\mathbb{R}^2 \\). We observe that \\( A + v = B = (2, 1) \\), or\n\n\\[\n\\vec{v} = \\vec{AB} = B - A = (1, -1).\n\\]\n\nAlso note that \\( \\vec{v} = \\vec{OP} \\) with \\( P = (1, -1) \\) and \\( O = (0, 0) \\).\n\nIn the Cartesian coordinate system, at each point \\( P \\) we associate the vector \\( \\vec{OP} = P - O = P \\) with the same coordinates of \\( P \\).\n\nA line in the Cartesian plane can be defined by:\n\n- Two points on the line;\n- A point on the line and its slope;\n- A point and a straight line vector.\n\nGiven two fixed points on the plane, \\( A = (a_1, a_2) \\) and \\( B = (b_1, b_2) \\), the line \\( AB \\) has the direction of the vector\n\n\\[\n\\vec{AB} = B - A = (b_1 - a_1, b_2 - a_2) = (v_1, v_2)\n\\]\n\nand has a slope\n\n\\[\nm = \\frac{b_2 - a_2}{b_1 - a_1} = \\frac{v_2}{v_1}.\n\\]\n\nThe vector equation of the line \\( AB \\) is\n\n\\[\n(x, y) = (a_1, a_2) + k(v_1, v_2), \\quad k \\in \\mathbb{R},\n\\]\nFrom the vector equation we deduce the Cartesian equation\n\\[\n\\frac{x - a_1}{v_1} = \\frac{y - a_2}{v_2}\n\\]\nor we can still transform into the reduced equation\n\\[\ny = \\frac{v_2}{v_1}x + b.\n\\]\n\n**Example:** The vector equation of the line \\(AB\\), with \\(A = (2, -2)\\) e \\(B = (-1, 1)\\), is given by\n\\[\n(x, y) = (-1, 1) + k(3, -3), \\quad k \\in \\mathbb{R}.\n\\]\nBy eliminating the parameter \\(k\\), we obtain the reduced equation\n\\[\ny = -x\n\\]\nThis line intersects the \\(Oy\\) axis at the origin and has a slope of \\(-1\\).\n\n**Lines and planes in \\(\\mathbb{R}^3\\)**\n\nWe represent elements in space using the three-dimensional Cartesian framework \\(Oxyz\\)\n\nFor example, to say that a point \\(P \\in \\mathbb{R}^3\\) has Cartesian coordinates \\((2, 3, 2)\\), means to say that \\(P\\) is written as the linear combination\n\\[\nP = 2i + 3j + 2k\n\\]\nof unit vectors\n\\[\ni = (1, 0, 0), \\quad j = (0, 1, 0), \\quad k = (0, 0, 1),\n\\]\noriented according to the reference axes, as the image suggests.\n\nLike the vector equation of the line in the plane, also if \\(A = (a_1, a_2, a_3), B = (b_1, b_2, b_3) \\in \\mathbb{R}^3\\), the line \\(AB\\) is the locus of the points \\(P = (x, y, z)\\) to the space, such that\n\\[\nP = A + k\\vec{AB}, \\quad \\text{to any} \\quad k \\in \\mathbb{R}.\n\\]\n\nFrom the vector equation of the line \\(AB\\),\n\\[\nAB : (x, y, z) = (a_1, a_2, a_3) + k(b_1 - a_1, b_2 - a_2, b_3 - a_3), \\quad k \\in \\mathbb{R}.\n\\]\nIf \\(\\vec{AB}(b_1 - a_1, b_2 - a_2, b_3 - a_3) = (v_1, v_2, v_3)\\) is such that \\(v_1, v_2, v_3 \\neq 0\\), eliminating the parameter \\(k\\), we obtain the Cartesian equation:\n\\[\nAB : \\frac{x - a_1}{v_1} = \\frac{y - a_2}{v_2} = \\frac{z - a_3}{v_3}.\n\\]\nIf, for example \\(v_1, v_3 \\neq 0\\), but \\(v_2 = 0\\) we have\n\\[\nAB : \\frac{x - a_1}{v_1} = \\frac{z - a_3}{v_3} \\land y = a_2.\n\\]\nExample: The equation \\( \\frac{x + 1}{2} = \\frac{y - 1}{3} = z \\) represents the line that contains \\( A = (-1, 1, 0) \\) and has the direction of \\( v = (2, 3, 1) \\).\n\nA plane in space can be defined by:\n\n- Three non-collinear points;\n- Two cross lines;\n- Two parallel lines;\n- A point and a vector perpendicular to the plane.\n\nDefinition: The plane containing \\( A \\) and is perpendicular to \\( v \\) is the locus of the points \\( P = (x, y, z) \\), such that the scalar product \\( \\vec{AP} \\cdot v \\) is zero,\n\n\\[ \\vec{AP} \\cdot v = 0. \\]\n\nExample: The plane containing \\( A = (1, 0, 2) \\) and is perpendicular to \\( v = (-1, 3, 2) \\) has the equation \\((x - 1, y, z - 2) \\cdot (-1, 3, 2) = 0 \\iff -x + 1 + 3y + 2z - 4 = 0 \\iff -x + 3y + 2z - 3 = 0.\\)\n\nThus, we obtained the general equation of the plane, that is, a linear equation in the variables \\( x, y, z \\),\n\n\\[ Ax + By + Cz + D = 0. \\]\n\nExample: Let’s determine the plane containing the points \\( A = (1, 1, -1), B = (2, 1, 0) \\) and \\( C = (3, 0, 1) \\). We start by calculating a vector orthogonal to the plane \\( ABC \\), considering\n\n\\[ \\vec{AB} \\times \\vec{AC} = \\begin{vmatrix} i & j & k \\\\ 1 & 0 & 1 \\\\ 2 & -1 & 2 \\end{vmatrix} = (1, 0, -1) \\]\n\nThen, the equation of \\( ABC \\) is \\((x - 1, y - 1, z + 1) \\cdot (1, 0, -1) = 0 \\iff x - z - 2 = 0.\\)", "id": "./materials/156.pdf" }, { "contents": "Relative position of straight lines and planes\n\nRelative position of straight lines in the plane\n\nTwo straight lines on the plane can be secant, parallel or coincidental.\n\nLet \\( r : y = ax + b \\) and \\( s : y = cx + d \\). Notice that:\n\n- If \\( a = b \\) and \\( b \\neq d \\) then \\( r \\) and \\( s \\) are parallel;\n- If \\( a = b \\) and \\( b = d \\) then \\( r \\) and \\( s \\) coincidental\n- If \\( a \\neq b \\) then \\( r \\) and \\( s \\) are secant;\n\nRelative position of straight lines in \\( \\mathbb{R}^3 \\)\n\nIn space one more case is added to the cases in the plane, if the lines \\( AB \\) and \\( CD \\) don’t lie in the same plane.\n\nSo, for the relative position of two straight lines in space, the following cases are possible:\n\n1. Lines lie in one plane and have no common points (parallel lines).\n2. Lines lie in the same plane and have a common point (intersecting lines).\n3. Lines do not lie in any plane (skew lines).\n\nTwo lines belong to the same plane if they are parallel lines or have a common point. Otherwise, they are skew lines.\n\nLet be the vectorial equations of the lines \\( r \\) and \\( s \\), respectively:\n\n\\[\n\\begin{align*}\n r : (x, y, z) &= (a_1, a_2, a_3) + k(v_1, v_2, v_3), \\quad k \\in \\mathbb{R} \\\\\n s : (x, y, z) &= (b_1, b_2, b_3) + t(u_1, u_2, u_3), \\quad t \\in \\mathbb{R}\n\\end{align*}\n\\]\n\nNote that:\n1. If \\( v = (v_1, v_2, v_3) \\) and \\( u = (u_1, u_2, u_3) \\) are collinear vectors then \\( r \\) and \\( s \\) are parallel lines.\n\n2. If \\( v = (v_1, v_2, v_3) \\) and \\( u = (u_1, u_2, u_3) \\) are not collinear vectors and \\( r \\cap s \\neq \\emptyset \\), then \\( r \\) and \\( s \\) are intersecting lines.\n\n3. If \\( v = (v_1, v_2, v_3) \\) and \\( u = (u_1, u_2, u_3) \\) are not collinear vectors and \\( r \\cap s \\neq \\emptyset \\), then \\( r \\) and \\( s \\) are not lines that lie in any plane.\n\nIn \\( \\mathbb{R}^3 \\), we consider the line \\( r : (x, y, z) = (a_1, a_2, a_3) + k(v_1, v_2, v_3), k \\in \\mathbb{R} \\) whose direction is that of the vector \\( v = (v_1, v_2, v_3) \\) and the plane \\( \\pi : Ax + By + C = 0 \\) orthogonal to the vector \\( n = (A, B, C) \\). Notice that \\( r \\) is either parallel to a plane \\( \\pi : Ax + By + C = 0 \\) or intersects it in a single point. Specifically:\n\n1. If \\( n \\cdot v = 0 \\), then \\( r \\) is parallel to \\( \\pi \\);\n\n2. If \\( n \\cdot v \\neq 0 \\), then \\( r \\) intersects \\( \\pi \\) in a single point\n\nEven more, if \\( r \\) is parallel to \\( \\pi \\) we have that \\( r \\subset \\pi \\) (if all points on the \\( r \\) belong to \\( \\pi \\)) or \\( r \\cap \\pi = \\emptyset \\)\n\n**Relative position of planes**\n\nLet two planes, \\( \\pi_1 \\) and \\( \\pi_2 \\), be given by their general equations:\n\n\\[\n\\pi_1 : A_1 x + B_1 y + C_1 z + D_1 = 0 \\quad \\text{and} \\quad \\pi_2 : A_2 x + B_2 y + C_2 z + D_2 = 0\n\\]\n\nConsider the system of two linear equations:\n\n\\[\nS : \\begin{cases}\nA_1 x + B_1 y + C_1 z + D_1 = 0 \\\\\nA_2 x + B_2 y + C_2 z + D_2 = 0\n\\end{cases}\n\\]\n\n- If the system \\( S \\) is inconsistent then the planes are parallel, and so the coordinates of the normal vectors \\( n_1 = (A_1, B_1, C_1) \\) and \\( n_2 = (A_2, B_2, C_2) \\) are proportional, that is, \\( \\frac{A_1}{A_2} = \\frac{B_1}{B_2} = \\frac{C_1}{C_2} \\neq \\frac{D_1}{D_2} \\).\n\n- If system \\( S \\) is consistent and the equations are proportional to each other, then \\( \\pi_1 \\) is just the same plane as \\( \\pi_2 \\), that is, \\( \\frac{A_1}{A_2} = \\frac{B_1}{B_2} = \\frac{C_1}{C_2} = \\frac{D_1}{D_2} \\).\n\n- If system \\( S \\) is consistent, and the rank of the coefficient matrix equals 2, then \\( \\pi_1 \\) and \\( \\pi_2 \\) are intersecting planes.\n\nThe locus of these distinct intersecting planes is exactly one line \\( r \\) whose direction is given by \\( v = n_1 \\times n_2 \\) (cross product of vectors \\( n_1 \\) and \\( n_2 \\) that are orthogonal to the planes \\( \\pi_1 \\) and \\( \\pi_2 \\), respectively), according to the image beside.", "id": "./materials/157.pdf" }, { "contents": "Distances\n\nDistance from a point to a line\n\nThe distance from a point $A$ to a line $r$ is equal to the distance from $A$ to its orthogonal projection $A'$ on the line $r$, according to the figure beside.\n\nWe calculate the distance $d$ by doing:\n\n1. Determine the line $PP'$ that is perpendicular to $r$ containing $P$;\n2. Determine $P' = PP' \\cap r$;\n3. Determine $d = PP'$.\n\nExample:\nConsider in $\\mathbb{R}^3$, $P = (2, 1, 1)$ and $r : (x, y, z) = (0, 0, -1) + k(1, -1, 1)$, $k \\in \\mathbb{R}$. Let us determine the distance from $P$ to $r$.\n\nFor example, $u = (1, 2, 1)$ is orthogonal to $v = (1, -1, 1)$ because $u \\cdot v = 0$.\n\nThen, $PP' : (x, y, z) = (2, 1, 1) + t(1, 2, 1), t \\in \\mathbb{R}$.\n\nBesides that $P' = (x, y, z) = PP' \\cap r$ is such that\n\n\\[\n\\begin{align*}\n x &= \\frac{y}{-1} = z + 1 \\\\\n x - 2 &= \\frac{y - 1}{2} = z - 1\n\\end{align*}\n\\]\n\n\\[\\Leftrightarrow \\begin{cases} \n x = 1 \\\\\n y = -1 \\\\\n z = 0\n\\end{cases}.\n\\]\n\nThat is $P' = (1, -1, 0)$.\n\nFinally $d = PP' = \\sqrt{(2 - 1)^2 + (1 + 1)^2 + (1 - 0)^2} = \\sqrt{6}$.\n\nDistance from a point to a plane:\n\nWe can determine the distance from point $P$ to plane $p$, by performing:\n\n- Calculate the line $r$ that contains the point $P$ and is normal to the plane $p$;\n- Calculate $P' = r \\cap p$;\n- Determine the distance from $P$ to $P'$.\n\nExample: To calculate the distance from $P = (1, 2, -1)$ to the plane $p : x - y + z = 0$, we can take\n\\[ n = (2, -2, 1) \\perp p \\] and the line \\( r \\) that contains \\( P \\) and is normal to the plane \\( p \\) is \\( r : \\frac{x - 1}{2} = \\frac{y - 2}{-2} = z + 1 \\).\n\n\\[\nP' = r \\cap p = \\begin{cases} x - 1 = -y + 2 \\\\ -y + 2 = z + 1 \\\\ x - y + z = 1 \\end{cases} \\iff \\begin{cases} x = 2 \\\\ y = 0 \\\\ z = 1 \\end{cases}\n\\]\n\nThen, \\( d(P, p) = d(P, P') = \\sqrt{(1 - 2)^2 + (2 - 0)^2 + (-1 - 1)^2} = 3 \\)\n\n**Distance from a straight line to a parallel plane:**\n\nGiven a line \\( r \\) parallel to a plane \\( p \\), the distance \\( d \\) from the line \\( r \\) is the distance from any point \\( p \\) on the line to the plane, that is,\n\n\\[ d(r, p) = d(P, p) \\]\n\n**Example:** To calculate the distance from \\( r : \\frac{x - 1}{2} = -y = \\frac{z + 1}{-3} \\) to the plane \\( p : x - y + z = 0 \\) is the distance from \\( P = (1, 0, -1) \\in r \\) to the plane \\( p \\).\n\n**Distance between two parallel planes:**\n\nTo calculate the distance between two planes \\( \\alpha \\) and \\( \\beta \\) parallel to each other, we can perform:\n\n- Calculate the line \\( r \\) that is normal to the planes \\( \\alpha \\) and \\( \\beta \\);\n- Calculate \\( A = r \\cap \\alpha \\);\n- Calculate \\( B = r \\cap \\beta \\);\n- Determine the distance from \\( A \\) to \\( B \\).", "id": "./materials/158.pdf" }, { "contents": "Tests of Significance\n\nDiana Mindrila, Ph.D.\nPhoebe Balentyne, M.Ed.\n\nBased on Chapter 15 of The Basic Practice of Statistics (6th ed.)\n\nConcepts:\n- The Reasoning of Tests of Significance\n- Stating Hypotheses\n- P-value and Statistical Significance\n- Tests for a Population Mean\n- Significance from a Table\n\nObjectives:\n- Define statistical inference.\n- Describe the reasoning of tests of significance.\n- Describe the parts of a significance test.\n- State hypotheses.\n- Define P-value and statistical significance.\n- Conduct and interpret a significance test for the mean of a Normal population.\n- Determine significance from a table.\n\nReferences:\nMoore, D. S., Notz, W. I, & Flinger, M. A. (2013). The basic practice of statistics (6th ed.). New York, NY: W. H. Freeman and Company.\nStatistical Inference\n\nConfidence intervals are one of the two most common types of statistical inference. Researchers use a confidence interval when their goal is to estimate a population parameter. The second common type of inference, called a test of significance, has a different goal: to assess the evidence provided by data about some claim concerning a population.\n\nA test of significance is a formal procedure for comparing observed data with a claim (also called a hypothesis), the truth of which is being assessed.\n\n- The claim is a statement about a parameter, like the population proportion $p$ or the population mean $\\mu$.\n- The results of a significance test are expressed in terms of a probability that measures how well the data and the claim agree.\nThe Reasoning of Tests of Significance\n\nIt is helpful to start with an example:\n\nPopulation Mean:\nIQ=100\n\nIs the sample mean significantly different than the population mean?\n\n- In order to determine if two numbers are *significantly different*, a statistical test must be conducted to provide evidence. Researchers cannot rely on subjective interpretations.\n- Researchers must collect statistical evidence to make a claim, and this is done by conducting a test of statistical significance.\nStating Hypotheses\n\nThe first step in conducting a test of statistical significance is to state the hypothesis.\n\nA significance test starts with a careful statement of the claims being compared. The claim tested by a statistical test is called the null hypothesis \\( (H_0) \\). The test is designed to assess the strength of the evidence against the null hypothesis. Often the null hypothesis is a statement of “no difference.”\n\nThe claim about the population that evidence is being sought for is the alternative hypothesis \\( (H_a) \\).\n\nThe alternative is one-sided if it states that a parameter is larger or smaller than the null hypothesis value. It is two-sided if it states that the parameter is different from the null value (it could be either smaller or larger).\n\n- When using logical reasoning, it is much easier to demonstrate that a statement is false, than to demonstrate that it is true. This is because proving something false only requires one counterexample. Proving something true, however, requires proving the statement is true in every possible situation.\n- For this reason, when conducting a test of significance, a null hypothesis is used. The term null is used because this hypothesis assumes that there is no difference between the two means or that the recorded difference is not significant. The notation that is typically used for the null hypothesis is \\( H_0 \\).\n- The opposite of a null hypothesis is called the alternative hypothesis. The alternative hypothesis is the claim that researchers are actually trying to prove is true. However, they prove it is true by proving that the null hypothesis is false. If the null hypothesis is false, then its opposite, the alternative hypothesis, must be true. The notation that is typically used for the alternative hypothesis is \\( H_a \\).\nIn the example above, the null hypotheses states: “the sample mean is equal to 100” or “there is no difference between the sample mean and the population mean.”\n\n- The sample mean will not be exactly equal to the population mean. This null hypothesis is stating that the recorded difference is not a significant one.\n- If researchers can demonstrate that this null hypothesis is false, then its opposite, the alternative hypothesis, must be true.\n\nIn the example above, the alternative hypothesis states: “the sample mean is significantly different than 100” or “there is a significant difference between the sample mean and the population mean.”\n\nIf researchers are trying to prove that the mean IQ in the sample will specifically be higher or lower (just one direction) than the population mean, this is a one-sided alternative hypothesis because they are only looking at one direction in which the mean may vary. They are not interested in the other direction.\n\nIf researchers suspect that the sample mean could be either lower or higher than 100, the alternative hypothesis would be two-sided because both directions in which mean IQ may vary are being tested.\n\nWhen conducting a significance test, the goal is to provide evidence to reject the null hypothesis. If the evidence is strong enough to reject the null hypothesis, then the alternative hypothesis can automatically be accepted. However, if the evidence is not strong enough, researchers fail to reject the null hypothesis.\nTest-Statistic (z score)\n\nExample:\n\nPopulation mean: IQ = 100\nPopulation st dev = 16\nSample mean: IQ = 108\nSample size: N = 16\n\n\\[ Z = \\frac{\\bar{X} - \\mu_0}{\\sigma / \\sqrt{n}} \\]\n\n\\[ z = \\frac{(108 - 100)}{(16 / \\sqrt{16})} \\Rightarrow z = 8 / (16 / 4) \\Rightarrow z = 8 / 4 \\Rightarrow z = 2 \\]\n\n- In the above example, the mean IQ score for the sample is 108. This is slightly higher than the population mean, which is 100. The sample mean is obviously different from the population mean, but tests of significance must be done to determine if the difference is statistically significant. The difference could possibly be attributed to chance or to sampling error.\n- The first step is to compute the test statistic. The test statistic is simply the z score for the sample mean. The only difference is that the population standard deviation is divided by the square root of N, just like when a confidence interval is computed.\n- To compute the test statistic, the population standard deviation must be known for the variable. The population standard deviation for IQ is 16.\n- To compute the test statistic, the sample size must also be known. In this case, it is 16. (In a real research scenario, the sample size would be larger. Small sample sizes are being use in this example to make calculations simpler).\n- After putting the needed information into the formula, the result is a z score of 2. This means that the sample mean is exactly two standard deviations above the population mean.\n**P-Value**\n\n- After computing the test statistic, the next step is to find out the probability of obtaining this score when the null hypothesis is true.\n- The Normal curve helps researchers determine the percentage of individuals in the population who are located within certain intervals or above or below a certain score.\n- To find this information, the score needs to be standardized. In the case of the example, this was already done by computing $z$, the test statistic.\n\nThe Normal distribution can be used to compute the probability of obtaining a certain $z$ score.\n\nAssuming that $H_0$ is true:\n\nArea to the left of $z$ = the probability of obtaining scores lower than $z$\n\nArea to the right of $z$ (p-value) = the probability of obtaining scores higher than $z$\n\nThe smaller the p-value, the stronger the evidence against $H_0$ provided by the data.\n\n- When testing hypotheses, researchers are interested in the area to the right of $z$, which is called the p-value.\n- The p-value represents the probability of obtaining scores that are at the $z$ level or higher when the null hypothesis is true. In other words, what percent chance exists of getting this specific sample mean score if it is actually no different from the population mean.\n- If $z$ is far away from the mean, the p-value is small. The larger the test statistic (the farther from the mean), the smaller the p-value.\n- When the p-value is very small, researchers can say they have strong evidence that the null hypothesis is false. This is because if the p-value is very small, it means that the probability of obtaining a score that is so extreme or even higher is very small.\n**P-Value**\n\nNow it is important to know how to find the p-value.\n\n- Statistical textbooks provide tables listing the area to the left of the curve for every possible z score.\n - The rows of this table represent the first two digits of the z score.\n - The columns of this table represent the last two digits of the z score.\n\n**Example:**\n\nTable A provides the area to the left of z:\n\n| Z | .00 | .01 | .02 |\n|----|-----|-----|-----|\n| 1.9| .13 | .9719| .9726|\n| 2.0| .9772| .9778| .9783|\n| 2.1| .9821| .9826| .9830|\n\n- The z score in the example is exactly 2, so all decimals are zero. The area to the left of the curve for this z score is 0.9772.\n- However, for hypothesis testing, the area to the right of z is needed. This is called the p-value. Since the entire area under the curve is equal to one, simply subtract the area to the left of the value from one to obtain the p-value.\n- In this example, 1 – 0.9772 = 0.0228.\n- This means that there is only a 2% chance that the null hypothesis is true. In other words, if the population mean is 100, then there is only a 2% chance of having a sample mean equal to 108.\n**P-Value and Statistical Significance**\n\nIt is important to know how small the p-value needs to be in order to reject the null hypothesis.\n\n- The cutoff value for p is called **alpha**, or the significance level.\n- The researcher establishes the value of alpha prior to beginning the statistical analysis.\n- In social sciences, alpha is typically set at 0.05 (or 5%). This represents the amount of acceptable error, or the probability of rejecting a null hypothesis that is in fact true. It is also called the probability of Type I error.\n\nOnce the alpha level has been selected and the p-value has been computed:\n- If the p-value is larger than alpha, accept the null hypothesis and reject the alternative hypothesis.\n- If the p-value is smaller than alpha, reject the null hypothesis and accept the alternative hypothesis.\n\nIn the above example, the p-value of approximately 0.02 is smaller than 0.05, so the null hypothesis can be rejected.\n- The test of significance showed that the difference between the sample mean and the population mean is statistically significant.\nTwo-Sided Alternative Hypotheses\n\n- A two-sided alternative hypothesis is used when there is no reason to believe that the sample mean can only be higher or lower than a given value. Researchers are only hypothesizing that the values are significantly different.\n- In the example, the alternative hypothesis was one-sided, so it was only necessary to look at the probability that the sample mean was larger than 100.\n- However, when the alternative hypothesis is two-sided, the sample mean can be higher or lower than the given value, so researchers must look for both extremely high and extremely low values. This means that alpha is located at both ends of the curve.\n- Half of alpha is located at the higher end, and half is located at the lower end. So, there is both a low cutoff value and a high cutoff value.\n- Therefore, in two-sided cases, the p-value is obtained by multiplying the area to the right of z by 2.\n- Only after doubling this value can the p-value be compared to alpha.\n\n\\[\n\\text{P-value} = \\text{Area to the right of the curve (from Table A)} \\times 2\n\\]\n\nExample:\nNull Hypothesis: Sample IQ=100\nAlternative Hypothesis: Sample IQ≠100\n\n\\[Z=2\\]\nArea to the right=.0228\n\\[P\\_Value=.0228 \\times 2 = .0456\\]\n\nExample:\n\\[\\alpha = .05\\]\n\\[p\\text{-value}=0.04\\]\n\n\\[p\\text{-value}<\\alpha \\Rightarrow \\text{Decision: Reject the null hypothesis & accept the alternative}\\]\n\\[\\text{Conclusion: “The test of significance provided evidence that the sample IQ is significantly different than 100.”}\\]\n\n- In the above example, the area to the right of z was 0.0228. So, if the alternative hypothesis were two-sided, the p-value would be 0.0456.\n- The p-value (0.0456) is smaller than the alpha level (0.05), so the null hypothesis is rejected and the alternative hypothesis is accepted.\nHow to Obtain the *P*-Value for the *z* Statistic\n\n1) A table in a statistical textbook\n\n2) Statistical software\n\n3) A calculator (these can be found online)\nTests of Significance: The Four-Step Process\n\n1. State the null and alternative hypotheses.\n2. Calculate the test statistic.\n3. Find the $P$-value (using a table or statistical software).\n4. Compare $P$-value with $\\alpha$ and decide whether the null hypothesis should be rejected or accepted.\n\n- The $z$ statistic is computed based on the properties of the Normal distribution, so it should only be used when the distribution of the variable is approximately normal. If this condition is not met, a different statistic must be used.", "id": "./materials/159.pdf" }, { "contents": "Null Hypothesis Significance Testing II\nClass 18, 18.05\nJeremy Orloff and Jonathan Bloom\n\n1 Learning Goals\n\n1. Be able to list the steps common to all null hypothesis significance tests.\n2. Be able to define and compute the probability of Type I and Type II errors.\n3. Be able to look up and apply one and two sample $t$-tests.\n\n2 Introduction\n\nWe continue our study of significance tests. In these notes we will introduce two new tests: one-sample $t$-tests and two-sample $t$-tests. You should pay careful attention to the fact that every test makes some assumptions about the data – often that is drawn from a normal distribution. You should also notice that all the tests follow the same pattern. It is just the computation of the test statistic and the type of the null distribution that changes.\n\n3 Review: setting up and running a significance test\n\nThere is a fairly standard set of steps one takes to set up and run a null hypothesis significance test.\n\n1. Design an experiment to collect data and choose a test statistic $x$ to be computed from the data. The key requirement here is to know the null distribution $f(x|H_0)$. To compute power, one must also know the alternative distribution $f(x|H_A)$.\n2. Decide if the test is one or two-sided based on $H_A$ and the form of the null distribution.\n3. Choose a significance level $\\alpha$ for rejecting the null hypothesis. If applicable, compute the corresponding power of the test.\n4. Run the experiment to collect data $x_1, x_2, \\ldots, x_n$.\n5. Compute the test statistic $x$.\n6. Compute the $p$-value corresponding to $x$ using the null distribution.\n7. If $p < \\alpha$, reject the null hypothesis in favor of the alternative hypothesis.\n\nNotes.\n1. Rather than choosing a significance level, you could instead choose a rejection region and reject $H_0$ if $x$ falls in this region. The corresponding significance level is then the probability that $x$ falls in the rejection region.\n2. The null hypothesis is often the ‘cautious hypothesis’. The lower we set the significance level, the more “evidence” we will require before rejecting our cautious hypothesis in favor of a more sensational alternative. It is standard practice to publish the $p$ value itself so that others may draw their own conclusions.\n\n3. **A key point of confusion:** A significance level of 0.05 does not mean the test only makes mistakes 5% of the time. It means that if the null hypothesis is true, then the probability the test will mistakenly reject it is 5%. The power of the test measures the accuracy of the test when the alternative hypothesis is true. Namely, the power of the test is the probability of rejecting the null hypothesis if the alternative hypothesis is true. Therefore the probability of falsely failing to reject the null hypothesis is 1 minus the power.\n\n**Errors.** We can summarize these two types of errors and their probabilities as follows:\n\n- **Type I error** = rejecting $H_0$ when $H_0$ is true.\n- **Type II error** = failing to reject $H_0$ when $H_A$ is true.\n\n\\[\nP(\\text{type I error}) = \\text{probability of falsely rejecting } H_0 = P(\\text{test statistic is in the rejection region } | H_0) = \\text{significance level of the test}\n\\]\n\n\\[\nP(\\text{type II error}) = \\text{probability of falsely not rejecting } H_0 = P(\\text{test statistic is in the acceptance region } | H_A) = 1 - \\text{power}.\n\\]\n\n**Helpful analogies.** In terms of medical testing for a disease, a Type I error is a false positive and a Type II error is a false negative. In terms of a jury trial, a Type I error is convicting an innocent defendant and a Type II error is acquitting a guilty defendant.\n\n4 **Understanding a significance test**\n\nQuestions to ask:\n\n1. How did they collect data? What is the experimental setup?\n\n2. What are the null and alternative hypotheses?\n\n3. What type of significance test was used?\n - Does their data match the criteria needed to use this type of test?\n - How robust is the test to deviations from this criteria.\n\n4. For example, some tests comparing two groups of data assume that the groups are drawn from distributions that have the same variance. This needs to be verified before applying the test. Often the check is done using another significance test designed to compare the variances of two groups of data.\n\n5. How is the $p$-value computed?\n - A significance test comes with a test statistic and a null distribution. In most tests the $p$-value is\n\n\\[\np = P(\\text{data at least as extreme as what we got } | H_0)\n\\]\nWhat does ‘data at least as extreme as the data we saw,’ mean? I.e. is the test one or two-sided.\n\n6. What is the significance level $\\alpha$ for this test? If $p < \\alpha$ then the experimenter will reject $H_0$ in favor of $H_A$.\n\n5 $t$ tests\n\nMany significance tests assume that the data are drawn from a normal distribution, so before using such a test you should examine the data to see if the normality assumption is reasonable. We will describe how to do this in more detail later, but plotting a histogram is a good start. Like the $z$-test, the one-sample and two-sample $t$-tests we’ll consider below start from this normality assumption.\n\nWe don’t expect you to memorize all the computational details of these tests and those to follow. In real life, you have access to textbooks, google, and wikipedia; on the exam, you’ll have your notecard. Instead, you should be able to identify when a $t$ test is appropriate and apply this test after looking up the details and using a table or software like R.\n\n5.1 $z$-test\n\nLet’s first review the $z$-test.\n\n- Data: we assume $x_1, x_2, \\ldots, x_n \\sim N(\\mu, \\sigma^2)$, where $\\mu$ is unknown and $\\sigma$ is known.\n- Null hypothesis: $\\mu = \\mu_0$ for some specific value $\\mu_0$\n- Test statistic: $z = \\frac{\\bar{x} - \\mu_0}{\\sigma/\\sqrt{n}} = \\text{standardized mean}$\n- Null distribution: $f(z \\mid H_0)$ is the pdf of $Z \\sim N(0,1)$\n- One-sided $p$-value (right side): $p = P(Z > z \\mid H_0)$\n One-sided $p$-value (left side): $p = P(Z < z \\mid H_0)$\n Two-sided $p$-value: $p = P(|Z| > |z|)$.\n\nExample 1. Suppose that we have data that follows a normal distribution of unknown mean $\\mu$ and known variance 4. Let the null hypothesis $H_0$ be that $\\mu = 0$. Let the alternative hypothesis $H_A$ be that $\\mu > 0$. Suppose we collect the following data:\n\n$$1, 2, 3, 6, -1$$\n\nAt a significance level of $\\alpha = 0.05$, should we reject the null hypothesis?\n\nanswer: There are 5 data points with average $\\bar{x} = 2.2$. Because we have normal data with a known variance we should use a $z$ test. Our $z$ statistic is\n\n$$z = \\frac{\\bar{x} - \\mu_0}{\\sigma/\\sqrt{n}} = \\frac{2.2 - 0}{2/\\sqrt{5}} = 2.460$$\nOur test is one-sided because the alternative hypothesis is one-sided. So (using R) our $p$-value is\n\n$$p = P(Z > z) = P(Z > 2.460) = 0.007$$\n\nSince $p < .05$, we reject the null hypothesis in favor of the alternative hypothesis $\\mu > 0$.\n\nWe can visualize the test as follows:\n\n5.2 The Student $t$ distribution\n\n‘Student’ is the pseudonym used by the William Gosset who first described this test and this test and distribution. See [http://en.wikipedia.org/wiki/Student’s_t-test](http://en.wikipedia.org/wiki/Student’s_t-test)\n\nThe $t$-distribution is symmetric and bell-shaped like the normal distribution. It has a parameter $df$ which stands for degrees of freedom. For $df$ small the $t$-distribution has more probability in its tails than the standard normal distribution. As $df$ increases $t(df)$ becomes more and more like the standard normal distribution.\n\nHere is a simple applet that shows $t(df)$ and compares it to the standard normal distribution: [http://mathlets.org/mathlets/t-distribution/](http://mathlets.org/mathlets/t-distribution/)\n\nAs usual in R, the functions pt, dt, qt, rt correspond to cdf, pdf, quantiles, and random sampling for a $t$ distribution. Remember that you can type ?dt in RStudio to view the help file specifying the parameters of dt. For example, pt(1.65,3) computes the probability that $x$ is less than or equal 1.65 given that $x$ is sampled from the $t$ distribution with 3 degrees of freedom, i.e. $P(x \\leq 1.65)$ given that $x \\sim t(3)$.\n\n5.3 One sample $t$-test\n\nFor the $z$-test, we assumed that the variance of the underlying distribution of the data was known. However, it is often the case that we don’t know $\\sigma$ and therefore we must estimate it from the data. In these cases, we use a one sample $t$-test instead of a $z$-test and the studentized mean in place of the standardized mean\n\n- Data: we assume $x_1, x_2, \\ldots, x_n \\sim N(\\mu, \\sigma^2)$, where both $\\mu$ and $\\sigma$ are unknown.\n- Null hypothesis: $\\mu = \\mu_0$ for some specific value $\\mu_0$\n- Test statistic:\n\n$$t = \\frac{\\bar{x} - \\mu_0}{s/\\sqrt{n}}$$\nwhere\n\\[ s^2 = \\frac{1}{n-1} \\sum_{i=1}^{n} (x_i - \\bar{x})^2. \\]\n\nHere \\( t \\) is called the Studentized mean and \\( s^2 \\) is called the sample variance. The latter is an estimate of the true variance \\( \\sigma^2 \\).\n\n- Null distribution: \\( f(t \\mid H_0) \\) is the pdf of \\( T \\sim t(n-1) \\), the \\( t \\) distribution with \\( n-1 \\) degrees of freedom.*\n- One-sided \\( p \\)-value (right side): \\( p = P(T > t \\mid H_0) \\)\n - One-sided \\( p \\)-value (left side): \\( p = P(T < t \\mid H_0) \\)\n - Two-sided \\( p \\)-value: \\( p = P(|T| > |t|) \\).\n\n*It’s a theorem (not an assumption) that if the data is normal with mean \\( \\mu_0 \\) then the Studentized mean follows a \\( t \\)-distribution. A proof would take us too far afield, but you can look it up if you want: [http://en.wikipedia.org/wiki/Student’s_t-distribution#Derivation](http://en.wikipedia.org/wiki/Student’s_t-distribution#Derivation)\n\n**Example 2.** Now suppose that in the previous example the variance is unknown. That is, we have data that follows a normal distribution of unknown mean \\( \\mu \\) and and unknown variance \\( \\sigma \\). Suppose we collect the same data as before:\n\n\\[ 1, 2, 3, 6, -1 \\]\n\nAs above, let the null hypothesis \\( H_0 \\) be that \\( \\mu = 0 \\) and the alternative hypothesis \\( H_A \\) be that \\( \\mu > 0 \\). At a significance level of \\( \\alpha = 0.05 \\), should we reject the null hypothesis?\n\n**answer:** There are 5 data points with average \\( \\bar{x} = 2.2 \\). Because we have normal data with unknown mean and unknown variance we should use a one-sample \\( t \\) test. Computing the sample variance we get\n\n\\[ s^2 = \\frac{1}{4} \\left( (1 - 2.2)^2 + (2 - 2.2)^2 + (3 - 2.2)^2 + (6 - 2.2)^2 + (-1 - 2.2)^2 \\right) = 6.7 \\]\n\nOur \\( t \\) statistic is\n\n\\[ t = \\frac{\\bar{x} - \\mu_0}{s/\\sqrt{n}} = \\frac{2.2 - 0}{\\sqrt{6.7}/\\sqrt{5}} = 1.901 \\]\n\nOur test is one-sided because the alternative hypothesis is one-sided. So (using R) the \\( p \\)-value is\n\n\\[ p = P(T > t) = P(T > 1.901) = 1 - pt(1.901, 4) = 0.065 \\]\n\nSince \\( p > .05 \\), we do not reject the null hypothesis.\n\nWe can visualize the test as follows:\n5.4 Two-sample t-test with equal variances\n\nWe next consider the case of comparing the means of two samples. For example, we might be interested in comparing the mean efficacies of two medical treatments.\n\n- Data: We assume we have two sets of data drawn from normal distributions\n\n\\[ x_1, x_2, \\ldots, x_n \\sim N(\\mu_1, \\sigma^2) \\]\n\\[ y_1, y_2, \\ldots, y_m \\sim N(\\mu_2, \\sigma^2) \\]\n\nwhere the means \\( \\mu_1 \\) and \\( \\mu_2 \\) and the variance \\( \\sigma^2 \\) are all unknown. Notice the assumption that the two distributions have the same variance. Also notice that there are \\( n \\) samples in the first group and \\( m \\) samples in the second.\n\n- Null hypothesis: \\( \\mu_1 = \\mu_2 \\) (the values of \\( \\mu_1 \\) and \\( \\mu_2 \\) are not specified)\n\n- Test statistic:\n\n\\[ t = \\frac{\\bar{x} - \\bar{y}}{s_p}, \\]\n\nwhere \\( s_p^2 \\) is the pooled variance\n\n\\[ s_p^2 = \\frac{(n-1)s_x^2 + (m-1)s_y^2}{n+m-2} \\left( \\frac{1}{n} + \\frac{1}{m} \\right) \\]\n\nHere \\( s_x^2 \\) and \\( s_y^2 \\) are the sample variances of the \\( x_i \\) and \\( y_j \\) respectively. The expression for \\( t \\) is somewhat complicated, but the basic idea remains the same and it still results in a known null distribution.\n\n- Null distribution: \\( f(t \\mid H_0) \\) is the pdf of \\( T \\sim t(n + m - 2) \\).\n\n- One-sided p-value (right side): \\( p = P(T > t \\mid H_0) \\)\n One-sided p-value (left side): \\( p = P(T < t \\mid H_0) \\)\n Two-sided p-value: \\( p = P(|T| > |t|) \\).\n\n**Note 1:** Some authors use a different notation. They define the pooled variance as\n\n\\[ s_{p\\text{-other-authors}}^2 = \\frac{(n-1)s_x^2 + (m-1)s_y^2}{n+m-2} \\]\n\nand what we called the pooled variance they point out is the estimated variance of \\( \\bar{x} - \\bar{y} \\). That is,\n\n\\[ s_p^2 = s_{p\\text{-other-authors}} \\times \\left( \\frac{1}{n} + \\frac{1}{m} \\right) \\approx s_{\\bar{x}-\\bar{y}}^2 \\]\n\n**Note 2:** There is a version of the two-sample t-test that allows the two groups to have different variances. In this case the test statistic is a little more complicated but R will handle it with equal ease.\n\n**Example 3.** The following data comes from a real study in which 1408 women were admitted to a maternity hospital for (i) medical reasons or through (ii) unbooked emergency\nadmission. The duration of pregnancy is measured in complete weeks from the beginning of the last menstrual period. We can summarize the data as follows:\n\nMedical: 775 observations with $\\bar{x}_M = 39.08$ and $s^2_M = 7.77$.\n\nEmergency: 633 observations with $\\bar{x}_E = 39.60$ and $s^2_E = 4.95$\n\nSet up and run a two-sample $t$-test to investigate whether the mean duration differs for the two groups.\n\nWhat assumptions did you make?\n\n**answer:** The pooled variance for this data is\n\n$$s_p^2 = \\frac{774(7.77) + 632(4.95)}{1406} \\left( \\frac{1}{775} + \\frac{1}{633} \\right) = .0187$$\n\nThe $t$ statistic for the null distribution is\n\n$$\\frac{\\bar{x}_M - \\bar{x}_E}{s_p} = -3.8064$$\n\nWe have 1406 degrees of freedom. Using R to compute the two-sided $p$-value we get\n\n$$p = P(|T| > |t|) = 2*dt(-3.8064, 1406) = 0.00015$$\n\n$p$ is very small, much smaller than $\\alpha = .05$ or $\\alpha = .01$. Therefore we reject the null hypothesis in favor of the alternative that there is a difference in the mean durations.\n\nRather than compute the two-sided $p$-value exactly using a $t$-distribution we could have noted that with 1406 degrees of freedom the $t$ distribution is essentially standard normal and 3.8064 is almost 4 standard deviations. So\n\n$$P(|t| \\geq 3.8064) \\approx P(|z| \\geq 3.8064) < .001$$\n\nWe assumed the data was normal and that the two groups had equal variances. Given the large difference between the sample variances this assumption may not be warranted.\n\nIn fact, there are other significance tests that test whether the data is approximately normal and whether the two groups have the same variance. In practice one might apply these first to determine whether a $t$ test is appropriate in the first place. We don’t have time to go into normality tests here, but we will see the $F$ distribution used for equality of variances next week.\n\nhttp://en.wikipedia.org/wiki/Normality_test\nhttp://en.wikipedia.org/wiki/F-test_of_equality_of_variances\n", "id": "./materials/161.pdf" }, { "contents": "Higher Order Linear Equations\n\nDefinition An $n$-th linear differential equation is an equation of the form\n\n$$P_0(t) \\frac{d^n y}{dt^n} + P_1(t) \\frac{d^{n-1} y}{dt^{n-1}} + \\cdots + P_{n-1}(t) \\frac{dy}{dt} + P_n(t)y = G(t) \\quad (3.1)$$\n\nWe assume that the functions $P_0, \\ldots, P_n$ and $G$ are continuous real-valued functions on some interval $I : \\alpha < t < \\beta$, and that $P_0$ is nowhere zero in this interval. Then, dividing Eq. (3.1) by $P_0(t)$, we obtain\n\n$$\\frac{d^n y}{dt^n} + p_1(t) \\frac{d^{n-1} y}{dt^{n-1}} + \\cdots + p_{n-1}(t) \\frac{dy}{dt} + p_n(t)y = g(t) \\quad (3.2)$$\n\nIn developing the theory of linear differential equations, it is helpful to introduce a differential operator notation. Let $p_1, \\ldots, p_n$ be continuous functions on an open interval $I$, that is, for $\\alpha < t < \\beta$. The cases $\\alpha = -\\infty$, or $\\beta = \\infty$, or both, are included. Then, for any function $f$ that is $n$-th differentiable on $I$, we define the differential operator $L$ by the equation\n\n$$L[y](t) = \\frac{d^n y}{dt^n} + p_1(t) \\frac{d^{n-1} y}{dt^{n-1}} + \\cdots + p_{n-1}(t) \\frac{dy}{dt} + p_n(t)y.$$\n\nFirst by direct calculation it can prove the next assertion\n\n**Proposition 3.1** The differential operator $L$ is linear transform, that is\n\n$$L[ay_1 + by_2] = aL[y_1] + bL[y_2]$$\n\nwhere $a, b \\in \\mathbb{R}$ and $y_1, y_2$ are continuous functions.\n\nSince Eq. (3.2) involves the $n$-th derivative of $y$ with respect to $t$, it will, so to speak, require $n$ integrations to solve Eq. (3.2). Each of these integrations introduces an arbitrary constant. Hence we can expect that, to obtain a unique solution, it is necessary to specify $n$ initial conditions,\n\n$$y(t_0) = y_0, \\quad y'(t_0) = y'_0, \\ldots, y^{(n-1)}(t_0) = y^{(n-1)}_0, \\quad (3.3)$$\n\nwhere $t_0$ may be any point in the interval $I$ and $y_0, y'_0, \\ldots, y^{(n-1)}_0$ is any set of prescribed real constants. That there does exist such a solution and that\nit is unique are assured by the following existence and uniqueness theorem.\n\n**Theorem 3.1** If the functions \\( p_1, p_2, \\ldots, p_n \\), and \\( g \\) are continuous on the open interval \\( I \\), then there exists exactly one solution \\( y = \\psi(t) \\) of the differential equation (3.2) that also satisfies the initial conditions (3.3). This solution exists throughout the interval \\( I \\).\n\nWe will not give a proof of this theorem here. However, if the coefficients \\( p_1, \\ldots, p_n \\) are constants, then we can construct the solution of the initial value problem (3.2), (3.3).\n\n**The Homogeneous Equation.** We first discuss the homogeneous equation\n\n\\[\nL[y] = y^{(n)} + p_1(t)y^{(n-1)} + \\ldots + p_{n-1}(t)y' + p_n(t)y = 0. \\tag{3.4}\n\\]\n\nIf we denote by \\( V_0 \\) the set of all solution of the eq.(3.4) then it is easy to see that \\( V_0 \\) is subset of the set \\( V \\) of \\( n \\)-th differentiable function which is a linear space. By direct calculation it can prove\n\n**Proposition 3.2** \\( V_0 \\) is a real linear subspace of \\( V \\) of dimension \\( n \\).\n\nIf the functions \\( y_1, y_2, \\ldots, y_n \\) are solutions of Eq. (3.4), then it follows by direct computation that the linear combination\n\n\\[\ny = c_1y_1(t) + c_2y_2(t) + \\ldots + c_ny_n(t), \\tag{3.5}\n\\]\n\nwhere \\( c_1, \\ldots, c_n \\) are arbitrary constants, is also a solution of Eq. (3.4). It is then natural to ask whether every solution of Eq. (3.4) can be expressed as a linear combination of \\( y_1, \\ldots, y_n \\). This will be true if, regardless of the initial conditions (3.3) that are prescribed, it is possible to choose the constants \\( c_1, \\ldots, c_n \\) so that the linear combination (3.5) satisfies the initial conditions. Specifically, for any choice of the point \\( t_0 \\) in \\( I \\), and for any choice of \\( y_0, y_0', \\ldots, y_0^{(n-1)} \\), we must be able to determine \\( c_1, \\ldots, c_n \\) so that the equations\n\\[\n\\begin{align*}\n\\begin{cases}\n c_1 y_1(t_0) + c_n y_n(t_0) &= y_0 \\\\\n c_1 y_1'(t_0) + c_n y_n'(t_0) &= y_0' \\\\\n \\vdots & \\vdots \\\\\n c_1 y_1^{(n-1)}(t_0) + c_n y_n^{(n-1)}(t_0) &= y_0^{(n-1)}.\n\\end{cases}\n\\end{align*}\n\\] (3.6)\n\nare satisfied. Equations (3.6) can be solved uniquely for the constants \\(c_1, \\ldots, c_n\\), provided that the determinant of coefficients is not zero. On the other hand, if the determinant of coefficients is zero, then it is always possible to choose values of \\(y_0, y_0', \\ldots, y_0^{(n-1)}\\) such that Eqs. (3.6) do not have a solution. Hence a necessary and sufficient condition for the existence of a solution of Eqs. (3.6) for arbitrary values of \\(y_0, y_0', \\ldots, y_0^{(n-1)}\\) is that the Wronskian\n\n\\[\nW(y_1, \\ldots, y_n) = \\begin{vmatrix}\n y_1 & y_2 & \\cdots & y_n \\\\\n y_1' & y_2' & \\cdots & y_n' \\\\\n \\vdots & \\vdots & \\ddots & \\vdots \\\\\n y_1^{(n-1)} & y_2^{(n-1)} & \\cdots & y_n^{(n-1)}\n\\end{vmatrix}\n\\] (3.7)\n\nis not zero at \\(t = t_0\\). Since \\(t_0\\) can be any point in the interval \\(I\\), it is necessary and sufficient that \\(W(y_1, y_2, \\ldots, y_n)\\) be nonzero at every point in the interval. It can be shown that if \\(y_1, y_2, \\ldots, y_n\\) are solutions of Eq. (3.4), then \\(W(y_1, y_2, \\ldots, y_n)\\) is either zero for every \\(t\\) in the interval \\(I\\) or else is never zero there. Hence we have the following theorem.\n\n**Theorem 3.2** If the functions \\(p_1, p_2, \\ldots, p_n\\) are continuous on the open interval \\(I\\), if the functions \\(y_1, y_2, \\ldots, y_n\\) are solutions of Eq. (3.4), and if for at least one point in \\(I\\), then every solution of Eq. (3.4) can be expressed as a linear combination of the solutions \\(y_1, y_2, \\ldots, y_n\\).\n\nA set of solutions \\(y_1, y_2, \\ldots, y_n\\) of Eq. (3.4) whose Wronskian is nonzero is referred to as a fundamental set of solutions. Since all solutions of Eq. (3.4) are of the form (3.5), we use the term general solution to refer to an arbitrary linear combination of any fundamental set of solutions of Eq. (3.4).\n\nThe discussion of linear dependence and independence can also be generalized. The functions \\(f_1, f_2, \\ldots, f_n\\) are said to be linearly dependent on \\(I\\) if\nthere exists a set of constants $k_1, k_2, ..., k_n$, not all zero, such that\n\n$$k_1 f_1 + k_2 f_2 + \\cdots + k_n f_n = 0$$ \\hspace{1cm} (3.8)\n\nfor all $t$ in $I$. The functions $f_1, f_2, ..., f_n$ are said to be linearly independent on $I$ if they are not linearly dependent there. If $y_1, ..., y_n$ are solutions of Eq. (4), then it can be shown that a necessary and sufficient condition for them to be linearly independent is that $W(y_1, y_2, ..., y_n)(t_0) \\neq 0$ for some $t_0$ in $I$. Hence a fundamental set of solutions of Eq. (3.4) is linearly independent, and a linearly independent set of $n$ solutions of Eq. (3.4) forms a fundamental set of solutions.\n\n**The Non homogeneous Equation.**\n\nNow consider the non homogeneous equation (3.2),\n\n$$L[y] = y^{(n)} + p_1(t)y^{(n-1)} + \\cdots + p_{n-1}(t)y' + p_n(t)y = g(t).$$\n\nIf $Y_1$ and $Y_2$ are any two solutions of Eq. (3.2), then it follows immediately from the linearity of the operator $L$ that\n\n$$L[Y_1 - Y_2](t) = L[Y_1](t) - L[Y_2](t) = g(t) - g(t) = 0.$$\n\nHence the difference of any two solutions of the non homogeneous equation (3.2) is a solution of the homogeneous equation (3.4). Since any solution of the homogeneous equation can be expressed as a linear combination of a fundamental set of solutions $y_1, y_2, ..., y_n$ it follows that any solution of Eq. (3.2) can be written as\n\n$$y(t) = c_1 y_1(t) + c_2 y_2(t) + \\cdots + c_n y_n(t) + Y(t),$$ \\hspace{1cm} (3.9)\n\nwhere $Y$ is some particular solution of the non homogeneous equation (3.2). The linear combination (3.9) is called the general solution of the non homogeneous equation (3.2). Thus the primary problem is to determine a fundamental set of solutions $y_1, y_2, ..., y_n$, of the homogeneous equation (3.4). If the coefficients are constants, this is a fairly simple problem.\n\n**Homogeneous Equations with Constant Coefficients**\nConsider the nth order linear homogeneous differential equation\n\n\\[ L[y] = a_0 y^{(n)} + a_1 y^{(n-1)} + \\ldots + a_{n-1} y' + a_n y = 0, \\quad (3.10) \\]\n\nwhere \\( a_0, a_1, \\ldots, a_n \\) are real constants. By using the Euler’s method we find a solution of the equations with constant coefficients of the form \\( y = e^{\\lambda t} \\), for suitable values of \\( \\lambda \\). Indeed,\n\n\\[ L[e^{\\lambda t}] = e^{\\lambda t}(a_0 \\lambda^n + a_1 \\lambda^{n-1} + \\ldots + a_{n-1} \\lambda + a_n) = e^{\\lambda t} P(\\lambda) \\quad (3.11) \\]\n\nfor all \\( t \\), where\n\n\\[ P(\\lambda) = a_0 \\lambda^n + a_1 \\lambda^{n-1} + \\ldots + a_{n-1} \\lambda + a_n. \\quad (3.12) \\]\n\nFor those values of \\( \\lambda \\) for which \\( P(\\lambda) = 0 \\), it follows that \\( L[e^{\\lambda t}] = 0 \\) and \\( y = e^{\\lambda t} \\) is a solution of Eq. (3.10). The polynomial \\( P(\\lambda) \\) is called the characteristic polynomial, and the equation \\( P(\\lambda) = 0 \\) is the characteristic equation of the differential equation (3.10). Next we have the next situation:\n\n**Real and Unequal Roots.** If the roots of the characteristic equation are real and no two are equal, then we have \\( n \\) distinct solutions \\( e^{\\lambda_1 t}, e^{\\lambda_2 t}, \\ldots, e^{\\lambda_n t} \\) of Eq. (3.10). If these functions are linearly independent, then the general solution of Eq. (3.10) is\n\n\\[ y = c_1 e^{\\lambda_1 t} + c_2 e^{\\lambda_2 t} + \\ldots + c_n e^{\\lambda_n t}. \\quad (3.13) \\]\n\nOne way to establish the linear independence of \\( e^{\\lambda_1 t}, e^{\\lambda_2 t}, \\ldots, e^{\\lambda_n t} \\) is to evaluate their Wronskian determinant.\n\n**Example** Find the general solution of\n\n\\[ y^{(4)} + y^{(3)} - 7y'' - y' + 6y = 0. \\quad (3.14) \\]\n\nAlso find the solution that satisfies the initial conditions\n\n\\[ y(0) = 1, \\quad y'(0) = 0, \\quad y''(0) = -2, \\quad y'''(0) = -1 \\quad (3.15) \\]\n\nThe characteristic equation of the differential equation (3.16) is the polynomial equation\n\n\\[ r^4 + r^3 - 7r^2 - r + 6 = 0. \\quad (3.16) \\]\n\nThe roots of this equation are \\( r_1 = 1, r_2 = -1, r_3 = 2, r_4 = -3 \\). Therefore the general solution of Eq. (3.15) is\n\n\\[ y = c_1 e^t + c_2 e^{-t} + c_3 e^{2t} + c_4 e^{-3t}. \\quad (3.17) \\]\nThe initial conditions (3.15) require that $c_1, \\ldots, c_4$ satisfy the four equations\n\\[\n\\begin{align*}\n c_1 + c_2 + c_3 + c_4 &= 1, \\\\\n c_1 - c_2 + 2c_3 - 3c_4 &= 0, \\\\\n c_1 + c_2 + 4c_3 + 9c_4 &= -2, \\\\\n c_1 - c_2 + 8c_3 - 27c_4 &= -1.\n\\end{align*}\n\\] (3.18)\n\nBy solving this system of four linear algebraic equations, we find that $c_1 = 11/8, c_2 = 5/12, c_3 = -2/3, c_4 = -1/8$. Therefore the solution of the initial value problem is\n\\[\ny = \\frac{11}{8} e^t + \\frac{5}{12} e^{-t} - 23e^{2t} - 18e^{-3t}.\n\\] (3.19)\n\n**Complex Roots.** If the characteristic equation has complex roots, they must occur in conjugate pairs, $\\lambda \\pm j\\mu$, $j^2 = -1$, since the coefficients $a_0, \\ldots, a_n$ are real numbers. Provided that none of the roots is repeated, the general solution of Eq. (3.10) is still of the form (3.15). However, we can replace the complex-valued solutions $e^{(\\lambda+j\\mu)t}$ and $e^{(\\lambda-j\\mu)t}$ by the real-valued solutions\n\\[\ne^{\\lambda t} \\cos \\mu t, \\quad e^{\\lambda t} \\sin \\mu t\n\\] (3.20)\n\nobtained as the real and imaginary parts of $e^{(\\lambda+j\\mu)t}$. Thus, even though some of the roots of the characteristic equation are complex, it is still possible to express the general solution of Eq. (3.10) as a linear combination of real-valued solutions.\n\n**Example.** Find the general solution of\n\\[\ny^{(iv)} - y = 0.\n\\] (3.21)\n\nAlso find the solution that satisfies the initial conditions\n\\[\ny(0) = 7/2, \\quad y'(0) = -4, \\quad y''(0) = 5/2, y'''(0) = -2\n\\] (3.22)\n\nWe find that the characteristic equation of the differential equation (3.21)\n\\[\nr^4 - 1 = (r^2 - 1)(r^2 + 1) = 0.\n\\]\n\nTherefore the roots are $r = 1, -1, j, -j$, and the general solution of Eq. (3.22)\n\\[\ny = c_1 e^t + c_2 e^{-t} + c_3 \\cos t + c_4 \\sin t.\n\\]\nIf we impose the initial conditions (3.22), we find that $c_1 = 0$, $c_2 = 3$, $c_3 = 1/2$, $c_4 = -1$, thus the solution of the given initial value problem is\n\n$$y = 3e^{-t} + 12 \\cos t - \\sin t. \\quad (3.23)$$\n\nObserve that the initial conditions (3.22) cause the coefficient $c_1$ of the exponentially growing term in the general solution to be zero. Therefore this term is absent in the solution (3.23), which describes an exponential decay to a steady oscillation. However, if the initial conditions are changed slightly, then $c_1$ is likely to be nonzero and the nature of the solution changes enormously. For example, if the first three initial conditions remain the same, but the value of $y'''(0)$ is changed from -2 to $-15/8$, then the solution of the initial value problem becomes\n\n$$y = \\frac{1}{32}e^t + \\frac{95}{32}e^{-t} + \\frac{1}{2} \\cos t - \\frac{17}{16} \\sin t. \\quad (3.24)$$\n\n**Repeated Roots.** If the roots of the characteristic equation are not distinct, that is, if some of the roots are repeated, then the solution (3.13) is clearly not the general solution of Eq. (3.10). Recall that if $r_1$, has multiplicity $s$ (where $s \\leq n$), then $e^{r_1 t}, te^{r_1 t}, t^2 e^{r_1 t}, \\ldots, t^{s-1} e^{r_1 t}$ (18) are corresponding solutions of Eq. (3.10). If a complex root $\\lambda + j\\mu$ is repeated $s$ times, the complex conjugate $\\lambda + j\\mu$ is also repeated $s$ times. Corresponding to these $2s$ complex-valued solutions, we can find $2s$ real-valued solutions by noting that the real and imaginary parts of $e^{(\\lambda+j\\mu)t}, te^{(\\lambda+j\\mu)t}, t^2 e^{(\\lambda+j\\mu)t}, \\ldots, t^{s-1} e^{(\\lambda+j\\mu)t}, e^{(\\lambda-j\\mu)t}, te^{(\\lambda-j\\mu)t}, t^2 e^{(\\lambda-j\\mu)t}, \\ldots, t^{s-1} e^{(\\lambda-j\\mu)t}$, are also linearly independent solutions: $e^{\\lambda t} \\cos \\mu t$, $e^{\\lambda t} \\sin \\mu t$, $te^{\\lambda t} \\cos \\mu t$, $te^{\\lambda t} \\sin \\mu t$, $t^2 e^{\\lambda t} \\cos \\mu t$, $t^2 e^{\\lambda t} \\sin \\mu t$, $\\ldots$, $t^{s-1} e^{\\lambda t} \\cos \\mu t$, $t^{s-1} e^{\\lambda t} \\sin \\mu t$. Hence the general solution of Eq. (3.10) can always be expressed as a linear combination of $n$ real-valued solutions. Consider the following example.\n\n**Example.** Find the general solution of\n\n$$y^{iv} + 2y'' + y = 0. \\quad (3.25)$$\n\nThe characteristic equation is $r^4 + 2r^2 + 1 = (r^2 + 1)^2 = 0$. The roots are $r = j, j, -j, -j$, and the general solution of Eq. (19) is\n\n$$y = c_1 \\cos t + c_2 \\sin t + c_3 t \\cos t + c_4 t \\sin t.$$\nExercises. Find the general solution of the given differential equation.\n\n1. \\( y'' - y' - y + y = 0 \\)\n2. \\( y''' - 3y'' + 3y' + y = 0 \\)\n3. \\( 2y''' - 4y'' - 4y' + 4y = 0 \\)\n4. \\( y''' - y'' - y' + y = 0 \\)\n5. \\( y^{(iv)} - 4y''' + 4y'' = 0 \\)\n6. \\( y^{(iv)} + y = 0 \\)\n7. \\( y^{(iv)} - 5y''' + 4y'' = 0 \\)\n8. \\( y^{(iv)} - 3y''' + 3y'' - y = 0 \\)\n9. \\( y^{(v)} - y'' = 0 \\)\n10. \\( y^{(v)} - 3y^{(iv)} + 3y''' - 3y'' + 2y' = 0 \\)\n11. \\( y^{(iv)} - 8y' = 0 \\)\n12. \\( y^{(6)} + 8y^{(4)} + 16y = 0 \\)\n13. \\( 18y''' + 21y'' + 14y' + 4y = 0 \\)\n14. \\( y^{(4)} - 7y''' + 6y'' + 30y' - 36y = 0 \\)\n15. \\( 12y^{(4)} + 31y''' + 75y'' + 37y' + 5y = 0 \\)\n\nIn each of the given initial value problem, find the particular solution.\n\n1. \\( y'' + y' = 0; \\quad y(0) = 0, \\quad y'(0) = 1, \\quad y''(0) = 2 \\)\n2. \\( y^{(4)} + y = 0; \\quad y(0) = 0, \\quad y'(0) = 0, \\quad y''(0) = -1, \\quad y'''(0) = 0 \\)\n3. \\( y^{(4)} - 4y''' + 4y'' = 0; \\quad y(1) = -1, \\quad y'(1) = 2, \\quad y''(1) = 0, \\quad y'''(1) = 0 \\)\n4. \\( y''' - y'' + y' - y = 0; \\quad y(0) = 2, \\quad y'(0) = -1, \\quad y''(0) = -2 \\)\n5. \\( 2y^{(4)} - y''' - 9y'' + 4y' + 4y = 0; \\quad y(0) = -2, \\quad y'(0) = 0, \\quad y''(0) = -2, \\quad y'''(0) = 0 \\)\n6. \\( 4y''' + y' + 5y = 0; \\quad y(0) = 2, \\quad y'(0) = 1, \\quad y''(0) = -1 \\)\n7. \\( 6y''' + 5y'' + y' = 0; \\quad y(0) = -2, \\quad y'(0) = 2, \\quad y''(0) = 0 \\)\n8. \\( y^{(4)} + 6y''' + 17y'' + 22y' + y = 0; \\quad y(0) = 1, \\quad y'(0) = -2, \\quad y''(0) = 0, \\quad y'''(0) = 0 \\)\n\nThe Method of Undetermined Coefficients\n\nA particular solution \\( y \\) of the non homogeneous nth order linear equation with constant coefficients\n\n\\[\nL[y] = a_0 y^{(n)} + a_1 y^{(n-1)} + \\ldots + a_{n-1} y' + a_n y = g(t)\n\\]\n\n(4.1)\n\ncan be obtained by the method of undetermined coefficients, provided that \\( g(t) \\) is of an appropriate form. While the method of undetermined coefficients is not as general as the method of variation of parameters described in the next section, it is usually much easier to use when applicable.\n\nWhen the constant coefficient linear differential operator \\( L \\) is applied to a polynomial \\( A_0 t^m + A_1 t^{m-1} + \\ldots + A_m \\), an exponential function \\( e^{\\alpha t} \\), a sine function \\( \\sin \\beta t \\), or a cosine function \\( \\cos \\beta t \\), the result is a polynomial, an exponential function, or a linear combination of sine and cosine functions, respectively. Hence, if \\( g(t) \\) is a sum of polynomials, exponentials, sines, and cosines, or products of such functions, we can expect that it is possible\nto find $Y(t)$ by choosing a suitable combination of polynomials, exponentials, and so forth, multiplied by a number of undetermined constants. The constants are then determined so that Eq. (4.1) is satisfied. More exactly if $g(t) = e^{at}(P_n(t) \\cos bt + Q_m(t) \\sin bt)$ and if $r = a + jb$ is a root of the characteristic polynomial of the eq. 4.1, of multiplicity $s$ then it can be determine a particular solution of the form $y(t) = t^s e^{at}(P_u(t) \\cos bt + Q_u(t) \\sin bt)$, where $u = \\max\\{m, n\\}$ and $P_u, Q_u$ are the undetermined polynomials which are determinate if $y(t)$ is a solution of eq.1\n\n**Example** Find the general solution of\n\n$$y''' - 3y'' + 3y' - y = 4e^t. \\quad (4.2)$$\n\nThe characteristic polynomial for the homogeneous equation corresponding to Eq. (4.2) is $P(r) = r^3 - 3r^2 + 3r - 1 = (r - 1)^3$, so the general solution of the homogeneous equation is\n\n$$y_0(t) = c_1 e^t + c_2 te^t + c_3 t^2 e^t. \\quad (4.3)$$\n\nTo find a particular solution $Y(t)$ of Eq. (4.2), we start by assuming that $Y(t) = Ae^t$. However, since $e^t, te^t, t^2 e^t$ are all solutions of the homogeneous equation, we must multiply this initial choice by $t^3$. Thus our final assumption is that $Y(t) = At^3 e^t$, where $A$ is an undetermined coefficient. To find the correct value for $A$, we differentiate $Y(t)$ three times, substitute for $y$ and its derivatives in Eq. (4.2), and collect terms in the resulting equation. In this way we obtain $6Ae^t = 4e^t$. Thus $A = 2/3$ and the particular solution is\n\n$$Y_p(t) = \\frac{2}{3} t^3 e^t. \\quad (4.4)$$\n\nThe general solution of Eq. (4.2) is the sum of $y_0(t)$ from Eq. (4.3) and $Y(t)$ from Eq. (4.4), that is\n\n$$y(t) = Y_0(t) + y_p(t) = c_1 e^t + c_2 te^t + c_3 t^2 e^t + \\frac{2}{3} t^3 e^t.$$\n\n**Example** Find a particular solution of the equation\n\n$$y^{(4)} + 2y'' + y = 3\\sin t - 5\\cos t. \\quad (4.5)$$\n\nThe characteristic polynomial for the homogeneous equation corresponding to eq. 4.5 is $r^4 + 2r^2 + 1 = (r^2 + 1)^2$, and the general solution of the\nhomogeneous equation is\n\\[ y_0(t) = c_1 \\cos t + c_2 \\sin t + c_3 t \\cos t + c_4 t \\sin t, \\quad (4.6) \\]\ncorresponding to the roots \\( r = j, j, -j, -j \\) of the characteristic equation. Our initial assumption for a particular solution is \\( y_p(t) = A \\sin t + B \\cos t \\), but we must multiply this choice by \\( t^2 \\) to make it different from all solutions of the homogeneous equation. Thus our final assumption is\n\\[ y_p(t) = At^2 \\sin t + Bt^2 \\cos t. \\]\nNext, we differentiate \\( y_p(t) \\) four times, substitute into the differential equation (4.5), and collect terms, obtaining finally \\(-8A \\sin t - 8B \\cos t = 3 \\sin t - 5 \\cos t \\). Thus \\( A = -38, B = 58 \\), and the particular solution of Eq. (4.5) is\n\\[ y_p(t) = -38t^2 \\sin t + 58t^2 \\cos t. \\quad (4.7) \\]\nThe general solution of Eq. (4.5) is the sum of \\( y_0(t) \\) from Eq. (4.6) and \\( Y(t) \\) from Eq. (4.6), that is\n\\[ y(t) = Y_0(t) + y_p(t) = c_1 \\cos t + c_2 \\sin t + c_3 t \\cos t + c_4 t \\sin t - 38t^2 \\sin t + 58t^2 \\cos t. \\]\nIf \\( g(t) \\) is a sum of several terms, it is often easier in practice to compute separately the particular solution corresponding to each term in \\( g(t) \\). As for the second order equation, the particular solution of the complete problem is the sum of the particular solutions of the individual component problems. This is illustrated in the following example.\n\n**Example** Find a particular solution of\n\\[ y''' - 4y' = t + 3 \\cos t + e^{-2t}. \\quad (4.8) \\]\nFirst we solve the homogeneous equation. The characteristic equation is \\( r^3 - 4r = 0 \\), and the roots are 0, \\( \\pm 2 \\), hence\n\\[ y_0(t) = c_1 + c_2 e^{2t} + c_3 e^{-2t}. \\]\nWe can write a particular solution of Eq. (4.8) as the sum of particular solutions of the differential equations\n\\[ y''' - 4y' = t, \\quad y''' - 4y' = 3 \\cos t, \\quad y''' - 4y' = e^{-2t}. \\]\nOur initial choice for a particular solution \\( y_1(t) \\) of the first equation is \\( A_0 t + A_1 \\), but since a constant is a solution of the homogeneous equation, we\nmultiply by \\( t \\). Thus \\( y_1(t) = t(A_0 t + A_1) \\). For the second equation we choose \\( y_2(t) = B \\cos t + C \\sin t \\), and there is no need to modify this initial choice since \\( \\cos t \\) and \\( \\sin t \\) are not solutions of the homogeneous equation. Finally, for the third equation, since \\( e^{-2t} \\) is a solution of the homogeneous equation, we assume that \\( y_3(t) = E t e^{-2t} \\). The constants are determined by substituting into the individual differential equations; they are \\( A_0 = -\\frac{1}{8}, \\ A_1 = 0, \\ B = 0, \\ C = -\\frac{3}{8}, \\ E = \\frac{1}{8} \\). Hence a particular solution of Eq. (4.8) is\n\n\\[\ny_p(t) = -\\frac{1}{8} t^2 - \\frac{3}{5} \\sin t + \\frac{1}{8} t e^{-2t}.\n\\]\n\n**Exercises** In each of the next problems determine the general solution of the given differential equation.\n\n1. \\( y''' - y'' - y' + y = 2e^{-t} + 3 \\)\n2. \\( y^{(4)} - y = 3t + \\cos t \\)\n3. \\( y''' + y'' + y' + y = e^{-t} + 4t \\)\n4. \\( y''' - y' = 2 \\sin t \\)\n5. \\( y^{(4)} - 4y'' = t^2 + e^t \\)\n6. \\( y^{(4)} + 2y'' + y = 3 + \\cos 2t \\)\n7. \\( y^{(4)} + y^{(3)} = t \\)\n8. \\( y^{(4)} + y^{(3)} = \\sin 2t \\)\n\nIn each of Problems which follows find the solution of the given initial value problem.\n\n9. \\( y^{(3)} + 4y' = t, \\ y(0) = y'(0) = 0, \\ y''(0) = 1 \\)\n10. \\( y^{(4)} + 2y'' + y = 3t + 4, \\ y(0) = y'(0) = 0, \\ y''(0) = y^{(3)}(0) = 1 \\)\n11. \\( y^{(3)} - 3y'' + 2y' = t + e^t, \\ y(0) = 1, \\ y'(0) = -1/4, \\ y''(0) = -32 \\)\n12. \\( y^{(4)} + 2y''' + y'' + 8y' - 12y = 12 \\sin t - e^{-t}, \\ y(0) = 3, \\ y'(0) = 0, \\ y''(0) = -1, \\ y'''(0) = 2 \\)\n\n**The Method of Undetermined Coefficients** In each of Problems determine a suitable form for \\( y(t) \\) if the method of undetermined coefficients is to be used. Do not evaluate the constants.\n\n13. \\( y^{(3)} - 2y'' + y' = t^3 + 2e^t \\)\n14. \\( y^{(3)} - y' = te^{-t} + 2 \\cos t \\)\n15. \\( y^{(4)} - 2y'' + y' = e^t + \\sin t \\)\n16. \\( y^{(4)} + 4y'' = \\sin 2t + te^t + 4 \\)\n17. \\( y^{(4)} - y^{(3)} - y'' + y' = t^2 + 4 + t \\sin t \\)\n18. \\( y^{(4)} + 2y'' = 3e^t + 2te^{-t} + e^{-t} \\sin t \\)\n\n**The Method of Variation of Parameters**\n\nThe method of variation of parameters for determining a particular solution of the non homogeneous \\( n \\)-th order linear differential equation\n\n\\[\nL[y] = y^{(n)} + p_1(t)y^{(n-1)} + \\cdots + p_{n-1}(t)y' + p_n(t)y = g(t) \\quad (4.9)\n\\]\n\nis known as the Lagrange method. As before, to use the method of variation of parameters, it is first necessary to solve the corresponding homogeneous\ndifferential equation. In general, this may be difficult unless the coefficients are constants. However, the method of variation of parameters is still more general than the method of undetermined coefficients in that it leads to an expression for the particular solution for any continuous function \\( g \\), whereas the method of undetermined coefficients is restricted in practice to a limited class of functions \\( g \\). Suppose then that we know a fundamental set of solutions \\( y_1, y_2, \\ldots, y_n \\) of the homogeneous equation. Then the general solution of the homogeneous equation is\n\n\\[\ny_0(t) = c_1 y_1(t) + c_2 y_2(t) + \\cdots + c_n y_n(t).\n\\] (4.10)\n\nThe method of variation of parameters for determining a particular solution of Eq. (4.9) rests on the possibility of determining \\( n \\) functions \\( u_1, u_2, \\ldots, u_n \\) such that \\( y(t) \\) is of the form\n\n\\[\ny(t) = u_1(t)y_1(t) + u_2(t)y_2(t) + \\cdots + u_n(t)y_n(t).\n\\] (4.11)\n\nSince we have \\( n \\) functions to determine, we will have to specify \\( n \\) conditions. One of these is clearly that \\( y \\) satisfy Eq. (4.9). The other \\( n - 1 \\) conditions are chosen so as to make the calculations as simple as possible. Since we can hardly expect a simplification in determining \\( y \\) if we must solve high order differential equations for \\( u_1, \\ldots, u_n \\), it is natural to impose conditions to suppress the terms that lead to higher derivatives of \\( u_1, \\ldots, u_n \\). From Eq. (4.11) we obtain\n\n\\[\ny' = (u_1 y_1' + u_2 y_2' + \\cdots + u_n y_n') + (u_1' y_1 + u_2' y_2 + \\cdots + u_n' y_n),\n\\] (4.12)\n\nwhere we have omitted the independent variable \\( t \\) on which each function in Eq. (4.12) depends. Thus the first condition that we impose is that\n\n\\[\nu_1' y_1 + u_2' y_2 + \\cdots + u_n' y_n = 0.\n\\] (4.13)\n\nContinuing this process in a similar manner through \\( n - 1 \\) derivatives of \\( Y \\) gives\n\n\\[\ny^{(m)} = u_1 y_1^{(m)} + u_2 y_2^{(m)} + \\cdots + u_n y_n^{(m)}, \\quad m = 0, 1, 2, \\ldots, n - 1,\n\\] (4.14)\n\nand the following \\( n - 1 \\) conditions on the functions \\( u_1, \\ldots, u_n \\),\n\n\\[\nu_1' y_1^{(m-1)} + u_2' y_2^{(m-1)} + \\cdots + u_n' y_n^{(m-1)} = 0, \\quad m = 1, 2, \\ldots, n - 1.\n\\] (4.15)\nThe $n$-th derivative of $y$ is\n\n$$y^{(n)} = (u_1 y_1^{(n)} + \\cdots + u_n y_n^{(n)}) + (u_1' y_1^{(n-1)} + \\cdots + u_n' y_n^{(n-1)}). \\quad (4.16)$$\n\nFinally, we impose the condition that $y$ must be a solution of Eq. (4.9). On substituting for the derivatives of $y$ from Eqs. (4.14) and (4.16), collecting terms, and making use of the fact that $L[y_i] = 0, i = 1, 2, \\cdots, n$, we obtain\n\n$$u_1' y_1^{(n-1)} + u_2' y_2^{(n-1)} + \\cdots + u_n' y_n^{(n-1)} = g. \\quad (4.17)$$\n\nEquation (4.17), coupled with the $n - 1$ equations (4.15), gives $n$ simultaneous linear non homogeneous algebraic equations for $u_1', u_2', \\cdots, u_n'$:\n\n$$\\begin{cases}\n u_1' y_1 + u_2' y_2 + \\cdots + u_n' y_n = 0 \\\\\n u_1' y'' + u_2' y'' + \\cdots + u_n' y'' = 0 \\\\\n u_1' y_1^{(3)} + u_2' y_2^{(3)} + \\cdots + u_n' y_n^{(3)} = 0 \\\\\n \\vdots \\\\\n u_1' y_1^{(n-1)} + u_2' y_2^{(n-1)} + \\cdots + u_n' y_n^{(n-1)} = g\n\\end{cases} \\quad (4.18)$$\n\nThe system (4.18) is a linear algebraic system for the unknown quantities $u_1', u_2', \\cdots, u_n'$. By solving this system and then integrating the resulting expressions, you can obtain the coefficients $u_1, \\cdots, u_n$. A sufficient condition for the existence of a solution of the system of equations (4.18) is that the determinant of coefficients is nonzero for each value of $t$. However, the determinant of coefficients is precisely $W(y_1, y_2, \\ldots, y_n)$, and it is nowhere zero since $y_1, \\ldots, y_n$ are linearly independent solutions of the homogeneous equation. Hence it is possible to determine $u_1', u_2', \\cdots, u_n'$. Using Cramers rule, we find that the solution of the system of equations (4.18).\n\n**Remark.** If when we determine, by integrating, the functions $u_i$ it is considered and the additional arbitrary constants, then we obtain the general solution of non homogeneous equation (4.9).\n\n**Example** Find the general solution for the next equation\n\n$$y^{(3)} - y'' - y' + y = 4e^t, \\quad (4.19)$$\n\nFirst we determine the general solution for the homogeneous equation corresponding to the eq.(4.19)\n\n$$y^{(3)} - y'' - y' + y = 0 \\quad (4.20)$$\nThe characteristic algebraic equation is \\( r^3 - r^2 - r + 1 = 0 \\), and its roots are \\( r_1 = -1, \\ r_2 = r_3 = 1 \\). The fundamental system of solutions is \\( y_1(t) = e^t, \\ y_2(t) = te^t, \\ y_3(t) = e^{-t} \\). The general solution of eq.(4.20) is\n\n\\[\ny_0(t) = c_1e^t + c_2te^t + c_3e^{-t}.\n\\]\n\nNext we find the particular (general) solution of eq (4.19) of the form\n\n\\[\ny_p(t) = u_1(t)e^t + u_2(t)te^t + u_3(t)e^{-t}, \\tag{4.21}\n\\]\n\nwhere the unknown functions \\( u_i \\) are solutions of the algebraic system\n\n\\[\n\\begin{align*}\nu_1'(t)e^t + u_2'(t)te^t + u_3'(t)e^{-t} &= 0 \\\\\nu_1'(t)e^t + u_2'(t)(t+1)e^t - u_3'(t)e^{-t} &= 0 \\\\\nu_1'(t)e^t + u_2'(t)(t+2)e^t + u_3'(t)e^{-t} &= 4e^t.\n\\end{align*}\n\\]\n\nBy solving this system we find\n\n\\[\nu_1'(t) = -2t - 1, \\quad u_2'(t) = 2, \\quad u_3'(t) = e^{2t}.\n\\]\n\nNow we integrate the resulting expressions and obtain the coefficients\n\n\\[\nu_1(t) = -\\int (2t+1)dt = -t^2 - t + c_1, \\quad u_2(t) = 2t + c_2, \\quad u_3(t) = \\frac{1}{2}e^{2t} + c_3.\n\\]\n\nNow from eq.(4.21) we obtain the general solution of non homogeneous equation (4.19)\n\n\\[\ny(t) = (-t^2 - t + c_1)e^t + (2t + c_2)te^t + \\frac{1}{2}e^{2t} + c_3)e^{-t} = c_1e^t + c_2te^t + c_3e^{-t} + (t^2 - t + 3/2)e^t.\n\\]\n\n**Exercises** In each of the next problems use the method of variation of parameters to determine the general solution of the given differential equation.\n\n1. \\( y''' + y' = \\tan t, \\ 0 < t < \\pi \\)\n2. \\( y''' - y' = t \\)\n3. \\( y''' - 2y'' + y' + 2y = e^{4t} \\)\n4. \\( y''' + y' = \\sec t, \\ -\\frac{\\pi}{2} < t < \\frac{\\pi}{2} \\)\n5. \\( y''' - y'' + y' - y = e^{-t}\\sin t \\)\n6. \\( y^{(4)} + 2y'' + y = \\sin t \\)\n\nFind the solution of the given initial value problem.\n\n7. \\( y''' + y' = \\sec t, \\ y(0) = 2, \\ y'(0) = 1, \\ y''(0) = .2 \\)\n8. \\( y^{(4)} + 2y'' + y = \\sin t, \\ y(0) = 2, \\ y'(0) = 0, \\ y''(0) = -1, \\ y'''(0) = 1 \\)\n9. \\( y''' - y'' + y' - y = \\sec t, \\ y(0) = 2, \\ y'(0) = -1, \\ y''(0) = 1 \\)\n10. \\( y''' - y' = \\csc t, \\ y(\\pi/2) = 2, \\ y'(\\pi/2) = 1, \\ y''(\\pi/2) = -1 \\)", "id": "./materials/163.pdf" }, { "contents": "Circle and spherical surface\n\nCircle\n\nA circumference is a two-dimensional shape made by drawing a curve that is the same distance all around from the center.\n\nThe circle centered in \\( C = (c_1, c_2) \\) with radius \\( r \\) is the set of points \\( P = (x, y) \\) (the locus) that are distant from \\( C \\) the measure \\( r \\), that is,\n\n\\[\n||\\overrightarrow{CP}|| = r \\iff (x - c_1)^2 + (y - c_2)^2 = r^2.\n\\]\n\nThe distance between the midpoint and the circle border is called the radius.\n\nExample: Let us consider, on the Cartesian plane, the circle that contains points \\( A = (-1, 4) \\) and \\( B(3, 1) \\) and whose diameter measures \\( AB = 5 \\). Then the midpoint of \\( [AB] \\), \\( M = (1, \\frac{5}{2}) \\), corresponds to the center of the circle and the radius is equal to \\( \\frac{AB}{2} = \\frac{5}{2} \\). Thus, the cartesian equation for this circle is as follows:\n\n\\[\n(x - 1)^2 + (y - \\frac{5}{2})^2 = \\frac{25}{4}.\n\\]\n\nSpherical surface\n\nA Spherical surface is a three-dimensional shape where any of its points is at the same distance from a fixed point, called the center of the spherical surface.\n\nThe Spherical surface centered in \\( C = (c_1, c_2, c_3) \\) with radius \\( r \\) is the set of points \\( P = (x, y, z) \\) (the locus) that are distant from \\( C \\) the measure \\( r \\), that is,\n\n\\[\n||\\overrightarrow{CP}|| = r \\iff (x - c_1)^2 + (y - c_2)^2 + (z - c_3)^2 = r^2.\n\\]", "id": "./materials/164.pdf" }, { "contents": "DUALITY THEORY\n\nC. B. Vaz\nInstituto Politécnico de Bragança\nGiven the standard form for the primal problem at the left, its dual problem has the form shown at the right.\n\n**Primal Problem**\n\nMax \\( Z = \\sum_{j=1}^{n} c_j x_j \\)\n\nsubject to\n\n\\[ \\sum_{j=1}^{n} a_{ij} x_j \\leq b_i, \\quad i = 1, \\ldots, m \\]\n\n\\[ x_j \\geq 0, \\quad j = 1, \\ldots, n. \\]\n\n**Dual Problem**\n\nMin \\( W = \\sum_{i=1}^{m} b_i y_i \\)\n\nsubject to\n\n\\[ \\sum_{i=1}^{m} a_{ij} y_i \\geq c_j, \\quad j = 1, \\ldots, n \\]\n\n\\[ y_i \\geq 0, \\quad i = 1, \\ldots, m. \\]\n\nThe primal problem is in maximization form while the dual problem is in minimization form.\nMoreover, the dual problem uses the same parameters as the primal problem, but in different positions, as described below:\n\n- The coefficients in the objective function $c_j, j = 1, \\ldots, n$ of the primal problem are the right-hand sides of the functional constraints in the dual problem.\n- The right-hand sides of the functional constraints $b_i, i = 1, \\ldots, m$ in the primal problem are the coefficients in the objective function of the dual problem.\n- The coefficient matrix of the functional constraints of the dual problem is the transpose of the coefficient matrix of the functional constraints of the primal problem.\nTo highlight the comparison between the primal and the dual problems, the matrix notation can be used to define them. Set \\( c = [c_1 \\ c_2 \\ \\ldots \\ c_n] \\) and \\( y = [y_1 \\ y_2 \\ \\ldots \\ y_m] \\) the row vectors, considering that \\( b, x \\) and \\( 0 \\) are column vectors while \\( A \\) is a matrix, as described below:\n\n\\[\nx = \\begin{bmatrix}\nx_1 \\\\\nx_2 \\\\\n\\vdots \\\\\nx_n\n\\end{bmatrix}, \\quad b = \\begin{bmatrix}\nb_1 \\\\\nb_2 \\\\\n\\vdots \\\\\nb_m\n\\end{bmatrix}, \\quad 0 = \\begin{bmatrix}\n0 \\\\\n0 \\\\\n\\vdots \\\\\n0\n\\end{bmatrix}\n\\]\n\n\\[\nA = \\begin{bmatrix}\na_{11} & a_{12} & \\ldots & a_{1n} \\\\\na_{21} & a_{22} & \\ldots & a_{2n} \\\\\n\\vdots & \\vdots & \\ddots & \\vdots \\\\\na_{m1} & a_{m2} & \\ldots & a_{mn}\n\\end{bmatrix}\n\\]\nMatrix notation\n\nPrimal Problem\n\nMaximize $Z = cx$\nsubject to\n$Ax \\leq b$\n$x \\geq 0.$\n\nDual Problem\n\nMinimize $W = yb$\nsubject to\n$A^T y \\geq c$\n$y \\geq 0.$\nExample\n\nPrimal Problem\n\nMax $Z = 8x_1 + 5x_2$\n\nsubject to\n\n$3x_1 + 6x_2 \\leq 20$\n\n$5x_1 + 2x_2 \\leq 40$\n\n$x_2 \\leq 60$\n\n$x_1, x_2 \\geq 0.$\n\nDual Problem\n\nMin $W = 20y_1 + 40y_2 + 60y_3$\n\nsubject to\n\n$3y_1 + 5y_2 \\geq 8$\n\n$6y_1 + 2y_2 + y_3 \\geq 5$\n\n$y_1, y_2, y_3 \\geq 0.$\nReference\n\nHillier F. S., & Lieberman G.R. (2010). Introduction to operations research (9th ed.). New York: McGraw-Hill.", "id": "./materials/165.pdf" }, { "contents": "LINEAR PROGRAMMING (LP): Formulation\n\nC. B. Vaz\nInstituto Politécnico de Bragança\n\"The development of linear programming (LP) has been ranked among the most important scientific advances of the mid-20th century\" (Hillier & Lieberman, 2010)\n\nToday, the LP is a standard tool that enables to save many thousands or millions of euros for many companies or businesses in the various industrialized countries of the world.\n\nBriefly, the most common type of application of LP involves selecting the optimum level of certain activities that compete for scarce resources that are necessary to perform those activities.\n\nThe large diversity of situations in which the LP applies:\n- allocation of production facilities to products\n- allocation of national resources to domestic needs\n- portfolio selection\n- selection of shipping patterns\n- agricultural planning\n- design of radiation therapy, ...\nLP uses a mathematical model to describe the problem to solve:\n\n- The adjective linear means that all the mathematical functions in this model are required to be linear functions.\n- The word programming is essentially a synonym for planning.\n\nThus, the LP involves the planning of activities to obtain an optimal solution among all feasible alternatives.\nSo, a LP problem includes the following components:\n\n- Decision variables which are the quantities to be determined.\n- Constraints which define the admissible (and not admissible) values for the decision variables, being defined by the various resources available and the technical limitations of the problem; the optimal solution will have to respect them.\n- Objective function defines the evaluation criteria for the various admissible solutions which should be minimized or maximized.\nConsider the LP problem to allocate $m$ resources to $n$ activities:\n\n- The amount available of each resource $i$ is limited by $b_i$ ($i = 1, \\ldots, m$), being necessary a careful allocation of resources to activities $j$. Set $a_{ij}$ the amount of resource $i$ consumed by each unit of activity $j$.\n\n- To achieve the best possible value of the overall measure of performance $Z$, it is necessary to chose the levels of the activities $x_j$ for $j = 1, \\ldots, n$. Set $c_j$ the increase in $Z$ that would result from each unit increase in level of activity $j$.\n\nThus, the standard formulation of LP model is the following:\n\n$$\\text{Max } Z = \\sum_{j=1}^{n} c_j x_j$$\n\nsubject to\n\n$$\\sum_{j=1}^{n} a_{ij} x_j \\leq b_i, \\ i = 1, \\ldots, m$$\n\n$$x_j \\geq 0, \\ j = 1, \\ldots, n.$$\nComponents of this standard formulation of LP model:\n\n- **Objective function**: \\( Z = \\sum_{j=1}^{n} c_j x_j \\) which should be maximized.\n\n- **Functional constraints** (or structural constraints):\n \\( \\sum_{j=1}^{n} a_{ij} x_j \\leq b_i, \\ i = 1, \\ldots, m \\). The restrictions normally are referred as constraints.\n\n- **Nonnegativity constraints** (or nonnegativity conditions):\n \\( x_j \\geq 0, \\ j = 1, \\ldots, n \\).\nAssumptions of LP:\n\n- **Proportionality**: it is an assumption about both the objective function and the functional constraints.\n\n- **Additivity**: every function in a LP model (the objective function or the function on the left-hand side of a functional constraint) is the sum of the individual contributions of the respective activities.\n\n- **Divisibility**: the decision variables in a LP model are allowed to have any values, including noninteger values, that satisfy the functional and nonnegativity constraints.\n\n- **Certainty**: the value assigned to each parameter of a LP model is assumed to be a known constant, being important to conduct the sensitivity analysis after an optimum solution is found.\nReference\n\nHillier F. S., & Lieberman G.R. (2010). Introduction to operations research (9th ed.). New York: McGraw-Hill.", "id": "./materials/166.pdf" }, { "contents": "LINEAR PROGRAMMING (LP): Formulation example\n\nC. B. Vaz\nInstituto Politécnico de Bragança\nExample of LP model:\n\nA company produces two products (A and B), using the resources R1 and R2. The time required for the production of each product is shown in the table (in hours):\n\n| | R1 | R2 |\n|-------|----|----|\n| Product A | 2 | 4 |\n| Product B | 4 | 4 |\n\nThe capacity (monthly availability) of the resources R1 and R2 is 720 hours and 880 hours, respectively. It is known that the maximum sales of the product A are limited to 160 monthly units.\n\nThe profit obtained from the sales of these products is 6€ per unit of product A and 3€ per unit of product B.\n\nDetermine the monthly production plan in order to maximize the profit obtained.\nFormulation of LP model\n\n1st: Choice the decision variables: The company produces products A and B. How many units of these products the company should produce per month to maximize the profit?\n ▶ Number of units of product A to be produced per month: $x_1$\n ▶ Number of units of product B to be produced per month: $x_2$\n\n2nd: Nonnegativity constraints: $x_1, x_2 \\geq 0$. \n2nd: Functional constraints:\n\n- **Resource 1:** \\(2x_1 + 4x_2 \\leq 720\\), where:\n - \\(2x_1\\) is the number of hours used to produce the product A in the resource R1;\n - \\(4x_2\\) is the number of hours used to produce the product B in the resource R1;\n - 720 is the number of hours available in the resource R1 per month.\n\n- **Resource 2:** \\(4x_1 + 4x_2 \\leq 880\\), where:\n - \\(4x_1\\) is the number of hours used to produce the product A in the resource R2;\n - \\(4x_2\\) is the number of hours used to produce the product B in the resource R2;\n - 880 is the number of hours available in the resource R2 per month.\n\n- **Market:** \\(x_1 \\leq 160\\) represents the number of units of the product A that are absorbed by the market.\n3rd: **Objective function** $= Z$: It is necessary to know the company’s monthly profit to solve a maximization problem.\n\n- $Z = 6x_1 + 3x_2$, where:\n - $6x_1$ is the profit (in €) obtained from the sale of $x_1$ units of product A;\n - $3x_2$ is the profit (in €) obtained from the sale of $x_2$ units of product B;\n\nThe standard formulation of this LP model is the following:\n\n$$\\text{Max } Z = 6x_1 + 3x_2$$\n\nsubject to\n\n- $2x_1 + 4x_2 \\leq 720$\n- $4x_1 + 4x_2 \\leq 880$\n- $x_1 \\leq 160$\n- $x_1, x_2 \\geq 0$. \n\n5\nReference\n\nHillier F. S., & Lieberman G.R. (2010). Introduction to operations research (9th ed.). New York: McGraw-Hill.", "id": "./materials/167.pdf" }, { "contents": "LINEAR PROGRAMMING (LP): Excel Solver Add-in\n\nC. B. Vaz\nInstituto Politécnico de Bragança\nThe Solver is an Add-in of Excel that is necessary to install:\n\n- Select File > Options;\n- Manage Excel Add-ins > Go;\n- Select the Solver Add-in.\n- After the Solver Add-in is available in Data.\nSolving a LP model using Excel Solver Add-in:\n\nConsider the following linear programming (LP) model to determine the production plan of three products (1, 2 and 3), using 110, 150 and 200 hours (h) in machines M1, M2 and M3, respectively.\n\nMax \\( Z = 3x_1 + 3x_2 + 2x_3 \\)\n\nsubject to\n\n\\[\n\\begin{align*}\n2x_1 + 3x_2 + 4x_3 & \\leq 110 \\\\\n3x_1 + 2x_2 + 3x_3 & \\leq 150 \\\\\n4x_1 + 2x_2 + 3x_3 & \\leq 200 \\\\\nx_1, x_2, x_3 & \\geq 0.\n\\end{align*}\n\\]\n\nSet the decision variables \\( x_1, x_2 \\) and \\( x_3 \\) which are the number of units produced of A, B and C products, respectively.\n1st step: To define the model in the workbook, including:\n\n- Decision variables will be determined through the cells B4 to D4 after run the Solver;\n- Coefficients of the objective function: insert them in the cells B8 to D8.\n- The matrix constraints coefficients: insert them in the cells B12 to D14; the constants of the right side of the constraints: insert them in the cells G12 to G14.\n2nd step: To define the formulas of the objective function and the left side of each constraint:\n\n- **Objective function (cell E8)** - this formula relates the coefficients of the objective function to the decision variables \\((B8\\times B4 + C8\\times C4 + D8\\times D4)\\), using the \"SUMPRODUCT\" function.\n\n- **Left side of each constraint (cells E12, E13 and E14)** - this formula relates the coefficients of each constraint to the decision variables (in E12 is \\(B12\\times B4 + C12\\times C4 + D12\\times D4\\)).\n3rd step: To open the Solver in Data to insert:\n\n- **Objective function**: select the cell E8\n- **By Changing Variable Cells**: select the cells B4 to D4\n- **Add for each constraint**: selecting the cell on the left side, the sign and the cell on right side. In the example, for the 1st to 3rd constraints: select the cells E12 to E14, the sign ≤ and the cells G12 to G14.\n- **Select a solving method**: chose the Simplex LP\n\n![Solver Parameters](image)\n4th step: To obtain the optimum solution: the optimum profit is 156€ which is achieved if the company produces 46 units of A and 6 units of B. There are 4 hours available in the machine M3.\n5th step: To obtain the Answer and the Sensitivity reports\n### Answer Report\n\n#### Objective Cell (Max)\n\n| Cell | Name | Original Value | Final Value |\n|--------|------|----------------|-------------|\n| $E$8 | Z | 0 | 156 |\n\n#### Variable Cells\n\n| Cell | Name | Original Value | Final Value | Integer |\n|--------|------|----------------|-------------|---------|\n| $B$4 | x1 | 0 | 46 | Contin |\n| $C$4 | x2 | 0 | 6 | Contin |\n| $D$4 | x3 | 0 | 0 | Contin |\n\n#### Constraints\n\n| Cell | Name | Cell Value | Formula | Status | Slack |\n|--------|-----------------------|------------|---------------|------------|-------|\n| $E$12 | Hours used in M1 | 110 | $E$12<=$G$12 | Binding | 0 |\n| $E$13 | Hours used in M2 | 150 | $E$13<=$G$13 | Binding | 0 |\n| $E$14 | Hours used in M3 | 196 | $E$14<=$G$14 | Not Binding| 4 |\n## Sensitivity Report\n\n### Variable Cells\n\n| Cell | Name | Final Value | Reduced Cost | Objective Coefficient | Allowable Increase | Allowable Decrease |\n|-------|------|-------------|--------------|------------------------|--------------------|--------------------|\n| $B$4 | $x_1$ | 46 | 0 | 3 | 1,5 | 1 |\n| $C$4 | $x_2$ | 6 | 0 | 3 | 1,5 | 1 |\n| $D$4 | $x_3$ | 0 | -2,2 | 2 | 2,2 | 1E+30 |\n\n### Constraints\n\n| Cell | Name | Final Value | Shadow Price | Constraint R.H. Side | Allowable Increase | Allowable Decrease |\n|-------|-----------------------|-------------|--------------|----------------------|--------------------|--------------------|\n| $E$12 | Hours used in M1 | 110 | 0,6 | 110 | 115 | 10 |\n| $E$13 | Hours used in M2 | 150 | 0,6 | 150 | 2,5 | 76,666666667 |\n| $E$14 | Hours used in M3 | 196 | 0 | 200 | 1E+30 | 4 |\nReference\n\nHillier F. S., & Lieberman G.R. (2010). Introduction to operations research (9th ed.). New York: McGraw-Hill.", "id": "./materials/168.pdf" }, { "contents": "LINEAR PROGRAMMING: Sensitivity analysis example\n\nC. B. Vaz\nInstituto Politécnico de Bragança\nExample of LP model:\n\nA company produces three products (1, 2 and 3), using 110, 150 and 200 hours (h) in machines M1, M2 and M3, respectively, according to the following linear programming (LP) model:\n\n\\[\n\\text{Max } Z = 3x_1 + 3x_2 + 2x_3\n\\]\n\nsubject to\n\n\\[\n\\begin{align*}\n2x_1 + 3x_2 + 4x_3 & \\leq 110 \\\\\n3x_1 + 2x_2 + 3x_3 & \\leq 150 \\\\\n4x_1 + 2x_2 + 3x_3 & \\leq 200 \\\\\nx_1, x_2, x_3 & \\geq 0.\n\\end{align*}\n\\]\n\nThe decision variables \\(x_1, x_2\\) and \\(x_3\\) represent the units produced of A, B and C products, respectively.\nThe *Answer Report* provided by the Excel Solver Add-in indicates that the optimum profit is 156€ which is achieved if the company produces 46 units of product A and 6 units of product B. There are 4 hours available in the machine M3.\n\n| Objective Cell (Max) |\n|----------------------|\n| **Cell** | **Name** | **Original Value** | **Final Value** |\n| $E$8 | Z | 0 | 156 |\n\n| Variable Cells |\n|----------------|\n| **Cell** | **Name** | **Original Value** | **Final Value** | **Integer** |\n| $B$4 | x1 | 0 | 46 | Contin |\n| $C$4 | x2 | 0 | 6 | Contin |\n| $D$4 | x3 | 0 | 0 | Contin |\n\n| Constraints |\n|--------------|\n| **Cell** | **Name** | **Cell Value** | **Formula** | **Status** | **Slack** |\n| $E$12 | Hours used in M1 | 110 | $E$12<=$G$12 | Binding | 0 |\n| $E$13 | Hours used in M2 | 150 | $E$13<=$G$13 | Binding | 0 |\n| $E$14 | Hours used in M3 | 196 | $E$14<=$G$14 | Not Binding | 4 |\nOptimum solutions of the primal and dual problems from the Answer and Sensitivity Reports:\n\n### Sensitivity Report\n\n#### Variable Cells\n\n| Cell | Name | Final Value | Reduced Cost | Objective Coefficient | Allowable Increase | Allowable Decrease |\n|--------|------|-------------|--------------|------------------------|--------------------|--------------------|\n| $B$4 | $x_1$| 46 | 0 | 3 | 1,5 | 1 |\n| $C$4 | $x_2$| 6 | 0 | 3 | 1,5 | 1 |\n| $D$4 | $x_3$| 0 | -2.2 | 2 | 2,2 | 1E+30 |\n\n#### Constraints\n\n| Cell | Name | Final Value | Shadow Price | Constraint R.H. Side | Allowable Increase | Allowable Decrease |\n|--------|--------------------|-------------|--------------|----------------------|--------------------|--------------------|\n| $E$12 | Hours used in M1 | 110 | 0.6 | 110 | 115 | 10 |\n| $E$13 | Hours used in M2 | 150 | 0.6 | 150 | 2.5 | 76,666,666,667 |\n| $E$14 | Hours used in M3 | 196 | 0 | 200 | 1E+30 | 4 |\n\n### Primal problem\n\n| Primal problem | $x_1^*$ | $x_2^*$ | $x_3^*$ | $x_4^*$ | $x_5^*$ | $x_6^*$ |\n|----------------|---------|---------|---------|---------|---------|---------|\n| Primal Solution| 46 | 6 | 0 | 0 | 0 | 4 |\n\n### Dual problem\n\n| Dual problem | $y_4^*$ | $y_5^*$ | $y_6^*$ | $y_1^*$ | $y_2^*$ | $y_3^*$ |\n|--------------|---------|---------|---------|---------|---------|---------|\n| Dual Solution| 0 | 0 | 2.2 | 0.6 | 0.6 | 0 |\nEconomic interpretation from the Sensitivity Report:\n\nReduced costs of the products produced:\n\nAs the products A and B are produced (46 units and 6 units, respectively), their reduced costs (1st table) are equal to 0. As the product C is not produced, it has a reduced cost of 2.2, indicating that for each unit produced of the product C there will be a reduction of 2.2€ in the optimum profit obtained.\n\nShadow prices of the available resources:\n\nThe shadow price (2nd table) of the machine M3 is 0 due to it available hours (only 196 h were used from the 200 h available). The shadow prices of the machines M2 and M3 are both equal to 0.6, indicating that for each extra hour available on that machines, the optimum profit has an increase of 0.6 €.\nSensitivity analysis of the unit profit of the products: to determine the allowable range to keep the original optimal solution (see the last two columns in the 1st table - Sensitivity report)\n\n- Profit of product A:\n - Current value of profit of product A: $c_1 = 3$\n - Allowable increase of profit of product A: 1.5, so $c_1 \\leq 3 + 1.5 = 4.5$\n - Allowable decrease of profit of product A: 1, so $c_1 \\geq 3 - 1 = 2$\n - Allowable range for the profit of product A: $2 \\leq c_1 \\leq 4.5$\n\n- Allowable range for the profit of product B: $2 \\leq c_2 \\leq 4.5$\n- Allowable range for the profit of product C: $-\\infty \\leq c_3 \\leq 4.2$\n\nCalculation of the allowable range in the total profit due to the allowable range on profit of each product. For the product A:\n\n- The allowable range on the profit of product A implies an allowable range in total profit equal to $156 - 1 \\times 46 \\leq z^* \\leq 156 + 1.5 \\times 46$. \nSensitivity analysis of the available capacity of resources: to determine the allowable range to keep the original optimal solution (see the last two columns in the 2nd table - Sensitivity report)\n\n- Available capacity of machine M3:\n - Current value of capacity of M3: $b_1 = 110$\n - Allowable increase of capacity of M3: 115, so $b_1 \\leq 110 + 115 = 225$\n - Allowable decrease of capacity of M3: 10, so $b_1 \\geq 110 - 10 = 100$\n - Allowable range for the capacity of M3: $100 \\leq b_1 \\leq 225$\n\n- Allowable range for the profit of product B: $73.33 \\leq b_2 \\leq 152.5$\n\n- Allowable range for the profit of product C: $196 \\leq b_3 \\leq +\\infty$\nReference\n\nHillier F. S., & Lieberman G.R. (2010). Introduction to operations research (9th ed.). New York: McGraw-Hill.", "id": "./materials/169.pdf" }, { "contents": "Exponentials\n\nGiven a real number $a > 0$\n\n$$a^n = a \\cdot a \\cdot \\ldots \\cdot a$$\n\n$$a^{n+m} = a^n \\cdot a^m$$\n\n$$(a^n)^m = a^{n \\cdot m}$$\n\nTrue for every $n, m \\in \\mathbb{N}$\n\\( a^\\frac{p}{q} \\) is the unique positive real number \\( b \\) such that \\( b^q = a^p \\).\n\n\\[\n\\left( a^p \\right)^\\frac{1}{q} = \\sqrt[q]{a^p}\n\\]\n\nHow to extend the exponential to exponents which are real numbers?\n\n(Ex: how can I define \\( 2^{\\sqrt{2}} \\)?)\n\nLet us consider \\( A, B \\subseteq \\mathbb{R} \\)\n\n\\[\nA = \\{ 2^r : r \\in \\mathbb{Q}, r > 0 \\}\n\\]\n\n\\[\nB = \\{ 2^r : r \\in \\mathbb{Q}, r > \\sqrt{2} \\}\n\\]\n\\[ A = \\left\\{ 2^1, 2^{\\frac{14}{10}}, 2^{\\frac{1}{2}}, \\ldots \\right\\} \\]\n\n\\[ B = \\left\\{ 2^2, 2^{\\frac{15}{10}}, 2^{1.42}, 2^{1.415}, \\ldots, 2^{100} \\right\\} \\]\n\n\\[ A \\text{ is on the left of } B \\]\n\n\\[ \\exists \\ c \\text{ s.t. } a \\leq c \\leq b \\quad A \\ni a \\in A, \\ b \\in B \\]\n\nand also \\( c \\) is unique \\( c = 2^{\\sqrt{2}} \\)\nThis defines $a^b$ for every $a > 0$, $b > 0$, moreover $a^{-b} = \\frac{1}{a^b}$. This will define the exp. for all real number $b$.\n\nIt can be seen that actually the algebraic properties of the exponential on $\\mathbb{N}$ carries on on $\\mathbb{R}$.\n\n1. $a^{b+c} = a^b \\cdot a^c \\quad \\forall \\ a > 0 \\ \\forall \\ b, c \\in \\mathbb{R}$\n\n2. $(a^b)^c = a^{b \\cdot c} \\quad \\forall \\ a > 0 \\ \\forall \\ b, c \\in \\mathbb{R}$\nIn particular for (1), using \\( c = -b \\)\n\n\\[\n\\begin{align*}\n\\frac{b + (-b)}{c} &= a^b \\cdot a^{-b} \\\\\n\\Rightarrow \\quad a^0 &= 1 = a^b \\cdot a^{-b} \\\\\n\\Rightarrow \\quad a^{-b} &= \\frac{1}{a^b}\n\\end{align*}\n\\]\n\n(3) \\( (a \\cdot b)^c = a^c \\cdot b^c \\)\n\n(3') \\( \\left( \\frac{a}{b} \\right)^c = \\frac{a^c}{b^c} \\)\nLogarithms.\n\nQ. What is the number $x$ that I have to put at the exponent of $a$ so in order to get $y$.\n\n$$a^x = y$$\n\nAnswer: This is possible only if $y > 0$.\n\nAnd this is a unique solution, that we call $x = \\log_a(y)$.\n(1) \\( \\log_a (b \\cdot c) = \\log_a (b) + \\log_a (c) \\quad \\forall a, b, c > 0 \\)\n\n(1') \\( \\log_a (b / c) = \\log_a (b) - \\log_a (c) \\quad \\forall a, b, c > 0 \\)\n\n(2) \\( \\log_a (b^c) = c \\cdot \\log_a (b) \\quad \\forall a, b > 0 \\)\n\n(3) \\( \\log_a (c) = \\frac{\\log_b (c)}{\\log_b (a)} \\quad \\forall a, b, c \\in \\mathbb{R} \\)\n\n**Warning:** It could be that \\( \\log_a (y) < 0 \\), but always we have to have \\( y > 0 \\).\n\n**Example:** \\( \\log_2 (2) = 1 \\) (since \\( 2^1 = 2 \\))\n\n\\( \\log_2 \\left( \\frac{1}{2} \\right) = -1 \\) (since \\( 2^{-1} = \\frac{1}{2} \\))\n(Tautology) \\[ a^{\\log_a(y)} = y \\quad \\forall \\ a, y > 0 \\]\n\n(Tautology II) \\[ \\log_a(a^y) = y \\quad \\forall \\ a, y > 0 \\]", "id": "./materials/17.pdf" }, { "contents": "Change of Basis\n\nIn a vector space, the coordinates of a vector is always with respect to a basis and if we omit the basis, we naturally assume it to be the standard basis.\n\nWithout loss of generality, consider $A = \\{v_1, v_2, v_3\\}$ and $B = \\{u_1, u_2, u_3\\}$ two bases of three-dimensional space $\\mathbb{R}^3$. For all $v \\in V$, $v_A = (k_1, k_2, k_3)$ means that $v = k_1v_1 + k_2v_2 + k_3v_3$ and $v_B = (t_1, t_2, t_3)$ means that $v = t_1u_1 + t_2u_2 + t_3u_3$.\n\nIn particular, we can write the vectors $u_1, u_2, u_3$ of $B$ in base $A$ as follows:\n\n$$\n\\begin{align*}\n u_1 &= a_{11}v_1 + a_{21}v_2 + a_{31}v_3 \\\\\n u_2 &= a_{12}v_1 + a_{22}v_2 + a_{32}v_3 \\\\\n u_3 &= a_{13}v_1 + a_{23}v_2 + a_{33}v_3\n\\end{align*}\n$$\n\nThen,\n\n$$\nt_1u_1 + t_2u_2 + t_3u_3 = t_1(a_{11}v_1 + a_{21}v_2 + a_{31}v_3) + t_2(a_{12}v_1 + a_{22}v_2 + a_{32}v_3) + t_3(a_{13}v_1 + a_{23}v_2 + a_{33}v_3)\n$$\n\nAssociating the terms in $v_i$, we have:\n\n$$\nt_1u_1 + t_2u_2 + t_3u_3 = (t_1a_{11} + t_2a_{12} + t_3a_{13})v_1 + (t_1a_{21} + t_2a_{22} + t_3a_{23})v_2 + (t_1a_{31} + t_2a_{32} + t_3a_{33})v_3,\n$$\n\nAs the coordinates in relation to a base are unique, we have\n\n$$\nt_1a_{11} + t_2a_{12} + t_3a_{13} = k_1, t_1a_{21} + t_2a_{22} + t_3a_{23} = k_2 \\quad \\text{and} \\quad t_1a_{31} + t_2a_{32} + t_3a_{33} = k_3.\n$$\n\nThat is:\n\n$$\n\\begin{bmatrix}\n a_{11} & a_{12} & a_{13} \\\\\n a_{21} & a_{22} & a_{23} \\\\\n a_{31} & a_{32} & a_{33}\n\\end{bmatrix}\n\\begin{bmatrix}\n t_1 \\\\\n t_2 \\\\\n t_3\n\\end{bmatrix}\n= \n\\begin{bmatrix}\n k_1 \\\\\n k_2 \\\\\n k_3\n\\end{bmatrix}\n$$\n\nIn short, we can write\n\n$$\nP_A^B \\cdot v_A = v_B,\n$$\n\nwhere $P_A^B$ is called the change matrix from $B$ to base $A$.\n\nIn particular, if $B = \\{v_1, \\ldots, v_n\\}$ is a basis of a vector space $V$ and the matrix whose columns are the vectors of $B$,\n\n$$\n\\begin{bmatrix}\n v_1 & v_2 & \\cdots & v_n\n\\end{bmatrix},\n$$\n\nis a square matrix, then its determinant is nonzero.\n\nRemember that:\n\n- The standard basis of $\\mathbb{R}^2$ is $\\{(1, 0), (0, 1)\\}$;\n- The standard basis of $\\mathbb{R}^3$ is $\\{(1, 0, 0), (0, 1, 0), (0, 0, 1)\\}$.\nFor example, in $\\mathbb{R}^2$, $v = (2, 3)$ means\n\n$$v = 2(1, 0) + 3(0, 1) = \\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix} \\cdot \\begin{bmatrix} 2 \\\\ 3 \\end{bmatrix} \\text{ or }$$\n\n$$v = 1(3, 1) - 1(1, -2) = \\begin{bmatrix} 3 & 1 \\\\ 1 & -2 \\end{bmatrix} \\cdot \\begin{bmatrix} 1 \\\\ -1 \\end{bmatrix} \\text{ or }$$\n\n$$v = -\\frac{4}{3}(1, -1) + \\frac{5}{3}(2, 1) = \\begin{bmatrix} 1 & 2 \\\\ -1 & 1 \\end{bmatrix} \\cdot \\begin{bmatrix} -\\frac{4}{3} \\\\ \\frac{5}{3} \\end{bmatrix}.$$ \n\nSo, the coordinates of $v$ with respect to:\n\n- the standard basis are $v = (2, 3)$;\n- the basis $A = \\{(3, 1), (1, -2)\\}$ are $v_A = (1, -1)$;\n- the basis $B = \\{(1, -1), (2, 1)\\}$ are $v_B = \\left(\\begin{array}{c} -\\frac{4}{3} \\\\ \\frac{5}{3} \\end{array}\\right)$.\n\nNotice that\n\n$$v_A = \\begin{bmatrix} 3 & 1 \\\\ 1 & -2 \\end{bmatrix}^{-1} \\cdot v \\quad \\text{and} \\quad v_B = \\begin{bmatrix} 1 & 2 \\\\ -1 & 1 \\end{bmatrix}^{-1} \\cdot v.$$ \n\nBesides that,\n\n$$\\begin{bmatrix} 3 & 1 \\\\ 1 & -2 \\end{bmatrix} \\cdot \\begin{bmatrix} 1 \\\\ -1 \\end{bmatrix} = \\begin{bmatrix} 1 & 2 \\\\ -1 & 1 \\end{bmatrix} \\cdot \\begin{bmatrix} -\\frac{4}{3} \\\\ \\frac{5}{3} \\end{bmatrix} \\Leftrightarrow \\begin{bmatrix} 1 \\\\ -1 \\end{bmatrix} = \\begin{bmatrix} 3 & 1 \\\\ 1 & -2 \\end{bmatrix}^{-1} \\cdot \\begin{bmatrix} 1 & 2 \\\\ -1 & 1 \\end{bmatrix} \\cdot \\begin{bmatrix} -\\frac{4}{3} \\\\ \\frac{5}{3} \\end{bmatrix}.$$ \n\nThat is,\n\n$$v_A = \\begin{bmatrix} 3 & 1 \\\\ 1 & -2 \\end{bmatrix}^{-1} \\cdot \\begin{bmatrix} 1 & 2 \\\\ -1 & 1 \\end{bmatrix} \\cdot v_B.$$ \n\nSo, if $A$ and $B$ are two bases of a $n$ dimensional vector space and the matrices $A = [a_{i,j}]_{n \\times n}$ and $B = [b_{i,j}]_{n \\times n}$, whose columns are the vectors of bases $A$ and $B$ (respectively) are square matrices, then the coordinates of any vector $v \\in V$ in bases $A$ and $B$ are related as follows:\n\n$$v_A = A^{-1} \\cdot B \\cdot v_B \\quad \\text{and} \\quad v_B = B^{-1} \\cdot A \\cdot v_A.$$ \n\nThe product $A^{-1} \\cdot B$ corresponds to the change matrix from $B$ to base $A$, that is:\n\n$$P_A^B = A^{-1} \\cdot B.$$ \n\nWe still have\n\n**Properties:** If $A$ and $B$ are basis of a $V$ vector space of $n$ dimension, then:\n\n1. $P_A^B = (P_B^A)^{-1}$.\n2. Given $v \\in V$, we have $[v]_A = P_A^B \\cdot [v]_B$;\n3. Given $v \\in V$, we have $[v]_B = (P_A^B)^{-1} \\cdot [v]_A$;\n4. $P_C^B = P_A^B \\cdot P_A^C$. \n\n---\n\nEdite Martins Cordeiro \nFlora Silva \nPaula Maria Barros", "id": "./materials/173.pdf" }, { "contents": "Concept of Linear Transformation\n\nDefinition: Let $U$ and $V$ be two real vector spaces. $T: U \\rightarrow V$ is a linear transformation if:\n\ni. $\\forall x, y \\in U$, $T(x + y) = T(x) + T(y)$\n\nii. $\\forall x \\in U$, $\\forall \\alpha \\in \\mathbb{R}$, $T(\\alpha x) = \\alpha T(x)$\n\n1. Prove that the transformation $T: \\mathbb{R}^2 \\rightarrow \\mathbb{R}^3$, $T(x, y) = (2x, y, -y)$ is linear.\n\ni. Considering $(x_1, y_1), (x_2, y_2) \\in \\mathbb{R}^2$, we have:\n\n$$T((x_1, y_1) + (x_2, y_2)) = T(x_1 + x_2, y_1 + y_2)$$\n$$= (2(x_1 + x_2), y_1 + y_2, -(y_1 + y_2))$$\n$$= (2x_1 + 2x_2, y_1 + y_2, -y_1 - y_2)$$\n\nOn the other side,\n\n$$T(x_1, y_1) + T(x_2, y_2) = (2x_1, y_1, -y_1) + (2x_2, y_2, -y_2)$$\n$$= (2x_1 + 2x_2, y_1 + y_2, -y_1 - y_2)$$\n\nWe concluded that,\n\n$$T((x_1, y_1) + (x_2, y_2)) = T(x_1, y_1) + T(x_2, y_2), \\forall (x_1, y_1), (x_2, y_2) \\in \\mathbb{R}^2$$\n\n😊 The first condition of linearity of a transformation is proved.\n\nii. Considering $(x_1, y_1) \\in \\mathbb{R}^2$ and $\\alpha \\in \\mathbb{R}$, we have:\n\n$$T(\\alpha(x_1, y_1)) = T(\\alpha x_1, \\alpha y_1) = (\\alpha 2x_1, \\alpha y_1, -\\alpha y_1)$$\n$$= \\alpha(2x_1, y_1, -y_1) = \\alpha T(x_1, y_1)$$\n\nWe concluded that,\n\n$$T(\\alpha(x_1, y_1)) = \\alpha T(x_1, y_1), \\forall (x_1, y_1) \\in \\mathbb{R}^2, \\forall \\alpha \\in \\mathbb{R}$$\n\n😊 The second condition of linearity is also verified.\n\nConclusion: Since both linearity conditions are verified, $T$ is a linear transformation.\n2. The transformation $T: \\mathbb{R}^2 \\to \\mathbb{R}^2$, $T(x, y) = (x, 1 + y)$ is linear?\n\ni. Considering $(x_1, y_1), (x_2, y_2) \\in \\mathbb{R}^2$, we have:\n\n$$T((x_1, y_1) + (x_2, y_2)) = T(x_1 + x_2, y_1 + y_2)$$\n\n$$= (x_1 + x_2, 1 + y_1 + y_2)$$\n\nOn the other side,\n\n$$T(x_1, y_1) + T(x_2, y_2) = (x_1, 1 + y_1) + (x_2, 1 + y_2)$$\n\n$$= (x_1 + x_2, 2 + y_1 + y_2)$$\n\nWe concluded that,\n\n$$\\exists (x_1, y_1), (x_2, y_2) \\in \\mathbb{R}^2: T((x_1, y_1) + (x_2, y_2)) \\neq T(x_1, y_1) + T(x_2, y_2)$$\n\nThe first condition of linearity of a transformation is not verified.\n\nConclusion: As the first linearity condition is not verified, we concluded that $T$ is not a linear transformation.", "id": "./materials/174.pdf" }, { "contents": "Homogeneous Equations. First we recall some properties of a homogeneous function.\n\nDefinition 1.5 A function $f(x, y)$ is called homogeneous of degree $\\alpha$ if we have\n\n$$f(tx, ty) = t^\\alpha f(x, y)$$\n\nAn important case is $\\alpha = 0$ because the above relation is of the form $f(tx, ty) = f(x, y)$. This property is equivalent with\n\n$$f(x, y) = f(1, \\frac{y}{x}) = f(\\frac{x}{y}, 1)$$\n\nDefinition 1.6 A differential equation of the first order is called homogeneous zero degree if it is of the form\n\n$$y' = f(x, y), \\quad (1)$$\n\nwhere $f$ is a function of zero degree.\n\nTo integrate the equation (1) it is interesting to see that it can always be transformed into separable equation by change of the dependent variable. More exactly we put $y(x) = xu(x)$ where $u(x)$ is a unknown function. Thus we have\n\n$$y = xu \\quad \\Rightarrow \\quad y' = xu' + u.$$ \n\nReplace $y$ and $y'$ in th eq. (1.12) we obtain\n\n$$xu' + u = f(1, u) = g(u), \\quad \\Rightarrow \\quad \\frac{dx}{x} = \\frac{du}{g(u) - u}, \\quad g(u) \\neq u.$$ \n\nThe last equation is a separable equation and its solution it can be deduce.\n\nFind the general solution of the equations\n\n1. $x^2dy = (x^2 + xy + y^2)dy$ \n2. $2xydy = (x^2 + 3y^2)dx$ \n3. $(2x - y)dy + (3x - 4y)dx = 0$, \n4. $(2x + y)dy + (4x - 3y)dx = 0$, \n\n5. \\((x + 3y)dx - (x - y)dy = 0\\), 6. \\((x^2 + 3xy + y^2)dx - (x^2 - xy + 2y^2)dy = 0\\),\n7. \\((x^2 - 3y^2)dx = 2xydy\\) 8. \\(2xydy = (3y^2 - x^2)dx\\)\n9. \\(xy' - y = \\sqrt{x^2 + y^2}\\) 10. \\(xy' - y = \\frac{x}{\\arctg y}\\)\n11. \\(ydx + (2\\sqrt{xy} - x)dx = 0\\).\n\nReducible equations to the homogeneous equations.\n\nA differentiable equation of the form\n\\[\ny' = f\\left(\\frac{a_1x + b_1y + c_1}{a_2x + b_2y + c_2}\\right)\n\\] \\hspace{1cm} (2)\n\nis called reducible equation to homogeneous equations.\n\nFor to find the general solution we can follow the next steps:\n\na) solve the next system\n\\[\n\\begin{cases}\n a_1x + b_1y + c_1 = 0 \\\\\n a_2x + b_2y + c_2 = 0\n\\end{cases}\n\\] \\hspace{1cm} (3)\n\nSuppose that \\(x_0, y_0\\) is a unique solution of the system (3) Then if we put\n\\(y(x) - y_0 = (x - x_0)u(x)\\) where \\(u(x)\\) is a unknown function, the equation (2) become a separable equation.\n\nIf the system (3) is incompatible then \\(\\frac{a_1}{a_2} = \\frac{b_1}{b_2} \\neq \\frac{c_1}{c_2}\\) the by substitution\n\\(u(x) = a_1x + b_1y + c_1\\), the equation (2) become a separable equation.\n\nSolve the next differentiable homogeneous equations\n1) \\(2x + 2y + 1 + (x + 2y - 1)y' = 0\\), 2) \\(2x + 2y - 1 + (x - 2y + 3)y' = 0\\),\n3) \\((2x - y + 5)dx + (2x - y + 4)dy = 0\\), 4) \\((x + y - 1)dx + (-x + y + 1)dy = 0\\),\n5) \\((2x + 2y + 5)dx + (2x + 2y - 6)dy = 0\\) 6) \\((3x - 4y + 10)dx + (x + y + 2)dy = 0\\).", "id": "./materials/175.pdf" }, { "contents": "Definition. Let \\( f_1, f_2 \\) defined on \\( a \\leq x \\leq b \\) and \\( g_1, g_2 \\) defined on \\( c \\leq y \\leq d \\), continue, such that \\( f_2(x) \\neq 0 \\), \\( g_2(y) \\neq 0 \\). A differential equation of the form\n\n\\[\nf_1(x)g_1(y) + f_2(x)g_2(y)y' = 0 \\iff f_1(x)g_1(y)dx + f_2(x)g_2(y)dy = 0. \\tag{1}\n\\]\ncalled equation with separable variable.\n\nFor this equation we have the next result\n\nProposition. The general solution of the equation (1) is given by an implicit function in the following form\n\n\\[\n\\int \\frac{f_1(x)}{f_2(x)} \\, dx + \\int \\frac{g_1(y)}{g_2(y)} \\, dy = C, \\quad C \\in \\mathbb{R}\n\\]\n\nProof. It is easy to see that the equation (1) can be rewrite as follows\n\n\\[\n\\frac{f_1(x)}{f_2(x)} = \\frac{g_1(y)}{g_2(y)},\n\\]\n\nand by integrating each side of the above equation we obtain the desired result.\n\nExample. Integrate the next equation\n\n\\[\n(1 + x^2)dy + ydx = 0\n\\]\n\nSolution. The equation above is of the separable variable, and thus we have\n\n\\[\n(1 + x^2)dy + ydx = 0, \\quad \\Rightarrow \\quad \\frac{dx}{1 + x^2} = -\\frac{dy}{y} \\quad \\Rightarrow \\quad \\int \\frac{dx}{1 + x^2} = -\\int \\frac{dy}{y} \\quad \\Rightarrow \\quad \\arctan x = -\\ln y + C \\quad \\Rightarrow \\quad \\arctan x + \\ln y = C\n\\]\n\nExercises. Solve the next differential equations\n\n1. \\( yy' = -2x \\csc y \\),\n2. \\( y' + \\cos(x + 2y) = \\cos(x - 2y) \\),\n3. \\( 2x(2 \\cos y - 1)dx = (x^2 - 2x + 3)dy \\),\n4. \\( y' = \\frac{\\cos y - \\sin y - 1}{\\cos x - \\sin x - 1} \\)\n5. \\( y y' + xe^y = 0, \\quad y(1) = 0 \\)\n\n6. \\( y' = e^{x+y} + e^{x-y}, \\quad y(0) = 0 \\)\n\n7. \\( \\frac{dx}{x(y-1)} + \\frac{dy}{y(x+2)}, \\quad y(1) = 1 \\)\n\n8. \\( \\frac{y^2 + 4}{\\sqrt{x^2 + 4x + 13}} = \\frac{3y + 2}{x + 1}, \\quad y(1) = 2 \\)", "id": "./materials/176.pdf" }, { "contents": "Definition. Let $D \\subseteq \\mathbb{R}^2$ be a connex set and $P, Q$ the differential equations so that\n\\[\n\\frac{\\partial P}{\\partial y} = \\frac{\\partial Q}{\\partial x}\n\\] (1.15)\nthen the equation\n\\[\nP(x, y)dx + Q(x, y)dy = 0\n\\] (1.16)\nis called the exact differential equation, and its solution is implicitly defined by the next equation\n\\[\n\\int_{x_0}^{x} P(t, y_0)dt + \\int_{y_0}^{y} P(x, t)dt = C, \\quad \\text{or equivalently}\n\\]\n\\[\n\\int_{y_0}^{y} Q(x_0, t)dt + \\int_{x_0}^{x} Q(t, y)dt = C\n\\]\nRemark. We now show that if $P$ and $Q$ satisfy relation (1.15) then we can find the solution of differentiable (1.16). More exactly let $\\psi(x, y) = C$ (this fact is ensured by condition (1.15)), so that\n\\[\n\\frac{\\partial \\psi}{\\partial x} = P \\quad \\text{and} \\quad \\frac{\\partial \\psi}{\\partial y} = Q.\n\\] (1.17)\nNext we have\n\\[\n\\frac{\\partial \\psi}{\\partial x} = P \\quad \\rightarrow \\quad \\psi(x, y) = \\int P(x, y)dx + f(y)\n\\] (1.18)\nTaking into account the second relation of (1.17), we obtain\n\\[\n\\frac{\\partial \\psi}{\\partial y} = Q \\quad \\rightarrow \\quad Q(x, y) = \\frac{\\partial}{\\partial y} \\left( \\int P(x, y)dx \\right) + f'(y) = \\int \\left( \\frac{\\partial}{\\partial y} P(x, y)dx \\right) + f'(y).\n\\]\nThis above relation determine the function $\\psi$ as follows\n\\[\nf'(y) = Q(x, y) - \\int \\left( \\frac{\\partial}{\\partial y} P(x, y)dx \\right).\n\\]\nTo determine $h(y)$, it is essential that, despite its appearance, the right side of above equation be a function of $y$ only. This fact, is ensured by condition (1.15), and it can by proved by direct calculation. It should be noted that this proof contains a method for the computation of $\\psi(x, y)$ and thus for solving the original differential equation (1.16). Note also that the\nsolution is obtained in implicit form; it may or may not be feasible to find the solution explicitly.\n\n**Example** Solve the differential equation\n\n\\[(y \\cos x + 2xe^y) + (\\sin x + x^2e^y - 1)y' = 0.\\] \\hspace{1cm} (1.19)\n\nIt is easy to see that\n\n\\[(y \\cos x + 2xe^y)dx + (\\sin x + x^2e^y - 1)dy = 0,\n\nand thus\n\n\\[P(x, y) = \\cos x + 2xe^y = y \\cos x + 2xe^y, \\quad Q(x, y) = \\sin x + x^2e^y - 1 \\rightarrow \\frac{\\partial P}{\\partial y} = 2xe^y = \\frac{\\partial Q}{\\partial x},\\]\n\nso the given equation is exact. Thus there is a \\(\\psi(x, y)\\) such that\n\n\\[\\psi_x(x, y) = y \\cos x + 2xe^y,\\]\n\n\\[\\psi_y(x, y) = \\sin x + x^2e^y - 1\\]\n\nIntegrating the first of these equations, we obtain\n\n\\[\\psi(x, y) = y \\sin x + x^2e^y + h(y)\\]\n\nSetting \\(\\psi_y = Q(x, y)\\) gives\n\n\\[\\psi_y(x, y) = \\sin x + x^2e^y + h'(y) = \\sin x + x^2e^y - 1, \\rightarrow h'(y) = -1 \\rightarrow h(y) = -y.\\]\n\nThe constant of integration can be omitted since any solution of the preceding differential equation is satisfactory; we do not require the most general one. Substituting for \\(h(y)\\) gives \\(\\psi(x, y) = y \\sin x + x^2e^y - y.\\) Hence solutions of Eq. (1.19) are given implicitly by\n\n\\[y \\sin x + x^2e^y - y = c.\\]", "id": "./materials/177.pdf" }, { "contents": "Equation Reducible to Exact Differential Equation. Integrating Factors.\n\n**Integrating Factors.** It is sometimes possible to convert a differential equation that is not exact into an exact equation by multiplying the equation by a suitable integrating factor. To investigate the possibility of implementing this idea more generally, let us multiply the equation\n\n\\[ P(x, y)dx + Q(x, y)dy = 0 \\] \\hspace{1cm} (1.20)\n\nby a function \\( \\mu(x, y) \\) and then try to choose \\( \\mu(x, y) \\) so that the resulting equation\n\n\\[ \\mu(x, y)P(x, y)dx + \\mu(x, y)Q(x, y)dy = 0 \\] \\hspace{1cm} (1.21)\n\nis exact, that is taking into account (1.15), eq. (23) is exact if and only if\n\n\\[ \\frac{\\partial(\\mu P)}{\\partial y} = \\frac{\\partial(\\mu Q)}{\\partial x}. \\] \\hspace{1cm} (1.22)\n\nSince \\( P \\) and \\( Q \\) are given functions, Eq. (1.22) states that the integrating factor \\( \\mu \\) must satisfy the first order partial differential equation\n\n\\[ P\\mu_y - Q\\mu_x + (P_y - Q_x)\\mu = 0. \\] \\hspace{1cm} (1.23)\n\nIf a function \\( \\mu \\) satisfying Eq. (1.25) can be found, then Eq. (1.21) will be exact.\n\nA partial differential equation of the form (1.23) may have more than one solution; if this is the case, any such solution may be used as an integrating factor of Eq. (1.20). This possible non uniqueness of the integrating factor is illustrated in further example.\n\nUnfortunately, Eq. (1.23), which determines the integrating factor \\( \\mu \\), is ordinarily at least as difficult to solve as the original equation (1.20). Therefore, while in principle integrating factors are powerful tools for solving differential equations, in practice they can be found only in special cases. The most important situations in which simple integrating factors can be\nfound occur when \\( \\mu \\) is a function of only one of the variables \\( x \\) or \\( y \\), instead of both. Let us determine necessary conditions on \\( P \\) and \\( Q \\) so that Eq. (1.20) has an integrating factor \\( \\mu \\) that depends on \\( x \\) only. Assuming that \\( \\mu \\) is a function of \\( x \\) only, we have\n\n\\[\n(\\mu P)_y = \\mu P_y, \\quad (\\mu Q)_x = \\mu Q_x + Q \\frac{d\\mu}{dx}.\n\\]\n\nThus, if \\((\\mu P)_y\\) is to equal \\((\\mu Q)_x\\), it is necessary that\n\n\\[\n\\frac{d\\mu}{dx} = \\frac{P_y - Q_x}{Q} \\mu.\n\\] \n\n(1.24)\n\nIf \\( \\frac{P_y - Q_x}{Q} \\) is a function of \\( x \\) only, then there is an integrating factor \\( \\mu \\) that also depends only on \\( x \\); further, \\( \\mu(x) \\) can be found by solving Eq. (26), which is both linear and separable.\n\nA similar procedure can be used to determine a condition under which Eq. (1.20) has an integrating factor depending only on \\( y \\).\n\n**Example.** Find an integrating factor for the equation\n\n\\[\n(3xy + y^2) + (x^2 + xy)y' = 0 \\quad (1.25)\n\\]\n\nand then solve the equation. We showed that this equation is not exact. Let us determine whether it has an integrating factor that depends on \\( x \\) only. On computing the quantity \\( \\frac{P_y - Q_x}{Q} \\), we find that\n\n\\[\n\\frac{P_y(x, y) - Q_x(x, y)}{Q(x, y)} = \\frac{(3x + 2y) - (2x + y)}{x^2 + xy} = \\frac{1}{x}.\n\\]\n\nThus there is an integrating factor \\( \\mu \\) that is a function of \\( x \\) only, and it satisfies the differential equation\n\n\\[\n\\frac{d\\mu}{dx} = \\frac{\\mu}{x}.\n\\]\n\nHence \\( \\mu(x) = x \\). Multiplying Eq. (1.25) by this integrating factor, we obtain\n\n\\[\n(3x^2y + xy^2) + (x^3 + x^2y)y' = 0.\n\\]\n\nThe latter equation is exact and it is easy to show that its solutions are given implicitly by\n\n\\[\nx^3y + 12x^2y^2 = C.\n\\]\nYou may also verify that a second integrating factor of Eq. (1.25) is \\( \\mu(x, y) = \\frac{1}{xy(2x+y)} \\), and that the same solution is obtained, though with much greater difficulty, if this integrating factor is used.\n\nDetermine whether or not each of the next equations is exact. If it is exact, find the solution.\n\n1. \\((2x + 3) + (2y - 2)y' = 0\\), 2. \\((2x + 4y) + (2x - 2y)y' = 0\\),\n3. \\((3x^2 - 2xy + 2)dx + (6y^2 - x^2 + 3)dy = 0\\), 4. \\((2xy^2 + 2y) + (2x^2y + 2x)y' = 0\\),\n5. \\((e^x \\sin y - 2y \\sin x)dx + (e^x \\cos y + 2 \\cos x)dy = 0\\)\n\n6. \\(\\frac{dy}{dx} = \\frac{ax + by}{bx + cy}\\), 7. \\(\\frac{dy}{dx} = -\\frac{ax - by}{bx - cy}\\), 8. \\((e^x \\sin y + 3y)dx - (3x - e^x \\sin y)dy = 0\\),\n9. \\((ye^{xy} \\cos 2x - 2e^{xy} \\sin 2x + 2x)dx + (xe^{xy} \\cos 2x - 3)dy = 0\\),\n10. \\(\\left(\\frac{y}{x} + 6x\\right)dx + (\\ln x - 2)dy = 0, x > 0\\),\n11. \\((x \\ln y + xy)dx + (y \\ln x + xy)dy = 0; x > 0, y > 0\\).\n\nIn each of the next equation solve the given initial value problem and determine at least approximately where the solution is valid.\n\n\\((2x - y)dx + (2y - x)dy = 0, y(1) = 3\\), \\((9x^2 + y - 1)dx - (4y - x)dy = 0, y(1) = 0\\).\n\nIn each of the next problems find the value of \\(b\\) for which the given equation is exact and then solve it using that value of \\(b\\).\n\n\\((xy^2 + bx^2y)dx + (x + y)x^2dy = 0\\), \\((ye^{2xy} + x)dx + bxe^{2xy}dy = 0\\).\n\nShow that the next equations are not exact, but become exact when multiplied by the given integrating factor. Then solve the equations.\n\n1. \\(x^2y^3 + x(1 + y^2)y' = 0\\), \\(\\mu(x, y) = \\frac{1}{xy^3}\\),\n2. \\(\\left(\\frac{\\sin y}{y} - 2e^{-x} \\sin x\\right)dx + \\left(\\frac{\\cos y + 2e^{-x} \\cos x}{y}\\right)dy = 0\\), \\(\\mu(x, y) = ye^x\\),\n3. \\(ydx + (2x - ye^y)dy = 0, \\mu(x, y) = y\\),\n4. \\((x + 2) \\sin ydx + x \\cos ydy = 0, \\mu(x, y) = xe^x\\)\nShow that the equation\n\\[ y' + f(x)y = 0 \\]\nhas an integrating factor of the form\n\\[ \\mu(x) = e^{-\\int f(x)dx} \\]\nShow that if \\((Q_x - P_y)/(xP - yQ) = R\\), where \\(R\\) depends on the quantity \\(xy\\) only, then the differential equation\n\\[ P(x, y) + Q(x, y)y' = 0 \\]\nhas an integrating factor of the form \\(\\mu(xy)\\). A general formula for this integrating factor is of the form\n\\[ \\mu(xy) = e^{\\int R(xy)d(xy)}. \\]\nIn each of the next problems find an integrating factor and solve the given equation.\n\n1. \\((3x^2y + 2xy + y^3)dx + (x^2 + y^2)dy = 0, \\quad 2. \\ y' = e^{2x} + y - 1, \\)\n3. \\(dx + (x/y - \\sin y)dy = 0, \\quad 4. ydx + (2xy - e^{-2y})dy = 0, \\)\n5. \\(e^x dx + (e^x \\cot y + 2y \\cosec y)dy = 0, \\quad 6. (4(x^3/y^2) + 3/y)dx + 3x/y^2 + 4y)dy = 0, \\)\n7. \\((3x + 6/y) + (x^2/y + 3y/x)(dy/dx) = 0 \\)", "id": "./materials/178.pdf" }, { "contents": "Intersection and sum of two vector spaces and the relationship between their dimensions\n\nIntersection and sum of two vector subspaces\n\nDefinition: Let $V$ be a vector space, and let $U$ and $W$ be subspaces of $V$. Then:\n\n1. $U + W = \\{u + w : u \\in U \\land w \\in W\\}$ and is called the sum of $U$ and $W$.\n2. $U \\cap W = \\{v : v \\in U \\land v \\in W\\}$ and is called the intersection of $U$ and $W$.\n\nExample: Consider the plans $P_1 = \\{(x, y, z) \\in \\mathbb{R}^3 : z = 0\\}$ and $P_2 = \\{(x, y, z) \\in \\mathbb{R}^3 : x - y + z = 0\\}$. These are both subspaces of $\\mathbb{R}^3$, that we can define by its generic vectors as:\n\n- $P_1 = \\{(x_1, y_1, 0) : x_1, y_1 \\in \\mathbb{R}\\}$;\n- $P_2 = \\{(x_2, y_2, y_2 - x_2) : x_2, y_2 \\in \\mathbb{R}\\}$.\n\nTheir intersection is the subspace\n\n$$P_1 \\cap P_2 = \\{(x, y, z) \\in \\mathbb{R}^3 : z = 0 \\land x - y = 0\\} = \\{(x, x, 0) : x \\in \\mathbb{R}\\}.$$\n\nThe sum,\n\n$$P_1 + P_2 = \\{(x_1 + x_2, y_1 + y_2, y_2 - x_2) : x_1, x_2, y_1, y_2 \\in \\mathbb{R}\\}$$\n\nis the vector space $\\mathbb{R}^3$.\n\nRelationship between the dimensions of the vector spaces sum and intersection of two vector spaces\n\nIn relation to the previous example, we easily determine the size of each of the spaces $P_1$, $P_2$, $P_1 \\cap P_2$ and $P_1 + P_2$.\n\nJust consider the number of free variables of the generic vector and we conclude that both subspaces $P_1 = \\{(x, y, 0) : x, y \\in \\mathbb{R}\\}$ and $P_2 = \\{(x, y, y - x) : x, y \\in \\mathbb{R}\\}$ of $\\mathbb{R}^3$ have dimension 2.\n\nTheir intersection is the subspace $P_1 \\cap P_2 = \\{(x, x, 0) : x \\in \\mathbb{R}\\}$ with dimension 1 and the subspace $P_1 + P_2$ has dimension 3, because $P_1 + P_2$ spans $\\mathbb{R}^3$. We have\n\n$$\\dim(P_1 + P_2) = \\dim(P_1) + \\dim(P_2) - \\dim(P_1 \\cap P_2).$$\n\nIt will be always like this? The answer is affirmative:\n\n**Theorem Dimension of sum:** Let $V$ be a vector space with subspaces $U$ and $W$, each one of them have finite dimension. Then $U + W$ also has finite dimension which is given by\n\n$$\\dim(U + W) = \\dim(U) + \\dim(W) - \\dim(U \\cap W).$$\n**Example:** Let the vector subspaces $S_1 = \\left\\{ \\begin{bmatrix} a & 0 \\\\ 3a & b \\end{bmatrix} : a, b \\in \\mathbb{R} \\right\\}$ and $S_2 = \\left\\{ \\begin{bmatrix} c & d \\\\ -d & e \\end{bmatrix} : c, d, e \\in \\mathbb{R} \\right\\}$ of the space of the square matrices of order 2.\n\n$S_1$ has dimension 2, $S_2$ has dimension 3 and their intersection, $S_1 \\cap S_2 = \\left\\{ \\begin{bmatrix} x & 0 \\\\ 0 & y \\end{bmatrix} : x, y \\in \\mathbb{R} \\right\\}$, has dimension 2.\n\nThen the sum, $S_1 + S_2$ is a subspace with dimension $2 + 3 - 2 = 3$.\n\nNotice that\n\n$$B = \\left\\{ \\begin{bmatrix} 1 & 0 \\\\ 3 & 0 \\end{bmatrix}, \\begin{bmatrix} 0 & 0 \\\\ 0 & 1 \\end{bmatrix} \\right\\}$$\n\nis a basis of $S_1$ and\n\n$$C = \\left\\{ \\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix}, \\begin{bmatrix} 0 & 1 \\\\ -1 & 0 \\end{bmatrix}, \\begin{bmatrix} 0 & 0 \\\\ 0 & 1 \\end{bmatrix} \\right\\}$$\n\nis a basis of $S_2$.\n\nThen\n\n$$D = \\left\\{ \\begin{bmatrix} 1 & 0 \\\\ 3 & 0 \\end{bmatrix}, \\begin{bmatrix} 0 & 0 \\\\ 0 & 1 \\end{bmatrix}, \\begin{bmatrix} 1 & 0 \\\\ 0 & 0 \\end{bmatrix}, \\begin{bmatrix} 0 & 1 \\\\ -1 & 0 \\end{bmatrix}, \\begin{bmatrix} 0 & 0 \\\\ 0 & 1 \\end{bmatrix} \\right\\}$$\n\nspans $S_1 + S_2$.\n\nAs the equality\n\n$$k_1 \\begin{bmatrix} 1 & 0 \\\\ 3 & 0 \\end{bmatrix} + k_2 \\begin{bmatrix} 0 & 0 \\\\ 0 & 1 \\end{bmatrix} + k_3 \\begin{bmatrix} 1 & 0 \\\\ 0 & 0 \\end{bmatrix} + k_4 \\begin{bmatrix} 0 & 1 \\\\ -1 & 0 \\end{bmatrix} + k_5 \\begin{bmatrix} 0 & 0 \\\\ 0 & 1 \\end{bmatrix} = \\begin{bmatrix} 0 & 0 \\\\ 0 & 0 \\end{bmatrix}$$\n\nrepresents a system, whose expanded matrix is\n\n$$\\begin{bmatrix} 1 & 0 & 1 & 0 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 0 \\\\ 3 & 0 & 0 & -1 & 0 & 0 & 0 \\\\ 0 & 1 & 0 & 0 & 1 & 0 & 0 \\end{bmatrix} \\rightarrow \\begin{bmatrix} 1 & 0 & 1 & 0 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 0 \\\\ 0 & 0 & -3 & -1 & 0 & 0 & 0 \\\\ 0 & 1 & 0 & 0 & 1 & 0 & 0 \\end{bmatrix} \\rightarrow \\begin{bmatrix} 1 & 0 & 1 & 0 & 0 & 0 & 0 \\\\ 0 & 1 & 0 & 0 & 1 & 0 & 0 \\\\ 0 & 0 & -3 & -1 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 0 \\end{bmatrix}$$\n\nWe can conclude that the system is doubly indeterminate, so $D$ is linearly dependent. Thus, the minimum set that generates $S_1 + S_2$ has cardinality 3.", "id": "./materials/179.pdf" }, { "contents": "Elementary operations with sets\n\nA, B two sets\n\n\\[ A \\cup B = \\text{\"union\"} = \\{ \\text{elements which belong to } A \\} \\]\n\nor B (or both)\n\n\\[ A \\cap B = \\text{\"intersection\"} = \\{ \\text{elements which belong to } A \\} \\]\n\nboth A and B\n\n\\[ A \\setminus B = \\text{\"difference\"} = \\{ \\text{elements which are in } A \\} \\]\n\nbut NOT in B\n\\[ A \\cup B = (A \\setminus B) \\cup (B \\setminus A) \\cup (A \\cap B) \\]\n\ngraphically\nA = B if every element in A belongs also to B and vice-versa.\n\nExercise\n\n\\[ A = \\{1, 2, 3\\} \\quad B = \\{2, 3, 7, 8\\} \\]\n\n\\[ A \\cup B = \\{1, 2, 3, 7, 8\\} \\]\n\n\\[ A \\cap B = \\{2, 3\\} \\]\n\n\\[ A \\setminus B = \\{1\\} \\]\n\n\\[ B \\setminus A = \\{7, 8\\} \\]\n\\[ A = \\{1, 2, 3\\} \\quad B = \\{1, 2, 3, \\star\\} = \\{1, 2, 3, \\star\\} \\]\n\n\\[ A \\cup B = B = \\{1, 2, 3, \\star\\} \\]\n\n\\[ A \\cap B = A = \\{1, 2, 3\\} \\]\n\n\\[ A \\setminus B = \\emptyset \\quad \\text{empty set} \\]\n\n\\[ B \\setminus A = \\{\\star\\} \\]\nProperty: \\( A \\subseteq B \\)\n\n- \\( A \\) is a subset of \\( B \\)\n- \\( A \\) is contained in \\( B \\)\n- \\( B \\) contains \\( A \\)\n\nIf each element of \\( A \\) is also an element of \\( B \\)\n\n\\[ \\forall a \\in A \\text{ we have } a \\in B \\]\n\n\\[ \\forall \\text{ for each} \\]\n\n\\[ \\exists \\text{ it exists} \\]\n\n\\[ (\\text{existential operator}) \\]\nPower of a set $A$\n\n$\\mathcal{P}(A) = \\{ B : B \\text{ is a subset of } A \\}$\n\n\"power of $A$\"\n\n\"l'insieme delle parti di $A$\"\n\nEx. 1\n\n$A = \\{0, 1\\}$\n\n$\\mathcal{P}(A) = \\{\\{\\}, \\{0\\}, \\{1\\}, \\{0, 1\\}\\}$\n\\[ B = \\{ \\Delta, 0, \\star \\} \\]\n\n\\[ \\mathcal{P}(B) = \\{ \\{\\Delta\\}, \\{0\\}, \\{\\star\\}, \\emptyset, \\{\\Delta, 0\\}, \\{0, \\star\\}, \\{\\Delta, \\star\\}, \\{\\Delta, 0, \\star\\} \\} \\]\n\n**Cartesian Product of Sets**\n\n\\[ A \\times B = \\{ (a, b) : a \\in A, b \\in B \\} \\]\n\n\\[ \\uparrow \\text{ ordered couple} \\]\n\\{1, 2\\} = \\{2, 1\\} \\quad (1, 2) \\neq (2, 1)\n\n\\text{Ex.} \\quad A = \\{0, 1, 2\\} \\quad B = \\{\\sqrt{2}, \\pi\\}\n\nA \\times B = \\{(0, \\sqrt{2}), (1, \\pi), (2, \\sqrt{2}), (0, \\pi), (1, \\sqrt{2}), (2, \\pi)\\}\nCARDINALITY OF A SET \\( |A| \\in \\mathbb{N} \\)\n\n\\( n \\in \\mathbb{N} \\mid |A| = \\text{the number of elements of } A \\)\n\n\\[ |\\{0, 1, 2\\}| = 3 \\]\n\n\\[ |\\{p \\in \\mathbb{N} \\mid p \\text{ prime}, p \\leq 10\\}| \\]\n\n\\[ = |\\{3, 2, 5, 7\\}| = 4 \\]\nEx. \\( A, B \\) two sets such that \\(|A| = 3\\) \\(\\text{and} \\ |B| = 10\\). What is the cardinality of \\(A \\times B\\)?\n\n\\[ 30 = 3 \\times 10 \\]\n\n\\[\n\\begin{array}{c|ccc}\n& b_1 & b_2 & \\cdots & b_{10} \\\\\n\\hline\na_1 & (a_1, b_1) & (a_1, b_2) & \\cdots & (a_1, b_{10}) \\\\\na_2 & (a_2, b_1) & (a_2, b_2) & \\cdots & (a_2, b_{10}) \\\\\na_3 & (a_3, b_1) & (a_3, b_2) & \\cdots & (a_3, b_{10}) \\\\\n\\end{array}\n\\]\n\n\\[ |\\mathbb{R}^2| = |\\mathbb{R}| \\times |\\mathbb{R}| = \\{ (x, y) : x, y \\in \\mathbb{R} \\} \\]\n\\[ A \\times B \\times C = \\{ (a, b, c) : a \\in A, b \\in B, c \\in C \\} \\]\n\n**Ex.**\n\n\\[ |A| = 10 \\quad |\\mathcal{P}(A)| = ? \\]\n\n\\[ A = \\{1, 2, 3, \\ldots, 10\\} \\]\n\n\\[ \\mathcal{P}(A) = \\left\\{ \\{1\\}, \\{2\\}, \\ldots, \\{10\\}, \\emptyset, A, A \\setminus \\{1\\}, A \\setminus \\{2\\}, \\ldots, A \\setminus \\{10\\} \\right\\} \\]\n\n\\[ \\{1, 2\\}, \\{1, 3\\}, \\{1, 4\\}, \\ldots, \\{1, 10\\} \\leq 9 \\]\n\n\\[ \\{2, 3\\}, \\{2, 4\\}, \\{2, 5\\}, \\ldots, \\{2, 10\\} \\leq 8 \\]\n\\[\n\\begin{align*}\n\\{8, 9\\}, \\{9, 10\\} &< 2 \\\\\n\\{9, 10\\} &< 1 \\\\\n&= 45\n\\end{align*}\n\\]\n\n45 subsets of \\( A \\), with cardinality 2\n\n45 \" \" \" 1\" \" 8\n\\[ |B(A)| = 2^{|A|} \\]\n\n\\[ \\{1, 2, 3, \\ldots, 10\\} \\]\n\n\\[ B \\subseteq A \\text{ then for each } x \\in B? \\]\n\n\\[ \\begin{array}{ccc}\n\\text{Yes} & \\text{Yes} & \\text{Yes} \\\\\n\\text{No} & \\text{No} & \\text{No} \\\\\n\\end{array} \\]\n\n\\[ 2 \\cdot 2 \\cdot 2 \\cdot \\ldots \\cdot 2 = 2^{|A|} \\]\n\n\\[ 1024 = 2^{10} \\]\n\nLet's define \\( C = \\{ B \\subseteq A : |B| = 3 \\} \\). \\( |C| = ? \\)\n\n\"How many subsets of \\( A \\) of cardinality 3 are there?\"\n\n\\[ |C| = \\binom{n}{3} \\quad (n = |A|) \\quad \\text{binomial coefficients} \\]\n\\[\n\\binom{n}{k} = \\left( \\text{number of subsets of a set of size } n \\right) \\text{ with } k \\text{ elements}\n\\]\n\n\\[\n= \\frac{n!}{k! \\cdot (n-k)!} = \\frac{n(n-1) \\cdots (n-k+1)}{1 \\cdot 2 \\cdots k}\n\\]\n\n\\[\n\\binom{n}{3} = \\frac{n(n-1)(n-2)}{6}\n\\]\n\n\\[B \\subseteq A \\quad B = \\{ b_1, b_2, b_3 \\}\\]\n\nIn how many ways can I select \\( b_3 \\)?\n\n\\[b_1, b_2, b_3\\]\n10.3.8. But in this way I have selected many times \\( \\{1, 2, 3\\} \\)\n\n\\[\n\\begin{align*}\n&b_1 = 1 & b_1 = 2 & 1, 2, 3 \\\\\n&b_2 = 2 & b_2 = 1 & 2, 1, 3 \\\\\n&b_3 = 3 & b_3 = 2 & 1, 3, 2 \\\\\n& & & 3, 1, 2 \\\\\n& & & 2, 3, 1 \\\\\n& & & 1, 2, 3\n\\end{align*}\n\\]\n\n\\[\n\\text{# subsets with 3 elements} = \\frac{10 \\cdot 3 \\cdot 8^4}{6} = 120.\n\\]\n\n(binomial theorem)", "id": "./materials/18.pdf" }, { "contents": "Vector spaces\n\n**Definition:** A vector space is a set $V$ on which two operations $+$ and $\\cdot$ are defined, called vector addition and scalar multiplication, respectively.\n\nThe operation $+$ (vector addition) must satisfy the following conditions:\n\n1. **Closure:** If $u$ and $v$ are any vectors in $V$, then the sum $u + v$ belongs to $V$;\n2. **Commutative law:** For all vectors $u, v \\in V$, $u + v = v + u$;\n3. **Associative law:** For all vectors $u, v, w \\in V$, $(u + v) + w = u + (v + w)$;\n4. **Additive identity:** The set $V$ contains an additive identity element, denoted by $0$, such that for any vector $v \\in V$, $0 + v = v$ and $v + 0 = v$.\n5. **Additive inverses:** For each vector $v \\in V$, the equations $v + x = 0$ and $x + v = 0$ have a solution $x \\in V$, called an additive inverse of $v$, and denoted by $-v$.\n\nThe operation $\\cdot$ (scalar multiplication) is defined between real numbers (or scalars) and vectors, and must satisfy the following conditions:\n\n1. **Closure:** If $v \\in V$, and $c \\in \\mathbb{R}$, then the product $c \\cdot v \\in V$.\n2. **Distributive law:** For all $c \\in \\mathbb{R}$ and all vectors $u, v \\in V$, $c \\cdot (u + v) = c \\cdot u + c \\cdot v$;\n3. **Distributive law:** For all $c, d \\in \\mathbb{R}$ and all vectors $v \\in V$, $(c + d) \\cdot v = c \\cdot v + d \\cdot v$;\n4. **Associative law:** For all real numbers $c, d$ and all vectors $v \\in V$, $c \\cdot (d \\cdot v) = (c \\cdot d) \\cdot v$;\n5. **Unitary law:** For all vectors $v \\in V$, $1 \\cdot v = v$.\n\n**Examples:** Some sets that equipped with scalar addition and multiplication have a structure of vector spaces:\n\n1. $A = \\{(x, y, z) \\in \\mathbb{R}^3 : 2x - y + 3z = 0\\}$;\n2. $\\mathbb{R}^n$, with $n \\in \\mathbb{N}$;\n3. $P(n) = a_n x^n + a_{n-1} x^{n-1} + \\cdots + a_1 x + a_0$, with $a_i \\in \\mathbb{R}$;\n4. $M = [a_{i,j}]_{m \\times n}$, with $a_{i,j} \\in \\mathbb{R}$, $i = 1, \\ldots, m$ and $j = 1, \\ldots, n$.\n\nNote that $B = \\{(x, y, z) \\in \\mathbb{R}^3 : x + y - z + 1 = 0\\}$ is not a vector space.\n\nIndeed, $u = (u_1, u_2, u_1 + u_2 + 1), v = (v_1, v_2, v_1 + v_2 + 1) \\in B$, but $u + v = (u_1 + v_1, u_2 + v_2, u_1 + u_2 + v_1 + v_2 + 2) \\notin B$.\n\nBesides that, $(0, 0, 0) \\notin B$. \n\n---\n\nEdite Martins Cordeiro \nFlora Silva \nPaula Maria Barros", "id": "./materials/180.pdf" }, { "contents": "Subspace of a vector space\n\n**Definition:** Let V be a vector space, and let W be a subset of V. If W is a vector space with respect to the operations in V, then W is called a subspace of V.\n\nFor example, the vector space \\( A = \\{(x, y, z) \\in \\mathbb{R}^3 : 2x - y + 3z = 0\\} \\) is a subspace of \\( \\mathbb{R}^3 \\).\n\n**Theorem:** Let V be a vector space, with operations + and ·, and let W be a subset of V. Then W is a subspace of V if and only if the following conditions hold.\n\n- W is nonempty;\n- If \\( u \\) and \\( v \\) are any vectors in W, then \\( u + v \\in W \\) (closure under +);\n- If \\( v \\in W \\), and \\( c \\in \\mathbb{R} \\), then \\( c \\cdot v \\in W \\) (closure under ·).\n\nNote that if W is a vector subspace, then the null vector must belong to W.\n\n**Example:**\n\\( A = \\{(x, y) \\in \\mathbb{R}^2 : x - 2y = 0\\} \\) is a subspace of \\( \\mathbb{R}^2 \\);\n\nIndeed, \\( A = \\{(2y, y) : y \\in \\mathbb{R}\\} \\) and we have:\n\n- \\((0, 0) \\in A\\);\n- If \\( u = (2u_2, u_2) \\) and \\( v = (2v_2, v_2) \\), then \\( u + v = (2u_2 + 2v_2, u_2 + v_2) = (2(u_2 + v_2), u_2 + v_2) \\in A\\);\n- If \\( u = (2u_2, u_2) \\) and \\( k \\in \\mathbb{R} \\), then \\( ku = (2(ku_2), ku_2) \\in A \\).\n\n**Example:** \\( S = \\left\\{ \\begin{bmatrix} 2a \\\\ 3a + b \\\\ 3b \\end{bmatrix} : a, b \\in \\mathbb{R} \\right\\} \\) is a subspace of \\( M_{2 \\times 2} \\).\n\nHowever, the set of polynomials \\( P_1 = \\{a_0 + a_1x + a_2x^2 : a_0 + a_1 - a_2 = 3\\} \\) is not a subspace of the vector space \\( P = \\{a_0 + a_1x + a_2x^2 : a_0, a_1, a_2 \\in \\mathbb{R}\\} \\). In fact, the null polynomial does not belong to \\( P_1 \\).\n\nAlso \\( A = \\{(x, y) \\in \\mathbb{R}^2 : y = x^2\\} \\) is not a subspace of \\( \\mathbb{R}^2 \\). In fact, \\( u = (u_1, u_2^2), v = (v_1, v_1^2) \\in A \\), but \\( u + v = (u_1 + v_1, u_1^2 + v_1^2) \\) does not always belong to A. There are vectors \\((u_1, u_2), (v_1, v_2) \\in \\mathbb{R}^2 \\) such that \\( u_1^2 + v_1^2 \\neq (u_1 + v_1)^2 \\).", "id": "./materials/181.pdf" }, { "contents": "Subspace spanned\n\nRecall that:\n\n**Definition (linear combination):** For vectors \\( v_1, v_2, \\ldots, v_k \\) in a vector space \\( V \\), the vector\n\n\\[\nv = a_1 v_1 + a_2 v_2 + \\cdots + a_k v_k\n\\]\n\nis called a linear combination of the vectors \\( v_1, v_2, \\ldots, v_k \\). The scalars \\( a_i \\) are called coefficients.\n\n**Example:** Consider the vector space \\( \\mathbb{R}^3 \\).\n\nThe vector \\( v = (1, 2, 0) \\) is a linear combination of the vector set \\( A = \\{(3, 1, 2), (2, -1, 2)\\} \\), because\n\n\\[\n(1, 2, 0) = (3, 1, 2) - (2, -1, 2).\n\\]\n\nAlso \\( u = (0, -5, 2) \\) is a linear combination of the vector set \\( A = \\{(3, 1, 2), (2, -1, 2)\\} \\), because\n\n\\[\n(0, -5, 2) = -2(3, 1, 2) + 3(2, -1, 2).\n\\]\n\nThe set \\( S \\) of all vectors that are a linear combination of \\( A = \\{(3, 1, 2), (2, -1, 2)\\} \\) are all vectors \\( (x, y, z) \\in \\mathbb{R}^3 \\) such that\n\n\\[\n(x, y, z) = k_1(3, 1, 2) + k_2(2, -1, 2), \\quad k_1, k_2 \\in \\mathbb{R}.\n\\]\n\nThis equality represents the system\n\n\\[\n\\begin{align*}\n3k_1 + 2k_2 &= x \\\\\nk_1 - k_2 &= y \\\\\n2k_1 + 2k_2 &= z\n\\end{align*}\n\\]\n\n\\[\n\\begin{align*}\n3k_1 + 2k_2 &= x \\\\\nk_1 &= k_2 + y \\\\\n2k_1 + 2k_2 &= z\n\\end{align*}\n\\]\n\n\\[\n\\begin{align*}\n3(k_2 + y) + 2k_2 &= x \\\\\nk_1 &= k_2 + y \\\\\n2(k_2 + y) + 2k_2 &= z\n\\end{align*}\n\\]\n\n\\[\n\\begin{align*}\nk_2 &= \\frac{x - 3y}{5} \\\\\nk_1 &= k_2 + y \\\\\nk_2 &= \\frac{z - 2y}{4}\n\\end{align*}\n\\]\n\nNote that this system is only possible if\n\n\\[\n\\frac{x - 3y}{5} = \\frac{z - 2y}{4}.\n\\]\n\nIn conclusion, only the vectors that check the condition \\( 4x - 2y - 5z = 0 \\) are a linear combination of \\( A \\).\n\n**Definition (linear span):** Let \\( V \\) be a vector space and \\( A = \\{v_1, v_2, \\ldots, v_k\\} \\subset V \\). The linear span of \\( A \\) is the set of all linear combinations of the vectors \\( v_1, v_2, \\ldots, v_k \\), denoted by \\( \\langle A \\rangle \\), that is:\n\n\\[\n\\langle A \\rangle = \\{a_1 v_1 + a_2 v_2 + \\cdots + a_k v_k : a_1, a_2, \\ldots, a_k \\in \\mathbb{R}\\}.\n\\]\n\n**Theorem (subspace spanned):** If \\( A = \\{v_1, v_2, \\ldots, v_k\\} \\) is a set of vectors of a vector space \\( V \\), then \\( \\langle A \\rangle \\) is a subspace of \\( V \\) and is also called the subspace spanned by \\( A \\). It is the smallest subspace containing the vectors \\( v_1, v_2, \\ldots, v_k \\).\n\nNote that any vector of \\( \\mathbb{R}^2 \\) spans a line of the plane that contains \\( (0, 0) \\).\nExample: If \\( A = \\{(2, 1)\\} \\), then \\( \\langle A \\rangle = \\{(2k, k) : k \\in \\mathbb{R}\\} = \\{(x, y) \\in \\mathbb{R}^2 : x - 2y = 0\\} \\).\n\nTwo vectors of \\( \\mathbb{R}^2 \\) can define a straight line of the plane or the entire plane \\( \\mathbb{R}^2 \\).\n\nExample:\n\n- \\( \\langle\\{(2, 1), (-1, -1/2)\\}\\rangle = \\langle\\{(2, 1)\\}\\rangle = \\{(2k, k) : k \\in \\mathbb{R}\\} \\)\n\nbecause \\( \\{(2, 1), (-1, -1/2)\\} \\) is linearly dependent. According to the figure, the two vectors are collinear.\n\n- If \\( A = \\{(2, 1), (1, 2)\\} \\), then \\( \\langle A \\rangle = \\mathbb{R}^2 \\), because \\( A \\) is linearly independent and has cardinality 2. According to the figure beside, \\( v_1 = (2, 1) \\) and \\( v_2 = (1, 2) \\) have different directions and any vector of \\( \\mathbb{R}^2 \\) can be written as the sum of a scalar multiple of \\( v_1 \\) with a scalar multiple of \\( v_2 \\).\n\nExample: The linear space \\( \\langle B \\rangle \\) such that \\( B = \\{(1, 0, 1), (1, 2, 0), (0, 1, 1)\\} \\) is the set\n\n\\[\nS = \\{(x, y, z) \\in \\mathbb{R}^3 : (x, y, z) = k_1(1, 0, 1) + k_2(1, 2, 0) + k_3(0, 2, -1), k_1, k_2, k_3 \\in \\mathbb{R}\\}.\n\\]\n\nThat is, the set of vectors \\( (x, y, z) \\in \\mathbb{R}^3 \\) such that the system\n\n\\[\n\\begin{align*}\nk_1 + k_2 &= x \\\\\n2k_2 + k_3 &= y \\\\\nk_1 + k_3 &= z\n\\end{align*}\n\\]\n\nis possible. Through its expanded matrix\n\n\\[\n\\begin{bmatrix}\n1 & 1 & 0 & | & x \\\\\n0 & 2 & 2 & | & y \\\\\n1 & 0 & -1 & | & z\n\\end{bmatrix} \\rightarrow \\begin{bmatrix}\n1 & 1 & 0 & | & x \\\\\n0 & 2 & 2 & | & y \\\\\n0 & -1 & -1 & | & z - x\n\\end{bmatrix} \\rightarrow \\begin{bmatrix}\n1 & 1 & 0 & | & x \\\\\n0 & 2 & 2 & | & y \\\\\n0 & 0 & 0 & | & y + 2z - 2x\n\\end{bmatrix}\n\\]\n\nwe can conclude that the system is possible if \\(-2x + y + 2z = 0\\). That is,\n\n\\[\nS = \\{(x, y, z) \\in \\mathbb{R}^3 : -2x + y + 2z = 0\\},\n\\]\n\nwhich represents a plan of \\( \\mathbb{R}^3 \\).", "id": "./materials/182.pdf" }, { "contents": "Change of Basis\n\nIn a vector space, the coordinates of a vector is always with respect to a basis and if we omit the basis, we naturally assume it to be the standard basis.\n\nWithout loss of generality, consider \\( A = \\{v_1, v_2, v_3\\} \\) and \\( B = \\{u_1, u_2, u_3\\} \\) two bases of three-dimensional space \\( \\mathbb{R}^3 \\). For all \\( v \\in V \\), \\( v_A = (k_1, k_2, k_3) \\) means that \\( v = k_1v_1 + k_2v_2 + k_3v_3 \\) and \\( v_B = (t_1, t_2, t_3) \\) means that \\( v = t_1u_1 + t_2u_2 + t_3u_3 \\).\n\nIn particular, we can write the vectors \\( u_1, u_2, u_3 \\) of \\( B \\) in base \\( A \\) as follows:\n\n\\[\n\\begin{align*}\n u_1 &= a_{11}v_1 + a_{21}v_2 + a_{31}v_3 \\\\\n u_2 &= a_{12}v_1 + a_{22}v_2 + a_{32}v_3 \\\\\n u_3 &= a_{13}v_1 + a_{23}v_2 + a_{33}v_3\n\\end{align*}\n\\]\n\nThen,\n\n\\[\nt_1u_1 + t_2u_2 + t_3u_3 = t_1(a_{11}v_1 + a_{21}v_2 + a_{31}v_3) + t_2(a_{12}v_1 + a_{22}v_2 + a_{32}v_3) + t_3(a_{13}v_1 + a_{23}v_2 + a_{33}v_3)\n\\]\n\nAssociating the terms in \\( v_i \\), we have:\n\n\\[\nt_1u_1 + t_2u_2 + t_3u_3 = (t_1a_{11} + t_2a_{12} + t_3a_{13})v_1 + (t_1a_{21} + t_2a_{22} + t_3a_{23})v_2 + (t_1a_{31} + t_2a_{32} + t_3a_{33})v_3,\n\\]\n\nAs the coordinates in relation to a base are unique, we have\n\n\\[\nt_1a_{11} + t_2a_{12} + t_3a_{13} = k_1, t_1a_{21} + t_2a_{22} + t_3a_{23} = k_2 \\quad \\text{and} \\quad t_1a_{31} + t_2a_{32} + t_3a_{33} = k_3.\n\\]\n\nThat is:\n\n\\[\n\\begin{bmatrix}\n a_{11} & a_{12} & a_{13} \\\\\n a_{21} & a_{22} & a_{23} \\\\\n a_{31} & a_{32} & a_{33}\n\\end{bmatrix}\n\\begin{bmatrix}\n t_1 \\\\\n t_2 \\\\\n t_3\n\\end{bmatrix}\n= \n\\begin{bmatrix}\n k_1 \\\\\n k_2 \\\\\n k_3\n\\end{bmatrix}\n\\]\n\nIn short, we can write\n\n\\[\nP_A^B \\cdot v_A = v_B,\n\\]\n\nwhere \\( P_A^B \\) is called the change matrix from \\( B \\) to base \\( A \\).\n\nIn particular, if \\( B = \\{v_1, \\ldots, v_n\\} \\) is a basis of a vector space \\( V \\) and the matrix whose columns are the vectors of \\( B \\),\n\n\\[\n\\begin{bmatrix}\n v_1 & v_2 & \\cdots & v_n\n\\end{bmatrix},\n\\]\n\nis a square matrix, then its determinant is nonzero.\n\nRemember that:\n\n- The standard basis of \\( \\mathbb{R}^2 \\) is \\( \\{(1, 0), (0, 1)\\} \\);\n- The standard basis of \\( \\mathbb{R}^3 \\) is \\( \\{(1, 0, 0), (0, 1, 0), (0, 0, 1)\\} \\).\nFor example, in $\\mathbb{R}^2$, $v = (2, 3)$ means\n\n$$v = 2(1, 0) + 3(0, 1) = \\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix} \\cdot \\begin{bmatrix} 2 \\\\ 3 \\end{bmatrix} \\text{ or }$$\n\n$$v = 1(3, 1) - 1(1, -2) = \\begin{bmatrix} 3 & 1 \\\\ 1 & -2 \\end{bmatrix} \\cdot \\begin{bmatrix} 1 \\\\ -1 \\end{bmatrix} \\text{ or }$$\n\n$$v = -\\frac{4}{3}(1, -1) + \\frac{5}{3}(2, 1) = \\begin{bmatrix} 1 & 2 \\\\ -1 & 1 \\end{bmatrix} \\cdot \\begin{bmatrix} -\\frac{4}{3} \\\\ \\frac{5}{3} \\end{bmatrix}.$$\n\nSo, the coordinates of $v$ with respect to:\n\n- the standard basis are $v = (2, 3)$;\n- the basis $A = \\{(3, 1), (1, -2)\\}$ are $v_A = (1, -1)$;\n- the basis $B = \\{(1, -1), (2, 1)\\}$ are $v_B = \\left(\\begin{array}{c} -\\frac{4}{3} \\\\ \\frac{5}{3} \\end{array}\\right)$.\n\nNotice that\n\n$$v_A = \\begin{bmatrix} 3 & 1 \\\\ 1 & -2 \\end{bmatrix}^{-1} \\cdot v \\text{ and } v_B = \\begin{bmatrix} 1 & 2 \\\\ -1 & 1 \\end{bmatrix}^{-1} \\cdot v.$$\n\nBesides that,\n\n$$\\begin{bmatrix} 3 & 1 \\\\ 1 & -2 \\end{bmatrix} \\cdot \\begin{bmatrix} 1 \\\\ -1 \\end{bmatrix} = \\begin{bmatrix} 1 & 2 \\\\ -1 & 1 \\end{bmatrix} \\cdot \\begin{bmatrix} -\\frac{4}{3} \\\\ \\frac{5}{3} \\end{bmatrix} \\Leftrightarrow \\begin{bmatrix} 1 \\\\ -1 \\end{bmatrix} = \\begin{bmatrix} 3 & 1 \\\\ 1 & -2 \\end{bmatrix}^{-1} \\cdot \\begin{bmatrix} 1 & 2 \\\\ -1 & 1 \\end{bmatrix} \\cdot \\begin{bmatrix} -\\frac{4}{3} \\\\ \\frac{5}{3} \\end{bmatrix}.$$\n\nThat is,\n\n$$v_A = \\begin{bmatrix} 3 & 1 \\\\ 1 & -2 \\end{bmatrix}^{-1} \\cdot \\begin{bmatrix} 1 & 2 \\\\ -1 & 1 \\end{bmatrix} \\cdot v_B.$$\n\nSo, if $A$ and $B$ are two bases of a $n$ dimensional vector space and the matrices $A = [a_{i,j}]_{n \\times n}$ and $B = [b_{i,j}]_{n \\times n}$, whose columns are the vectors of bases $A$ and $B$ (respectively) are square matrices, then the coordinates of any vector $v \\in V$ in bases $A$ and $B$ are related as follows:\n\n$$v_A = A^{-1} \\cdot B \\cdot v_B \\text{ and } v_B = B^{-1} \\cdot A \\cdot v_A.$$\n\nThe product $A^{-1} \\cdot B$ corresponds to the change matrix from $B$ to base $A$, that is:\n\n$$P_A^B = A^{-1} \\cdot B.$$\n\nWe still have\n\n**Properties:** If $A$ and $B$ are basis of a $V$ vector space of $n$ dimension, then:\n\n1. $P_A^B = (P_B^A)^{-1}$.\n2. Given $v \\in V$, we have $[v]_A = P_A^B \\cdot [v]_B$;\n3. Given $v \\in V$, we have $[v]_B = (P_A^B)^{-1} \\cdot [v]_A$;\n4. $P_C^B = P_A^B \\cdot P_A^C$. \n\n---\n\nEdite Martins Cordeiro \nFlora Silva \nPaula Maria Barros", "id": "./materials/183.pdf" }, { "contents": "Bases and dimension of a vector space\n\nBases of a vector space\n\nDefinition (basis): Let $V$ be a vector space. Then the subset $A = \\{v_1, v_2, \\ldots, v_t\\}$ of $V$ is said to be a basis for $V$ if:\n\n1. $A$ is a linearly independent set of vectors;\n2. $A$ spans $V$, that is, $V = \\langle A \\rangle$.\n\nExample: The set $A = \\{(1, 1), (1, -2)\\}$ is a basis of $\\mathbb{R}^2$.\n\nIn fact:\n\n1. $k_1(1, 1) + k_2(1, -2) = (0, 0) \\iff \\begin{cases} k_1 + k_2 = 0 \\\\ k_1 - 2k_2 = 0 \\end{cases} \\iff \\begin{cases} k_2 = 0 \\\\ k_1 = 0 \\end{cases}$, that is, $A$ is linearly independent;\n\n2. $A$ spans $\\mathbb{R}^2$, because for any $(x, y) \\in \\mathbb{R}^2$ there are $k_1, k_2 \\in \\mathbb{R}$ such that $k_1(1, 1) + k_2(1, -2) = (x, y)$.\n\nIndeed $\\begin{cases} k_1 + k_2 = x \\\\ k_1 - 2k_2 = y \\end{cases} \\iff \\begin{cases} k_2 = (x + y)/3 \\\\ k_1 = (2x - y)/3 \\end{cases}$.\n\nExample: The set $A = \\{(1, 1, 2), (1, -2, 0), (2, -1, 2)\\}$ is not a basis of $\\mathbb{R}^3$, because $A$ is linearly dependent.\n\nIn fact, $(1, 1, 2) = -(1, -2, 0) + (2, -1, 2)$, that is, the first is a linear combination of the others.\n\nTheorem: If $V$ is a vector space, then a smallest spanning set is a basis of $V$.\n\nExample: The set $B = \\{(1, 1, 2), (1, -2, 0), (1, -1, 1)\\}$ is a basis of $\\mathbb{R}^3$.\n\nIn fact, $k_1(1, 1, 2) + k_2(1, -2, 0) + k_3(1, -1, 1) = (0, 0, 0) \\iff \\begin{cases} k_1 + k_2 + k_3 = 0 \\\\ k_1 - 2k_2 - k_3 = 0 \\\\ 2k_1 + k_3 = 0 \\end{cases} \\iff \\begin{cases} -k_1 + k_2 = 0 \\\\ 3k_1 - 2k_2 = 0 \\\\ k_3 = -2k_1 \\end{cases} \\iff \\begin{cases} k_1 = k_2 \\\\ k_1 = 0 \\\\ k_3 = -2k_1 \\end{cases}$.\n\nThat is, the vector equation has the unique solution $k_1 = k_2 = k_3 = 0$.\n\nIn addition, $B$ spans $\\mathbb{R}^3$. In fact, any $(x, y, z) \\in \\mathbb{R}^3$ is a linear combination of the vectors of $B$. That is, the linear system $k_1(1, 1, 2) + k_2(1, -2, 0) + k_3(1, -1, 1) = (x, y, z)$ in variables $k_1, k_2, k_3$ is possible for any values of $x, y, z \\in \\mathbb{R}$.\n\nNote that $B \\cup \\{v\\}$, for any $v \\in \\mathbb{R}^3$, is not a basis of $\\mathbb{R}^3$. In fact, $B \\cup \\{v\\}$ is linearly dependent.\n**Dimension of a vector space**\n\n**Theorem:** Consider $V$ a vector space with a basis $B = \\{v_1, v_2, \\ldots, v_n\\}$ of $n$ vectors. Then any set of $n + 1$ vectors is linearly dependent. Besides that, any basis of $V$ of $n$ vectors is linearly independent, that is, there are other solutions besides the null solution.\n\n**Example:** The set $C = \\{(1, 0, 0), (0, 1, 0), (0, 0, 1)\\}$ is a basis of $\\mathbb{R}^3$, because $C$ is linearly independent and any $(x, y, z) = x(1, 0, 0) + y(0, 1, 0) + z(0, 0, 1)$.\n\nThe set $D = C \\cup \\{(-1, 2, 0)\\}$ is not a basis of $\\mathbb{R}^3$. In fact,\n\n$$k_1(1, 0, 0) + k_2(0, 1, 0) + k_3(0, 0, 1) + k_4(-1, 2, 0) = (0, 0, 0)$$\n\nrepresents a system whose expanded matrix is\n\n$$\\begin{bmatrix}\n1 & 0 & 0 & -1 & | & 0 \\\\\n0 & 1 & 0 & 2 & | & 0 \\\\\n0 & 0 & 1 & 0 & | & 0\n\\end{bmatrix}$$\n\nwhich is linearly dependent, that is, there are other solutions besides the null solution.\n\nAlso the space generated for $S = \\{(1, 0, 0), (0, 1, 0)\\} \\subset C$ is its own subspace that represents a plane, it is not $\\mathbb{R}^3$. Indeed, the vectors $(x, y, z)$ that $(x, y, z) = k_1(1, 0, 0) + k_2(0, 1, 0)$ are those for which the system whose expanded matrix\n\n$$\\begin{bmatrix}\n1 & 0 & | & x \\\\\n0 & 1 & | & y \\\\\n0 & 0 & | & z\n\\end{bmatrix}$$\n\nis possible. This happens if $z = 0$, that is, $\\langle S \\rangle = \\{(x, y, 0) : x, y \\in \\mathbb{R}\\}$.\n\n**Definition (dimension):** The number $n$ of vectors of any basis of vector space $V$ is called the dimension of $V$ and is denoted by $\\dim(V)$.\n\n**Examples:**\n\n- We say that $\\mathbb{R}^2$ is a space two-dimensional because any of its bases has 2 vectors of $\\mathbb{R}^2$;\n- We say that $\\mathbb{R}^3$ is a space three-dimensional because any of its bases has 3 vectors of $\\mathbb{R}^3$.\n\n**Example:** The set $A = \\{(1, 1), (1, -2)\\}$ is a basis of $\\mathbb{R}^2$ and the set $C = \\{(1, 0), (0, 1)\\}$ is other basis of $\\mathbb{R}^2$, called the canonical basis of $\\mathbb{R}^2$.\n\nFurthermore, all bases of $\\mathbb{R}^2$ are sets of two linearly independent vectors of $\\mathbb{R}^2$, so $\\mathbb{R}^2$ has dimension 2.\n\nNote that:\n\n- $\\mathbb{R}^2 = \\{(x, y) : x, y \\in \\mathbb{R}\\}$ has dimension 2 and the generic vector $(x, y)$ has 2 free variables.\n- $\\mathbb{R}^3 = \\{(x, y, z) : x, y, z \\in \\mathbb{R}\\}$ has dimension 3 and the generic vector $(x, y, z)$ has 3 free variables.\n\nGenerally, the dimension of a vector space is equal to the number of free variables in its generic vector.\n\n**Example:** The set $A = \\{(1, -1, 0), (2, 0, 1)\\}$ spans\n\n$$\\{(x, y, z) \\in \\mathbb{R}^3 : (x, y, z) = k_1(1, -1, 0) + k_2(2, 0, 1)\\}.$$ \n\nIn fact, the condition that defines $\\langle A \\rangle$ represents a system whose expanded matrix is:\n\n$$\\begin{bmatrix}\n1 & 2 & | & x \\\\\n-1 & 0 & | & y \\\\\n0 & 1 & | & z\n\\end{bmatrix} \\rightarrow \\begin{bmatrix}\n1 & 2 & | & x \\\\\n0 & 2 & | & y + x \\\\\n0 & 1 & | & z\n\\end{bmatrix} \\rightarrow \\begin{bmatrix}\n1 & 2 & | & x \\\\\n0 & 2 & | & y + x \\\\\n0 & 0 & | & 2z - x - y\n\\end{bmatrix}$$\n\nThis system is possible, if $x + y - 2z = 0$. This means that $\\langle A \\rangle = \\{(2z - y, y, z) : y, z \\in \\mathbb{R}\\}$. The generic vector has 2 free variables and therefore $\\langle A \\rangle$ has dimension 2.", "id": "./materials/184.pdf" }, { "contents": "Orthogonal sets and basis\n\nRecall that the dot product:\n\n\\[ u \\cdot v = ||u|| ||v|| \\cos(\\theta), \\quad \\theta = \\hat{u} \\hat{v} \\in [0, \\pi]. \\]\n\nIf, for example, \\( u = (u_1, u_2, u_3), v = (v_1, v_2, v_3) \\in \\mathbb{R}^3 \\), we also have\n\n\\[ u \\cdot v = u_1v_1 + u_2v_2 + u_3v_3. \\]\n\nAn inner product of 2 vectors is a generalization of the dot product, it is a way to multiply vectors, whose product being a scalar. More precisely, an inner product, \\( \\langle \\cdot, \\cdot \\rangle \\), in a real vector space \\( V \\) is any operator that satisfies the following properties:\n\n1. \\( (u + v)/w = (u/w) + (v/w), \\forall u, v, w \\in V. \\)\n2. \\( (ku)/v = k(u/v)u/w + v/w, \\forall u, v \\in V, \\forall k \\in \\mathbb{R}. \\)\n3. \\( u/v = v/u, \\forall u, v \\in V. \\)\n4. \\( u/u \\leq 0 \\) if and only if \\( u = 0. \\)\n\nThe vector space \\( V \\) together with \\( \\langle \\cdot, \\cdot \\rangle \\) is called an inner product space.\n\nAn inner product space \\( V \\) induces a norm, that is, a notion of length of a vector:\n\n\\[ ||v|| = \\sqrt{v \\cdot v}. \\]\n\nIn particular, the usual inner product (scalar product) in the vector space \\( \\mathbb{R}^2 \\) induces a norm\n\n\\[ ||v|| = \\sqrt{v \\cdot v}. \\]\n\nThat is, if \\( v = (v_1, v_2) \\), then \\( v \\cdot v = v_1^2 + v_2^2 \\) and \\( ||v|| = \\sqrt{v_1^2 + v_2^2} \\)\n\n**Definition:** The set \\( A = \\{v_1, v_2, \\ldots, v_k\\} \\in V \\setminus \\{0\\} \\) is an orthogonal set if each vector of \\( A \\) is orthogonal to each of the other vectors in the set, that is,\n\n\\[ v_i \\cdot v_j = 0 \\quad \\text{for} \\quad i \\neq j. \\]\n\nIf, in addition, all vectors are of unit norm, \\( ||v_i|| = 1 \\), then \\( \\{v_1, v_2, \\ldots, v_k\\} \\) is called an orthonormal set.\n\n**Examples:**\n\n1. The set \\( A = \\{(1, -2), (4, 2)\\} \\subset \\mathbb{R}^2 \\) is orthogonal but is not orthonormal.\n \n In fact \\( (1, -2) \\cdot (4, 2) = 1 \\times 4 - 2 \\times 2 = 0 \\), but \\( ||(1, -2)|| = \\sqrt{(1, -2) \\cdot (1, -2)} = \\sqrt{5} \\neq 1. \\)\n\n2. The set \\( B = \\{(1, 0, 0), (0, 1, 0)\\} \\subset \\mathbb{R}^3 \\) is orthonormal.\n**Theorem:** Any orthogonal set is linearly independent.\n\n**Definition:** An orthonormal basis for an inner product space $V$ with finite dimension is a basis for $V$ whose vectors are orthonormal to each other and are all unit vectors.\n\nThe following are examples of orthonormal bases:\n- The standard basis of $\\mathbb{R}^2$, $A = \\{(1, 0), (0, 1)\\}$;\n- The standard basis of $\\mathbb{R}^3$, $B = \\{(1, 0, 0), (0, 1, 0), (0, 0, 1)\\}$;\n- The basis $C = \\left\\{ \\left( \\frac{3}{5}, \\frac{4}{5} \\right), \\left( \\frac{4}{5}, -\\frac{3}{5} \\right) \\right\\}$ of $\\mathbb{R}^2$;\n\nLet $V$ an inner product space with an inner product, “$\\cdot$”. The following properties give us the coordinates of $V$ vectors with respect to orthogonal bases.\n\n**Theorem:** If $B = \\{w_1, w_2, \\ldots, w_n\\}$ is an orthonormal basis of $V$, then for any $v \\in V$ we have\n\n$$v = (v \\cdot w_1) w_1 + (v \\cdot w_2) w_2 + \\cdots + (v \\cdot w_n) w_n.$$ \n\n**Example:** $B = \\left\\{ \\left( \\frac{3}{5}, \\frac{4}{5} \\right), \\left( \\frac{4}{5}, -\\frac{3}{5} \\right) \\right\\}$ is an orthonormal basis of $\\mathbb{R}^2$. We can write any vector $v = (x, y)$ as\n\n$$v_B = \\left( \\frac{3x + 4y}{5}, \\frac{4x - 3y}{5} \\right).$$\n\n**Theorem:** Let $A = \\{w_1, w_2, \\ldots, w_r\\} \\subset \\mathbb{R}^n \\setminus \\{0\\}$ such that $w_i \\cdot w_j = 0$, for $i \\neq j$ and $i, j \\in \\{1, 2, \\ldots, r\\}$. Then:\n\n1. $A$ is a basis of $\\langle A \\rangle$;\n2. For any $v \\in \\langle A \\rangle$, we have $v = k_1 w_1 + k_2 w_2 + \\cdots + k_r w_r$, with\n\n$$k_i = \\frac{v \\cdot w_i}{||w_i||^2}.$$ \n\n**Example:** $A = \\{(1, -1, 0), (0, 0, 2)\\} \\subset \\mathbb{R}^3$ is an orthogonal basis, because $(1, -1, 0) \\cdot (0, 0, 2) = 0$. The norms of vectors are $||(1, -1, 0)|| = \\sqrt{2}$ and $||(0, 0, 2)|| = 2$.\n\nBesides that, $\\langle A \\rangle = \\{(x, -x, z) : x, z \\in \\mathbb{R}\\}$ and any $v = (x, -x, z) \\in \\langle A \\rangle$ is such that\n\n$$(x, -x, z) = \\frac{(x, -x, z) \\cdot (1, -1, 0)}{||(1, -1, 0)||^2} (1, -1, 0) + \\frac{(x, -x, z) \\cdot (0, 0, 2)}{||(0, 0, 2)||^2} (0, 0, 2) = \\frac{2x}{2} (1, -1, 0) + \\frac{2z}{4} (0, 0, 2).$$", "id": "./materials/185.pdf" }, { "contents": "Linear differential equation of first order\n\nLet’s see how to solve a linear first order differential equation. The goal is to obtain a solution in the form \\( x = x(t) \\).\n\nThe general form of a linear first order differential equation is:\n\n\\[\nx'(t) + a(t)x(t) = b(t), \\quad t \\in I, I \\subset \\mathbb{R},\n\\]\n\nwhere both \\( a(t) \\) and \\( b(t) \\) are continuous functions on the interval \\( I \\).\n\nIf the differential equation is not in this form then the process we are going to use will not work.\n\nWe assume that there is some function called an integrating factor, \\( \\mu(t) \\), to multiply equation (1) such that\n\n\\[\n\\mu(t)x'(t) + \\mu(t)a(t)x(t) = \\mu(t)b(t),\n\\]\n\nIf \\( \\mu(t) \\neq 0, t \\in I, \\) exist, it will satisfy the following relation:\n\n\\[\n\\mu(t)a(t) = \\mu'(t)\n\\]\n\nSo substituting (3) into (2) we now arrive at\n\n\\[\n\\mu(t)x'(t) + \\mu'(t)x(t) = \\mu(t)b(t).\n\\]\n\nBut \\( \\mu(t)x'(t) + \\mu'(t)x(t) = (\\mu(t)x(t))' \\) and we replace at (4) so we obtain\n\n\\[\n(\\mu(t)x(t))' = \\mu(t)b(t)\n\\]\n\nWe integrate both sides:\n\n\\[\n\\mu(t)x(t) = \\int \\mu(t)b(t) + C\n\\]\n\nwhere \\( C \\) is a real constant of integration.\n\nWe obtain the general solution\n\n\\[\nx(t) = \\mu^{-1}(t) \\left( \\int \\mu(t)b(t)dt + C \\right), t \\in I.\n\\]\n\nWe need to determine the function \\( \\mu(t) \\). We start relation (3). Divide both sides by \\( \\mu(t) \\) and integrate\n\n\\[\n\\frac{\\mu'(t)}{\\mu(t)} = a(t) \\Leftrightarrow (\\ln \\mu(t))' = a(t) \\Leftrightarrow \\ln \\mu(t) + C_1 = \\int a(t)dt\n\\]\n\n\\[\n\\ln \\mu(t) = \\int a(t)dt - C_1 \\Leftrightarrow \\mu(t) = e^{\\int a(t)dt - C_1}\n\\]\n\nIt is inconvenient to have the \\( C_1 \\) in the exponent so we’re going to get it out of the exponent in the following way.\n\n\\[\n\\mu(t) = e^{\\int a(t)dt - C_1} = e^{-C_1}e^{\\int a(t)dt},\n\\]\nwhere $e^{-C_1}$ is a constant. Because we need a function, not all the function with this propriety, we can choose $C_1 = 0$, so\n\n$$\\mu(t) = e^{\\int a(t)dt}. \\quad (7)$$\n\nSo substituting (7) into (6) we now arrive at\n\n$$x(t) = e^{-\\int a(t)dt} \\left( \\int b(t)e^{\\int a(t)dt}dt + C \\right). \\quad (8)$$\n\n**Solution Process**\n\nThe solution process for linear differential equation of first order is as follows:\n\n1. Put the differential equation in the correct initial form, (1).\n2. Find the function $\\mu(t)$, using (7).\n3. Multiply the both sides of differential equation by $\\mu(t)$ and verify that the left side becomes the product rule $(\\mu(t)x(t))'$ and write it as such (5).\n4. Integrate both sides and take care to the constant of integration.\n5. Solve for the solution $x(t)$.\n\n**Example 1** Find the solution to the following differential equation\n\n$$x(t) + 2tx(t) = 4t, t \\in R.$$ \n\n**Solution.**\n\n1. We observe that the differential equation is in the correct form.\n2. Find the integrating factor, $\\mu(t)$.\n \n $a(t) = 2t, \\mu(t) = e^{\\int 2tdt} = e^{t^2}.$\n3. Multiply the both sides of the differential equation by $\\mu(t) = e^{t^2}$,\n \n $x'(t)e^{t^2} + 2te^{t^2}x(t) = 4te^{t^2}.$\n \n Verify that the left side becomes the product rule $(\\mu(t)x(t))'$,\n \n $$\\left( x(t)e^{t^2} \\right)' = x'(t)e^{t^2} + 2te^{t^2}x(t),$$\n \n and write it as such\n \n $$\\left( x(t)e^{t^2} \\right)' = 4te^{t^2}.$$\n4. Integrate both sides\n \n $x(t)e^{t^2} = 2e^{t^2} + C.$\n5. Solve for the solution $x(t)$\n \n $$x(t) = \\left( 2e^{t^2} + C \\right) e^{-t^2}, x(t) = 2 + Ce^{-t^2}.$$\nExercise 2 Find the solution to the following differential equation\n\n\\[ tx'(t) + x(t) = 3t^2, \\quad t \\neq 0. \\]\n\nSolution.\n\n1. We convert this equation into the standard form. So we divide both part by \\( t \\):\n\n\\[ x'(t) + \\frac{1}{t} x(t) = 3t, \\quad t \\neq 0. \\]\n\n2. Find the integrating factor, \\( \\mu(t) \\).\n\n\\[ a(t) = \\frac{1}{t}, \\quad \\mu(t) = e^{\\int \\frac{1}{t} dt} = e^{\\ln t} = t. \\]\n\n3. Multiply both sides of the differential equation by \\( \\mu(t) = t \\),\n\n\\[ x'(t)t + x(t) = 3t^2. \\]\n\nVerify that the left side becomes the product rule \\((\\mu(t)x(t))'\\), \\((x(t)t)' = x'(t)t + x(t)\\), and write it as such\n\n\\[ (x(t)t)' = 3t^2. \\]\n\n4. Integrate both sides\n\n\\[ x(t)t = t^3 + C. \\]\n\n5. Solve for the solution \\( x(t) = t^2 + \\frac{C}{t} \\).\n\nExercise 3 Find the solution to the following differential equation\n\n\\[\n\\begin{cases}\n x'(t) \\cos t + x(t) \\sin t + 4 \\cos^3 t = 0, \\quad t \\in \\left(-\\frac{\\pi}{2}, \\frac{\\pi}{2}\\right), \\\\\n x(0) = 1\n\\end{cases}\n\\]\n\nSolution.\n\n1. We convert this equation into the standard form. So we divide both part by \\( \\cos t \\):\n\n\\[ x'(t) + \\frac{\\sin t}{\\cos t} x(t) = -4 \\cos^2 t, \\quad t \\in \\left(-\\frac{\\pi}{2}, \\frac{\\pi}{2}\\right). \\]\n\n2. Find the integrating factor, \\( \\mu(t) \\).\n\n\\[ a(t) = \\frac{\\sin t}{\\cos t}, \\quad \\mu(t) = e^{\\int \\frac{\\sin t}{\\cos t} dt} = e^{-\\ln \\cos t} = \\frac{1}{\\cos t}. \\]\n\n3. Multiply both sides of the differential equation by \\( \\mu(t) = \\frac{1}{\\cos t} \\),\n\n\\[ x'(t) \\frac{1}{\\cos t} + \\frac{\\sin t}{\\cos^2 t} x(t) = -4 \\cos t, \\quad t \\in \\left(-\\frac{\\pi}{2}, \\frac{\\pi}{2}\\right). \\]\n\nVerify that the left side becomes the product rule \\((\\mu(t)x(t))'\\), \\((x(t)\\frac{1}{\\cos t})' = x'(t)\\frac{1}{\\cos t} + \\frac{\\sin t}{\\cos^2 t} x(t)\\), and write it as such\n\n\\[ (x(t)\\frac{1}{\\cos t})' = -4 \\cos t. \\]\n\n4. Integrate both sides\n\n\\[ x(t)\\frac{1}{\\cos t} = -4 \\sin t + C. \\]\n\n5. Solve for the solution \\( x(t) = (-4 \\sin t + C) \\cos t \\).\n\nUse condition to find the constant \\( C \\),\n\n\\[ x(0) = C = 1 \\Rightarrow x(t) = (-4 \\sin t + 1) \\cos t. \\]\n\nAuthor: Ariadna Lucia Pletea", "id": "./materials/186.pdf" }, { "contents": "Bernoulli differential equation\n\nIn this section we’ll see how to solve the Bernoulli differential equation. The general form of Bernoulli differential equation is,\n\n\\[ x'(t) + a(t)x(t) = b(t)x^n(t), \\quad t \\in I, I \\subset \\mathbb{R}, \\quad (1) \\]\n\nwhere both \\( a(t) \\) and \\( b(t) \\) are continuous functions on the interval \\( I \\).\n\n**Remark.** If \\( n = 0 \\) or \\( n = 1 \\) then the equation is linear and we already know how to solve it in these case. Therefore, in this section we’re going to obtain solutions for values of \\( n \\) other than these two.\n\nIn order to solve these we’ll first divide the differential equation by \\( x^n(t) \\) to get,\n\n\\[ x'(t)x^{-n}(t) + a(t)x^{1-n}(t) = b(t). \\quad (2) \\]\n\nWe use the substitution \\( y = x^{1-n} \\) to convert this into a differential equation in terms of \\( y \\). As we will see this will lead to a differential equation that we can solve.\n\nWe must be careful when we determine \\( y' \\). We’ll need to use the chain rule for differentiation,\n\n\\[ y'(t) = (x^{1-n}(t))' = (1-n)x^{-n}(t)x'(t). \\]\n\nWe obtain\n\n\\[ x^{-n}(t)x'(t) = \\frac{1}{1-n}y'(t). \\]\n\nNow, using substitution into the differential equation, we\n\n\\[ \\frac{1}{1-n}y'(t) + a(t)y(t) = b(t). \\quad (3) \\]\n\nThis is a linear differential equation that we can solve for \\( y \\) and once we have this in hand we can also get the solution to the original differential equation by plugging \\( y \\) back into our substitution and solving for \\( x \\).\n\n**Solution Process**\n\nThe solution process for Bernoulli differential equation is as follows.\n\n1. We observe if the differential equation is in the correct form (1). If the differential equation is not in the correct form we do it.\n2. Observe that \\( n \\) must be different of 0 or 1.\n3. We divide the differential equation by \\( x^n(t) \\) to get (2).\n4. We use the substitution \\( y = x^{1-n} \\). We determine \\( y' \\),\n\n\\[\ny'(t) = (1 - n) x^{-n}(t)x'(t).\n\\] (4)\n\n5. We make the substitution (4) on equation (2) and obtain (3).\n\n6. This is a linear differential equation.\n\n**Example 1** Find the solution to the following differential equation:\n\n\\[\n\\begin{align*}\nx'(t) + \\frac{4}{t}x(t) &= t^3x^2(t), \\quad t > 0, \\\\\nx(2) &= -1.\n\\end{align*}\n\\]\n\n**Solution.**\n\n1. We observe if the differential equation is in the correct form (1).\n\n2. Observe that \\( n = 2 \\).\n\n3. We divide the differential equation by \\( x^2(t) \\)\n\n\\[\nx'(t)x^{-2}(t) + \\frac{4}{t}x^{-1}(t) = t^3.\n\\]\n\n4. We determine \\( y' \\).\n\n\\[\ny'(t) = -x^{-2}(t)x'(t).\n\\]\n\n5. We use the substitution \\( y = x^{-1} \\). We make the substitution on equation\n\n\\[\nx'(t)x^{-2}(t) + \\frac{4}{t}x^{-1}(t) = t^3\n\\]\n\nand obtain\n\n\\[\n-y'(t) + \\frac{4}{t}y(t) = t^3\n\\]\n\n6. This is a linear differential equation.\n\n\\[\ny'(t) - \\frac{4}{t}y(t) = -t^3.\n\\]\n\nThe solution is\n\n\\[\ny(t) = Ct^4 - t^4 \\ln t \\Rightarrow x^{-1}(t) = Ct^4 - t^4 \\ln t.\n\\]\n\nNow we need to determine the constant of integration.\n\n\\[\nx^{-1}(2) = C2^4 - 2^4 \\ln 2 \\Rightarrow -1 = C2^4 - 2^4 \\ln 2 \\Rightarrow C = \\ln 2 - \\frac{1}{16}\n\\]\n\n\\[\nx^{-1}(t) = (\\ln 2 - \\frac{1}{16}) t^4 - t^4 \\ln t \\Rightarrow x(t) = \\frac{1}{(\\ln 2 - \\frac{1}{16} - \\ln t)t^4}.\n\\]\n\n**Example 2** Find the solution to the following differential equation:\n\n\\[\n\\begin{align*}\ntx'(t) + x(t) + 3t (\\ln t) x^2(t), \\quad t > 0, \\\\\nx(1) &= 5.\n\\end{align*}\n\\]\n\n**Solution.**\n1. We observe if the differential equation is not in the correct form (1). So we divide both part by $t$:\n\n$$x'(t) + \\frac{1}{t}x(t) = -3(\\ln t) x^2(t)$$\n\n2. Observe that $n = 2$.\n\n3. We divide the differential equation by $x^2(t)$\n\n$$x'(t)x^{-2}(t) + \\frac{1}{t}x^{-1}(t) = -3(\\ln t).$$\n\n4. We determine $y'$.\n\n$$y'(t) = -x^{-2}(t)x'(t).$$\n\n5. We use the substitution $y = x^{-1}$. We make the substitution on equation\n\n$$x'(t)x^{-2}(t) + \\frac{1}{t}x^{-1}(t) = -3(\\ln t)$$\n\nand obtain\n\n$$-y'(t) + \\frac{1}{t}y(t) = -3(\\ln t).$$\n\n6. This is a linear differential equation.\n\n$$y'(t) - \\frac{1}{t}y(t) = 3(\\ln t).$$\n\nThe solution is\n\n$$y(t) = Ct + \\frac{3}{2}t \\ln t \\Rightarrow x^{-1}(t) = Ct + \\frac{3}{2}t \\ln t.$$ \n\nNow we need to determine the constant of integration.\n\n$$x^{-1}(1) = C \\Rightarrow C = \\frac{1}{5}$$\n\n$$x^{-1}(t) = \\frac{1}{5}t + \\frac{3}{2}t \\ln t \\Rightarrow x(t) = \\frac{10}{(2+15 \\ln t)t}, \\ t > 0.$$ \n\nAuthor: Ariadna Lucia Pletea", "id": "./materials/187.pdf" }, { "contents": "Riccati differential equation\n\nThe Riccati equation is one of the most interesting nonlinear differential equations of first order. The general form is:\n\n\\[ x'(t) = a(t)x(t) + b(t)x^2(t) + c(t), \\] \n\n(1)\n\nwhere \\( a(t), b(t) \\) and \\( c(t) \\) are continuous functions of \\( t \\in I \\subset \\mathbb{R} \\). If \\( b \\equiv 0 \\) equation (1) is linear differential equation of first order and for \\( c \\equiv 0 \\) the equation (1) is a Bernoulli equation with \\( n = 2 \\).\n\nIt can be solved if a particular solution \\( x_1(t) \\) of a Riccati equation is known. Unfortunately, there is no strict algorithm to find the particular solution, which depends on the types of the functions \\( a(t), b(t) \\) and \\( c(t) \\).\n\nIf a particular solution \\( x_1(t) \\) of a Riccati equation is known, the general solution of the equation is given by\n\n\\[ x(t) = x_1(t) + y(t). \\] \n\n(2)\n\nIndeed, substituting the solution into Riccati equation, we have\n\n\\[ (x_1(t) + y(t))' = a(t)(x_1(t) + y(t)) + b(t)(x_1(t) + y(t))^2 + c(t) \\]\n\n\\[ x_1'(t) + y'(t) = a(t)x_1(t) + a(t)y(t) + b(t)x_1^2(t) + 2b(t)x_1(t)y(t) + b(t)y(t)^2 + c(t) \\]\n\nThe underlined terms in the left and in the right side can be canceled because \\( x_1 \\) is a particular solution satisfying the equation. As a result we obtain the differential equation for the function\n\n\\[ y'(t) = (a(t) + 2b(t)x_1(t))y(t) + b(t)y^2(t) \\] \n\n(3)\n\nwhich is a Bernoulli equation.\n\nSubstitution of \\( z(t) = y^{-1}(t) \\) converts the given Bernoulli equation into a linear differential equation that allows integration.\n\nBesides the general Riccati equation, there is an infinite number of particular cases of Riccati equation at certain coefficients of \\( a(t), b(t) \\) and \\( c(t) \\). Many of these particular cases have integrable solutions.\n\nSolution Process\n\nThe solution process for linear differential equation of first order is as follows:\n\n1. Put the differential equation in the correct initial form, (1).\n2. If a particular solution \\( x_1(t) \\) of a Riccati equation is known, we use the substitution (2).\n3. We obtain the Bernoulli equation (3) with \\( n = 2 \\).\n4. Solve for the Bernoulli equation (3).\n\n**Example 1** Find the solution to the following differential equation:\n\n\\[ tx'(t) - 3x(t) + 2x^2(t) = 2, \\quad t \\neq 0. \\]\n\nA particular solution \\( x_1(t) = 2 \\) of a Riccati equation is known.\n\n1. We convert this equation into the standard form:\n \\[ x'(t) = \\frac{3}{t} x(t) - \\frac{2}{t} x^2(t) + \\frac{2}{t}. \\]\n It is a Riccati equation.\n\n2. We use the particular solution \\( x_1(t) = 2 \\) in the substitution (2),\n \\[ x(t) = 2 + y(t). \\]\n\n3. We get the following differential equation for the new function\n \\[ (2 + y(t))' = \\frac{3}{t} (2 + y(t)) - \\frac{2}{t} (t^2 + y(t))^2 + \\frac{2}{t}, \\]\n \\[ y'(t) = \\frac{3}{t} - \\frac{1}{t} y(t) + t + \\frac{2}{t} y(t) + \\frac{1}{t^2} y^2(t) + 2t, \\]\n \\[ y'(t) = \\frac{1}{t} y(t) + \\frac{1}{t^2} y^2(t). \\]\n It is a Bernoulli equation with \\( n = 2 \\).\n\n4. Solve for the Bernoulli equation. The substitution \\( z(t) = y^{-1}(t) \\) converts it to a linear differential equation.\n We divide the differential equation by \\( y^2(t) \\)\n \\[ y^{-2}(t)y'(t) = \\frac{1}{t} y^{-1}(t) + \\frac{1}{t^3}. \\] (4)\n We make the substitution \\( z(t) = y^{-1}(t) \\) and \\( z'(t) = -y^{-2}(t)y'(t) \\) on equation (4) and obtain\n \\[ -z'(t) = \\frac{1}{t} z(t) + \\frac{1}{t^2} \\Rightarrow z'(t) + \\frac{1}{t} z(t) = -\\frac{1}{t^2} \\]\n a linear differential equation.\n\n5. We solve it and \\( z(t) = \\frac{1 + ct}{t^2} \\Rightarrow y(t) = \\frac{t^2}{1 + ct} \\Rightarrow x(t) = t^2 + \\frac{t^2}{1 + ct}. \\)\n\n**Example 2** Find the solution to the following differential equation:\n\n\\[ tx'(t) + x(t) - \\frac{1}{t^2} x^2(t) = 2t^2, \\quad t \\neq 0. \\]\n\n1. We convert this equation into the standard form:\n \\[ x'(t) = -\\frac{1}{t} x(t) + \\frac{1}{t^2} x^2(t) + 2t. \\]\n It is a Riccati equation.\n2. Try to find a particular solution in the form \\( x_1(t) = ct^2 \\). Substituting this into the Riccati equation, we can determine the coefficient \\( c \\).\n\n(a) \\( 2ct = -ct + c^2t + 2t \\Rightarrow c \\in \\{1, 2\\} \\).\n\nThus, there are even two particular solutions. However, we need only one of them. So we choose, for example, \\( c = 1 \\), \\( x_1(t) = t^2 \\). We use the substitution (2),\n\n\\[\nx(t) = t^2 + y(t).\n\\]\n\nWe get the following differential equation for the new function\n\n\\[\n(t^2 + y(t))' = -\\frac{1}{t} (t^2 + y(t)) + \\frac{1}{t^2} (t^2 + y(t))^2 + 2t,\n\\]\n\n\\[\n2t + y'(t) = -t - \\frac{1}{t} y(t) + t + \\frac{2}{t} y(t) + \\frac{1}{t^2} y^2(t) + 2t,\n\\]\n\n\\[\ny'(t) = \\frac{1}{t} y(t) + \\frac{1}{t^2} y^2(t).\n\\]\n\n3. It is a Bernoulli equation with Bernoulli equation \\( n = 2 \\). The substitution \\( z(t) = y^{-1}(t) \\) converts it to a linear differential equation\n\n\\[\nz'(t) = -y^{-2}(t)y'(t).\n\\]\n\nWe divide the differential equation by \\( y^2(t) \\)\n\n\\[\ny^{-2}(t)y'(t) = \\frac{1}{t} y^{-1}(t) + \\frac{1}{t^2}.\n\\]\n\n4. We make the substitution \\( z(t) = y^{-1}(t) \\) and \\( z'(t) = -y^{-2}(t)y'(t) \\) on equation (5) and obtain\n\n\\[\n-z'(t) = \\frac{1}{t} z(t) + \\frac{1}{t^2} \\Rightarrow z'(t) + \\frac{1}{t} z(t) = -\\frac{1}{t^2} \\text{ a linear differential equation.}\n\\]\n\nWe solve it and \\( z(t) = \\frac{1 + ct}{t^2} \\Rightarrow y(t) = \\frac{t^2}{1 + ct} \\Rightarrow x(t) = t^2 + \\frac{t^2}{1 + ct} \\).\n\nAuthor: Ariadna Lucia Pletea", "id": "./materials/188.pdf" }, { "contents": "Differential equations\n\nIn this section some of the common definitions and concepts in a differential equations course are introduced including order, linear vs. nonlinear, initial conditions, initial value problem and interval of validity.\n\nA differential equation is any equation which contains derivatives, either ordinary derivatives or partial derivatives.\n\nExamples: Newton’s second law of motion is written as a differential equation in terms of either the velocity, \\( v \\), or the position, \\( u \\), of the object as follows\n\n\\[\nm \\frac{dv}{dt} = F(t, v),\n\\]\n\n\\[\nm \\frac{d^2u}{dt^2} = F \\left( t, u, \\frac{du}{dt} \\right).\n\\]\n\nMore examples of differential equations:\n\n\\[\nx''(t) + x(t) = \\sin t,\n\\]\n\n\\[\nx'(t) = tx^2(t) + 2tx(t),\n\\]\n\n\\[\nx'''(t) + t^2x(t) - x'(t) = 2t^4,\n\\]\n\n\\[\nx(t) - tx'(t) = \\sqrt{1 + (x'(t))^2}, t \\in \\mathbb{R},\n\\]\n\n\\[\na^2 \\frac{\\partial^2 u}{\\partial x^2} = \\frac{\\partial^2 u}{\\partial t^2}, u = u(t, x),\n\\]\n\n\\[\n\\frac{\\partial u}{\\partial x} + \\frac{\\partial u}{\\partial y} = 0, u = u(x, y).\n\\]\n\nThe order of a differential equation is the largest derivative present in the differential equation.\n\nSo equations (1), (4), (6), (8) are of first order, equations (2), (3), (7) are of second order and equation (5) is of third order.\n\nOrdinary and Partial Differential Equations\n\nA differential equation is called an ordinary differential equation, abbreviated by ODE, if it has ordinary derivatives in it, equations (1), (2), (3), (4), (5) and (6). Likewise, a differential\nequation is called a **partial differential equation**, abbreviated by PDE, if it has partial derivatives in it, equations (7) and (8).\n\n**Linear Differential Equations**\n\nA **linear differential equation** is any differential equation that can be written in the following form\n\n\\[ x^{(n)}(t) + a_1(t)x^{(n-1)}(t) + \\cdots + a_{n-1}(t)x'(t) + a_n(t)x(t) = f(t) \\] \n\n(9)\n\nThe important thing to note about linear differential equations is that there are no products of the function, \\( x(t) \\), and its derivatives and neither the function or its derivatives occur to any power other than the first power. Also note that neither the function or its derivatives are “inside” another function, for example, \\( \\sqrt{x(t)} \\) or \\( \\ln x(t) \\).\n\nThe coefficients \\( a_1(t), a_2(t), \\ldots, a_n(t) \\) and \\( f(t) \\) can be zero or non-zero functions, constant or non-constant functions, linear or non-linear functions.\n\nOnly the function, \\( x(t) \\) and its derivatives are used in determining if a differential equation is linear.\n\nIf a differential equation cannot be written in the form (9) then it is called a non-linear differential equation. Equations (4) and (6) are nonlinear.\n\nA solution of a differential equation on an interval \\( t \\in (a, b) \\) is any function \\( x(t) \\) which satisfies the differential equation in question on the interval \\( (a, b) \\). It is important to note that solutions are often accompanied by intervals and these intervals can give some important information about the solution.\n\n**General Solution**\n\nThe general solution to a differential equation is the most general form that the solution can take and doesn’t take any initial conditions into account.\n\nThe solution \\( x(t) = 2 + Ce^{-t^2} \\) is the general solution of equation\n\n\\[ x(t) + 2tx(t) = 4t. \\] \n\n(10)\n\nThe solution to an ordinary differential equation is a family of parametric curve. So there are an infinite number of solutions to the differential equation. The set of solutions that we’ve graphed below are for some values of \\( C, C = -1, 0, 1, 2, 3, \\)\n\n![Graph of solutions](image)\n\n(11)\n\nWe can ask a natural question. Which is the solution that we want or does it matter which solution we use? This question leads us to the next ideas,\n**Initial Condition(s)**\n\nInitial Condition(s) are a condition, or set of conditions, on the solution that will allow us to determine which solution that we are after. Initial conditions are of the form,\n\n\\[ x(t_0) = x_0 \\text{ or } x^{(k)}(t_0) = x_{0,k}. \\]\n\nSo, in other words, initial conditions are values of the solution and/or its derivative(s) at specific points.\n\nThe number of initial conditions that are required for a given differential equation will depend upon the order of the differential equation.\n\nIf we consider the initial condition \\( x(0) = 3 \\) for equation (10) the solution will be \\( x(t) = 2 + e^{-t^2} \\). It is a particular solution. This solution is colored on the figure (11) red.\n\n**Initial Value Problem**\n\nAn Initial Value Problem (or IVP) is a differential equation along with an appropriate number of initial conditions.\n\nThe following problems are IVP\n\n\\[\n\\begin{align*}\n\\begin{cases}\n x(t) + 2tx(t) = 4t, \\\\\n x(0) = 3,\n\\end{cases}\n\\end{align*}\n\\]\n\nand\n\n\\[\n\\begin{align*}\n\\begin{cases}\n x''' + 3x'' - x' - 3x = 0, \\\\\n x(0) = 0, x'(0) = 1, x''(0) = -1.\n\\end{cases}\n\\end{align*}\n\\]\n\nAs I noted earlier the number of initial condition required will depend on the order of the differential equation.\n\n**Interval of Validity**\n\nThe interval of validity for an IVP with initial condition(s) is the largest possible interval on which the solution is valid and contains \\( t_0 \\).\n\nA Cauchy problem can be an Initial Value Problem.\n\nThe simplest Cauchy problem is to find a function \\( x(t) \\) defined on the half-line \\( t \\geq t_0 \\), satisfying a first-order ordinary differential equation\n\n\\[\n\\frac{dx}{dt} = f(x, t)\n\\]\n\n(\\( f \\) is a given function) and taking a specified value \\( x_0 \\) at \\( t = t_0 \\):\n\n\\( x(t_0) = x_0 \\).\n\nWe write this\n\n\\[\n\\begin{align*}\n\\begin{cases}\n \\frac{dx}{dt} = f(x, t), \\\\\n x(t_0) = x_0.\n\\end{cases}\n\\end{align*}\n\\]\n\nIn geometrical terms this means that, considering the family of integral curves of equation (12) in the \\((t, x)\\)-plane, one wishes to find the curve passing through the point \\((t_0, x_0)\\).\n\nAuthor: Ariadna Lucia Pletea", "id": "./materials/189.pdf" }, { "contents": "I numeri reali\n\nNote per il corso di Analisi Matematica 1\n\nG. Mauceri\n\na.a. 2003-04\nContents\n\n1 Introduzione 3\n2 Gli assiomi di campo 3\n3 Gli assiomi dell’ordine 4\n4 Valore assoluto 5\n5 I numeri naturali e gli interi 6\n6 I razionali 11\n7 La rappresentazione geometrica dei razionali 12\n8 L’assioma di completezza 12\n9 Conseguenze della completezza 14\n10 Archimedea dei reali 16\n11 Altre conseguenze della completezza 17\n12 Allineamenti decimali 19\n13 Esistenza e unicità del campo reale 24\n14 I tagli di Dedekind 25\n1 Introduzione\n\nLa nozione intuitiva del sistema dei numeri reali è legata al concetto di misura per classi di grandezze omogenee, come ad esempio la classe delle lunghezze, la classe delle aree, la classe dei tempi. Anche se questa nozione intuitiva è stata usata dai matematici almeno dai tempi degli antichi geometri Greci, soltanto verso la fine del diciannovesimo secolo si è arrivati a una costruzione formale rigorosa del sistema dei numeri reali a partire dai numeri razionali.\n\nDal punto di vista fondazionale un metodo abbastanza soddisfacente di introdurre i numeri reali consiste nel prendere i numeri naturali 0, 1, 2, 3... come concetti primitivi, enunciando un sistema appropriato di assiomi per essi, e di usare poi i numeri naturali e la teoria degli insiemi per costruire il sistema dei numeri razionali. A loro volta i numeri razionali possono essere usati come base per la costruzione dei numeri reali. Poiché questo procedimento è piuttosto lungo e laborioso e noi siamo interessati esclusivamente nello sviluppo dell’analisi matematica a partire dalle proprietà dei numeri reali, seguiremo un procedimento diverso, che consiste nell’assumere i numeri reali come oggetti primitivi, indefiniti, che soddisfano certi assiomi. Supporremo quindi che esista un insieme \\( \\mathbb{R} \\), i cui elementi sono detti numeri reali, che soddisfa i nove assiomi enunciati nelle prossime sezioni. Gli assiomi possono essere riuniti in modo naturale in tre gruppi, che indicheremo con i nomi di assiomi di campo, assiomi dell’ordine e assioma di completezza. A partire da questi assiomi dedurremo tutte le altre proprietà dei numeri reali.\n\n2 Gli assiomi di campo\n\nSu \\( \\mathbb{R} \\) sono definite due operazioni che ad ogni coppia di numeri reali \\((x, y)\\) associano i numeri reali \\(x + y\\) e \\(xy\\), detti rispettivamente somma e prodotto di \\(x\\) e \\(y\\), che soddisfano gli assiomi seguenti. Per ogni \\(x, y\\) e \\(z\\) in \\(\\mathbb{R}\\)\n\n**Assioma 1 (proprietà associative)** \\((x + y) + z = x + (y + z), \\quad (xy)z = x(yz)\\)\n\n**Assioma 2 (proprietà commutative)** \\(x + y = y + x, \\quad xy = yx\\)\n\n**Assioma 3 (proprietà distributiva)** \\(x(y + z) = xy + xz\\)\n\n**Assioma 4 (esistenza degli elementi neutri)** Esistono due numeri reali distinti, che indicheremo con 0 e 1, tali che per ogni \\(x\\) in \\(\\mathbb{R}\\), si ha \\(x + 0 = x\\) e \\(1x = x\\).\n\n**Assioma 5 (esistenza dell’opposto)** Per ogni numero reale \\(x\\) esiste un numero reale \\(y\\) tale che \\(x + y = 0\\).\n\n**Assioma 6 (esistenza del reciproco)** Per ogni numero reale \\(x \\neq 0\\) esiste un numero reale \\(y\\) tale che \\(xy = 1\\).\n\n---\n\n1. Due diverse costruzioni dei numeri reali vennero date da Richard Dedekind e George Cantor nel 1872.\n2. Il concetto di operazione su \\(\\mathbb{R}\\) può essere ricondotto a quello di funzione definita sul prodotto cartesiano \\(\\mathbb{R} \\times \\mathbb{R}\\) a valori in \\(\\mathbb{R}\\).\nOsserviamo che il numero reale $y$ la cui esistenza è garantita dall’Assioma 5 è unico. Infatti, se $x + y = x + z = 0$, allora $y = y + 0 = y + (x + z) = z + (x + y) = z + 0 = z$. Il numero reale $y$ tale che $x + y = 0$ si dice l’opposto di $x$ e si denota con $-x$. La somma $x + (-y)$ si denota anche con $x - y$ ed è detta la differenza di $x$ e $y$. Analogamente si dimostra l’unicità del reciproco di un numero reale $x \\neq 0$. Il reciproco di $x$ si denota con $x^{-1}$ o con $1/x$. Il prodotto $yx^{-1}$ si denota anche con $y/x$ ed è detto anche il rapporto di $y$ e $x$. Da questi assiomi possiamo dedurre tutte le abituali leggi dell’aritmetica; ad esempio:\n\n$$-(-x) = x, \\quad (x^{-1})^{-1} = x, \\quad -(x - y) = y - x, \\quad x + (-y) = x - y, \\quad 0x = x0 = 0, \\quad x/y + a/b = (xb + ya)/yb, \\ldots$$\n\nSi noti anche che se $xy=0$ allora $x = 0$ oppure $y = 0$. La dimostrazione di tutte queste proprietà a partire dagli assiomi è un utile esercizio per il lettore.\n\n### 3 Gli assiomi dell’ordine\n\nSu $\\mathbb{R}$ si introduce una relazione d’ordine a partire dal concetto non definito di positività.\n\nEsiste un sottoinsieme $\\mathbb{R}_+$ di $\\mathbb{R}$, detto insieme dei numeri reali positivi, che soddisfa i due assiomi seguenti.\n\n**Assioma 7** Per ogni $x$ e $y$ in $\\mathbb{R}_+$ anche $x + y$ e $xy \\in \\mathbb{R}_+$.\n\n**Assioma 8** Per ogni numero reale $x$ si verifica una e una sola delle tre alternative: $x = 0$, oppure $x \\in \\mathbb{R}_+$, oppure $-x \\in \\mathbb{R}_+$.\n\n**Definizione.** Definiamo una relazione $<$ in $\\mathbb{R}$ ponendo $x < y$ (leggi $x$ minore di $y$) se e solo se $y - x \\in \\mathbb{R}_+$. Scriveremo anche $y > x$ ($y$ maggiore di $x$) se $x < y$; $x \\leq y$ ($x$ minore o uguale a $y$) se $x < y$ oppure $x = y$; $y \\geq x$ ($y$ maggiore o uguale di $x$) se $x \\leq y$. Se $x < 0$ diciamo che $x$ è negativo. Denoteremo con $\\mathbb{R}_-$ l’insieme dei numeri reali negativi. Se $x \\geq 0$ diremo anche che $x$ è non negativo.\n\n**Teorema 3.1** La relazione $\\leq$ è una relazione d’ordine totale su $\\mathbb{R}$. In altri termini essa soddisfa le seguenti proprietà:\n\na) riflessività: per ogni $x$ in $\\mathbb{R}$ si ha $x \\leq x$,\n\nb) antisimmetria: per ogni $x$, $y$ in $\\mathbb{R}$, $x \\leq y$ e $y \\leq x$ implica $x = y$,\n\nc) transitività: per ogni $x$, $y$ e $z$ in $\\mathbb{R}$, $x \\leq y$ e $y \\leq z$ implica $x \\leq z$,\n\nd) totalità: per ogni coppia di numeri reali $x$ e $y$ vale sempre una delle relazioni $x \\leq y$ oppure $y \\leq x$.\n\nLa relazione $<$ soddisfa le seguenti proprietà di compatibilità con la somma e il prodotto:\n\ne) monotonia rispetto alla somma: per ogni $x$, $y$ e $z$ in $\\mathbb{R}$ se $x < y$ allora $x + z < y + z$,\nf) monotonia rispetto al prodotto: per ogni \\( x, y, z \\) in \\( \\mathbb{R} \\) se \\( x < y \\) e \\( z \\) è positivo allora \\( zx < zy \\).\n\nLasciamo al lettore la cura di dimostrare il Teorema 3.1 a partire dagli Assiomi 7 e 8. Elenchiamo alcune altre proprietà utili della relazione d’ordine.\n\n- Se \\( x \\neq 0 \\) allora \\( x^2 > 0 \\).\n- Per ogni \\( x, y, z \\) in \\( \\mathbb{R} \\) se \\( x < y \\) e \\( z \\) è negativo allora \\( zx > zy \\).\n- Se \\( x < y \\) allora \\( -x > -y \\).\n- Se \\( xy > 0 \\) allora \\( x \\) e \\( y \\) sono entrambi positivi o entrambi negativi.\n\n**Definizione.** Siano \\( a \\) e \\( b \\) due numeri reali tali che \\( a < b \\). L’intervallo aperto di estremi \\( a \\) e \\( b \\) è l’insieme \\( (a, b) = \\{ x \\in \\mathbb{R} : a < x < b \\} \\); \\( a \\) è il primo estremo, \\( b \\) il secondo estremo. L’intervallo chiuso di estremi \\( a \\) e \\( b \\) è l’insieme \\( [a, b] = \\{ x \\in \\mathbb{R} : a \\leq x \\leq b \\} \\). In maniera analoga si definiscono l’intervallo semiaperto a destra \\( [a, b) \\) e l’intervallo semiaperto a sinistra \\( (a, b] \\) mediante le disuguaglianze \\( a \\leq x < b \\) e \\( a < x \\leq b \\). Questi intervalli sono detti limitati. Gli intervalli semilititati sono gli insiemi\n\n\\[\n\\begin{align*}\n(a, +\\infty) : &= \\{ x \\in \\mathbb{R} : a < x \\} \\quad \\text{intervallo aperto semilititato a sinistra} \\\\\n[a, +\\infty) : &= \\{ x \\in \\mathbb{R} : a \\leq x \\} \\quad \\text{intervallo chiuso semilititato a sinistra} \\\\\n(-\\infty, a) : &= \\{ x \\in \\mathbb{R} : x < a \\} \\quad \\text{intervallo aperto semilititato a destra} \\\\\n(-\\infty, a] : &= \\{ x \\in \\mathbb{R} : x \\leq a \\} \\quad \\text{intervallo chiuso semilititato a destra}\n\\end{align*}\n\\]\n\nSono intervalli anche l’insieme \\( \\mathbb{R} \\) (denotato talvolta con \\( (-\\infty, +\\infty) \\)), l’insieme vuoto \\( \\emptyset \\) e l’insieme costituito da un solo elemento \\( \\{a\\} \\). Gli ultimi due sono detti intervalli degeneri.\n\n### 4 Valore assoluto\n\n**Definizione.** Sia \\( x \\) un numero reale. Si dice valore assoluto o modulo di \\( x \\) il numero reale \\( |x| \\) definito da\n\n\\[\n|x| = \\begin{cases} \n x & \\text{se } x \\geq 0 \\\\\n -x & \\text{se } x < 0 \n\\end{cases}\n\\]\n\n**Proposizione 4.1** Il valore assoluto soddisfa le seguenti proprietà\n\na) \\( |x| \\geq 0 \\) per ogni \\( x \\) in \\( \\mathbb{R} \\) e \\( |x| = 0 \\) se e solo se \\( x = 0 \\);\n\nb) \\( |x| = |-x| \\);\nc) $-|x| \\leq x \\leq |x|$\n\nd) se $a$ è un numero reale maggiore o uguale a 0 allora $|x| \\leq a$ se e solo se $-a \\leq |x| \\leq a$.\n\ne) $|xy| = |x| |y|$\n\nf) $|x + y| \\leq |x| + |y|$\n\ng) $||x| - |y|| \\leq |x - y|$\n\nper ogni $x$ e $y$ in $\\mathbb{R}$.\n\n**Dimostrazione.** Le prime quattro proprietà sono conseguenze immediate della definizione di valore assoluto. Per dimostrare la quarta si possono considerare separatamente i quattro casi $x \\geq 0$ e $y \\geq 0$; $x \\geq 0$ e $y < 0$; $x < 0$ e $y \\geq 0$; $x < 0$ e $y < 0$. Per dimostrare la f) si sommano le disuguaglianze $-|x| \\leq x \\leq |x|$ e $-|y| \\leq y \\leq |y|$ ottenendo\n\n$$-(|x| + |y|) \\leq x + y \\leq |x| + |y|.$$ \n\nQuindi, per la d), $|x + y| \\leq |x| + |y|$. Per dimostrare la g) osserviamo che\n\n$$|x| = |(x - y) + y| \\leq |x - y| + |y|$$\n\ne quindi\n\n$$|x| - |y| \\leq |x - y|.$$ \n\nScambiando $x$ e $y$ tra loro si ha anche\n\n$$|y| - |x| \\leq |y - x| = |x - y|.$$ \n\nLa conclusione segue dalla d) perché $|x| - |y|$ e $|y| - |x|$ sono opposti l’uno dell’altro. □\n\nLe disuguaglianze f) e g) sono dette *disuguaglianze triangolari*, perché quando vengono generalizzate ai vettori esprimono il fatto che la lunghezza di un lato di un triangolo è minore o uguale alla somma ed è maggiore o uguale alla differenza delle lunghezze degli altri due.\n\n## 5 I numeri naturali e gli interi\n\nIn questa sezione descriviamo due sottoinsiemi particolari di $\\mathbb{R}$, l’insieme $\\mathbb{N}$ dei *numeri naturali* e l’insieme $\\mathbb{Z}$ degli *interi*.\n\n**Definizione.** Un sottoinsieme $A$ di $\\mathbb{R}$ si dice *induttivo* se ha le seguenti due proprietà:\n\na) $0 \\in A$;\n\nb) per ogni $x$, se $x \\in A$ allora $x + 1 \\in A$. \n\nI numeri naturali e gli interi\n\nSono esempi di insiemi induttivi $\\mathbb{R}$ stesso e $\\mathbb{R}_+ \\cup \\{0\\}$.\n\n**Definizione.** L’insieme $\\mathbb{N}$ dei numeri naturali è l’intersezione di tutti i sottoinsiemi induttivi di $\\mathbb{R}$.\n\nL’insieme $\\mathbb{N}$ è non vuoto, perché $0 \\in \\mathbb{N}$. Inoltre è facile vedere che $\\mathbb{N}$ stesso è induttivo. Pertanto $1 = 0 + 1 \\in \\mathbb{N}$, $2 = 1 + 1 \\in \\mathbb{N}$, $3 = 2 + 1 \\in \\mathbb{N}$. Poiché $\\mathbb{N}$ è contenuto in ogni insieme induttivo, esso è il più piccolo insieme induttivo. Precisamente si ha il seguente teorema.\n\n**Teorema 5.1 (teorema di induzione)** Sia $A$ un sottoinsieme di $\\mathbb{N}$ tale che\n\na) $0 \\in A$,\n\nb) per ogni $n$ in $\\mathbb{N}$, $n \\in A$ implica $n + 1 \\in A$,\n\nallora $A = \\mathbb{N}$.\n\n**Dimostrazione.** L’insieme $A$ è induttivo. Quindi $\\mathbb{N} \\subseteq A$ e, poiché $A \\subseteq \\mathbb{N}$, si ha $A = \\mathbb{N}$. □\n\nIl Teorema di induzione è alla base di uno schema dimostrativo detto *principio di induzione*. Prima di enunciare il principio di induzione dobbiamo definire alcune nozioni logiche.\n\n**Definizione.** Diremo *enunciato* o *formula* un’espressione sintatticamente ben formata, come ad esempio: “Parigi è in Inghilterra”, “$x$ è un uomo”, “$x+y=4$”, “se $x \\in \\mathbb{R}$ allora $x^2 \\neq -1$”. Denoteremo gli enunciati con lettere come $P$, $Q$, ecc.\n\nCome si vede un enunciato può contenere delle variabili che possono assumere valori in un dato insieme. Le variabili che sono precedute dalle parole *per ogni* o *esiste* si dicono legate, le altre variabili si dicono libere. Così nell’enunciato “per ogni $x$ in $\\mathbb{R}$ esiste un $y$ in $\\mathbb{R}$ tale che $x + y = z$”, le variabili $x$ e $y$ sono legate, mentre la variabile $z$ è libera. Gli enunciati si dicono *aperti* se contengono delle variabili libere, *chiusi* se non contengono variabili o se tutte le variabili che appaiono in essi sono legate. Un *enunciato aperto* è detto anche un *predicato*. Se $P$ è un predicato, scriveremo $P[x]$ tutte le volte che vorremo indicare che $x$ è una delle variabili libere di $P$.\n\nLa negazione di un enunciato si ottiene premettendo la parola *non* all’enunciato stesso. Un enunciato $P$ si dice *vero* se è un assioma o è deducibile dagli assiomi, si dice *falso* (o *refutabile*) se non–$P$ è vero.\\(^3\\) Si osservi che ‘falso’ non significa ‘non vero’. Ad esempio l’enunciato “$x = 0$” non è né vero né falso nella teoria i cui assiomi sono A.1–A.8. Infatti in ogni campo vi sono due elementi distinti 0 e 1, uno solo dei quali soddisfa il predicato.\n\n\\(^3\\)Nel definire i termini ‘vero’ e ‘falso’ abbiamo seguito l’uso prevalente tra i matematici. Avvertiamo però il lettore che i logici matematici preferiscono dire $P$ è dimostrabile e non–$P$ è dimostrabile, riservando l’uso dei termini ‘vero’ e ‘falso’ ad un altro contesto.\nSi noti che vi sono degli enunciati aperti che sono veri, come ad esempio “se \\( x \\in \\mathbb{R} \\) allora \\( x^2 \\geq 0 \\).”\n\n**Teorema 5.2 (principio di induzione)** Sia \\( P[n] \\) un predicato in cui la variabile \\( n \\) è un numero naturale. Supponiamo che i seguenti due enunciati\n\na) \\( P[0] \\),\n\nb) \\( P[n] \\) implica \\( P[n + 1] \\),\n\nsiano veri. Allora \\( P[n] \\) è vero per ogni \\( n \\) in \\( \\mathbb{N} \\).\n\n**Dimostrazione.** Proviamo che l’insieme \\( A = \\{ n \\in \\mathbb{N} : P[n] \\text{ è vero} \\} \\) è induttivo. Infatti \\( 0 \\in A \\) perché \\( P[0] \\) è vero. Se \\( n \\in A \\) allora \\( P[n] \\) è vero e quindi per b) è vero anche \\( P[n + 1] \\). Quindi \\( n + 1 \\in A \\). Poiché \\( n \\) è arbitrario, abbiamo dimostrato che, per ogni \\( n \\), \\( n \\in A \\) implica \\( n + 1 \\in A \\). Pertanto \\( A \\) è induttivo e, quindi, \\( A = \\mathbb{N} \\) per il Teorema di induzione. Questo significa che \\( P[n] \\) è vero per ogni \\( n \\in \\mathbb{N} \\).\n\nÈ essenziale che il lettore comprenda perfettamente il significato del principio di induzione. Essa afferma che, se vogliamo dimostrare che una proprietà vale per tutti i numeri naturali, non è necessario dimostrarla per ogni numero. Basta dimostrare che essa vale per 0 e che, se vale per un numero naturale \\( n \\), vale anche per \\( n + 1 \\). Il principio di induzione non è l’unico modo per dimostrare che una proprietà vale per tutti i numeri naturali. Per il principio di generalizzazione (vedi nota a piè di pagina), per dimostrare che \\( P[n] \\) è vero per ogni \\( n \\) basta dimostrarlo per un \\( n \\) generico. Tuttavia in molti casi è più facile dimostrare che \\( P[0] \\) è vero e che “\\( P[n] \\) implica \\( P[n + 1] \\)” è vero per un generico \\( n \\), piuttosto che dimostrare direttamente che \\( P[n] \\) è vero per un generico \\( n \\).\n\nEsaminiamo ora alcune proprietà dei numeri naturali.\n\n**Proposizione 5.3** I numeri naturali sono maggiori o uguali a 0. Se \\( m \\) e \\( n \\) sono numeri naturali allora anche \\( m + n \\) e \\( mn \\) sono numeri naturali. Se \\( m > n \\) allora \\( m - n \\) è un numero naturale.\n\n**Dimostrazione.** Tutte queste proprietà si dimostrano facilmente mediante il principio di induzione. A titolo d’esempio dimostriamo l’ultima, cioè che se \\( m > n \\) allora \\( m - n \\) è un numero naturale. Osserviamo dapprima che, poiché l’insieme \\( \\{0\\} \\cup \\{k \\in \\mathbb{N} : k = q+1, q \\in \\mathbb{N}\\} \\) è induttivo, per ogni numero naturale \\( k \\neq 0 \\) esiste un numero naturale \\( q \\) tale \\( k = q + 1 \\). Fissato un numero naturale arbitrario \\( m > 0 \\), indichiamo con \\( P[n] \\) il predicato: “se \\( m > n \\)\n\n---\n\n4 Si potrebbe credere che un enunciato chiuso debba sempre essere dimostrabile o refutabile. Tuttavia non è così: un risultato profondo di logica, dovuto a K. Gödel (1931), asserisce che, se l’aritmetica non è contradditoria, vi sono in essa enunciati chiusi \\( P \\) che sono indecidibili, cioè tali che \\( P \\) non è dimostrabile e non–\\( P \\) non è dimostrabile.\n\n5 Questo è un esempio di applicazione di un metodo di dimostrazione detto principio di generalizzazione: sia \\( P[x] \\) un enunciato che contiene la variabile libera \\( x \\); se \\( P[x] \\) è vero allora “per ogni \\( x \\) \\( P[x] \\)” è vero. In linguaggio meno formale: per dimostrare che \\( P[x] \\) è vero per ogni \\( x \\) basta dimostrare che \\( P[x] \\) è vero per un “generico” \\( x \\).\nallora $m - n$ è un numero naturale”. Se $m > 0$ allora $m - 0 = m \\in \\mathbb{N}$. Quindi $P[0]$ è vero. Dimostriamo ora che $P[n]$ implica $P[n+1]$. Se $m > n + 1$ allora si ha anche $m > n$ e quindi, per $P[n]$, si ha che $m - n \\in \\mathbb{N}$. Poiché $m - n > 1$, $m - n$ è un numero naturale diverso da 0 e quindi esiste un numero naturale $q$ tale che $m - n = q + 1$. Quindi $m - (n + 1) = q \\in \\mathbb{N}$ e $P[n + 1]$ è così dimostrato.\n\n**Proposizione 5.4** Se $n$ è un numero naturale non esistono numeri naturali maggiori di $n$ e minori di $n + 1$.\n\n**Dimostrazione.** Non vi sono numeri naturali maggiori di 0 e minori di 1, perché l’insieme \\{0\\} ∪ \\{n ∈ \\mathbb{N} : n ≥ 1\\} è induttivo. Se $m$ e $n$ fossero due numeri naturali tali che $n < m < n + 1$, allora $m - n$ sarebbe un numero naturale maggiore di 0 e minore di 1. Ma questo contraddice quanto abbiamo appena provato.\n\nUna variante del principio di induzione che è utile in talune applicazioni è la seguente.\n\n**Corollario 5.5 (Principio di induzione generalizzato)** Siano $P[n]$ un predicato in cui la variabile $n$ è un numero naturale e $n_0$ un numero naturale. Supponiamo che i seguenti due enunciati\n\na) $P[n_0]$,\n\nb) $P[n]$ implica $P[n + 1]$,\n\nsiano veri. Allora $P[n]$ è vero per ogni $n ≥ n_0$.\n\n**Dimostrazione.** Se $n ≥ n_0$ allora, per la Proposizione 5.3, esiste un numero naturale $k$ tale che $n = n_0 + k$. Basta quindi osservare che l’insieme \\{k ∈ \\mathbb{N} : P[k + n_0] è vero\\} coincide con $\\mathbb{N}$ perché è induttivo.\n\n**Definizione.** Sia $X$ un insieme. Una *successione a valori in $X$* è una funzione $a$ definita su $\\mathbb{N}$ a valori in $X$. Se $X = \\mathbb{R}$ si dice che $a$ è una successione a *valori reali*. Se $a : \\mathbb{N} → X$ è una successione è uso comune scrivere $a_n$ per indicare il valore $a(n)$ di $a$ in $n$. Si dice anche che $a_n$ è il *termine n-esimo* della successione $(a_n)$. Una successione può essere definita mediante una formula esplicita che permette di calcolare il valore $a_n$ per ogni numero naturale $n$. Ad esempio $a_n = n^3 + 1$. Vi è però un altro modo di definire una successione mediante un procedimento noto come *definizione per ricorrenza*. Grosso modo il metodo di ricorrenza consiste nel definire una successione definendo $a_0$ e assegnando una procedura per calcolare $a_{n+1}$, dopo aver supposto che $a_n$ sia già stato definito. La giustificazione di questo metodo si basa sul seguente *teorema di ricorrenza*.\n\n**Teorema 5.6** Se $x$ è un elemento di un insieme $X$ e $φ$ è una funzione da $X$ in $X$, esiste un’unica successione $(a_n)$ tale che $a_0 = x$ e $a_{n+1} = φ(a_n)$, per ogni $n$ in $\\mathbb{N}$.\n\n---\n\n6Questo è un primo esempio di *dimostrazione per contraddizione* o *per assurdo*: se aggiungendo l’enunciato non-$Q$ agli assiomi si ottiene una contraddizione, cioè si dimostra sia $R$ che non-$R$, allora $Q$ è vero.\nDimostrazione. La dimostrazione del teorema di ricorrenza si basa sul principio di induzione. L’unicità della successione \\((a_n)\\) è facile da dimostrare. Per dimostrare l’esistenza proviamo dapprima che per ogni intero \\(n \\geq 1\\) esiste una funzione \\(g_n\\), definita su \\(\\{0, 1, \\ldots, n\\}\\) a valori in \\(X\\), tale che\n\n\\[\ng_n(0) = x, \\quad g_n(k + 1) = \\phi(g_n(k)), \\quad k = 0, 1, \\ldots, n - 1. \\tag{5.1}\n\\]\n\nSe \\(n = 1\\) poniamo \\(g_1(0) = x, g_1(1) = \\phi(g_1(0))\\). Supponiamo ora che l’assunto sia vero per un intero \\(n \\geq 1\\) arbitrario e definiamo \\(g_{n+1}\\) su \\(\\{0, 1, \\ldots, n + 1\\}\\) ponendo\n\n\\[\ng_{n+1}(k) = g_n(k), \\quad \\text{se } k = 0, 1, \\ldots, n, \\quad g_{n+1}(n + 1) = \\phi(g_{n+1}(n)).\n\\]\n\nÈ immediato verificare che \\(g_{n+1}\\) soddisfa la relazione\n\n\\[\ng_{n+1}(0) = x, \\quad g_{n+1}(k + 1) = \\phi(g_{n+1}(k)), \\quad k = 0, 1, \\ldots, n.\n\\]\n\nQuindi, per il principio di induzione, abbiamo dimostrato l’esistenza della funzione \\(g_n\\) per ogni intero \\(n \\geq 1\\). A questo punto è immediato verificare che la successione \\((a_n)\\) definita da\n\n\\[\na_0 = x, \\quad a_n = g_n(n), \\quad n \\geq 1,\n\\]\n\nsoddisfa la relazione ricorsiva \\(a_{n+1} = \\phi(a_n)\\). \\(\\square\\)\n\nIllustriamo ora l’uso del teorema di ricorrenza con un esempio. Consideriamo la funzione \\(\\phi\\) di \\(\\mathbb{R}\\) in \\(\\mathbb{R}\\), definita da \\(\\phi(x) = x^3 + 1\\). Per il teorema di ricorrenza, fissato un numero reale \\(x\\), è possibile definire una successione \\(a : \\mathbb{N} \\to \\mathbb{R}\\) ponendo\n\n\\[\na_0 = x, \\quad a_{n+1} = a_n^3 + 1.\n\\]\n\nUtilizzando la formula ricorsiva possiamo calcolare \\(a_n\\) qualunque sia il numero naturale \\(n\\). Ad esempio, prendendo \\(x = 1\\), si ha\n\n\\[\na_0 = 1, \\quad a_1 = a_0^3 + 1 = 2, \\quad a_2 = a_1^3 + 1 = 9, \\quad \\ldots,\n\\]\n\nSi noti che il teorema di ricorrenza garantisce che la successione \\((a_n)\\) è definita anche se non è data una formula esplicita per calcolare \\(a_n\\) a partire da \\(n\\).\n\nNon sempre il rapporto di due numeri naturali è un numero naturale. Tuttavia in \\(\\mathbb{N}\\) è possibile effettuare la divisione con resto, come dimostra il seguente teorema.\n\n**Teorema 5.7** Siano \\(n, d\\) in \\(\\mathbb{N}\\), con \\(d > 0\\). Esistono due numeri naturali \\(q\\) e \\(r\\), univocamente determinati, tali che \\(n = qd + r\\) e \\(0 \\leq r < d\\).\n\n**Dimostrazione.** Per dimostrare l’esistenza procediamo per induzione su \\(n\\). Se \\(n = 0\\) basta prendere \\(q = r = 0\\). Supponiamo che per un numero naturale \\(n\\) arbitrario si abbia \\(n = qd + r\\), con \\(0 \\leq r < d\\) e dimostriamo che esistono \\(q'\\) e \\(r'\\), con \\(0 \\leq r' < d\\), tali che \\(n + 1 = q'd + r'\\). Se \\(r < d - 1\\) allora \\(n + 1 = qd + r + 1\\) e \\(0 \\leq r + 1 < d\\). Quindi \\(q' = q\\) e \\(r' = r + 1\\). Se \\(r = d - 1\\) allora \\(n + 1 = qd + d = (q + 1)d\\). Quindi \\(q' = q + 1\\) e \\(r' = 0\\).\nPer dimostrare l’unicità supponiamo che \\( n = q_0d + r_0 = q_1d + r_1 \\) con \\( 0 \\leq r_0, r_1 < d \\). Allora \\((q_1 - q_0)d = r_1 - r_0\\). Se, per assurdo, \\( q_0 \\neq q_1 \\) possiamo supporre che \\( q_0 < q_1 \\). Allora si ottiene la contraddizione \\( d \\leq (q_1 - q_0)d = r_1 - r_0 < d \\). Pertanto \\( q_1 = q_0 \\) e \\( r_1 = r_0 \\).\n\nI numeri \\( q \\) e \\( r \\) si dicono il quoziente e il resto della divisione di \\( n \\) per \\( d \\). Se \\( r = 0 \\) si dice che \\( d \\) è un divisore di \\( n \\) o che \\( n \\) è un multiplo di \\( d \\) e si scrive \\( d|n \\) (leggi: \\( d \\) divide \\( n \\)).\n\n**Definizione.** Un numero naturale \\( p > 1 \\) si dice primo se i suoi divisori sono solo \\( p \\) e 1. L’intero 1 non è un numero primo.\n\nUn risultato fondamentale della teoria dei numeri interi è il seguente **teorema di fattorizzazione unica**.\n\n**Teorema 5.8** Ogni numero naturale \\( > 1 \\) è primo o si fattorizza nel prodotto di numeri primi. La fattorizzazione è unica a parte l’ordine dei fattori.\n\n**Dimostrazione.** Ci limitiamo a dimostrare che ogni intero \\( n > 1 \\) è primo o si fattorizza nel prodotto di fattori primi, rinviando il lettore al corso di algebra per la dimostrazione dell’unicità della fattorizzazione. Sia \\( P[n] \\) il predicato: tutti gli interi \\( k \\) tali che \\( 2 \\leq k \\leq n \\) sono primi o si fattorizzano nel prodotto di numeri primi. Allora \\( P[2] \\) è vero. Supponiamo che \\( P[n] \\) sia vero. Per dimostrare che anche \\( P[n+1] \\) è vero è sufficiente provare che anche \\( n+1 \\) si fattorizza nel prodotto di numeri primi. Se \\( n+1 \\) stesso non è primo esistono due interi \\( k \\leq n \\) e \\( m \\leq m \\) tali che \\( n = km \\). Poiché \\( k \\) e \\( m \\) sono entrambi minori o uguali di \\( n \\), essi sono primi o si fattorizzano in prodotto di numeri primi; quindi anche \\( n+1 \\) è il prodotto di primi.\n\n**Definizione.** L’insieme \\( \\mathbb{Z} = \\{ k \\in \\mathbb{R} : k \\in \\mathbb{N} \\text{ oppure } -k \\in \\mathbb{N} \\} \\) è detto insieme degli interi. È facile vedere che se \\( k \\) è un intero anche \\( k-1 \\) e \\( k+1 \\) sono interi e che non vi sono interi maggiori di \\( k \\) e minori di \\( k+1 \\). Gli interi \\( k-1 \\) e \\( k+1 \\) sono detti rispettivamente l’antecedente e il successivo di \\( k \\). L’insieme degli interi positivi si denota con \\( \\mathbb{Z}_+ \\) o con \\( \\mathbb{N}_+ \\).\n\n### 6 I razionali\n\nI quozienti di interi \\( m/n \\), dove \\( n \\neq 0 \\) sono detti **numeri razionali**. L’insieme dei numeri razionali si denota con \\( \\mathbb{Q} \\) e contiene l’insieme degli interi \\( \\mathbb{Z} \\). A differenza degli interi, dati due numeri razionali \\( r \\) e \\( s \\) con \\( r < s \\), esiste sempre un altro numero razionale \\( t \\) tale che \\( r < t < s \\): basta prendere per \\( t \\) la media \\((r+s)/2\\). Questa proprietà di \\( \\mathbb{Q} \\) si esprime dicendo che \\( \\mathbb{Q} \\) è **denso in sé**. Iterando il procedimento si può intuire che tra due numeri razionali vi sono sempre infiniti numeri razionali.\\(^7\\) Non ha quindi senso parlare di antecedente o successivo di un numero razionale.\n\nIl lettore può verificare facilmente che l’insieme \\( \\mathbb{Q} \\) soddisfa gli assiomi di campo e gli assiomi dell’ordine. Per questa ragione si dice che \\( \\mathbb{Q} \\) è un **campo ordinato**. L’insieme dei razionali positivi si denota con \\( \\mathbb{Q}_+ \\).\n\n\\(^7\\)La definizione rigorosa di insieme infinito verrà precisata in seguito.\n7 La rappresentazione geometrica dei razionali\n\nI numeri razionali si possono rappresentare geometricamente come punti su una retta euclidea. Sia $\\mathcal{R}$ una retta orientata. Scegliamo su di essa un punto $O$, detto origine, un punto $U$ distinto da $O$ sulla semiretta positiva e assumiamo il segmento $OU$ come unità di misura delle lunghezze. Con semplici operazioni geometriche è possibile costruire multipli e sottomultipli del segmento $OU$. Quindi assegniamo al numero razionale $r = \\frac{m}{n}$ il punto $P$ sulla semiretta positiva tale che $OP = \\frac{m}{n}OU$ se $m, n > 0$, il punto $P$ sulla semiretta negativa tale che $PO = -\\frac{m}{n}OU$ se $m < 0$, $n > 0$. Al numero 0 assegniamo il punto $O$.\n\nIl numero razionale $r$ si dice ascissa del punto $P$. L'applicazione di $Q$ in $\\mathcal{R}$ così definita è iniettiva ma non surgettiva. L'iniettività segue dal fatto che se $r < s$ l'immagine di $r$ sulla retta precede l'immagine di $s$ nell'orientazione fissata. Per dimostrare che non tutti i punti della retta hanno un'ascissa razionale premettiamo il seguente lemma.\n\nLemma 7.1 Non esiste alcun numero razionale $r$ tale che $r^2 = 2$.\n\nDimostrazione. Supponiamo per assurdo che esista un numero razionale $r$ tale che $r^2 = 2$. Poiché $r^2 = (-r)^2$, possiamo supporre che $r$ sia positivo. Sia quindi $r = \\frac{m}{n}$ con $m, n > 0$. Allora $m^2 = 2n^2$ e questo contraddice l'unicità della decomposizione dei numeri naturali in fattori primi, perché il fattore 2 compare con un esponente pari nel primo membro e con un esponente dispari nel secondo membro.\n\nProposizione 7.2 Sia $\\overline{OP}$ il segmento di lunghezza uguale alla diagonale del quadrato di lato $\\overline{OU}$. Allora il punto $P$ non corrisponde ad alcun numero razionale.\n\nDimostrazione. Se $r$ fosse un numero razionale tale che $\\overline{OP} = r\\overline{OU}$, per il teorema di Pitagora si avrebbe $r^2 = 2$. Ma questo contraddice il Lemma 7.1.\n\nL'insieme dei numeri razionali si rivela quindi insufficiente per risolvere il problema della misura delle grandezze omogenee. Infatti i numeri razionali permettono di misurare solo quelle grandezze che sono commensurabili con l'unità di misura scelta. Vedremo invece che questo inconveniente non si verifica con l'insieme dei numeri reali perché $\\mathbb{R}$ soddisfa un ulteriore assioma: l'assioma di completezza.\n\n8 L'assioma di completezza\n\nPrima di enunciare l'assioma di completezza è necessario introdurre alcune definizioni e notazioni. Sia $X$ un insieme totalmente ordinato, cioè un insieme su cui è definita una relazione $\\leq$ riflessiva, antisimmetrica, transitiva e che soddisfa la condizione di totalità del Teorema 3.1.\n\nDefinizione. Sia $S$ un sottoinsieme non vuoto di $X$. Diremo che un elemento $a$ di $X$ è un minorante di $S$ se $a \\leq x$ per ogni $x \\in S$. Analogamente diremo che elemento $b$ di $X$ è un maggiorante di $S$ se $x \\leq b$ per ogni $x \\in S$. Un insieme $S$ si dice inferiormente limitato se esiste almeno un minorante di $S$, inferiormente illimitato se l'insieme dei minoranti di\nL’assioma di completezza\n\n$S$ è vuoto. Le definizioni di insieme superiormente limitato, superiormente illimitato sono analoghe. Un insieme si dice limitato se è superiormente e inferiormente limitato. Un minorante di $S$ che appartiene a $S$ si dice minimo di $S$; un maggiorante di $S$ che appartiene a $S$ si dice massimo di $S$. Un insieme può avere al più un minimo e al più un massimo. Infatti, se $a$ e $a'$ sono minoranti di $S$ e appartengono entrambi a $S$, si ha $a \\leq a'$ e $a' \\leq a$ e quindi $a = a'$. L’unicità del massimo si prova in modo analogo. Le notazioni min $S$ e max $S$ denotano il minimo e il massimo di $S$.\n\nDiamo alcuni esempi per illustrare questi concetti.\n\n**Esempio 1.** Sia $X = \\mathbb{R}$. L’intervallo $[0, 1)$ è limitato. L’insieme dei minoranti è l’intervallo $(-\\infty, 0]$. L’insieme dei maggioranti è l’intervallo $[1, +\\infty)$. Poiché $0$ è un minorante e appartiene all’insieme $[0, 1)$ si ha $0 = \\min[0, 1)$. L’insieme è privo di massimo.\n\n**Esempio 2.** Sia $X = \\mathbb{Q}$, $S = \\{1/n : n \\in \\mathbb{Z}_+\\}$. Allora l’insieme dei maggioranti di $S$ è l’insieme dei razionali maggiori o uguali a $1$. Poiché $1 \\in S$ si ha che $1 = \\max S$. Ogni numero razionale $\\leq 0$ è un minorante di $S$ e non vi sono altri minoranti, perché, se $r = m/n$ è un numero razionale positivo, $1/(n+1) < r$ e quindi $r$ non è un minorante di $S$. Pertanto $0$ è il massimo dell’insieme dei minoranti di $S$, ma $S$ è privo di minimo.\n\n**Definizione.** Se l’insieme dei minoranti di $S$ ha massimo $a$ si dice che $a$ è l’estremo inferiore di $S$. Analogamente, se l’insieme dei maggioranti di $S$ ha minimo $b$ si dice che $b$ è l’estremo superiore di $S$.\n\nL’estremo inferiore e l’estremo superiore di un insieme, se esistono, sono necessariamente unici. Essi si denotano rispettivamente con $\\inf S$ e $\\sup S$. Si noti che se $S$ ha minimo esso coincide con l’estremo inferiore. Tuttavia, come mostra l’Esempio 2, un insieme può avere estremo inferiore senza avere minimo. Viceversa se l’estremo inferiore di $S$ appartiene a $S$, esso è il minimo di $S$. Considerazioni analoghe valgono per estremo superiore e massimo.\n\nPossiamo ora formulare l’assioma di completezza.\n\n**Assioma 9** Ogni sottoinsieme non vuoto superiormente limitato di $\\mathbb{R}$ ha estremo superiore.\n\nIl fatto che ogni sottoinsieme non vuoto e inferiormente limitato di $\\mathbb{R}$ ha estremo inferiore è una conseguenza del seguente risultato generale valido in ogni insieme totalmente ordinato.\n\n**Proposizione 8.1** Sia $X$ un insieme totalmente ordinato. Ogni sottoinsieme non vuoto inferiormente limitato di $X$ ha estremo inferiore se e solo se ogni sottoinsieme non vuoto superiormente limitato di $X$ ha estremo superiore.\n\n**Dimostrazione.** Proviamo la sufficienza della condizione, lasciando la dimostrazione della necessità al lettore per esercizio. Sia $S$ un sottoinsieme non vuoto inferiormente limitato di $X$ e sia $S_*$ l’insieme dei minoranti di $S$. Allora $S_*$ è non vuoto e superiormente limitato (tutti gli elementi di $S$ sono maggioranti di $S_*$). Pertanto per l’assioma di completezza $S_*$ ha estremo superiore $a$. Poiché $a$ è il minimo dei maggioranti di $S_*$ e gli elementi di $S$ sono maggioranti di $S_*$, si ha che $a \\leq x$ per ogni $x$ in $S$. Quindi $a$ è un minorante di $S$ e perciò $a \\in S_*$. Questo prova che $a = \\max S_* = \\inf S$. □\nIl lettore può verificare facilmente che, se $S$ è un sottoinsieme non vuoto e superiormente limitato di $\\mathbb{R}$ e $a = \\sup S$, allora l’insieme dei maggioranti di $S$ è l’intervallo chiuso semilimitato a sinistra $[a, +\\infty)$. Analogamente, se $S$ è un sottoinsieme non vuoto e inferiormente limitato di $\\mathbb{R}$ e $b = \\inf S$, allora l’insieme dei minoranti di $S$ è l’intervallo chiuso semilimitato a destra $(-\\infty, b]$. Queste osservazioni forniscono un criterio utile per determinare l’estremo superiore e l’estremo inferiore degli insiemi. Ad esempio per determinare l’estremo superiore di un insieme $S$ si determina dapprima l’insieme dei maggioranti di $S$ risolvendo le disequazioni $x \\geq s$ per ogni $s$ in $S$. L’insieme delle soluzioni è un intervallo chiuso semilimitato a sinistra $[a, +\\infty)$. Quindi $\\sup S = \\min [a, +\\infty) = a$.\n\nUn altro criterio utile per determinare l’estremo inferiore e l’estremo superiore di un insieme si basa sulla seguente proprietà: se l’insieme $S$ ha estremo superiore vi sono elementi in $S$ “arbitrariamente vicini” a $\\sup S$; se $S$ ha estremo inferiore vi sono in $S$ elementi “arbitrariamente” vicini a $\\inf S$.\n\n**Proposizione 8.2** Sia $S$ un sottoinsieme non vuoto di $\\mathbb{R}$. Un numero reale $a$ è l’estremo superiore di $S$ se e solo se $a$ è un maggiorante di $S$ e per ogni $\\epsilon > 0$ esiste un elemento $s$ in $S$ tale che $a - \\epsilon < s$. Analogamente $a$ è l’estremo inferiore di $S$ se e solo se $a$ è un minorante di $S$ e per ogni $\\epsilon > 0$ esiste un elemento $s$ in $S$ tale che $s < a + \\epsilon$.\n\n**Dimostrazione.** Sia $a = \\sup S$. Se esistesse un $\\epsilon > 0$ tale che per ogni $s$ in $S$ si avesse $s \\leq a - \\epsilon$, allora $a - \\epsilon$ sarebbe un maggiorante di $S$ minore di $a$. Questo contraddice il fatto che $a$ è il minimo dei maggioranti. Viceversa, supponiamo che $a$ sia un maggiorante di $S$ e che per ogni $\\epsilon > 0$ esista un $s$ in $S$ tale che $a - \\epsilon < s$. Dimostriamo che $a$ è il minimo dei maggioranti di $S$. Se $b < a$, posto $\\epsilon = b - a > 0$, si avrebbe che esiste un elemento $s$ in $S$ tale che $a - \\epsilon = b < s$. Pertanto $b$ non sarebbe un maggiorante di $S$. La dimostrazione della caratterizzazione dell’estremo inferiore è simile. \\(\\square\\)\n\n### 9 Conseguenze della completezza\n\nIn questa sezione esponiamo alcune proprietà dei numeri reali che derivano dall’assioma di completezza. Cominciamo con il dimostrare che l’insieme dei numeri reali soddisfa una proprietà analoga al postulato di continuità della retta.\\(^8\\)\n\n**Definizione.** Una coppia $(A, B)$ di sottoinsiemi non vuoti di $\\mathbb{R}$ si dice *separata* se per ogni $a$ in $A$ e $b$ in $B$ si ha $a \\leq b$. Una coppia separata si dice *contigua* se per ogni numero reale $\\epsilon > 0$ esistono due elementi $a$ in $A$ e $b$ in $B$ tali che $b - a < \\epsilon$. Un numero reale $x$ si dice un *elemento di separazione* della coppia $(A, B)$ se $a \\leq x \\leq b$ per ogni $a$ in $A$ e $b$ in $B$.\n\n\\(^8\\)Il postulato di continuità della retta, nella formulazione di Cantor, è il seguente: siano $A$ e $B$ sono due sottoinsiemi della retta orientata tali che:\n\n(i) ogni punto di $A$ precede ogni punto di $B$ nell’orientazione fissata;\n\n(ii) comunque fissato un numero razionale $r > 0$ esistono due punti $P$ in $A$ e $Q$ in $B$ tali che la lunghezza del segmento $PQ$ è minore di $r$.\n\nAllora esiste uno e un solo punto della retta compreso tra $A$ e $B$. \n\nTeorema 9.1 Se \\((A, B)\\) è una coppia separata di sottoinsiemi non vuoti di \\(\\mathbb{R}\\) esiste almeno un elemento di separazione. Se la coppia è contigua l’elemento di separazione è unico.\n\nDimostrazione. Ogni elemento di \\(B\\) è un maggiorante di \\(A\\). Quindi \\(A\\) è superiormente limitato e, per l’assioma di completezza, esiste sup \\(A\\). Poiché sup \\(A\\) è il minimo dei maggioranti di \\(A\\), si ha sup \\(A \\leq b\\) per ogni \\(b\\) in \\(B\\). Quindi sup \\(A\\) è un minorante di \\(B\\) e pertanto sup \\(A \\leq \\inf B\\). Tutti gli elementi dell’intervallo \\([\\sup A, \\inf B]\\) sono elementi di separazione della coppia \\((A, B)\\). Supponiamo ora che \\((A, B)\\) sia contigua. Osserviamo che se \\(x\\) è un elemento di separazione allora deve essere sup \\(A \\leq x \\leq \\inf B\\). Quindi, per dimostrare l’unicità dell’elemento di separazione, basta provare che sup \\(A = \\inf B\\). Se, per assurdo, fosse sup \\(A < \\inf B\\), preso \\(\\epsilon = (\\inf B - \\sup A)/2 > 0\\) esisterebbero due elementi \\(a\\) in \\(A\\) e \\(b\\) in \\(B\\), tali che \\(b - a < (\\inf B - \\sup A)/2\\). Ma questo è impossibile perché \\(a \\leq \\sup A < \\inf B \\leq b\\) e quindi \\(\\inf B - \\sup A \\leq b - a\\). Pertanto sup \\(A = \\inf B\\) e il teorema è dimostrato. \\(\\square\\)\n\nUtilizzando il Teorema 9.1 e il postulato di continuità della retta, si ottiene la rappresentazione geometrica dei numeri reali. Precisamente si può dimostrare il seguente teorema.\n\nTeorema 9.2 La corrispondenza che ad ogni numero razionale associa un punto sulla retta \\(\\mathbb{R}\\), definita nella Sezione 6, si estende in modo unico a una corrispondenza biunivoca di \\(\\mathbb{R}\\) su \\(\\mathbb{R}\\) che preserva l’ordine. In altri termini esiste un’unica funzione bigettiva \\(\\phi : \\mathbb{R} \\rightarrow \\mathbb{R}\\) tale che\n\n- \\(\\phi(r) = r\\overrightarrow{OU}\\) per ogni \\(r\\) in \\(\\mathbb{Q}\\),\n- se \\(x\\) e \\(y\\) sono due numeri reali tali che \\(x < y\\) allora \\(\\phi(x)\\) precede \\(\\phi(y)\\) nell’orientamento fissato sulla retta.\n\nLa corrispondenza tra \\(\\mathbb{R}\\) e la retta euclidea risulta univocamente determinata, una volta che siano stati fissati un orientamento sulla retta, un’origine e un’unità di misura. Essa viene detta un riferimento cartesiano su \\(\\mathbb{R}\\). Il numero reale \\(x\\) che corrisponde al punto \\(P\\) è detto ascissa di \\(P\\). L’insieme dei numeri reali viene quindi rappresentato geometricamente come l’insieme dei punti della retta. Questa rappresentazione geometrica ci consente di utilizzare un linguaggio geometrico nel parlare dei numeri reali. Così spesso parleremo del punto \\(x\\) anziché del punto di ascissa \\(x\\); il valore assoluto \\(|x - y|\\) verrà detto distanza dei punti \\(x\\) e \\(y\\), e diremo che il punto \\(x\\) si trova tra \\(a\\) e \\(b\\) se \\(a \\leq x \\leq b\\). Conseguentemente le espressioni retta reale o asse reale verranno utilizzati indifferenmente per denotare l’insieme dei numeri reali o una retta dotata di un riferimento cartesiano. Tuttavia noi utilizzeremo questa identificazione dell’insieme dei numeri reali con la retta reale soltanto come stampella per la nostra intuizione, senza mai fare riferimento a considerazioni geometriche per la dimostrazione delle proprietà di \\(\\mathbb{R}\\), che verranno dedotte esclusivamente dagli assiomi.\\(^9\\)\n\n\\(^9\\)L’identificazione dei numeri reali con i punti della retta euclidea orientata, dopo aver scelto un’origine e un’unità di misura delle lunghezze, è alla base della costruzione geometrica del campo dei numeri reali data nel XVI secolo da R. Bombelli nel IV libro della sua Algebra. L’operazione di somma di due punti \\(P\\) e \\(Q\\) “positivi” (cioè che seguono l’origine nell’orientazione della retta) è definita geometricamente come somma dei segmenti \\(\\overrightarrow{OP}\\) e \\(\\overrightarrow{OQ}\\). Per definire il prodotto di \\(P\\) e \\(Q\\) si utilizza l’assioma di esistenza e unicità del\n10 Archimedeanità dei reali\n\nIn questa sezione esaminiamo alcune altre proprietà dei numeri reali che derivano dalla completezza.\n\nProposizione 10.1 Ogni sottoinsieme non vuoto e superiormente limitato di \\( \\mathbb{Z} \\) ha massimo.\n\nDimostrazione. Sia \\( A \\) un sottoinsieme non vuoto e superiormente limitato di \\( \\mathbb{Z} \\). Per l’assioma di completezza esiste \\( a \\in \\mathbb{R} \\) tale che \\( a = \\sup A \\). Supponiamo, per assurdo, che \\( a \\) non sia il massimo di \\( A \\), cioè che \\( a \\notin A \\). Per la caratterizzazione dell’estremo superiore (Proposizione 8.2) dovrebbe esistere un elemento \\( k \\) di \\( A \\) tale che \\( a - 1 < k < a \\). Poiché abbiamo supposto che \\( A \\) non ha massimo, esiste un altro elemento \\( h \\) in \\( A \\) tale che \\( a - 1 < k < h < a \\). Pertanto \\( k - h \\) sarebbe un intero strettamente compreso tra 0 e 1. Questo contraddice la Proposizione 5.4. \\( \\square \\)\n\nIn modo analogo si dimostra che ogni sottoinsieme non vuoto inferiormente limitato di \\( \\mathbb{Z} \\) ha minimo. In particolare ogni sottoinsieme non vuoto di \\( \\mathbb{N} \\) ha minimo. Un insieme ordinato che gode della proprietà che ogni sottoinsieme non vuoto ha minimo si dice bene ordinato.\n\nCorollario 10.2 L’insieme dei numeri naturali è illimitato superiormente.\\(^{10}\\)\n\nDimostrazione. Se, per assurdo, \\( \\mathbb{N} \\) fosse superiormente limitato esso avrebbe massimo \\( m \\) per la Proposizione 10.1. Poiché \\( m + 1 \\) è un numero naturale maggiore di \\( m \\) si ha una contraddizione. \\( \\square \\)\n\nDefinizione. Un campo ordinato \\( K \\) si dice archimedeano \\(^{11}\\) se per ogni coppia di elementi positivi \\( \\epsilon \\) e \\( a \\) esiste un intero positivo \\( n \\) tale che \\( n\\epsilon > a \\).\n\nTeorema 10.3 Il campo reale è archimedeano.\n\nDimostrazione. Se \\( n\\epsilon \\leq a \\) per ogni intero positivo \\( n \\), l’insieme dei numeri naturali sarebbe superiormente limitato da \\( a/\\epsilon \\), in contraddizione con il Corollario 10.2. \\( \\square \\)\n\nOsservazione. Anche il campo \\( \\mathbb{Q} \\), pur non essendo completo, è archimedeano. Lasciamo al lettore la facile verifica di questo fatto. Tuttavia la proprietà archimedea non è una conseguenza dei soli assiomi di campo, perché esistono dei campi ordinati non archimedei \\(^{12}\\) (necessariamente non completi).\n\n\\(^{10}\\)È opportuno osservare che questo corollario enuncia anche una proprietà di \\( \\mathbb{R} \\), che possiamo esprimere in modo vago e impreciso, ma suggestivo, dicendo che non esistono numeri reali “infiniti”.\n\n\\(^{11}\\)In termini geometrici la proprietà archimedea significa che, date due grandezze, esiste sempre un multiplo della prima che è maggiore della seconda. In questa forma esso venne enunciato da Archimede, che ne fece uno degli assiomi della geometria e che lo utilizzò per il calcolo delle aree e dei volumi. In realtà Archimede stesso afferma che questo assioma era già stato impiegato dai suoi predecessori.\n\n\\(^{12}\\)Ad esempio il campo dei numeri reali non standard costruito da A. Robinson nel 1960.\nProposizione 10.4 Siano $a$ e $x$ due numeri reali tali che\n\\[ a \\leq x \\leq a + \\frac{1}{n} \\] (10.1)\nper ogni intero positivo $n$. Allora $a = x$.\n\nDimostrazione. Se $a < x$ allora, per l’archimedea dei reali esisterebbe un intero positivo $n$ tale che $n(x - a) > 1$. Quindi $a + \\frac{1}{n} < x$, contraddicendo la (10.1). Ne segue che non può essere $a < x$ e quindi $a = x$. □\n\nDefinizione. Un sottoinsieme $A$ di $\\mathbb{R}$ si dice denso in $\\mathbb{R}$ se per ogni coppia di numeri reali $x$ e $y$ con $x < y$ esiste un elemento $a$ in $A$ tale che $x < a < y$.\n\nProposizione 10.5 L’insieme dei numeri razionali $\\mathbb{Q}$ è denso in $\\mathbb{R}$.\n\nDimostrazione. Siano $x$ e $y$ numeri reali con $x < y$. Poniamo $\\epsilon = y - x$. Allora per l’archimedea di $\\mathbb{R}$ esiste un intero positivo $n$ tale che $n\\epsilon > 1$, cioè $1/n < \\epsilon$. Per la Proposizione 10.1 esiste $m = \\max\\{k \\in \\mathbb{Z} : k \\leq nx\\}$. Allora $m \\leq nx < m + 1$ e $m/n \\leq x < (m + 1)/n = m/n + 1/n < x + \\epsilon = y$. □\n\n11 Altre conseguenze della completezza\n\nEsaminiamo ora un’altra conseguenza importante dell’assioma di completezza.\n\nDefinizione. Una successione di intervalli $(I_n)$ si dice monotona se $I_n \\supseteq I_{n+1}$ per ogni $n$.\\(^{13}\\) Si ha quindi\n\\[ I_0 \\supseteq I_1 \\supseteq \\ldots \\supseteq I_n \\supseteq \\ldots \\]\nPuò accadere che l’intersezione $\\cap_{n \\in \\mathbb{N}} I_n$ di una successione monotona di intervalli sia vuota. Ad esempio se $I_n = (0, 1/n)$ è facile verificare, usando la proprietà di archimedea dei reali, che $\\cap_{n \\in \\mathbb{N}} I_n = \\emptyset$. Un altro esempio è dato dalla successione di intervalli illimitati $I_n = [n, +\\infty)$. Tuttavia, l’intersezione di una successione monotona di intervalli chiusi e limitati non è mai vuota. Si ha infatti il seguente\n\nTeorema 11.1 Se $(I_n)$ è una successione monotona di intervalli chiusi e limitati esiste un elemento comune a tutti gli intervalli. Tale elemento è unico se l’estremo inferiore delle lunghezze degli intervalli è 0.\n\nDimostrazione. Sia $I_n = [a_n, b_n]$. La lunghezza di $I_n$ è allora $\\ell(I_n) = b_n - a_n$. Consideriamo gli insiemi $A = \\{a_n : n \\in \\mathbb{N}\\}$ e $B = \\{b_n : n \\in \\mathbb{N}\\}$. Dimostriamo che la coppia $(A, B)$ è separata. Si ha che $a_n \\leq b_n$ per ogni $n$, perché $a_n$ e $b_n$ sono, rispettivamente, primo e secondo estremo dello stesso intervallo. Sia $n \\leq m$; allora poiché $I_m \\subseteq I_n$ si ha che $a_n \\leq a_m$ e $b_m \\leq b_n$. Siano ora $a_k$ un generico elemento di $A$ e $b_k$ un generico elemento di $B$. Se $k \\leq h$ si ha $a_k \\leq a_h \\leq b_h$. Se $h \\leq k$ si ha $a_h \\leq a_k \\leq b_k$. In ogni caso $a_k \\leq b_h$ e, quindi,\n\n\\(^{13}\\)Se $(I_n)$ è una successione monotona di intervalli si dice anche che gli intervalli sono *incapsulati*. \n(A, B) è separata. Per il Teorema 9.1 esiste un numero reale \\( x \\) tale che \\( a_n \\leq x \\leq b_n \\) per ogni \\( n \\). Pertanto \\( x \\in I_n \\) per ogni \\( n \\).\n\nPer dimostrare l’unicità osserviamo che un numero reale appartiene a tutti gli intervalli \\( I_n \\) se e solo se è un elemento di separazione della coppia \\( (A, B) \\). L’elemento di separazione è unico se la coppia \\( (A, B) \\) è contigua e questo avviene se \\( \\inf\\{b_n - a_n : n \\in \\mathbb{N}\\} = \\inf\\{\\ell(I_n) : n \\in \\mathbb{N}\\} = 0 \\).\n\nIl Teorema 11.1 può essere utilizzato per dimostrare l’esistenza delle radici \\( n \\)-esime dei numeri reali positivi.\n\n**Teorema 11.2** Sia \\( y \\) un numero reale positivo. Per ogni intero \\( n \\geq 2 \\) esiste un unico numero reale positivo \\( x \\) tale che \\( x^n = y \\).\n\n**Dimostrazione.** Dimostriamo dapprima l’unicità di \\( x \\). Siano \\( x_1 \\) e \\( x_2 \\) sono due numeri reali positivi distinti. Supponiamo che \\( x_1 < x_2 \\). Allora \\( x_1^n < x_2^n \\) e quindi non può essere \\( x_1^n = x_2^n \\). Dimostriamo ora l’esistenza di \\( x \\). Se \\( y = 1 \\) si ha \\( x = 1 \\). Supponiamo quindi che \\( y < 1 \\). Costruiremo una successione monotona di intervalli chiusi e limitati \\( (I_k) \\) tali che, posto \\( I_k = [a_k, b_k] \\), si ha\n\n\\[\n\\begin{align*}\n&\\text{a)} \\quad a_k^n \\leq y \\leq b_k^n, \\\\\n&\\text{b)} \\quad \\ell(I_k) = b_k - a_k \\leq 2^{-k}(1 - y).\n\\end{align*}\n\\]\n\nLa successione \\( (I_n) \\) è definita ricorsivamente nel modo seguente. Definiamo \\( I_0 = [y, 1] \\). Supponendo quindi di aver già definito \\( I_k \\), consideriamo il punto medio \\( c = (a_k + b_k)/2 \\) di \\( I_k \\). Se \\( c^n < y \\) definiamo \\( I_{k+1} = [c, b_k] \\), cioè \\( a_{k+1} = c \\), \\( b_{k+1} = b_k \\); altrimenti definiamo \\( I_{k+1} = [a_k, c] \\), cioè \\( a_{k+1} = a_k \\), \\( b_{k+1} = c \\). Pertanto \\( I_{k+1} \\subset I_k \\) e \\( \\ell(I_{k+1}) = \\ell(I_k)/2 \\). La successione \\( (I_k) \\) è quindi monotona ed è facile dimostrare per induzione che essa soddisfa le condizioni a) e b). Per il Teorema 11.1 esiste un elemento \\( x \\) comune a tutti gli intervalli \\( I_k \\). Dimostriamo che \\( x^n = y \\). Per la condizione a) si ha \\( a_k^n \\leq y \\leq b_k^n \\) per ogni \\( n \\) in \\( \\mathbb{N} \\). Inoltre, poiché \\( a_k \\leq x \\leq b_k \\) per ogni \\( k \\), si ha anche \\( a_k^n \\leq x^n \\leq b_k^n \\) per ogni \\( k \\) in \\( \\mathbb{N} \\). Quindi\n\n\\[\n|y - x^n| \\leq b_k^n - a_k^n.\n\\]\n\nPoiché \\( a_k \\) e \\( b_k \\) sono minori o uguali a 1,\n\n\\[\nb_k^n - a_k^n = (b_k - a_k)(b_k^{n-1} + a_k b_k^{n-2} + \\cdots + b_k a_k^{n-2} + a_k^{n-1}) \\\\\n\\leq n(b_k - a_k) \\\\\n= \\frac{n(1 - y)}{2^k}\n\\]\n\nper ogni \\( k \\) in \\( \\mathbb{N} \\). Pertanto\n\n\\[\n|y - x^n| \\leq \\inf\\left\\{\\frac{n(1 - y)}{2^k} : k \\in \\mathbb{N}\\right\\} = 0.\n\\]\n\nQuesto dimostra che \\( x^n = y \\) nel caso \\( 0 < y < 1 \\). Se \\( y > 1 \\) si consideri \\( z = 1/y \\). Allora \\( 0 < z < 1 \\) e, per quanto abbiamo appena dimostrato, esiste un numero reale positivo \\( u \\) tale che \\( u^n = z \\). Ma allora basta porre \\( x = 1/u \\) per avere \\( x^n = y \\). \\( \\square \\)\nDefinizione. Se $y$ è un numero reale positivo, l’unico numero reale positivo $x$ tale che $x^n = y$ si dice radice $n$-esima di $y$ e si denota con $\\sqrt[n]{y}$ ($\\sqrt{y}$ se $n = 2$).\n\nIn particolare in $\\mathbb{R}$ esiste un numero $x$ tale che $x^2 = 2$. Poiché $x \\notin \\mathbb{Q}$ per il Lemma 7.1, l’insieme $\\mathbb{R} \\setminus \\mathbb{Q}$ è non vuoto. Gli elementi di $\\mathbb{R} \\setminus \\mathbb{Q}$ si dicono numeri irrazionali.\n\nProposizione 11.3 Il campo $\\mathbb{Q}$ non è completo.\n\nDimostrazione. Se $\\mathbb{Q}$ fosse completo, potremmo ripetere le dimostrazioni dei Teoremi 11.1 e 11.2 in $\\mathbb{Q}$ anziché in $\\mathbb{R}$, giungendo alla conclusione che l’equazione $x^2 = 2$ ha soluzione in $\\mathbb{Q}$. Ma questo contraddice il Lemma 7.1. □\n\nProposizione 11.4 L’insieme dei numeri irrazionali $\\mathbb{R} \\setminus \\mathbb{Q}$ è denso in $\\mathbb{R}$.\n\nDimostrazione. Siano $x$ e $y$ numeri reali con $x < y$. Per la Proposizione 10.5 esiste un numero razionale $r$ tale che $x - \\sqrt{2} < r < y - \\sqrt{2}$. Allora $x < r + \\sqrt{2} < y$ e $r + \\sqrt{2}$ è irrazionale. □\n\n12 Allineamenti decimali\n\nIn questa sezione dimostreremo che a ogni numero reale è possibile associare un allineamento decimale limitato o illimitato. I numeri razionali sono esattamente quei numeri che hanno allineamenti decimali limitati o periodici. Anche se gli allineamenti decimali non giocheranno alcun ruolo nel resto del corso, abbiamo voluto trattare questo argomento per convincere il lettore che fosse rimasto scettico di fronte a una presentazione assiomatica, in cui il concetto di numero reale rimane indefinito, che gli assiomi di campo ordinato completo conducono alla rappresentazione dei numeri reali che gli è più familiare.\n\nAd ogni numero razionale maggiore o uguale a 0 è possibile associare un allineamento decimale nel modo seguente. Sia $r = p/q$ con $p$ e $q$ interi, $p \\geq 0$ e $q > 0$. Effettuando successivamente la divisione (Teorema 5.7) si definiscono per ricorrenza due successioni di numeri naturali $(p_n)$ e $(r_n)$ tali che\n\n\\[\n\\begin{align*}\np &= p_0q + r_0, & 0 \\leq r_0 < q, \\\\\n10r_0 &= p_1q + r_1, & 0 \\leq r_1 < q, \\\\\n&\\vdots \\\\\n10r_n &= p_{n+1}q + r_{n+1}, & 0 \\leq r_{n+1} < q \\\\\n&\\vdots\n\\end{align*}\n\\]\n\n(12.1)\n\nIl numero naturale $p_0$ è detto parte intera del numero razionale $r$ e i numeri naturali $p_1, p_2, p_3, \\ldots$ sono le cifre decimali. Le cifre decimali sono numeri naturali $\\leq 9$. Infatti dalla (12.1) si vede che\n\n\\[\np_{n+1} = \\frac{10r_n - r_{n+1}}{q} < 10,\n\\]\nperché $r_n < q$ e $r_{n+1} \\geq 0$. Poiché i resti delle divisioni successive sono numeri naturali compresi tra 0 e $q - 1$, dopo $q$ divisioni al massimo uno dei resti $r_k$ si presenterà per la seconda volta. Ma, allora, tutti i resti seguenti si ripeteranno nello stesso ordine in cui si presentavano dopo la prima apparizione di $r_k$. Questo mostra che la successione delle cifre decimali di un numero razionale è periodica, cioè esiste una cifra o un gruppo di cifre che, dopo essere apparso la prima volta, si ripete infinite volte. Questo gruppo di cifre è detto periodo del numero razionale $r$. Le cifre decimali che precedono il periodo sono dette antiperiodo. La successione $(p_n)$ viene detta allineamento decimale di $r$. Per indicare che $(p_n)$ è l’allineamento decimale di $r$ si scrive\n\n$$r = p_0.p_1p_2p_3\\ldots p_n\\ldots$$\n\nSe il periodo è 0 l’allineamento si dice limitato e il periodo non si scrive (ad esempio $3/5=0.6$). In caso contrario l’allineamento si dice illimitato periodico e si scrivono soltanto le cifre decimali fino al periodo, che viene segnalato con una sopralineatura (ad esempio $3422/990 = 3.456$). Diamo ora una diversa caratterizzazione dell’allineamento decimale dei numeri razionali.\n\n**Proposizione 12.1** Una successione $(p_n)$ di numeri naturali tale che $p_n \\leq 9$ per $n \\geq 1$ è l’allineamento decimale del numero razionale $r$ se e solo se per ogni $n$ essa soddisfa le disuguaglianze\n\n$$p_0 + \\frac{p_1}{10} + \\frac{p_2}{10^2} + \\frac{p_3}{10^3} + \\cdots + \\frac{p_n}{10^n} \\leq r < p_0 + \\frac{p_1}{10} + \\frac{p_2}{10^2} + \\frac{p_3}{10^3} + \\cdots + \\frac{p_n}{10^n} + \\frac{1}{10^n}. \\quad (12.2)$$\n\n**Dimostrazione.** È facile dimostrare per induzione che per ogni $n$ in $\\mathbb{N}$\n\n$$r = p_0 + \\frac{p_1}{10} + \\frac{p_2}{10^2} + \\frac{p_3}{10^3} + \\cdots + \\frac{p_n}{10^n} + \\frac{r_n}{10^n q}. \\quad (12.3)$$\n\nPertanto, tenendo conto del fatto che $0 \\leq r_n < q$, si vede che le disuguaglianze (12.2) sono soddisfatte. Viceversa, sia $r = p/q$ un razionale maggiore o uguale a 0 e supponiamo che le (12.2) siano soddisfatte per ogni $n$. Allora $p_0 \\leq p/q < p_0 + 1$. Pertanto $p_0 q \\leq p < (p_0 + 1)q$ e quindi, posto $r_0 = p - p_0 q$, si ha che $p = p_0 q + r_0$ con $0 \\leq r_0 < q$. Analogamente, se\n\n$$p_0 + \\frac{p_1}{10} \\leq p/q < p_0 + \\frac{p_1}{10} + \\frac{1}{10},$$\n\nposto $r_1 = p - p_0 q - p_1 q/10$, si ha che $10 r_0 = p_1 q + r_1$ e $0 \\leq r_1 < q$. Lasciamo al lettore il compito di completare la dimostrazione provando per induzione che $10 r_n = p_{n+1} q + r_{n+1}$ con $0 \\leq r_{n+1} < q$. \\[\\square\\]\n\n\\[14\\] Per rendere questa argomentazione rigorosa basta osservare che, se i resti $r_n$ e $r_{n+k}$ sono uguali, le due successioni dei quozienti e dei resti definite ricorsivamente a partire da $r_n$ e $r_{n+k}$ sono uguali per il teorema di ricorrenza (Teorema 5.6).\nLe disuguaglianze (12.2) hanno una semplice interpretazione geometrica. Poiché la retta reale è l’unione degli intervalli disgiunti \\([n, n + 1)\\), \\(n\\) in \\(\\mathbb{Z}\\), il punto \\(r\\) appartiene a uno e uno solo di tali intervalli, diciamo \\([x_0, x_0 + 1)\\). Questo determina la parte intera \\(x_0\\) di \\(r\\). Successivamente dividiamo l’intervallo \\([x_0, x_0 + 1)\\) nei dieci sottointervalli disgiunti\n\\[\n[x_0, x_0 + \\frac{1}{10}), [x_0 + \\frac{1}{10}, x_0 + \\frac{2}{10}), \\ldots, [x_0 + \\frac{9}{10}, x_0 + 1),\n\\]\nciascuno dei quali ha lunghezza \\(1/10\\). Scegliamo l’intervallo \\([x_0 + \\frac{x_1}{10}, x_0 + \\frac{x_1+1}{10})\\) in cui cade \\(r\\). Questo determina la prima cifra decimale \\(x_1\\) di \\(r\\). Le cifre decimali successive vengono determinate ricorsivamente, suddividendo ogni volta l’intervallo precedente in dieci sottointervalli semichiusi a sinistra di uguale ampiezza e scegliendo tra essi quello in cui cade il punto \\(r\\). L’allineamento decimale è limitato se e solo se \\(r\\) cade in uno degli estremi degli intervalli semichiusi così costruiti. Lo stesso procedimento può essere utilizzato per generare l’allineamento decimale di un qualunque numero reale.\n\nSia \\(x\\) un numero reale maggiore o uguale a 0. La rappresentazione decimale \\((x_n)\\) di \\(x\\) è definita ricorsivamente come segue. Si pone\n\\[\nx_0 = \\max\\{k \\in \\mathbb{Z} : k \\leq x\\}.\n\\]\nSi noti che il massimo esiste per il Lemma 10.1, perché l’insieme è non vuoto \\((0 \\leq x)\\) e superiormente limitato da \\(x\\). Successivamente si definisce\n\\[\nx_1 = \\max\\{k \\in \\mathbb{Z} : x_0 + \\frac{k}{10} \\leq x\\}\n\\]\n\\[\nx_2 = \\max\\{k \\in \\mathbb{Z} : x_0 + \\frac{x_1}{10} + \\frac{k}{10^2} \\leq x\\}\n\\]\n\\[\n\\ldots\n\\]\nIn generale, supposto di aver già definito \\(x_0, x_1, x_2, \\ldots, x_n\\) si definisce\n\\[\nx_{n+1} = \\max\\{k \\in \\mathbb{Z} : x_0 + \\frac{x_1}{10} + \\cdots + \\frac{x_n}{10^n} + \\frac{k}{10^{n+1}} \\leq x\\}.\n\\]\nDalla definizione della successione \\((x_n)\\) discende immediatamente che \\(0 \\leq x_n \\leq 9\\) se \\(n \\geq 1\\) e\n\\[\nx_0 + \\frac{x_1}{10} + \\frac{x_2}{10^2} + \\frac{x_3}{10^3} + \\cdots + \\frac{x_n}{10^n} \\leq x < x_0 + \\frac{x_1}{10} + \\frac{x_2}{10^2} + \\frac{x_3}{10^3} + \\cdots + \\frac{x_n}{10^n} + \\frac{1}{10^n}. \\quad (12.4)\n\\]\nper ogni \\(n\\) in \\(\\mathbb{N}\\). Pertanto, se \\(x\\) è razionale, la successione \\((x_n)\\) coincide con l’allineamento decimale di \\(x\\) ottenuto mediante divisioni successive, per la Proposizione 12.1.\n\nAbbiamo così definito un’applicazione \\(x \\mapsto (x_n)\\) dell’insieme dei numeri reali maggiori o uguali a 0 nell’insieme degli allineamenti decimali. L’applicazione è iniettiva, perché, se \\(x\\) e \\(x’\\) sono due numeri reali che generano lo stesso allineamento decimale \\((x_n)\\), per le disuguaglianze (12.4), si ha \\(|x - x'| < 1/10^n\\) per ogni \\(n\\) in \\(\\mathbb{N}\\). Pertanto \\(x = x’\\). È naturale chiedersi se questa applicazione è surgettiva. La risposta è negativa.\nProposizione 12.2 \\textsuperscript{15} Nessun numero reale genera un allineamento decimale con periodo 9.\n\nDimostrazione. Dimostriamo dapprima che non esiste un numero reale che abbia 0.\\overline{9} come allineamento decimale. Infatti, se 0.\\overline{9} fosse l’allineamento decimale del numero reale \\( x \\), per la (12.4) si avrebbe per ogni \\( n \\)\n\\[\n\\frac{9}{10} \\left( 1 + \\frac{1}{10} + \\cdots + \\frac{1}{10^{n-1}} \\right) \\leq x < \\frac{9}{10} \\left( 1 + \\frac{1}{10} + \\cdots + \\frac{1}{10^{n-1}} \\right) + \\frac{1}{10^n}. \\tag{12.5}\n\\]\nDalla (12.5), utilizzando l’identità \\( 1 + t + t^2 + \\cdots + t^{n-1} = (1 - t^n)/(1 - t) \\) per \\( t = 1/10 \\), si ottiene\n\\[\n1 - \\frac{1}{10^n} \\leq x < 1.\n\\]\nPertanto \\( 0 \\leq 1 - x < 1/10^n \\) per ogni \\( n \\). Poiché \\( \\inf \\{1/10^n : n \\in \\mathbb{N} \\} = 0 \\), ne segue che \\( x = 1 \\) e questo contraddice la disuguaglianza \\( x < 1 \\).\n\nPer passare al caso generale osserviamo che è facile dimostrare per induzione che, se \\( y_0.y_1y_2 \\cdots y_n \\cdots \\) è l’allineamento decimale del numero reale \\( y \\), allora l’allineamento decimale di \\( 10^n y \\) è\n\\[\n10^n y_0 + 10^{n-1} y_1 + \\cdots + y_n y_{n+1} y_{n+2} \\cdots\n\\]\nPertanto, se \\( x_0.x_1x_2 \\cdots x_n \\overline{9} \\) fosse l’allineamento decimale di un numero reale \\( x \\), il numero reale \\( 10^n x - (10^n x_0 + 10^{n-1} x_1 + \\cdots + 10 x_{n-1} + x_n) \\) avrebbe allineamento decimale 0.\\overline{9}, ma questo è impossibile, come abbiamo appena dimostrato. \\( \\square \\)\n\nTuttavia, se si escludono gli allineamenti decimali con periodo 9, tutti gli altri allineamenti decimali, periodici o no, sono generati da un numero reale maggiore o uguale a 0.\n\nProposizione 12.3 Ogni allineamento decimale che non ha periodo 9 è generato da un numero reale maggiore o uguale a 0.\n\nDimostrazione. Sia \\( (x_n) \\) una allineamento decimale che non ha periodo 9. La successione di intervalli \\( I_n = [a_n, b_n] \\), dove\n\\[\na_n = x_0 + \\frac{x_1}{10} + \\frac{x_2}{10^2} + \\frac{x_3}{10^3} + \\cdots + \\frac{x_n}{10^n},\n\\]\n\\[\nb_n = x_0 + \\frac{x_1}{10} + \\frac{x_2}{10^2} + \\frac{x_3}{10^3} + \\cdots + \\frac{x_n}{10^n} + \\frac{1}{10^n}\n\\]\nè monotona e l’ampiezza di \\( I_n \\) è \\( 10^{-n} \\). Pertanto, per la Proposizione 11.1, esiste un unico numero reale \\( x \\) comune a tutti gli \\( I_n \\). Il punto \\( x \\) soddisfa le disuguaglianze \\( a_n \\leq x \\leq b_n \\) per ogni \\( n \\). Per dimostrare che \\( (x_n) \\) è l’allineamento decimale generato da \\( x \\) basta provare che si ha \\( x < b_n \\) per ogni \\( n \\). Se, per assurdo, esistesse un numero naturale \\( m \\) tale che\n\\[\nx = b_m = x_0 + \\frac{x_1}{10} + \\frac{x_2}{10^2} + \\frac{x_3}{10^3} + \\cdots + \\frac{x_m}{10^m} + \\frac{1}{10^m},\n\\]\n\\textsuperscript{15}Il lettore può verificare che l’impossibilità di allineamenti decimali con periodo 9 è una conseguenza dell’aver scelto il segno \\( \\leq \\) a sinistra e il segno \\( < \\) a destra nelle disuguaglianze (12.4). Se scambiamo tra loro i due segni \\( \\leq \\) e \\( < \\) si ottiene una definizione diversa di allineamento decimale, per cui gli allineamenti con periodo 9 sono consentiti, ma non quelli con periodo 0.\nsi avrebbe che \\( b_m = x \\leq b_{m+1} \\). Pertanto\n\\[\n\\frac{1}{10^m} \\leq \\frac{x_{m+1}}{10^{m+1}} + \\frac{1}{10^{m+1}}\n\\]\ne quindi \\( x_{m+1} \\geq 1/10^m - 1/10^{m+1} = 9 \\). In modo analogo si dimostra che, se \\( x_{m+k} = 9 \\) per \\( k = 1, 2, 3, \\ldots, n \\), allora anche \\( x_{m+n+1} = 9 \\). Ne consegue, per induzione, che tutte le cifre a partire da \\( x_{m+1} \\) sono uguali a 9. Pertanto il periodo di \\( (x_n) \\) sarebbe 9. □\n\n**Definizione.** Diciamo che un allineamento decimale è **ammissibile** se non ha periodo 9.\n\nIndichiamo con \\( \\mathcal{D} \\) l’insieme degli allineamenti decimali ammissibili. Abbiamo visto che l’applicazione \\( x \\mapsto (x_n) \\) è una corrispondenza biunivoca dell’insieme \\( \\mathbb{R}_+ \\cup \\{0\\} \\) sull’insieme \\( \\mathcal{D} \\). Questa corrispondenza ci permette di trasportare su \\( \\mathcal{D} \\) la struttura algebrica di \\( \\mathbb{R}_+ \\cup \\{0\\} \\), definendo la somma e il prodotto di allineamenti decimali nel modo seguente: se \\( (x_n) \\) e \\( (y_n) \\) sono gli allineamenti associati ai numeri reali \\( x \\) e \\( y \\) rispettivamente, la loro somma è l’allineamento associato al numero \\( x + y \\). Il prodotto di due decimali si definisce in modo analogo come l’allineamento associato al prodotto dei numeri reali corrispondenti. Possiamo quindi identificare l’insieme dei numeri reali maggiori o uguali a 0 con i loro allineamenti decimali e scrivere\n\\[\nx = x_0.x_1x_2x_3\\cdots x_n\\cdots\n\\]\nper indicare che \\( (x_n) \\) è l’allineamento decimale che corrisponde al numero reale \\( x \\).\n\nIn generale non è possibile dare una formula esplicita per il calcolo delle cifre della somma e del prodotto di due decimali illimitati. Tuttavia, usando la (12.4) si può facilmente dimostrare per induzione che, se uno dei decimali è limitato e ha solo \\( n \\) cifre non nulle dopo il punto, mentre le prime \\( n \\) cifre decimali dell’altro sono nulle, la loro somma si ottiene aggiungendo le cifre del secondo subito dopo quelle del primo. In simboli:\n\\[\nx_0.x_1x_2\\cdots x_n + 0.\\underbrace{0\\cdots0}_{n \\text{ cifre}} y_{n+1} y_{n+2} \\cdots y_{n+k} \\cdots\n\\]\n\\[\n= x_0.x_1x_2\\cdots x_n y_{n+1} y_{n+2} \\cdots y_{n+k} \\cdots\n\\]\n(12.6)\n\nInoltre, le cifre del prodotto di un decimale \\( 0.x_1x_2\\cdots \\) per \\( 10^{-m} \\) si ottengono spostando a destra di \\( m \\) posti le cifre del primo fattore e inserendo \\( m \\) zeri dopo il punto:\n\\[\n10^{-m} 0.x_1x_2\\cdots x_n \\cdots = 0.\\underbrace{0\\cdots0}_{m \\text{ cifre}} x_1 \\cdots x_n \\cdots\n\\]\n(12.7)\n\nPossiamo ora dimostrare che un allineamento decimale periodico corrisponde sempre a un numero razionale.\n\n**Proposizione 12.4** L’allineamento decimale periodico \\( c.a_1a_2\\ldots a_k\\overline{b_1b_2\\cdots b_m} \\) corrisponde al numero razionale\n\\[\nc + \\frac{a_1}{10} + \\frac{a_2}{10^2} + \\cdots + \\frac{a_k}{10^k} + \\frac{1}{1 - 10^{-m}} \\left( \\frac{b_1}{10^{k+1}} + \\frac{b_2}{10^{k+2}} + \\cdots + \\frac{b_m}{10^{k+m}} \\right).\n\\]\n**Dimostrazione.** Sia \\( d = c.a_1a_2 \\ldots a_k \\overline{b_1b_2 \\ldots b_m} \\). Allora, per la (12.6)\n\n\\[\nd = c.a_1a_2 \\ldots a_k + 0. \\overline{0 \\ldots 0 b_1b_2 \\ldots b_m}\n\\]\n\nIl decimale limitato \\( c.a_1a_2 \\ldots a_k \\) corrisponde al numero razionale\n\n\\[\na = c + \\frac{a_1}{10} + \\frac{a_2}{10^2} + \\cdots + \\frac{a_k}{10^k}.\n\\]\n\nInoltre\n\n\\[\n0. \\overline{0 \\ldots 0 b_1b_2 \\ldots b_m} = 0. \\overline{0 \\ldots 0 b_1 \\ldots b_m b_1b_2 \\ldots b_m}\n\\]\n\n\\[\n= 0. \\overline{0 \\ldots 0 b_1 \\ldots b_m} + 0. \\overline{0 \\ldots 0 b_1b_2 \\ldots b_m}\n\\]\n\n(12.8)\n\nSiano\n\n\\[\nx = 0. \\overline{0 \\ldots 0 b_1b_2 \\ldots b_m}\n\\]\n\n\\[\nb = 0. \\overline{0 \\ldots 0 b_1 \\ldots b_m}.\n\\]\n\nPoiché l’allineamento decimale di \\( b \\) è limitato, \\( b \\) è il numero razionale\n\n\\[\n\\frac{b_1}{10^{k+1}} + \\frac{b_2}{10^{k+2}} + \\cdots + \\frac{b_m}{10^{k+m}}.\n\\]\n\nPer la (12.8) si ha che \\( x = b + 10^{-m}x \\) e, quindi, risolvendo rispetto a \\( x \\) si ottiene\n\n\\[\nx = \\frac{1}{1 - 10^{-m}} b.\n\\]\n\nPertanto \\( d = a + \\frac{1}{1 - 10^{-m}} b \\), come volevamo dimostrare.\n\nPer estendere la rappresentazione decimale ai numeri negativi si associa a ogni numero reale negativo l’allineamento decimale del suo valore assoluto preceduto dal segno \\(-\\). In questo modo si ottiene una corrispondenza biunivoca di tutto \\( \\mathbb{R} \\) sull’insieme degli allineamenti decimali ammissibili con segno \\( \\mathcal{D}_\\pm \\).\n\n### 13 Esistenza e unicità del campo reale\n\nIn un approccio costruttivo alla teoria dei numeri reali, partendo dall’insieme \\( \\mathbb{N} \\) dei numeri naturali si può definire l’insieme dei numeri reali come l’insieme di tutti gli allineamenti decimali ammissibili con segno. Questo approccio però si scontra con una difficoltà tecnica: è difficile definire la somma e il prodotto di decimali illimitati, perché vi è la possibilità di\ninfiniti “riporti”. Anche se questa difficoltà può essere superata, definendo prima la somma e il prodotto di allineamenti limitati e estendendo poi le definizioni ai decimali illimitati con un procedimento di approssimazione, rimane la necessità di dimostrare che le operazioni così definite soddisfano gli assiomi di campo ordinato. La trattazione risulta così decisamente più laboriosa e pesante dell’approccio assiomatico. D’altra parte la trattazione assiomatica lascia aperto il problema dell’“esistenza” di $\\mathbb{R}$, cioè di un insieme che verifica gli assiomi di campo ordinato completo. La questione dell’esistenza, ossia, in linguaggio moderno, la non-contradditorietà della teoria, viene risolta costruendo un “modello” di $\\mathbb{R}$, come ad esempio la retta reale, con le operazione definite geometricamente, o l’insieme degli allineamenti decimali o ancora l’insieme dei “tagli” di Dedekind, a cui accenneremo brevemente nella sezione 14. Naturalmente la costruzione di un modello non fa altro che spostare la questione, riconducendola alla non-contradditorietà degli assiomi della geometria euclidea o dei numeri naturali (e a quelli della teoria degli insiemi).\n\nL’esistenza di diversi modelli per gli assiomi di campo ordinato completo pone il problema dell’unicità del sistema dei numeri reali. In altri termini ci si chiede: il modello basato sugli allineamenti decimali è “sostanzialmente diverso” da quello basato sulla retta reale o da quello basato sui tagli di Dedekind? La questione viene risolta mediante la nozione di isomorfismo di campi ordinati.\n\nSiano $K$ e $K'$ due campi ordinati.\n\n**Definizione.** Un isomorfismo tra $K$ e $K'$ è un’applicazione iniettiva e surgettiva $\\phi : K \\to K'$, che preserva la struttura: per ogni $x$ e $y$ in $K$ si ha $\\phi(x + y) = \\phi(x) + \\phi(y)$ e $\\phi(xy) = \\phi(x)\\phi(y)$. Inoltre se $x < y$ allora $\\phi(x) < \\phi(y)$.\n\n**Teorema 13.1** Se $K$ e $K'$ sono due campi ordinati completi esiste un unico isomorfismo di campi ordinati $\\phi : K \\to K'$.\n\nIl Teorema 13.1 ci dice che esiste sostanzialmente un unico campo ordinato completo, perché tutti i modelli che soddisfano gli assiomi A1–A9 sono tra loro isomorfi. In questo senso si può parlare di unicità dei numeri reali. Rinviamo il lettore a [Bu, Capitolo 7] per la dimostrazione del Teorema 13.1.\n\n### 14 I tagli di Dedekind\n\nIn questa sezione descriviamo per sommi capi la costruzione di Dedekind di un campo ordinato completo a partire dal campo dei numeri razionali. Supponendo di aver già costruito l’insieme dei numeri reali, osserviamo che ogni numero reale $x$ individua un “taglio” dell’insieme $\\mathbb{Q}$ dei numeri razionali in due sottoinsiemi: l’insieme dei numeri razionali maggiori di $x$ e l’insieme dei numeri razionali minori di $x$. Denotiamo con $S(x) = \\{ r \\in \\mathbb{Q} : r > x \\}$ l’insieme dei numeri razionali maggiori di $x$.\n\nL’insieme $S(x)$ gode delle seguenti proprietà: è non vuoto e diverso da $\\mathbb{Q}$; se $r$ è un numero razionale in $S(x)$ ogni numero razionale maggiore di $r$ è ancora in $S(x)$. Inoltre $S(x)$ è privo di minimo. Viceversa, ogni sottoinsieme $S$ di $\\mathbb{Q}$ tale che\n1. $\\emptyset \\neq S \\neq \\mathbb{Q}$;\n\n2. se $r \\in S$ e $q > r$ allora $q \\in S$;\n\n3. $S$ non ha minimo,\n\ne della forma $S = S(x)$ per qualche numero reale $x$ (basta prendere per $x$ l’estremo inferiore di $S$). È evidente che l’insieme $S(x)$ è sufficiente per individuare il numero reale $x$. Dedekind ebbe l’idea di definire numero reale un qualunque sottoinsieme $S$ di $\\mathbb{Q}$ che soddisfa le proprietà 1.–3.\n\nIn altri termini, supponiamo di conoscere solamente l’insieme dei numeri razionali $\\mathbb{Q}$. Diremo che un sottoinsieme $S$ di $\\mathbb{Q}$ che soddisfa le proprietà 1.–3. è un taglio di Dedekind. Definiamo l’insieme $T$ dei numeri reali di Dedekind come l’insieme dei tagli di Dedekind.\n\nPer ogni numero numero razionale $r$ l’insieme $S(r) = \\{q \\in \\mathbb{Q} : q > r\\}$ è un taglio di Dedekind e quindi è un elemento di $T$. L’applicazione $r \\mapsto S(r)$ è chiaramente iniettiva. Quindi l’insieme dei numeri razionali si immerge in $T$. Chiameremo i tagli di questo tipo tagli razionali. In particolare sono razionali il taglio zero $S(0)$ e il taglio unità $S(1)$.\n\nCi proponiamo ora di definire su $T$ una struttura di campo ordinato completo. Cominciamo dalla relazione d’ordine ponendo $S_1 \\leq S_2$ se $S_2$ è un sottoinsieme di $S_1$. Per definire la somma di due tagli di Dedekind $S_1$ e $S_2$ osserviamo che anche l’insieme $S_1 + S_2 = \\{r + q : r \\in S_1, q \\in S_2\\}$ è un taglio di Dedekind. Definiamo ora l’opposto del taglio $S$. Sia $S'$ l’insieme dei minoranti di $S$ in $\\mathbb{Q}$, da cui è stato tolto l’estremo inferiore di $S$ se $S$ è un taglio razionale. Allora $-S = \\{-r : r \\in S'\\}$ è un taglio, che chiameremo l’opposto di $S$. È facile vedere che $S \\leq S(0)$ se e solo se $S(0) \\leq -S$. Diremo che il taglio $S$ è non negativo se $S(0) \\leq S$. Per definire il prodotto in $T$ consideriamo dapprima il caso di due tagli non negativi. Se $S_1$ e $S_2$ sono due tagli non negativi il loro prodotto è l’insieme $S_1S_2 = \\{rq : r \\in S_1, q \\in S_2\\}$. Naturalmente occorre verificare che l’insieme $S_1S_2$ è un taglio. Se $S_1$ è non negativo e $S_2$ è negativo, poniamo $S_1S_2 = -S_1(-S_2)$. Se $S_1$ e $S_2$ sono entrambi negativi il loro prodotto è $S_1S_2 = (-S_1)(-S_2)$. Non è difficile verificare che sui tagli razionali le operazioni di somma e prodotto e la relazione d’ordine così definite coincidono con la somma e il prodotto in $\\mathbb{Q}$. In altri termini, se $S(r)$ e $S(q)$ sono due tagli razionali, si ha $S(r) + S(q) = S(r + q)$ e $S(r)S(q) = S(rq)$. Inoltre $S(r) \\leq S(q)$ se e solo se $r \\leq q$.\n\n**Teorema 14.1** Con le operazioni e la relazione d’ordine definite sopra l’insieme $T$ è un campo ordinato completo.\n\nOmettiamo la dimostrazione che è piuttosto lunga e laboriosa (vedi [Bu, Capitolo 8]). Lasciamo al lettore come esercizio il compito di verificare che alcune delle proprietà che definiscono la struttura di campo ordinato completo sono soddisfatte. Ci limitiamo a osservare che la completezza di $T$ discende dal fatto che, se $A$ è un sottoinsieme inferiormente limitato di $T$, il taglio $\\bigcup_{S \\in A} S$ è l’estremo inferiore di $A$. \n\nReferences\n\n[Ha] P. R. Halmos, *Teoria elementare degli insiemi*, Feltrinelli Editore, Milano, 1970.\n\n[Bu] C. W. Burrill, *Foundations of Real Numbers*, McGraw-Hill Book Company, New York St. Louis San Francisco Toronto London Sydney, 1967.", "id": "./materials/19.pdf" }, { "contents": "Concept of Linear Transformation\n\n**Definition:** Let $U$ and $V$ be two real vector spaces. $T: U \\rightarrow V$ is a linear transformation if:\n\n(i) $\\forall x, y \\in U$, $T(x + y) = T(x) + T(y)$\n(ii) $\\forall x \\in U$, $\\forall \\alpha \\in \\mathbb{R}$, $T(\\alpha x) = \\alpha T(x)$\n\n1. Prove that the transformation $T: \\mathbb{R}^2 \\rightarrow \\mathbb{R}^3$, $T(x, y) = (2x, y, -y)$ is linear.\n\n(i) Considering $(x_1, y_1), (x_2, y_2) \\in \\mathbb{R}^2$, we have:\n\n$$T((x_1, y_1) + (x_2, y_2)) = T(x_1 + x_2, y_1 + y_2)$$\n$$= (2(x_1 + x_2), y_1 + y_2, -(y_1 + y_2))$$\n$$= (2x_1 + 2x_2, y_1 + y_2, -y_1 - y_2)$$\n\nOn the other side,\n\n$$T(x_1, y_1) + T(x_2, y_2) = (2x_1, y_1, -y_1) + (2x_2, y_2, -y_2)$$\n$$= (2x_1 + 2x_2, y_1 + y_2, -y_1 - y_2)$$\n\nWe concluded that,\n\n$$T((x_1, y_1) + (x_2, y_2)) = T(x_1, y_1) + T(x_2, y_2), \\forall (x_1, y_1), (x_2, y_2) \\in \\mathbb{R}^2$$\n\n😊 The first condition of linearity of a transformation is proved.\n\n(ii) Considering $(x_1, y_1) \\in \\mathbb{R}^2$ and $\\alpha \\in \\mathbb{R}$, we have:\n\n$$T(\\alpha(x_1, y_1)) = T(\\alpha x_1, \\alpha y_1) = (\\alpha 2x_1, \\alpha y_1, -\\alpha y_1)$$\n$$= \\alpha(2x_1, y_1, -y_1) = \\alpha T(x_1, y_1)$$\n\nWe concluded that,\n\n$$T(\\alpha(x_1, y_1)) = \\alpha T(x_1, y_1), \\forall (x_1, y_1) \\in \\mathbb{R}^2, \\forall \\alpha \\in \\mathbb{R}$$\n\n😊 The second condition of linearity is also verified.\n\n**Conclusion:** Since both linearity conditions are verified, $T$ is a linear transformation.\n2. The transformation $T: \\mathbb{R}^2 \\to \\mathbb{R}^2$, $T(x, y) = (x, 1 + y)$ is linear?\n\n(i) Considering $(x_1, y_1), (x_2, y_2) \\in \\mathbb{R}^2$, we have:\n\n$$T((x_1, y_1) + (x_2, y_2)) = T(x_1 + x_2, y_1 + y_2) = (x_1 + x_2, 1 + y_1 + y_2)$$\n\nOn the other side,\n\n$$T(x_1, y_1) + T(x_2, y_2) = (x_1, 1 + y_1) + (x_2, 1 + y_2) = (x_1 + x_2, 2 + y_1 + y_2)$$\n\nWe concluded that,\n\n$$\\exists (x_1, y_1), (x_2, y_2) \\in \\mathbb{R}^2: T((x_1, y_1) + (x_2, y_2)) \\neq T(x_1, y_1) + T(x_2, y_2)$$\n\n⚠️ The first condition of linearity of a transformation is not verified.\n\n**Conclusion:** As the first linearity condition is not verified, we concluded that $T$ is not a linear transformation.", "id": "./materials/194.pdf" }, { "contents": "**Isomorphism**\n\n**Definition:** Let $U$ and $V$ be vector spaces and let the linear transformation $T: U \\rightarrow V$. Then:\n\n(i) $T$ is an **injection** if and only if:\n\\[\n\\forall x, y \\in U, T(x) = T(y) \\Rightarrow x = y \\text{ or } \\forall x, y \\in U, x \\neq y \\Rightarrow T(x) \\neq T(y)\n\\]\n\n(ii) $T$ is a **sobrejection** if $\\text{Range } (T) = V$\n\n**Examples:**\n\n- $T$ is a **monomorphism** if it is an injection;\n- $T$ is an **epimorphism** if it is a sobrejection;\n- $T$ is an **isomorphism** if it is a bijection (an injection and a sobrejection);\n- $T$ is an **endomorphism** if $U = V$;\n- $T$ is an **automorphism** if it is also an isomorphism and an endomorphism.\n\nThe following statements are equivalent:\n\n- $T$ is an injection;\n- $\\ker(T) = \\{0_U\\}$. \n1. The linear transformation $T: \\mathbb{R}^3 \\to \\mathbb{R}^3$ defined by $T(x, y, z) = (x + 2y + 3z, y + 2z, -z)$ is an endomorphism. Verify it is an isomorphism.\n\nWe will determine the kernel to verify if the linear transformation is an injection:\n\n$$\\ker(T) = \\{(x, y, z) \\in \\mathbb{R}^3 : T(x, y, z) = (0, 0, 0)\\}$$\n\nThen,\n\n$$T(x, y, z) = (0, 0, 0) \\iff (x + 2y + 3z, y + 2z, -z) = (0, 0, 0)$$\n\n$$\\iff \\begin{cases} x + 2y + 3z = 0 \\\\ y + 2z = 0 \\\\ -z = 0 \\end{cases} \\iff \\begin{cases} x = 0 \\\\ y = 0 \\\\ z = 0 \\end{cases}$$\n\nLike $\\ker(T) = \\{(0, 0, 0)\\}$, $T$ is an injection. Consequently $T$ is a monomorphism.\n\nLet us now determine the range to verify if the linear transformation is a surjection:\n\n$$\\text{range}(T) = \\{(a, b, c) \\in \\mathbb{R}^3 : T(x, y, z) = (a, b, c) \\text{ with } (x, y, z) \\in \\mathbb{R}^3\\}$$\n\nWe have:\n\n$$T(x, y, z) = (a, b, c) \\iff (x + 2y + 3z, y + 2z, -z) = (a, b, c)$$\n\n$$\\iff \\begin{cases} x + 2y + 3z = a \\\\ y + 2z = b \\\\ -z = c \\end{cases}$$\n\nThe matrix of the system is:\n\n$$\\begin{bmatrix} 1 & 2 & 3 & | & a \\\\ 0 & 1 & 2 & | & b \\\\ 0 & 0 & -1 & | & c \\end{bmatrix}$$\n\nConsidering that $A$ is the matrix of the coefficients, $A|B$ is the augmented matrix of the system and $n$ is the number of unknowns, we observed that:\n\n$$\\text{rank}(A) = 3; \\text{rank}(A|B) = 3; n = 3$$\n\nAs $\\text{rank}(A) = \\text{rank}(A|B) = n$, the system is possible (and determined).\nTherefore, there are no restrictions to be imposed on variables $a$ and $b$, so we conclude that the range($T$) = $\\mathbb{R}^3$.\n\nAs range ($T$) = $\\mathbb{R}^3 = V$, then $T$ is a **sobrejection**, this is, $T$ is an **epimorphism**.\n\n**Conclusion:** As $T$ is a **bijection** (an injection and a sobrejection), so it is an **isomorphism**.\n\n2. Verify if the linear transformation $T$: $\\mathbb{R}^2 \\rightarrow \\mathbb{R}^2$ defined by $T(x, y) = (x + y, x + y)$ is an automorphism.\n\n- We must verify if $T$ is an **endomorphism** and an **isomorphism**.\n\n $T$ is a **endomorphism** because the starting set and the finishing set are the same, this is, $\\mathbb{R}^2$.\n\nRemember $T$ is an **isomorphism**, if $T$ is a monomorphism and an epimorphism.\n\nLet’s see if $T$ is a monomorphism:\n\nConsidering the vectors $u = (2,3)$ and $v = (3,2)$:\n\n$$T(2,3) = (5,5) \\text{ and } T(3,2) = (5,5)$$\n\nWe have $(2,3) \\neq (3,2)$, but $T(2,3) = T(3,2) = (5,5)$. Then, $T$ is not a monomorphism.\n\n**Consequently, $T$ is not an isomorphism.**\n\n**Note:** To verify if $T$ is a monomorphism, alternatively we can determine the kernel:\n\n$$\\ker(T) = \\{(x, y) \\in \\mathbb{R}^2: T(x, y) = (0,0)\\}$$\n\nThen,\n\n$$T(x, y) = (0,0) \\iff (x + y, x + y) = (0,0)$$\n\n$$\\iff \\begin{cases} x + y = 0 \\\\ x + y = 0 \\end{cases} \\iff \\begin{cases} x = -y \\\\ -y + y = 0 \\end{cases} \\iff \\begin{cases} x = -y \\\\ 0 = 0 \\end{cases}$$\nLike \\( \\ker(T) = \\{(-y, y) : y \\in \\mathbb{R}\\} \\), we concluded again that \\( T \\) is not a monomorphism.\n\nAlternatively, we can begin to verify if \\( T \\) is an epimorphism. For this we determine the range of \\( T \\):\n\n\\[\n\\text{range}(T) = \\{(a, b) \\in \\mathbb{R}^2 : T(x, y) = (a, b) \\text{ with } (x, y) \\in \\mathbb{R}^2\\}\n\\]\n\nWe have:\n\n\\[\nT(x, y) = (a, b) \\iff (x + y, x + y) = (a, b)\n\\]\n\n\\[\n\\iff \\begin{cases} \n x + y = a \\\\\n x + y = b\n\\end{cases}\n\\]\n\nThe matrix of the system is:\n\n\\[\n\\begin{bmatrix}\n 1 & 1 & | & a \\\\\n 1 & 1 & | & b\n\\end{bmatrix} \\rightarrow \\begin{bmatrix}\n 1 & 1 & | & a \\\\\n 0 & 0 & | & -a + b\n\\end{bmatrix}\n\\]\n\n\\[\nL_2 \\leftarrow -L_1 + L_2\n\\]\n\nConsidering that \\( A \\) is the matrix of the coefficients, \\( A|B \\) is the augmented matrix of the system and \\( n \\) is the number of unknowns, we observed that:\n\nIf, \\(-a + b = 0\\):\n\n\\[\n\\text{rank}(A) = 1; \\text{rank}(A|B) = 1; n = 2\n\\]\n\nAs \\( \\text{rank}(A) = \\text{rank}(A|B) < n \\), the system is possible (and indeterminate).\n\nIf, \\(-a + b \\neq 0\\), the system is impossible (\\( \\text{rank}(A) \\neq \\text{rank}(A|B) \\)).\n\nTherefore:\n\n\\[\n\\text{range}(T) = \\{(a, b) \\in \\mathbb{R}^2 : a = b\\}\n\\]\n\nLike \\( \\text{range}(T) \\neq \\mathbb{R}^2 \\), \\( T \\) is not an epimorphism. And we can conclude that \\( T \\) is not an isomorphism.\n\n**Note:** Is sufficient to fail one condition (to be a monomorphism or to be an epimorphism) for we conclude that \\( T \\) is not an isomorphism.\n\n**Conclusion:** Despite \\( T \\) being an endomorphism is not an isomorphism, so \\( T \\) is not an automorphism.", "id": "./materials/195.pdf" }, { "contents": "Kernel and Range of a Linear Transformation\n\n**Definition:** Let $T: U \\rightarrow V$ be a linear transformation.\n\nThe kernel of $T$ ($\\ker(T)$) is the set of vectors of $U$ that $T$ transforms into the null element of $V$:\n\n$$\\ker(T) = \\{ u \\in U : T(u) = 0_V \\}$$\n\nThe range of $T$ ($\\text{range}(T)$) is the set of vectors of $V$ that are image by $T$ of at least one vector of $U$:\n\n$$\\text{range}(T) = \\{ v \\in V : T(u) = v, u \\in U \\}$$\n\n1. Determine the kernel and the range of the linear transformation $T: \\mathbb{R}^4 \\rightarrow \\mathbb{R}^2$ defined by $T(x, y, z, w) = (x - z, y + 2w)$.\n\nLet us first determine the **kernel** of the transformation $T$. By definition we have:\n\n$$\\ker(T) = \\{(x, y, z, w) \\in \\mathbb{R}^4 : T(x, y, z, w) = (0, 0)\\}$$\nThen,\n\n\\[ T(x, y, z, w) = (0,0) \\iff (x - z, y + 2w) = (0,0) \\]\n\n\\[ \\iff \\begin{cases} x - z = 0 \\\\ y + 2w = 0 \\end{cases} \\iff \\begin{cases} x = z \\\\ y = -2w \\end{cases} \\]\n\nTherefore,\n\n\\[ \\ker(T) = \\{(x, y, z, w) \\in \\mathbb{R}^4: x = z \\land y = -2w\\} \\]\n\n\\[ = \\{(z, -2w, z, w): z, w \\in \\mathbb{R}\\} \\]\n\nLet us now determine the range of the transformation \\( T \\):\n\n\\[ \\text{range}(T) = \\{(a, b) \\in \\mathbb{R}^2: T(x, y, z, w) = (a, b) \\text{ with } (x, y, z, w) \\in \\mathbb{R}^4\\} \\]\n\nWe have:\n\n\\[ T(x, y, z, w) = (a, b) \\iff (x - z, y + 2w) = (a, b) \\iff \\begin{cases} x - z = a \\\\ y + 2w = b \\end{cases} \\]\n\nThe matrix of the system is:\n\n\\[\n\\begin{bmatrix}\n1 & 0 & -1 & 0 & | & a \\\\\n0 & 1 & 0 & 2 & | & b\n\\end{bmatrix}\n\\]\n\nConsidering that \\( A \\) is the matrix of the coefficients, \\( A|B \\) is the augmented matrix of the system and \\( n \\) the number of unknowns, we observed that:\n\n\\[ \\text{rank}(A) = 2; \\text{rank}(A|B) = 2; n = 4 \\]\n\nAs \\( \\text{rank}(A) = \\text{rank}(A|B) < n \\), the system is possible (and indeterminate).\n\nTherefore, there are no restrictions to be imposed on variables \\( a \\) and \\( b \\).\n\n**Conclusion:** \\( \\text{range}(T) = \\mathbb{R}^2 \\).\n\n**Note:** \\( \\ker(T) \\) is a vectorial subspace of \\( \\mathbb{R}^4 \\) (starting set) and \\( \\text{range}(T) \\) is a vectorial subspace of \\( \\mathbb{R}^2 \\) (finishing set).", "id": "./materials/196.pdf" }, { "contents": "Linear Transformation and matrices\n\nIn a linear application the coordinates of the image vector are a linear combination of the coordinates of the object vector.\n\nFor example, a linear application $T: \\mathbb{R}^3 \\rightarrow \\mathbb{R}^2$ defined by\n\n$$T(x, y, z) = (a_{11}x + a_{12}y + a_{13}z, a_{21}x + a_{22}y + a_{23}z)$$\n\ncan be represented in matrix form in the following way:\n\n$$T(x, y, z) = \\begin{bmatrix} a_{11} & a_{12} & a_{13} \\\\ a_{21} & a_{22} & a_{23} \\end{bmatrix} \\cdot \\begin{bmatrix} x \\\\ y \\\\ z \\end{bmatrix}$$\n\nSpecifically, if $T(x, y, z) = (x - y, 2y + z)$, then:\n\n$$T(x, y, z) = \\begin{bmatrix} 1 & -1 & 0 \\\\ 0 & 2 & 1 \\end{bmatrix} \\cdot \\begin{bmatrix} x \\\\ y \\\\ z \\end{bmatrix}$$\n\nAny matrix $A = [a_{ij}]_{m \\times n}$ represents an application of $\\mathbb{R}^n$ in $\\mathbb{R}^m$, which depends on the bases considered for $\\mathbb{R}^n$ and $\\mathbb{R}^m$, respectively. If those are the canonical bases, then $T$ is defined by:\n\n$$T_A: \\mathbb{R}^n \\rightarrow \\mathbb{R}^m$$\n\n$$v \\rightarrow A \\cdot v$$\n\nExample:\n\nThe matrix\n\n$$A = \\begin{bmatrix} 1 & 0 & 2 & 3 \\\\ 0 & -1 & 1 & 0 \\\\ 1 & 0 & -1 & 2 \\end{bmatrix}$$\n\ninduces a linear application $T_A: \\mathbb{R}^4 \\rightarrow \\mathbb{R}^3$, defined by:\n\n$$T_A \\left( \\begin{bmatrix} x \\\\ y \\\\ z \\\\ w \\end{bmatrix} \\right) = \\begin{bmatrix} 1 & 0 & 2 & 3 \\\\ 0 & -1 & 1 & 0 \\\\ 1 & 0 & -1 & 2 \\end{bmatrix} \\cdot \\begin{bmatrix} x \\\\ y \\\\ z \\\\ w \\end{bmatrix} = \\begin{bmatrix} x + 2z + 3w \\\\ -y + z \\\\ x - z + 2w \\end{bmatrix}$$\n\nThis is, $A$ defines the application $T(x, y, z, w) = (x + 2z + 3w, -y + z, x - z + 2w)$, when considering the canonical bases of $\\mathbb{R}^4$ and $\\mathbb{R}^3$, respectively. In fact, considering other bases of the vector spaces involved, matrix $A$ would define another linear transformation.\nIf \\( T: U \\to V \\) be a linear transformation and \\( U = \\{ u_1, u_2, \\ldots, u_n \\} \\) be a base of \\( U \\) and \\( V = \\{ v_1, v_2, \\ldots, v_m \\} \\) be a base of \\( V \\), the following procedure allows to determine the matrix of the \\( T \\) transformation from the base \\( U \\) to the base \\( V \\), denoted by \\( M(T, U, V) \\):\n\n(i) Determine \\( T(u_1), T(u_2), \\ldots, T(u_n) \\);\n\n(ii) Determine the coordinates of \\( T(u_1), T(u_2), \\ldots, T(u_n) \\) in the base \\( V \\):\n\n\\[\nT(u_1) = a_{11}v_1 + \\cdots + a_{m1}v_m \\\\\nT(u_2) = a_{12}v_1 + \\cdots + a_{m2}v_m \\\\\n\\vdots \\\\\nT(u_n) = a_{1n}v_1 + \\cdots + a_{mn}v_m\n\\]\n\n(iii) Write these coordinates as columns of a matrix (which will be of the type \\( m \\times n \\)):\n\n\\[\nM(T, U, V) = \\begin{bmatrix}\na_{11} & \\cdots & a_{1n} \\\\\na_{21} & \\cdots & a_{2n} \\\\\n\\vdots & \\cdots & \\vdots \\\\\na_{m1} & \\cdots & a_{mn}\n\\end{bmatrix}\n\\]\n\n1. Consider the linear transformation \\( T: \\mathbb{R}^3 \\to \\mathbb{R}^2 \\) defined by \\( T(x, y, z) = (x - 3y, 2z) \\). Determine the matrix of \\( T \\) from the base \\( A = \\{(1, 2, 4), (0, 3, 0), (3, 0, 0)\\} \\) of \\( \\mathbb{R}^3 \\) to the base \\( B = \\{(1, 2), (0, 3)\\} \\) of \\( \\mathbb{R}^2 \\).\n\n(i) Calculate \\( T(1,2,4), T(0,3,0) \\) and \\( T(3,0,0) \\):\n\n\\[\nT(1,2,4) = (1 - 3 \\times 2, 2 \\times 4) = (-5, 8) \\\\\nT(0,3,0) = (0 - 3 \\times 3, 2 \\times 0) = (-9, 0) \\\\\nT(3,0,0) = (3 - 3 \\times 0, 2 \\times 0) = (3, 0)\n\\]\n\n(ii) Write \\( T(1,2,4), T(0,3,0) \\) and \\( T(3,0,0) \\) as a linear combination of the \\( B \\) vectors:\n\n\\[\n(-5,8) = c_1(1,2) + c_2(0,3)\n\\]\n\n\\[\n\\begin{align*}\n2c_1 + 3c_2 &= -5 \\\\\n2c_1 + 3c_2 &= 8\n\\end{align*} \\Rightarrow \\begin{cases} c_1 = -5 \\\\ c_2 = 6 \\end{cases}\n\\]\n\n\\[\n(-9,0) = c_1(1,2) + c_2(0,3)\n\\]\n\n\\[\n\\begin{align*}\n2c_1 + 3c_2 &= -9 \\\\\n2c_1 + 3c_2 &= 0\n\\end{align*} \\Rightarrow \\begin{cases} c_1 = -9 \\\\ c_2 = 6 \\end{cases}\n\\]\n(3,0) = c_1(1,2) + c_2(0,3)\n\n\\[\n\\begin{align*}\n c_1 + 0c_2 &= 3 \\\\\n 2c_1 + 3c_2 &= 0\n\\end{align*}\n\\]\n\n\\[\\begin{align*}\n c_1 &= 3 \\\\\n c_2 &= -2\n\\end{align*}\\]\n\n(iii) Write the coefficients of each of the previous linear combinations as columns of a matrix:\n\n\\[\nM(T, A, B) = \\begin{bmatrix}\n-5 & -9 & 3 \\\\\n6 & 6 & -2\n\\end{bmatrix}\n\\]", "id": "./materials/197.pdf" }, { "contents": "Determine a basis for a subspace\n\nFind one basis for the subspace \\( F = \\{(x, y, x) \\in \\mathbb{R}^3 : y - 3x + 5z = 0\\} \\).\n\nRemember: A subset \\( A \\) of a vector space \\( V \\) is a basis of \\( V \\) if \\( A \\) is a linearly independent set and \\( A \\) spans \\( V \\).\n\nWe have \\( y - 3x + 5z = 0 \\iff y = 3x - 5z \\). So, \\((x, 3x - 5z, z)\\) represents any vector of \\( F \\).\n\nLike \\((x, 3x - 5z, z) = x(1,3,0) + z(0,-5,1)\\), we conclude the vectors \\((1,3,0)\\) and \\((0,-5,1)\\) spans \\( F \\).\n\nNow, we must verify if \\((1,3,0)\\) and \\((0,-5,1)\\) are linearly independents.\n\n\\[\n\\begin{align*}\n c_1(1,3,0) + c_2(0,-5,1) &= (0,0,0) \\\\\n \\begin{cases}\n c_1 = 0 \\\\\n 3c_1 - 5c_2 = 0 \\\\\n c_2 = 0\n \\end{cases} &\\iff \\begin{cases}\n c_1 = 0 \\\\\n 0 = 0 \\\\\n c_2 = 0\n \\end{cases}\n\\end{align*}\n\\]\n\nWe conclude the vectors \\((1,3,0)\\) and \\((0,-5,1)\\) are linearly independents.\n\nThus, \\{\\((1,3,0), (0,-5,1)\\)\\} spans \\( F \\) and is a linearly independent set.\n\nConclusion: \\{\\((1, 3, 0), (0, -5, 1)\\)\\} is a basis of \\( F \\).\n\nWe can say, \\( \\text{dim}(F) = 2 \\).\n\nFind one basis for the subspace \\( H = \\left\\{ \\begin{bmatrix} a + b & -3b \\\\ 2c - 4a & c \\end{bmatrix} \\in M(\\mathbb{R})_{2 \\times 2} \\right\\} \\).\n\nNotice that\n\n\\[\n\\begin{bmatrix} a + b & -3b \\\\ 2c - 4a & c \\end{bmatrix} = \\begin{bmatrix} a & 0 \\\\ -4a & 0 \\end{bmatrix} + \\begin{bmatrix} b & -3b \\\\ 0 & 0 \\end{bmatrix} + \\begin{bmatrix} 0 & 0 \\\\ 2c & c \\end{bmatrix}\n\\]\n\n\\[\n= a \\begin{bmatrix} 1 & 0 \\\\ -4 & 0 \\end{bmatrix} + b \\begin{bmatrix} 1 & -3 \\\\ 0 & 0 \\end{bmatrix} + c \\begin{bmatrix} 0 & 0 \\\\ 2 & 1 \\end{bmatrix}\n\\]\n\nWe conclude \\( \\begin{bmatrix} 1 & 0 \\\\ -4 & 0 \\end{bmatrix}, \\begin{bmatrix} 1 & -3 \\\\ 0 & 0 \\end{bmatrix} \\) and \\( \\begin{bmatrix} 0 & 0 \\\\ 2 & 1 \\end{bmatrix} \\) spans \\( H \\).\nWe must verify if \\[\n\\begin{bmatrix}\n1 & 0 \\\\\n-4 & 0\n\\end{bmatrix},\n\\begin{bmatrix}\n1 & -3 \\\\\n0 & 0\n\\end{bmatrix}\n\\text{ and }\n\\begin{bmatrix}\n0 & 0 \\\\\n2 & 1\n\\end{bmatrix}\n\\] are linearly independents:\n\n\\[\nc_1 \\begin{bmatrix}\n1 & 0 \\\\\n-4 & 0\n\\end{bmatrix} + c_2 \\begin{bmatrix}\n1 & -3 \\\\\n0 & 0\n\\end{bmatrix} + c_3 \\begin{bmatrix}\n0 & 0 \\\\\n2 & 1\n\\end{bmatrix} = \\begin{bmatrix}\n0 & 0 \\\\\n0 & 0\n\\end{bmatrix}\n\\]\n\nMultiplying by the scalar and adding the matrices we have,\n\n\\[\n\\begin{bmatrix}\nc_1 + c_2 & -3c_2 \\\\\n-4c_1 + 2c_3 & c_3\n\\end{bmatrix} = \\begin{bmatrix}\n0 & 0 \\\\\n0 & 0\n\\end{bmatrix}\n\\]\n\nSince the matrices are equal if their corresponding entries are equal, we have\n\n\\[\n\\begin{cases}\nc_1 + c_2 = 0 \\\\\n-3c_2 = 0 \\\\\n-4c_1 + 2c_3 = 0 \\\\\nc_3 = 0\n\\end{cases}\n\\]\n\nSolving the system, we obtain \\(c_1 = c_2 = c_3 = 0\\). Thus, \\[\n\\begin{bmatrix}\n1 & 0 \\\\\n-4 & 0\n\\end{bmatrix},\n\\begin{bmatrix}\n1 & -3 \\\\\n0 & 0\n\\end{bmatrix}\n\\text{ and }\n\\begin{bmatrix}\n0 & 0 \\\\\n2 & 1\n\\end{bmatrix}\n\\] are linearly independents.\n\nThus, \\(\\left\\{\\begin{bmatrix}\n1 & 0 \\\\\n-4 & 0\n\\end{bmatrix},\n\\begin{bmatrix}\n1 & -3 \\\\\n0 & 0\n\\end{bmatrix}\n\\text{ and }\n\\begin{bmatrix}\n0 & 0 \\\\\n2 & 1\n\\end{bmatrix}\\right\\}\\) spans \\(H\\) and is a linearly independent set.\n\n**Conclusion:** \\(\\left\\{\\begin{bmatrix}\n1 & 0 \\\\\n-4 & 0\n\\end{bmatrix},\n\\begin{bmatrix}\n1 & -3 \\\\\n0 & 0\n\\end{bmatrix}\n\\text{ and }\n\\begin{bmatrix}\n0 & 0 \\\\\n2 & 1\n\\end{bmatrix}\\right\\}\\) is a basis of \\(H\\).\n\nWe can say, \\(\\text{dim}(H) = 3\\).\n\n**To think:** Find another bases for \\(F\\) and \\(H\\) subspaces!\n\nRemember that any basis of a subspace always has the same number of vectors.", "id": "./materials/198.pdf" }, { "contents": "Subsets that span \\( \\mathbb{R}^2 \\)\n\n- The subset \\( A = \\{(2, -8), (-1, 4)\\} \\) spans \\( \\mathbb{R}^2 \\)?\n\nAttend to the\n\nDefinition: Let \\( V \\) a vector space. Consider \\( A = \\{v_1, v_2, \\ldots, v_j\\} \\) a subset of \\( V \\). \\( A \\) spans \\( V \\) if\n\n\\[\n\\forall u \\in V \\exists c_1, c_2, \\ldots, c_j \\in \\mathbb{R}: c_1 v_1 + c_2 v_2 + \\cdots + c_j v_j = u\n\\]\n\nand applying it to our question, we have to check if\n\n\\[\n\\forall (x, y) \\in \\mathbb{R}^2 \\exists c_1, c_2 \\in \\mathbb{R}: c_1 (2, -8) + c_2 (-1, 4) = (x, y)\n\\]\n\nSolving the system resulting from this expression:\n\n\\[\n\\begin{align*}\n2c_1 - c_2 &= x \\\\\n-8c_1 + 4c_2 &= y\n\\end{align*}\n\\]\n\n\\[\n\\begin{align*}\nc_2 &= -x + 2c_1 \\\\\n-8c_1 + 4(-x + 2c_1) &= y \\\\\n-8c_1 - 4x + 8c_1 &= y \\\\\nc_2 &= -x + 2c_1 \\\\\n-4x &= y\n\\end{align*}\n\\]\n\nConclusion: For \\( y \\neq -4x \\), the system doesn’t have any solution, therefore \\( A \\) does not span \\( \\mathbb{R}^2 \\).\n\nIn this case, we can conclude that \\( A \\) spans the subset \\( \\{(x, y) \\in \\mathbb{R}^2: y = -4x\\} \\).\n\n- The subset \\( B = \\{(2, -10), (0, 2), (4, -1)\\} \\) spans \\( \\mathbb{R}^2 \\)?\n\nAs in the previous case, we must check if\n\n\\[\n\\forall (x, y) \\in \\mathbb{R}^2 \\exists c_1, c_2, c_3 \\in \\mathbb{R}: c_1 (2, 10) + c_2 (0, 2) + c_3 (4, -1) = (x, y)\n\\]\n\nSolving the system resulting from this expression:\n\n\\[\n\\begin{align*}\n2c_1 + 4c_3 &= x \\\\\n-10c_1 + 2c_2 - c_3 &= y\n\\end{align*}\n\\]\n\n\\[\n\\begin{align*}\nc_1 &= \\frac{x - 4c_3}{2} \\\\\n-10c_1 + 2c_2 - c_3 &= y\n\\end{align*}\n\\]\n\\[\n\\begin{align*}\n\\iff & \\quad \\begin{cases} \n c_1 = \\frac{x - 4c_3}{2} \\\\\n -10 \\left( \\frac{x - 4c_3}{2} \\right) + 2c_2 - c_3 = y \n\\end{cases} \\\\\n\\iff & \\quad \\begin{cases} \n c_1 = \\frac{x - 4c_3}{2} \\\\\n -5x + 20c_3 + 2c_2 - c_3 = y \n\\end{cases} \\\\\n\\iff & \\quad \\begin{cases} \n c_1 = \\frac{x - 4c_3}{2} \\\\\n -5x + 19c_3 + 2c_2 = y \n\\end{cases} \\\\\n\\iff & \\quad \\begin{cases} \n c_1 = \\frac{x - 4c_3}{2} \\\\\n c_2 = \\frac{5x - 19c_3 + y}{2} \n\\end{cases}\n\\end{align*}\n\\]\n\nConclusion: For all \\((x, y) \\in \\mathbb{R}^2\\), the system has always a solution. Therefore \\(B\\) spans \\(\\mathbb{R}^2\\).\n\nAlternatively, we can use the Gaussian elimination method to solve the system:\n\n\\[\n\\begin{bmatrix}\n2 & 0 & 4 & | & x \\\\\n-10 & 2 & -1 & | & y\n\\end{bmatrix}\n\\xrightarrow{L_2 \\leftarrow 5L_1 + L_2}\n\\begin{bmatrix}\n2 & 0 & 4 & | & x \\\\\n0 & 2 & -19 & | & 5x + y\n\\end{bmatrix}\n\\]\n\nObserve that for all \\(x, y \\in \\mathbb{R}\\) the system is always possible. Note that, considering that \\(A\\) is the matrix of the coefficients, \\(A|B\\) the augmented matrix of the system and \\(n\\) the number of unknowns, we have\n\n\\(\\text{rank}(A) = 2; \\text{rank}(A|B) = 2; n = 3\\), this is, \\(\\text{rank}(A) = \\text{rank}(A|B) < n\\).\n\nAs previously, we can conclude that \\(B\\) spans \\(\\mathbb{R}^2\\).\n\nTo think:\n\nNote that the system has an infinite number of solutions.\n\n\\textbf{Couldn't two vectors of }B\\textbf{ be enough to generate }\\mathbb{R}^2\\textbf{?}", "id": "./materials/199.pdf" }, { "contents": "Change of variables\n\nDefinition of Jacobian\n\nLet $T : D^* \\subset \\mathbb{R}^2 \\rightarrow \\mathbb{R}^2$ of class $C^1$ defined by $x = x(u, v)$ and $y = y(u, v)$. The Jacobian of $T$, denoted by $J = \\frac{\\partial(x, y)}{\\partial(u, v)}$, is the determinant of the matrix $DT(u, v)$:\n\n$$J = \\left| \\frac{\\partial(x, y)}{\\partial(u, v)} \\right| = \\left| \\begin{array}{cc} \\frac{\\partial x}{\\partial u} & \\frac{\\partial x}{\\partial v} \\\\ \\frac{\\partial y}{\\partial u} & \\frac{\\partial y}{\\partial v} \\end{array} \\right|$$\nChange of variables\n\n**Theorem**\n\nLet $D$ and $D^*$ be elementary regions (in the plane) and $T : D^* \\rightarrow D$ a function of $C^1$ class and such that $D = T(D^*)$. Then, for all integrable function $f : D \\rightarrow \\mathbb{R}$:\n\n$$\\int\\int_D f(x, y) \\, dx \\, dy = \\int\\int_{D^*} f(x(u, v), y(u, v)) \\times |J| \\, du \\, dv$$\nChanging to polar coordinates\n\nPolar Coordinates\n\nA coordinate system represents a point in the plane by a pair of real numbers denominated coordinates.\n\nExamples\n\n- Cartesian coordinates: \\((x, y)\\)\n- Polar coordinates:\n\n- We must choose a point in the plane for origin or polo; let it be \\(O\\).\n- We draw a line (from left to right) with origin in \\(O\\) — the polar axis.\n- \\(P\\) is a point in the plane with \\(\\rho\\) the distance from \\(O\\) to \\(P\\) and \\(\\theta\\) the angle between the polar axis and the line \\(\\overline{OP}\\).\nPolar Coordinates\n\nChanging to Polar Coordinates\n\nWe have:\n\n- \\( \\text{polo} = (0, 0) \\).\n- \\( x = \\rho \\cos(\\theta) \\)\n- \\( y = \\rho \\sin(\\theta) \\)\n\nor, generally,\n\n- \\( \\text{polo} = (x_0, y_0) \\).\n- \\( x - x_0 = \\rho \\cos(\\theta) \\)\n- \\( y - y_0 = \\rho \\sin(\\theta) \\)\nElementary Polar Region\n\n**Theorem**\n\nIf $f$ is continuous in the polar rectangle $R_p$ defined by: $\\rho \\in [a, b]$ and $\\theta \\in [\\alpha, \\beta]$, with $0 \\leq \\beta - \\alpha \\leq 2\\pi$ then,\n\n$$\\int\\int_{R_p} f(x, y) \\, dA = \\int_{\\alpha}^{\\beta} \\int_{a}^{b} f(\\rho \\cos(\\theta), \\rho \\cos(\\theta)) \\cdot \\rho \\, d\\rho \\, d\\theta$$\nGeneral polar region\n\nTheorem\n\nIf \\( f \\) is continuous in the polar rectangle \\( D \\) defined by:\n\\( \\theta \\in [\\alpha, \\beta] \\) and \\( h_1(\\theta) \\leq \\rho \\leq h_2(\\theta) \\) then,\n\n\\[\n\\iint_D f(x, y) \\, dA = \\int_{\\alpha}^{\\beta} \\int_{h_1(\\theta)}^{h_2(\\theta)} f(\\rho \\cos(\\theta), \\rho \\cos(\\theta)) \\cdot \\rho \\, d\\rho \\, d\\theta\n\\]", "id": "./materials/20.pdf" }, { "contents": "Subspace spanned by a subset of vectors\n\nWhich is the subspace spanned by the subset\n\n\\[ A = \\{(-6, 4), (9, -6)\\} \\text{ of } \\mathbb{R}^2? \\]\n\nMultiply each vector by a scalar and sum the resultants vectors. What kind of vectors do we get?\n\nFor example:\n\n\\[\n\\begin{align*}\n2 \\times (-6, 4) + 1 \\times (9, -6) &= (-3, 2) \\\\\n-3 \\times (-6, 4) + 0 \\times (9, -6) &= (18, -12) \\\\\n\\frac{1}{2} \\times (-6, 4) - \\frac{5}{6} \\times (9, -6) &= \\left(-\\frac{21}{2}, 7\\right) \\\\\n\\sqrt{3} \\times (-6, 4) + \\frac{1}{3} \\times (9, -6) &= (-6\\sqrt{3} + 3, 4\\sqrt{3} - 2)\n\\end{align*}\n\\]\n\nWe say that the vectors \\((-3, 2), (18, -12), \\left(-\\frac{21}{2}, 7\\right)\\) and \\((-6\\sqrt{3} + 3, 4\\sqrt{3} - 2)\\) belong to the subspace spanned by \\(A\\) [denoted by \\(< A >\\)].\n\nHow do you can meet all vectors of the \\(< A >\\)?\n\nLet \\(V\\) a vector space. Consider \\(A = \\{v_1, v_2, \\ldots, v_j\\}\\) a subset of \\(V\\) and \\(c_1, c_2, \\ldots, c_j \\in \\mathbb{R}\\). We can meet all vectors of the \\(< A >\\) if we determine all vectors resulting from the linear combination of the elements of \\(A\\), this is\n\n\\[ c_1 v_1 + c_2 v_2 + \\cdots + c_j v_j, \\forall c_1, c_2, \\ldots, c_j \\in \\mathbb{R} \\]\n\nAttend to this, and considering \\(\\alpha, \\beta \\in \\mathbb{R}\\) we have:\n\n\\[ \\alpha(-6, 4) + \\beta(9, -6) = (-6\\alpha + 9\\beta, 4\\alpha - 6\\beta) \\]\n\nFor each achievement of \\(\\alpha\\) and \\(\\beta\\), we have a vector belong to the \\(< A >\\), so \\((-6\\alpha + 9\\beta, 4\\alpha - 6\\beta)\\) represents a general vector of the subspace spanned by \\(A\\). Analysing the vector, we can see that their coordinates depend on each other. For determining the relationship between its coordinates we can consider\n\n\\[ (-6\\alpha + 9\\beta, 4\\alpha - 6\\beta) = (x, y) \\]\n\nand solve the resultant system.\n\n\\[\n\\begin{align*}\n-6\\alpha + 9\\beta &= x \\\\\n4\\alpha - 6\\beta &= y\n\\end{align*}\n\\]\n\n\\[\n\\begin{align*}\n\\alpha &= \\frac{6\\beta + y}{4} \\\\\nx &= \\frac{-6\\alpha + 9\\beta}{4}\n\\end{align*}\n\\]\n\\[\n\\begin{align*}\n\\begin{cases}\n-6 \\frac{6\\beta + y}{4} + 9\\beta &= x \\\\\n\\alpha &= \\frac{6\\beta + y}{4}\n\\end{cases} &\\iff \\\\\n\\begin{cases}\n-36\\beta - 6y + 36\\beta &= x \\\\\n\\alpha &= \\frac{6\\beta + y}{4}\n\\end{cases} &\\iff \\\\\n\\begin{cases}\ny &= -\\frac{2}{3}x \\\\\n\\alpha &= \\frac{6\\beta + y}{4}\n\\end{cases}\n\\end{align*}\n\\]\n\nConclusion:\nWe say \\( = \\{(x, y) \\in \\mathbb{R}^2 : y = -\\frac{2}{3}x \\} \\) or \\( A \\) spans \\( \\{(x, y) \\in \\mathbb{R}^2 : y = -\\frac{2}{3}x \\} \\).\n\nGeometrically the subspace is a line that passes through the origin.\nThe vectors \\( v_1 = (-6, 4) \\) and \\( v_2 = (9, -6) \\) are collinear, so all linear combination of this vectors give rise to a vector contained on the same line, this is \\( y = -\\frac{2}{3}x \\).", "id": "./materials/200.pdf" }, { "contents": "Linear Independence versus Linear Dependence\n\nThe vectors \\((2, 3)\\) and \\((1, -4)\\) are linearly independents?\n\nAttend of the\n\n**Definition:** Consider \\(u_1, u_2, u_3, \\ldots, u_n\\) vectors of vectorial space \\(V\\), and \\(c_1, c_2, c_3, \\ldots, c_n \\in \\mathbb{R}\\). The vectors \\(u_1, u_2, u_3, \\ldots, u_n\\) are **linearly independents** if\n\n\\[c_1 u_1 + c_2 u_2 + c_3 u_3 + \\cdots + c_n u_n = 0_K \\Rightarrow c_1 = c_2 = c_3 = \\cdots = c_n = 0.\\]\n\nApplying this definition to the example, and solving the system, we have:\n\n\\[\n\\begin{align*}\n2c_1 - c_2 &= 0 \\\\\n3c_1 + 4c_2 &= 0\n\\end{align*}\n\\]\n\n\\[\n\\begin{align*}\n\\Rightarrow \\begin{cases} c_2 = 2c_1 \\\\\n3c_1 + 4c_2 = 0 \\Rightarrow 3c_1 + 8c_1 = 0 \\Rightarrow c_1 = 0 \\Rightarrow c_2 = 0\n\\end{cases}\n\\]\n\n**Conclusion:** The vectors \\((2, 3)\\) and \\((1, -4)\\) are linearly independents.\n\nThe vectors \\((-4, 3)\\) and \\((12, -9)\\) are linearly independents?\n\nCan we find \\(c_1\\) and \\(c_2\\) not simultaneously null, that\n\n\\[c_1(-4,3) + c_2(12, -9) = (0,0)\\]?\n\nYes. If we consider \\(c_1 = 2\\) and \\(c_2 = 1\\), for example. Meet another values!\n\nBut, solving the system, how many solutions we meet?\n\n\\[\n\\begin{align*}\n-4c_1 + 12c_2 &= 0 \\\\\n3c_1 - 9c_2 &= 0\n\\end{align*}\n\\]\n\n\\[\n\\begin{align*}\n\\Rightarrow \\begin{cases} c_1 = \\frac{-12}{-4}c_2 \\\\\n3c_1 - 9c_2 = 0 \\Rightarrow 3c_1 - 9c_2 = 0 \\Rightarrow c_1 = 3c_2\n\\end{cases}\n\\]\n\n**Conclusion:** The system has an infinite number of solutions. So the vectors \\((-4, 3)\\) and \\((12, -9)\\) are linearly dependents.\n\nNote that \\((12, -9) = -3(-4,3)\\).\nWhat happens geometrically?\n\nConsider \\( u_1 = (2,3) \\), \\( u_2 = (-1,4) \\), \\( v_1 = (-4,3) \\) and \\( v_2 = (12,-9) \\).\n\nThe vectors \\( u_1 \\) and \\( u_2 \\) aren’t on the same line, they are linearly independents.\n\nThe vectors \\( v_1 \\) and \\( v_2 \\) are both in the same line: \\( y = -\\frac{3}{4}x \\), they are linearly dependents.\n\nInvestigate what happen with three vectors of \\( \\mathbb{R}^2 \\)!\n\nThe vectors \\( u_1 \\), \\( u_2 \\) and \\( v_1 \\) are linearly dependents or independents?", "id": "./materials/201.pdf" }, { "contents": "Linear Combination\n\nThe vector \\((-10, 13, -14)\\) is a linear combination of the vectors \\((1, 5, -7)\\) and \\((4, -1, 0)\\)?\n\nCan we write \\((-10, 13, -14)\\) as the sum resulting from the product of scalars by the vectors \\((1, 5, -7)\\) and \\((4, -1, 0)\\)? This is, there will be \\(c_1, c_2 \\in \\mathbb{R}\\), such that\n\n\\[\n(-10, 13, -14) = c_1(1, 5, -7) + c_2(4, -1, 0)\n\\]\n\nWe must write the system and solve it.\n\n\\[\n\\begin{align*}\n5c_1 - c_2 &= 13 \\\\\n-7c_1 &= -14\n\\end{align*}\n\\]\n\n\\[\n\\begin{align*}\nc_1 + 4c_2 &= -10 \\\\\nc_2 &= -13 + 5 \\times 2 \\\\\nc_1 &= 2\n\\end{align*}\n\\]\n\n\\[\n\\begin{align*}\nc_1 + 4c_2 &= -10 \\\\\nc_2 &= -3 \\\\\nc_1 &= 2\n\\end{align*}\n\\]\n\nConclusion: \\((-10, 13, -14) = 2(1, 5, -7) - 3(4, -1, 0)\\), so \\((-10, 13, -14)\\) is a linear combination of \\((1, 5, -7)\\) and \\((4, -1, 0)\\).\n\nGeometrically, we can observe the way to obtain the vector \\(u = (-10, 13, -14)\\) from the vectors \\(v_1 = (1, 5, -7)\\) and \\(v_2 = (4, -1, 0)\\).\nThe vector \\((5, 6)\\) is a linear combination of the vectors \\((1, 2)\\), \\((2, 4)\\) and \\((-1, -2)\\)?\n\nThere will be \\(c_1, c_2, c_3 \\in \\mathbb{R}\\), such that\n\n\\[\n(5, 6) = c_1 (1, 2) + c_2 (2, 4) + c_3 (-1, -2)\\]\n\n\\[\n\\begin{align*}\n c_1 + 2c_2 - c_3 &= 5 \\\\\n 2c_1 + 4c_2 - 2c_3 &= 6\n\\end{align*}\n\\]\n\n\\[\n\\begin{align*}\n 2c_1 + 4c_2 - 2c_3 &= 6 \\\\\n 2(-2c_2 + c_3) + 4c_2 - 2c_3 &= 6 \\\\\n -4c_2 + 2c_3 + 4c_2 - 2c_3 &= 6 \\\\\n c_1 &= -2c_2 + 3c_3 \\\\\n 0 &= 6\n\\end{align*}\n\\]\n\nThe system doesn’t have any solution.\n\n**Conclusion:** \\((5, 6)\\) is not a linear combination of the vectors \\((1, 2)\\), \\((2, 4)\\) and \\((-1, -2)\\).\n\nGeometrically, representing the vectors, we can see that it is not possible to obtain the vector \\(u = (5, 6)\\) from a linear combination of the vectors \\(v_1 = (1, 2)\\), \\(v_2 = (2, 4)\\) and \\(v_3 = (-1, -2)\\). Note that the vectors \\(v_1\\), \\(v_2\\) and \\(v_3\\) have the same direction, so any linear combination of one or more of these vectors is a vector in that direction.\nObservation:\nThe vector \\( v_2 = (2,4) \\) is a linear combination of \\( v_1 = (1,2) \\) and vice versa, since\n\\[\n(2,4) = 2(1,2) \\quad \\text{and} \\quad (1,2) = \\frac{1}{2} (2,4)\n\\]\nSimilarly \\( v_3 = (-1,-2) \\) is a linear combination of \\( v_1 = (1,2) \\), of \\( v_2 = (2,4) \\), or of \\( v_1 \\) and \\( v_2 \\). In this case, we can write this linear combination in several ways:\n\\[\n(-1,-2) = -1(1,2) + 0(2,4) \\\\\n(-1,-2) = (1,2) - (2,4) \\\\\n(-1,-2) = -4(1,2) + \\frac{3}{2} (2,4)\n\\]\n(…)\nGeneralizing, \\( (-1,-2) = (-1 - 2b)(1,2) + b(2,4) \\), for \\( b \\in \\mathbb{R} \\).\n\nIn summary, to analyze whether a vector is a linear combination of other vectors, we can use the\n\n**Definition:** The vector \\( u \\in \\mathbb{R}^n \\) is a **linear combination** of the vectors \\( v_1, v_2, ..., v_j \\in \\mathbb{R}^n \\) if\n\\[\n\\exists \\ c_1, c_2, ..., c_j \\in \\mathbb{R}: \\ u = c_1 v_1 + c_2 v_2 + \\cdots + c_j v_j.\n\\]", "id": "./materials/202.pdf" }, { "contents": "MathE project\n\nLimit for real functions of several variables\n\n1 The $\\varepsilon - \\delta$ characterization\n\nLet $E \\subseteq \\mathbb{R}^k$ be a nonempty set, let $a$ be a cluster point for $E$ and let us consider a real function $f : E \\to \\mathbb{R}$.\n\n**Definition 1.1** (with neighbourhoods) One says that $\\ell \\in \\mathbb{R}$ is the limit of $f$ at the point $a$ if for any $U \\in \\mathcal{V}(\\ell)$ from $\\mathbb{R}$, there exists $V \\in \\mathcal{V}(a)$ from $\\mathbb{R}^k$, such that for any $x \\in V \\cap E$ with $x \\neq a$, we have $f(x) \\in U$. We denote this by\n\n$$\\ell = \\lim_{x \\to a} f(x).$$\n\n**Proposition 1.1** (with $\\varepsilon - \\delta$)\n\n(i) Let $\\ell \\in \\mathbb{R}$. The limit of $f$ at the point $a$ is $\\ell$ if and only if for any $\\varepsilon > 0$, there exists $\\delta = \\delta(\\varepsilon) > 0$ such that for $x \\in E$ with $x \\neq a$ and $\\|x - a\\| < \\delta$ we have $|f(x) - \\ell| < \\varepsilon$.\n\n(ii) The limit of $f$ at the point $a$ is $+\\infty$ if and only if for any $\\varepsilon > 0$, there exists $\\delta = \\delta(\\varepsilon) > 0$ such that for $x \\in E$ with $x \\neq a$ and $\\|x - a\\| < \\delta$ we have $f(x) > \\varepsilon$.\n\n(iii) The limit of $f$ at the point $a$ is $-\\infty$ if and only if for any $\\varepsilon > 0$, there exists $\\delta = \\delta(\\varepsilon) > 0$ such that for $x \\in E$ with $x \\neq a$ and $\\|x - a\\| < \\delta$ we have $f(x) < -\\varepsilon$.\n\nFor a two-variables function $f : E \\subseteq \\mathbb{R}^2 \\to \\mathbb{R}$, $f = f(x, y)$ we obtain:\n\n**Proposition 1.2** (with $\\varepsilon - \\delta$)\n\n(i) Let $\\ell \\in \\mathbb{R}$. The limit of $f$ at the point $(a, b)$ is $\\ell$ if and only if for any $\\varepsilon > 0$, there exists $\\delta = \\delta(\\varepsilon) > 0$ such that for $(x, y) \\in E$, $(x, y) \\neq (a, b)$ with $|x - a| < \\delta$ and $|y - b| < \\delta$ we have $|f(x, y) - \\ell| < \\varepsilon$.\n\n(ii) The limit of $f$ at the point $(a, b)$ is $+\\infty$ if and only if for any $\\varepsilon > 0$, there exists $\\delta = \\delta(\\varepsilon) > 0$ such that for $(x, y) \\in E$, $(x, y) \\neq (a, b)$ with $|x - a| < \\delta$ and $|y - b| < \\delta$ we have $f(x, y) > \\varepsilon$.\n\n(iii) The limit of $f$ at the point $(a, b)$ is $-\\infty$ if and only if for any $\\varepsilon > 0$, there exists $\\delta = \\delta(\\varepsilon) > 0$ such that for $(x, y) \\in E$, $(x, y) \\neq (a, b)$ with $|x - a| < \\delta$ and $|y - b| < \\delta$ we have $f(x, y) < -\\varepsilon$.\n\n**Example 1.1** Using the $\\varepsilon - \\delta$ criterion of the limit, show that\n\n$$\\lim_{(x,y)\\to(3,1)} (2x - y) = 5.$$\n\n**Solution.** Consider $\\varepsilon > 0$, there is $\\delta = \\delta(\\varepsilon) > 0$ such that for any $(x, y) \\in \\mathbb{R}^2$ with $(x, y) \\neq (3, 1)$, $|x - 3| < \\delta$ and $|y - 1| < \\delta$ we have $|2x - y - 5| < \\varepsilon$. Indeed, we can write\n\n$$|2x - y - 5| = |2(x - 3) - (y - 1)| \\leq 2|x - 3| + |y - 1| < 4\\delta$$\n\nand for $\\delta \\leq \\frac{\\varepsilon}{4}$ the inequality is fulfilled.\nExample 1.2 Using the $\\varepsilon - \\delta$ criterion of the limit, show that\n\n$$\\lim_{(x,y) \\to (4, +\\infty)} \\frac{xy - 1}{y + 2} = 4.$$ \n\nSolution. Consider $\\varepsilon > 0$, there is $\\delta = \\delta(\\varepsilon) > 0$ such that for any $(x, y) \\in \\mathbb{R}^2$ with $|x - 4| < \\delta$ and $y > \\frac{1}{\\delta}$ we have\n\n$$\\left| \\frac{xy - 1}{y + 2} - 4 \\right| < \\varepsilon.$$\n\nIndeed, we can write\n\n$$\\left| \\frac{xy - 1}{y + 2} - 4 \\right| = \\left| \\frac{(x - 4)y - 9}{y + 2} \\right| \\leq |x - 4| \\cdot \\frac{y}{y + 2} + 9 \\cdot \\frac{1}{y} < \\delta + 9\\delta = 10\\delta$$\n\nand for $\\delta \\leq \\frac{\\varepsilon}{10}$ the inequality is fulfilled.\n\nExample 1.3 Find the limit of the following functions:\n\na) $\\lim_{(x,y) \\to (2,0)} \\frac{\\sin(xy)}{y}$\n\nb) $\\lim_{(x,y) \\to (+\\infty, 2)} \\left(1 + \\frac{y^2}{x}\\right)^{xy}$\n\nc) $\\lim_{(x,y) \\to (+\\infty, +\\infty)} \\left(\\frac{xy}{x^2 + y^2}\\right)^{y^2}$\n\nd) $\\lim_{(x,y) \\to (+\\infty, +\\infty)} \\frac{x + y}{x^2 - xy + y^2}$\n\nSolution.\n\na) Since $\\lim_{t \\to 0} \\frac{\\sin t}{t} = 1$, we have $\\lim_{(x,y) \\to (2,0)} \\frac{\\sin(xy)}{y} = \\lim_{(x,y) \\to (2,0)} \\frac{\\sin(xy)}{xy} \\cdot x = 2$.\n\nb) Using the fundamental limit $\\lim_{t \\to 0} (1 + t)^{\\frac{1}{t}} = e$, we deduce\n\n$$\\lim_{(x,y) \\to (+\\infty, 2)} \\left(1 + \\frac{y^2}{x}\\right)^{xy} = \\lim_{(x,y) \\to (+\\infty, 2)} \\left[\\left(1 + \\frac{y^2}{x}\\right)^{\\frac{y^2}{x}}\\right]^{y^2} = e^{\\lim_{y \\to 2} y^3} = e^8.$$\n\nc) We have $0 < \\left(\\frac{xy}{x^2 + y^2}\\right)^{y^2} \\leq \\left(\\frac{1}{2}\\right)^{y^2}$ for each $x, y > 0$, so we get\n\n$$0 \\leq \\lim_{(x,y) \\to (+\\infty, +\\infty)} \\left(\\frac{xy}{x^2 + y^2}\\right)^{y^2} \\leq \\lim_{(x,y) \\to (+\\infty, +\\infty)} \\left(\\frac{1}{2}\\right)^{y^2} = 0,$$\n\nthus\n\n$$\\lim_{(x,y) \\to (+\\infty, +\\infty)} \\left(\\frac{xy}{x^2 + y^2}\\right)^{y^2} = 0.$$\n\nd) Since\n\n$$\\frac{x + y}{x^2 - xy + y^2} \\leq \\frac{1}{x} + \\frac{1}{y}$$\n\nfor each $x, y > 0$, and\n\n$$\\lim_{(x,y) \\to (+\\infty, +\\infty)} \\left(\\frac{1}{x} + \\frac{1}{y}\\right) = 0$$\n\nit follows\n\n$$\\lim_{(x,y) \\to (+\\infty, +\\infty)} \\frac{x + y}{x^2 - xy + y^2} = 0.$$", "id": "./materials/205.pdf" }, { "contents": "Points of extremum for functions of several variables\n\nOne of the main uses of ordinary derivatives is in finding maximum and minimum values (extreme values). We see how to use partial derivatives to locate maximum and minimum of functions of two variables.\n\nMaximum, minimum and saddle points\n\nLook at the hills and valleys in the graph of shown in Figure 1.\n\n![Figure 1](image)\n\nThere are two points \\((a_1, a_2)\\) where \\(f\\) has a local maximum, that is, where \\(f(a_1, a_2)\\) is larger than nearby values of \\(f(x, y)\\). The larger of these two values is the absolute maximum. Likewise, \\(f\\) has two local minimum, where \\(f(x, y)\\) is smaller than nearby values. The smaller of these two values is the absolute minimum.\n\n**Definition 1** Let \\(D \\subset \\mathbb{R}^n\\) and \\(f : D \\subset \\mathbb{R}\\). The point \\(a = (a_1, a_2, ..., a_n) \\in D\\) whose coordinates verify the equations\n\n\\[\n\\frac{\\partial f}{\\partial x_i}(a_1, a_2, ..., a_n) = 0, i = 1, n\n\\]\n\nis called a **critical** or **stationary point** for \\(f\\).\n\n**Definition 2** Let \\(D \\subset \\mathbb{R}^n\\) and \\(f : D \\subset \\mathbb{R}\\). The point \\(a \\in D\\) is said to be:\n\n1. a **local maximum** if \\(f(x) \\leq f(a)\\) for all points \\(x\\) sufficiently close to \\(a\\);\n2. a **local minimum** if \\(f(x) \\geq f(a)\\) for all points \\(x\\) sufficiently close to \\(a\\);\n3. a **global** (or absolute) **maximum** if \\(f(x) \\leq f(a)\\) for all points \\(x \\in D\\);\n(4) a **global** (or absolute) **minimum** if \\( f(x) \\geq f(a) \\) for all points \\( x \\in D \\);\n(5) a local or global extremum if it is a local or global maximum or minimum.\n\n**Definition 3** A critical point \\( a \\) which is neither a local maximum nor minimum is called a **saddle point**.\n\nThere are three types of stationary points possible, these being a maximum point, a minimum point, and a saddle point.\n\nAn analogous of Fermat Theorem is the following:\n\n**Theorem 4** A point of local extremum for a function \\( f \\), belonging to interior of the domain, is a stationary point for \\( f \\).\n\nFermat’s theorem gives only a necessary condition for extreme function values.\n\n**Procedure to determine maxima, minima and saddle points for functions of several variables**\n\nWe can located the point of extremum. Then we must classifying them in points of maximum, points of minimum or saddle point.\n\n**Theorem 5** Let be a stationary point \\( a = (a_1, a_2, \\ldots, a_n) \\in D \\) for \\( f : D \\subset \\mathbb{R}^n \\rightarrow \\mathbb{R} \\) and suppose \\( f \\) has continuous partial derivatives of second order in a neighborhood of \\( a \\)\n\n1. if \\( d^2f(a) \\) is a positive quadric form, then \\( a \\) is a local minimum point,\n2. if \\( d^2f(a) \\) is a negative quadric form, then \\( a \\) is a local maximum point,\n3. if \\( d^2f(a) \\) is a undefined quadric form, then \\( a \\) is not a point of extremum.\n\nIn order to establish if a quadric form is positive, negative or undefined we turn it to canonical expression, by using an algebraic method. While using the method of Jacobi (when is possible) the above theorem can be rewritten as follows:\n\n**Definition 6** If \\( f : D \\subset \\mathbb{R}^2 \\rightarrow \\mathbb{R} \\) is a function of two variables such that all second order partial derivatives exist at the point \\( (a_1, a_2) \\), then the Hessian matrix of \\( f \\) at \\( (a_1, a_2) \\) is the matrix\n\n\\[\nH_2 = \\begin{pmatrix}\n\\frac{\\partial^2 f}{\\partial x^2} & \\frac{\\partial^2 f}{\\partial x \\partial y} \\\\\n\\frac{\\partial^2 f}{\\partial y \\partial x} & \\frac{\\partial^2 f}{\\partial y^2}\n\\end{pmatrix}\n\\]\n\nwhere the derivatives are evaluated at \\( (a_1, a_2) \\).\n\n**Definition 7** If \\( f : D \\subset \\mathbb{R}^3 \\rightarrow \\mathbb{R} \\) is a function of three variables such that all second order partial derivatives exist at the point \\( (a_1, a_2, a_3) \\), then the Hessian\nof $f$ at $(a_1, a_2, a_3)$ is the matrix\n\n$$H_3 = \\begin{pmatrix}\n\\frac{\\partial^2 f}{\\partial x^2} & \\frac{\\partial^2 f}{\\partial x \\partial y} & \\frac{\\partial^2 f}{\\partial x \\partial z} \\\\\n\\frac{\\partial^2 f}{\\partial y \\partial x} & \\frac{\\partial^2 f}{\\partial y^2} & \\frac{\\partial^2 f}{\\partial y \\partial z} \\\\\n\\frac{\\partial^2 f}{\\partial z \\partial x} & \\frac{\\partial^2 f}{\\partial z \\partial y} & \\frac{\\partial^2 f}{\\partial z^2}\n\\end{pmatrix}$$\n\n(3)\n\nwhere the derivatives are evaluated at $(a_1, a_2, a_3)$.\n\nWe note $H_1 = \\left( \\frac{\\partial^2 f}{\\partial x^2} \\right)$.\n\n**Theorem 8** Let $f : D \\subset \\mathbb{R}^2 \\to \\mathbb{R}$ is a function of two variables having partial derivatives of order three on $D$ and consider $(a_1, a_2)$ a stationary point of $f$ and $H_2$ the Hessian of $f$ at $(a_1, a_2)$. If $\\det(H_2) \\neq 0$, then $(a_1, a_2)$ is:\n\n1. a local maximum if $\\frac{\\partial^2 f}{\\partial x^2} < 0$ and $\\det(H_2) > 0$.\n2. a local minimum if $\\frac{\\partial^2 f}{\\partial x^2} > 0$ and $\\det(H_2) > 0$.\n3. a saddle point if neither of the above hold, where the partial derivatives are evaluated at $(a_1, a_2)$.\n\n**Theorem 9** If $f : D \\subset \\mathbb{R}^3 \\to \\mathbb{R}$ is a function of three variables having partial derivatives of order three on $D$ and consider $(a_1, a_2, a_3)$ a stationary point of $f$ and $H_3$ the Hessian of $f$ at $(a_1, a_2, a_3)$. If $\\det(H_3) \\neq 0$, then $(a_1, a_2, a_3)$ is:\n\n1. a local maximum if $\\det(H_1) < 0$, $\\det(H_2) > 0$ and $\\det(H_3) < 0$;\n2. a local minimum if $\\det(H_1) > 0$, $\\det(H_2) > 0$ and $\\det(H_3) > 0$;\n3. a saddle point if neither of the above hold, where the partial derivatives are evaluated at $(a_1, a_2, a_3)$.\n\nIn each case, if $\\det(H_i) = 0$, $i = 1, 2$, then $a$ can be either a local extremum or a saddle point.\n\n**Example 10** Find the points of extremum for the function\n\n$$f(x, y, z) = y + \\frac{z^2}{4y} + \\frac{x^2}{z} + \\frac{2}{x}, x \\neq 0, y \\neq 0, z \\neq 0.$$ \n\n**Solution.**\n\nFind the stationary (critical) point.\n\n$$\\frac{\\partial f}{\\partial x} = \\frac{2x}{z} - \\frac{2}{x^2}, \\quad \\frac{\\partial f}{\\partial y} = 1 - \\frac{z^2}{4y^2}, \\quad \\frac{\\partial f}{\\partial z} = \\frac{z}{2y} - \\frac{x^2}{z^2}$$\n\n$$\\begin{cases}\n\\frac{2x}{z} - \\frac{2}{x^2} = 0 \\\\\n1 - \\frac{z^2}{4y^2} = 0 \\\\\n\\frac{z}{2y} - \\frac{x^2}{z^2} = 0\n\\end{cases}$$\nThe solutions are: \\([x = 1, y = \\frac{1}{2}, z = 1], [x = -1, y = -\\frac{1}{2}, z = -1]\\).\n\nEstablish the sign of the quadric form \\(d^2 f (1, \\frac{1}{2}, 1)\\) and \\(d^2 f (-1, -\\frac{1}{2}, -1)\\).\n\n\\[\n\\frac{\\partial^2 f}{\\partial x^2} = \\frac{2}{z} + \\frac{4}{x^2}; \\quad \\frac{\\partial^2 f}{\\partial y^2} = \\frac{z^2}{2y^3}; \\quad \\frac{\\partial^2 f}{\\partial z^2} = \\frac{1}{2y} + \\frac{2x^2}{z^3};\n\\]\n\n\\[\n\\frac{\\partial^2 f}{\\partial x \\partial y} = 0; \\quad \\frac{\\partial^2 f}{\\partial y \\partial z} = -\\frac{z}{2y^2}; \\quad \\frac{\\partial^2 f}{\\partial x \\partial z} = -\\frac{2x}{z^2};\n\\]\n\n\\[\n\\frac{\\partial^2 f}{\\partial x^2} (1, \\frac{1}{2}, 1) = 6; \\quad \\frac{\\partial^2 f}{\\partial y^2} (1, \\frac{1}{2}, 1) = 4; \\quad \\frac{\\partial^2 f}{\\partial z^2} (1, \\frac{1}{2}, 1) = 3;\n\\]\n\n\\[\n\\frac{\\partial^2 f}{\\partial x \\partial y} (1, \\frac{1}{2}, 1) = 0; \\quad \\frac{\\partial^2 f}{\\partial y \\partial z} (1, \\frac{1}{2}, 1) = -2; \\quad \\frac{\\partial^2 f}{\\partial x \\partial z} (1, \\frac{1}{2}, 1) = -2.\n\\]\n\n\\(d^2 f (1, \\frac{1}{2}, 1) = 6dx^2 + 4dy^2 + 3dz^2 - 4dydz - 4dxdz,\n\\]\n\n\\(H_3 = \\begin{pmatrix} 6 & 0 & -2 \\\\ 0 & 4 & -2 \\\\ -2 & -2 & 3 \\end{pmatrix},\n\\]\n\n\\(\\det (H_3) = 32 \\neq 0\\), so we can apply Theorem 9.\n\n\\(\\det (H_1) = 6, \\det (H_2) = \\det \\begin{pmatrix} 6 & 0 \\\\ 0 & 4 \\end{pmatrix} = 24, \\det (H_3) = 32.\n\\]\n\nSince \\(\\det (H_1) > 0, \\det (H_2) > 0, \\det (H_3) > 0\\) then \\((1, \\frac{1}{2}, 1)\\) is a maximum point.\n\nSimilarly\n\n\\[\n\\frac{\\partial^2 f}{\\partial x^2} (-1, -\\frac{1}{2}, -1) = 2; \\quad \\frac{\\partial^2 f}{\\partial y^2} (-1, -\\frac{1}{2}, -1) = -4; \\quad \\frac{\\partial^2 f}{\\partial z^2} (-1, -\\frac{1}{2}, -1) = -3;\n\\]\n\n\\[\n\\frac{\\partial^2 f}{\\partial x \\partial y} (-1, -\\frac{1}{2}, -1) = 0; \\quad \\frac{\\partial^2 f}{\\partial y \\partial z} (-1, -\\frac{1}{2}, -1) = 2; \\quad \\frac{\\partial^2 f}{\\partial x \\partial z} (-1, -\\frac{1}{2}, -1) = 2.\n\\]\n\n\\(H_3 = \\begin{pmatrix} 2 & 0 & 2 \\\\ 0 & -4 & 2 \\\\ 2 & 2 & -3 \\end{pmatrix},\n\\]\n\n\\(\\det (H_3) = 32 \\neq 0\\), so we can apply Theorem 9.\n\n\\(\\det (H_1) = 2, \\det (H_2) = \\det \\begin{pmatrix} 2 & 0 \\\\ 0 & -4 \\end{pmatrix} = -8, \\det (H_3) = 32.\n\\]\n\nSince \\(\\det (H_1) > 0, \\det (H_2) < 0, \\det (H_3) > 0\\) then \\((-1, -\\frac{1}{2}, -1)\\) is a saddle point.\n\n**Example 11** Find the points of extremum for the function \\(f : \\mathbb{R}^2 \\to \\mathbb{R},\n\\]\n\n\\(f(x, y) := x^4 + y^4 - x^2 - 2xy - y^2.\n\\]\n\n**Solution.**\n\nFind the stationary (critical) point.\n\n\\[\n\\begin{cases}\n\\frac{\\partial f}{\\partial x}(x, y) = 0 \\\\\n\\frac{\\partial f}{\\partial y}(x, y) = 0\n\\end{cases} \\iff \\begin{cases}\n4x^3 - 2x - 2y = 0 \\\\\n4y^3 - 2x - 2y = 0.\n\\end{cases}\n\\]\nThe solutions are: \\([x = 1, y = 1], [x = 0, y = 0], [x = -1, y = -1]\\).\n\nNext, we will establish the sign of the quadratic forms \\(d^2 f(0, 0), d^2 f(1, 1), d^2 f(-1, -1),\\)\n\n\\[\n\\frac{\\partial^2 f}{\\partial x^2}(x, y) = 12x^2 - 2; \\quad \\frac{\\partial^2 f}{\\partial y^2}(x, y) = 12y^2 - 2; \\quad \\frac{\\partial^2 f}{\\partial x \\partial y}(x, y) = -2.\n\\]\n\n\\[\n\\frac{\\partial^2 f}{\\partial x^2}(0, 0) = -2; \\quad \\frac{\\partial^2 f}{\\partial y^2}(0, 0) = -2; \\quad \\frac{\\partial^2 f}{\\partial x \\partial y}(x, y) = -2.\n\\]\n\n\\[\n\\frac{\\partial^2 f}{\\partial x^2}(1, 1) = 10; \\quad \\frac{\\partial^2 f}{\\partial y^2}(1, 1) = 10; \\quad \\frac{\\partial^2 f}{\\partial x \\partial y}(x, y) = -2.\n\\]\n\n\\[\n\\frac{\\partial^2 f}{\\partial x^2}(-1, -1) = 10; \\quad \\frac{\\partial^2 f}{\\partial y^2}(-1, -1) = 10; \\quad \\frac{\\partial^2 f}{\\partial x \\partial y}(-1, -1) = -2.\n\\]\n\nFor \\((0, 0)\\) we have\n\n\\[\nH_2 = \\begin{pmatrix} -2 & -2 \\\\ -2 & -2 \\end{pmatrix}, \\quad \\det(H_2) = 0\n\\]\n\nand we cannot use Theorem 8. We construct the quadratic form. Thus,\n\n\\[\nd^2 f(0, 0) = -2dx^2 - 4dxdy - 2dy^2 = -2(dx + dy)^2.\n\\]\n\nThe quadratic form is negative semidefinite, so we cannot use Theorem 5.\n\nWe observe that, for \\(\\varepsilon > 0\\) small, we have \\(f(\\varepsilon, 0) = \\varepsilon^4 - \\varepsilon^2 < 0 = f(0, 0),\\) and \\(f(\\varepsilon, -\\varepsilon) = 2\\varepsilon^4 > 0 = f(0, 0),\\) so the point \\((0, 0)\\) is not extremum point.\n\nFor \\((1, 1)\\) we have\n\n\\[\nH_2 = \\begin{pmatrix} 10 & -2 \\\\ -2 & 10 \\end{pmatrix}, \\quad \\det(H_2) \\neq 0.\n\\]\n\nWe can apply Theorem 8. So \\(\\frac{\\partial^2 f}{\\partial x^2}(1, 1) = 10 > 0\\) and \\(\\det(H_2) = 36 > 0.\\)\n\nThe point \\((1, 1)\\) is a minimum point. Similarly for \\((-1, -1),\\)\n\nAuthor: Ariadna Lucia Pletea", "id": "./materials/206.pdf" }, { "contents": "Limit for real functions of several variables\n\n1 The characterization with sequences\n\nLet $E \\subseteq \\mathbb{R}^k$ be a nonempty set, let $a$ be a cluster point for $E$ and let us consider a real function $f : E \\to \\mathbb{R}$.\n\n**Definition 1.1** (with neighbourhoods) One says that $\\ell \\in \\mathbb{R}$ is the limit of $f$ at the point $a$ if for any $U \\in \\mathcal{V}(\\ell)$ from $\\mathbb{R}$, there exists $V \\in \\mathcal{V}(a)$ from $\\mathbb{R}^k$, such that for any $x \\in V \\cap E$ with $x \\neq a$, we have $f(x) \\in U$. We denote this by\n\n$$\\ell = \\lim_{x \\to a} f(x).$$\n\n**Proposition 1.1** (with sequences) Let $\\ell \\in \\mathbb{R}$. The limit of $f$ at the point $a$ is $\\ell$ if and only if for any sequence $(x_n)_n \\subset E$ with $x_n \\neq a$ and $\\lim_{n \\to +\\infty} x_n = a$, we have that\n\n$$\\lim_{n \\to +\\infty} f(x_n) = \\ell.$$\n\nFor a two-variables function $f : E \\subseteq \\mathbb{R}^2 \\to \\mathbb{R}$, $f = f(x, y)$ the above proposition is:\n\n**Proposition 1.2** (with sequences) Let $\\ell \\in \\mathbb{R}$. The limit of $f$ at the point $(a, b)$ is $\\ell$ if and only if for any sequence $(x_n, y_n)_n \\subset E$ with $(x_n, y_n) \\neq (a, b)$, $\\lim_{n \\to +\\infty} x_n = a$ and $\\lim_{n \\to +\\infty} y_n = b$, we have that\n\n$$\\lim_{n \\to +\\infty} f(x_n, y_n) = \\ell.$$\n\n**Remark 1.1** If there are two sequences $(x_n, y_n)_n, (x'_n, y'_n)_n \\subset E$ with $(x_n, y_n) \\neq (a, b), (x'_n, y'_n) \\neq (a, b)$ and $\\lim_{n \\to +\\infty} (x_n, y_n) = \\lim_{n \\to +\\infty} (x'_n, y'_n) = (a, b)$ such that\n\n$$\\lim_{n \\to +\\infty} f(x_n, y_n) = \\ell \\quad \\text{and} \\quad \\lim_{n \\to +\\infty} f(x'_n, y'_n) = \\ell', \\quad \\ell \\neq \\ell',$$\n\nthen the limit of the function $f$ does not exist.\n\n**Example 1.1** Find $\\ell_1 = \\lim_{x \\to 0} (\\lim_{y \\to 0} f(x, y))$, $\\ell_2 = \\lim_{y \\to 0} (\\lim_{x \\to 0} f(x, y))$ and $\\ell = \\lim_{(x, y) \\to (0, 0)} f(x, y)$, where\n\n$$f(x, y) = \\frac{x + y}{x - y}, \\quad x \\neq y.$$\nSolution.\n\n\\[ \\ell_1 = \\lim_{x \\to 0} \\left( \\lim_{y \\to 0} \\frac{x + y}{x - y} \\right) = \\lim_{x \\to 0} \\frac{x}{x} = 1 \\]\n\nand\n\n\\[ \\ell_2 = \\lim_{y \\to 0} \\left( \\lim_{x \\to 0} \\frac{x + y}{x - y} \\right) = \\lim_{y \\to 0} \\frac{y}{y} = -1. \\]\n\nSince the both limits \\( \\ell_1 \\) and \\( \\ell_2 \\) exist and \\( \\ell_1 \\neq \\ell_2 \\) it follows that the global limit \\( \\ell \\) of the function doesn’t exist. Indeed, we may choose the next two sequences\n\n\\[ (x_n, y_n) = \\left( \\frac{1}{n}, 0 \\right) \\to (0, 0) \\quad \\text{and} \\quad (x'_n, y'_n) = \\left( 0, \\frac{1}{n} \\right) \\to (0, 0), \\]\n\nfor which we have\n\n\\[ \\lim_{n \\to +\\infty} f(x_n, y_n) = 1 \\quad \\text{and} \\quad \\lim_{n \\to +\\infty} f(x'_n, y'_n) = -1. \\]\n\n**Example 1.2** Find \\( \\ell_1 = \\lim_{x \\to 0} \\left( \\lim_{y \\to 0} f(x, y) \\right) \\), \\( \\ell_2 = \\lim_{y \\to 0} \\left( \\lim_{x \\to 0} f(x, y) \\right) \\) and \\( \\ell = \\lim_{(x,y) \\to (0,0)} f(x, y) \\), where\n\n\\[ f(x, y) = \\frac{2xy}{x^2 + y^2}, \\quad (x, y) \\neq (0, 0). \\]\n\nSolution.\n\n\\[ \\ell_1 = \\lim_{x \\to 0} \\left( \\lim_{y \\to 0} \\frac{2xy}{x^2 + y^2} \\right) = \\lim_{x \\to 0} 0 = 0 \\]\n\nand\n\n\\[ \\ell_2 = \\lim_{y \\to 0} \\left( \\lim_{x \\to 0} \\frac{2xy}{x^2 + y^2} \\right) = \\lim_{y \\to 0} 0 = 0. \\]\n\nIn spite of the fact that the both limits \\( \\ell_1 \\) and \\( \\ell_2 \\) exist and \\( \\ell_1 = \\ell_2 \\), the global limit \\( \\ell \\) of the function doesn’t exist. Indeed, we may choose the next two sequences\n\n\\[ (x_n, y_n) = \\left( \\frac{1}{n}, \\frac{1}{n} \\right) \\to (0, 0) \\quad \\text{and} \\quad (x'_n, y'_n) = \\left( \\frac{2}{n}, \\frac{1}{n} \\right) \\to (0, 0), \\]\n\nfor which we have\n\n\\[ \\lim_{n \\to +\\infty} f(x_n, y_n) = \\lim_{n \\to +\\infty} \\frac{2}{n^2} = 1 \\quad \\text{and} \\quad \\lim_{n \\to +\\infty} f(x'_n, y'_n) = \\lim_{n \\to +\\infty} \\frac{4}{n^2} = \\frac{4}{5}. \\]\n\n**Example 1.3** Find \\( \\ell_1 = \\lim_{x \\to 0} \\left( \\lim_{y \\to 0} f(x, y) \\right) \\), \\( \\ell_2 = \\lim_{y \\to 0} \\left( \\lim_{x \\to 0} f(x, y) \\right) \\) and \\( \\ell = \\lim_{(x,y) \\to (0,0)} f(x, y) \\), where\n\n\\[ f(x, y) = \\frac{xy^2}{2x^2 + y^4}, \\quad (x, y) \\neq (0, 0). \\]\n\nSolution.\n\n\\[ \\ell_1 = \\lim_{x \\to 0} \\left( \\lim_{y \\to 0} \\frac{xy^2}{2x^2 + y^4} \\right) = \\lim_{x \\to 0} 0 = 0 \\]\n\nand\n\n\\[ \\ell_2 = \\lim_{y \\to 0} \\left( \\lim_{x \\to 0} \\frac{xy^2}{2x^2 + y^4} \\right) = \\lim_{y \\to 0} 0 = 0. \\]\nIn spite of the fact that the both limits $\\ell_1$ and $\\ell_2$ exist and $\\ell_1 = \\ell_2$, the global limit $\\ell$ of the function doesn’t exist. Indeed, we may choose the next two sequences\n\n$$(x_n, y_n) = \\left(\\frac{1}{n^2}, \\frac{1}{n}\\right) \\to (0, 0) \\quad \\text{and} \\quad (x'_n, y'_n) = \\left(\\frac{2}{n^2}, \\frac{1}{n}\\right) \\to (0, 0),$$\n\nfor which we have\n\n$$\\lim_{n \\to +\\infty} f(x_n, y_n) = \\lim_{n \\to +\\infty} \\frac{1}{n^4} = \\frac{1}{3} \\quad \\text{and} \\quad \\lim_{n \\to +\\infty} f(x'_n, y'_n) = \\lim_{n \\to +\\infty} \\frac{2}{n^4} = \\frac{2}{9}.$$\n\n**Example 1.4** Consider the function\n\n$$f(x, y) = x \\cos \\frac{1}{y}, \\quad y \\neq 0.$$\n\nStudy the existence of the limits $\\ell_1 = \\lim_{x \\to 0} (\\lim_{y \\to 0} f(x, y))$, $\\ell_2 = \\lim_{y \\to 0} (\\lim_{x \\to 0} f(x, y))$ and $\\ell = \\lim_{(x, y) \\to (0, 0)} f(x, y)$.\n\n**Solution.** As $\\lim_{y \\to 0} \\cos \\frac{1}{y}$ does not exist, it follows that the limit $\\ell_1 = \\lim_{x \\to 0} (\\lim_{y \\to 0} f(x, y))$ does not exist, while the limit $\\ell_2 = 0$. Indeed\n\n$$\\ell_2 = \\lim_{y \\to 0} \\left(\\lim_{x \\to 0} x \\cos \\frac{1}{y}\\right) = \\lim_{y \\to 0} \\cos \\frac{1}{y} \\cdot 0 = 0.$$\n\nFor the study of the global limit $\\ell$, let us observe that we have\n\n$$\\left| \\cos \\frac{1}{y} \\right| \\leq 1$$\n\nfor each $y \\neq 0$, hence $|f(x, y)| \\leq |x|$ and, because $\\lim_{(x, y) \\to (0, 0)} |x| = 0$, we deduce that the limit $\\ell$ exists and $\\ell = \\lim_{(x, y) \\to (0, 0)} f(x, y) = 0$.\n\n**Example 1.5** Consider the function\n\n$$f(x, y) = (x + y) \\sin \\frac{1}{x} \\sin \\frac{1}{y}, \\quad x \\neq 0, y \\neq 0.$$\n\nShow that the limits $\\ell_1 = \\lim_{x \\to 0} (\\lim_{y \\to 0} f(x, y))$ and $\\ell_2 = \\lim_{y \\to 0} (\\lim_{x \\to 0} f(x, y))$ do not exist, but the global limit $\\ell = \\lim_{(x, y) \\to (0, 0)} f(x, y)$ exists.\n\n**Solution.** As $\\lim_{y \\to 0} \\sin \\frac{1}{y}$ and $\\lim_{x \\to 0} \\sin \\frac{1}{x}$ do not exist, it follows that the both limits $\\ell_1$ and $\\ell_2$ do not exist. Regarding to the global limit $\\ell$, let us observe that we have\n\n$$\\left| \\sin \\frac{1}{x} \\sin \\frac{1}{y} \\right| \\leq 1$$\n\nfor each $x \\neq 0, y \\neq 0$, hence $|f(x, y)| \\leq |x| + |y|$ and, because $\\lim_{(x, y) \\to (0, 0)} (|x| + |y|) = 0$ we deduce that the global limit $\\ell$ exists and $\\ell = \\lim_{(x, y) \\to (0, 0)} f(x, y) = 0$. \n\n3", "id": "./materials/207.pdf" }, { "contents": "Constrained Extrema and Lagrange Multipliers\n\n**Definition 1** Let $D \\subset \\mathbb{R}^n$ and $f, g : D \\subset \\mathbb{R}$. An extreme value of $f$ subject to the condition $g(x) = 0$, is called a **constrained extreme value** and $g(x) = 0$ is called the constraint.\n\n**Definition 2** Let $D \\subset \\mathbb{R}^n$ and $f, g : D \\subset \\mathbb{R}$. The **Lagrangian function** of $f$ subject to the constraint $g(x) = 0$ is the function of $n + 1$ variables\n\n$$L(x; \\lambda) = f(x) + \\lambda g(x),$$\n\nwhere $\\lambda$ is known as the **Lagrange multiplier**.\n\n**Definition 3** The **Lagrangian function** of $f$ subject to the $k$ constraints $g_i(x) = 0$, $i = 1, k$ is the function with $k$ Lagrange multipliers, $\\lambda_i$, $i = 1, k$,\n\n$$L(x; \\lambda) = f(x) + \\sum_{i=1}^{k} \\lambda_i g_i(x).$$\n\n**Theorem 4** *(The Extreme Value Theorem for Functions of $n$ Variables)* Let $D \\subset \\mathbb{R}^n$ and $f : D \\subset \\mathbb{R}$ be a continuous $n$ variable real-valued function. If $D(f)$ is a closed and bounded set in $\\mathbb{R}^n$ then $R(f)$ is a closed and bounded set in $\\mathbb{R}$ and there exists $x, y \\in D(f)$ such that $f(x)$ is an absolute maximum value of $f$ and $f(y)$ is an absolute minimum value of $f$.\n\nTo find the extreme values of $f$ subject to the constraint $g(x) = 0$:\n\n1. calculate $\\frac{\\partial L}{\\partial x_1}, ..., \\frac{\\partial L}{\\partial x_n}, \\frac{\\partial L}{\\partial \\lambda}$ remembering that $L$ it is a function of the $n + 1$ variables $x = (x_1, ..., x_n)$ and $\\lambda$,\n2. solve the equations $\\frac{\\partial L}{\\partial x_1} = 0, ..., \\frac{\\partial L}{\\partial x_n} = 0$ and $g(x) = 0$,\n3. evaluate $f$ at these points to find the required extrema.\n\n**Example 5** Find the extreme values of\n\n$$f : \\mathbb{R}^2 \\to \\mathbb{R}, f(x, y) = x^2 + y^2 - x - y + 1$$\n\non the set $S = \\{(x, y) : x^2 + y^2 = 1\\}$. \n\n1\nSolution.\n\nLet \\( g(x, y) = x^2 + y^2 - 1 \\), \\( L(x, y; \\lambda) = x^2 + y^2 - x - y + 1 + \\lambda (x^2 + y^2 - 1) \\).\n\nCompute\n\\[\n\\frac{\\partial L}{\\partial x} = 2x - 1 + \\lambda 2x, \\quad \\frac{\\partial L}{\\partial y} = 2y - 1 + \\lambda 2y, \\quad \\frac{\\partial L}{\\partial \\lambda} = x^2 + y^2 - 1\n\\]\n\nSolve the system\n\\[\n\\begin{align*}\n2x - 1 + \\lambda 2x &= 0 \\\\\n2y - 1 + \\lambda 2y &= 0 \\\\\nx^2 + y^2 - 1 &= 0\n\\end{align*}\n\\]\n\nHence \\( x = y \\).\n\nFrom the last equation, it now follows that \\( 2x^2 = 1 \\Rightarrow x = \\pm \\frac{1}{\\sqrt{2}} \\). Thus we have two points to consider for extreme values: \\( \\left( \\frac{1}{\\sqrt{2}}, \\frac{1}{\\sqrt{2}} \\right), \\left( -\\frac{1}{\\sqrt{2}}, -\\frac{1}{\\sqrt{2}} \\right) \\).\n\nSince \\( S \\) is closed and bounded, we know from the Extreme Value Theorem that one of these values is an absolute maximum of \\( f \\) on \\( S \\) and the other an absolute minimum of \\( f \\) on \\( S \\). Now\n\\[\nf \\left( \\frac{1}{\\sqrt{2}}, \\frac{1}{\\sqrt{2}} \\right) = 2 - \\sqrt{2}\n\\]\nand\n\\[\nf \\left( -\\frac{1}{\\sqrt{2}}, -\\frac{1}{\\sqrt{2}} \\right) = 2 + \\sqrt{2}\n\\]\nso \\( f \\) has an absolute maximum value of \\( 2 + \\sqrt{2} \\) at \\( \\left( -\\frac{1}{\\sqrt{2}}, -\\frac{1}{\\sqrt{2}} \\right) \\) and an absolute minimum value of \\( 2 - \\sqrt{2} \\) at \\( \\left( \\frac{1}{\\sqrt{2}}, \\frac{1}{\\sqrt{2}} \\right) \\).\n\nExample 6 Find the shortest distance from the origin to the curve \\( x^6 + 3y^2 = 1 \\).\n\nSolution.\n\nWe extremize \\( f(x, y) = x^2 + y^2 \\) under the constraint \\( g(x, y) = x^6 + 3y^2 - 1 \\).\n\n\\( L(x, y; \\lambda) = x^2 + y^2 + \\lambda (x^6 + 3y^2 - 1) \\)\n\nCompute\n\\[\n\\frac{\\partial L}{\\partial x} = 2x + 6\\lambda x^5, \\quad \\frac{\\partial L}{\\partial y} = 2y + \\lambda 6y, \\quad \\frac{\\partial L}{\\partial \\lambda} = x^6 + 3y^2 - 1\n\\]\n\nSolve the system\n\\[\n\\begin{align*}\n2x + 6\\lambda x^5 &= 0 \\\\\n2y + \\lambda 6y &= 0 \\\\\nx^6 + 3y^2 - 1 &= 0\n\\end{align*}\n\\]\n\nSolutions are: \\( [x = 0, y = \\frac{1}{3}\\sqrt{3}, \\lambda = -\\frac{1}{3}] \\), \\( [x = 0, y = -\\frac{1}{3}\\sqrt{3}, \\lambda = -\\frac{1}{3}] \\), \\( [x = -1, y = 0, \\lambda = -\\frac{1}{3}] \\), \\( [x = 1, y = 0, \\lambda = -\\frac{1}{3}] \\).\n\nSo, we have the solutions \\( (0, \\pm \\sqrt{1/3}) \\) and \\( (1, 0), (-1, 0) \\). To see which is the minimum, just evaluate \\( f \\) on each of the points. We see that \\( (0, \\pm \\sqrt{1/3}) \\) are the minima.\n\nExample 7 Find the rectangular box with the largest volume that fits inside the ellipsoid \\( \\frac{x^2}{a^2} + \\frac{y^2}{b^2} + \\frac{z^2}{c^2} = 1 \\), given that it sides are parallel to the axes.\nSolution.\n\nClearly the box will have the greatest volume if each of its corners touch the ellipsoid. Let one corner of the box be point \\((x, y, z)\\) in the positive octant, then the box has corners \\((\\pm x; \\pm y; \\pm z)\\) and its volume is \\(V = 8xyz\\).\n\nWe want to maximize \\(V\\) given that \\(\\frac{x^2}{a^2} + \\frac{y^2}{b^2} + \\frac{z^2}{c^2} - 1 = 0\\). (Note that since the constraint surface is bounded the max does exist). The Lagrangian is\n\n\\[\nL(x, y, z; \\lambda) = 8xyz + \\lambda \\left( \\frac{x^2}{a^2} + \\frac{y^2}{b^2} + \\frac{z^2}{c^2} - 1 \\right).\n\\]\n\nThe critical points are solutions of the system\n\n\\[\n\\begin{align*}\n\\frac{\\partial L}{\\partial x} &= 0 \\\\\n\\frac{\\partial L}{\\partial y} &= 0 \\\\\n\\frac{\\partial L}{\\partial z} &= 0 \\\\\n\\frac{\\partial L}{\\partial \\lambda} &= 0\n\\end{align*}\n\\]\n\n(Note that \\(\\frac{\\partial L}{\\partial \\lambda}\\) will always be the constraint equation.) As we want to maximize \\(V\\) we can assume that \\(xyz \\neq 0\\) so that \\(x, y, z \\neq 0\\). Hence, eliminating \\(\\lambda\\), we get\n\n\\[\n\\lambda = -\\frac{4yz}{x}, \\quad \\lambda = -\\frac{4xz}{y}, \\quad \\lambda = -\\frac{4xy}{z}\n\\]\n\nso that\n\n\\[\n\\frac{4yz}{x} = \\frac{4xz}{y} = \\frac{4xy}{z} \\Rightarrow y^2a^2 = x^2b^2, \\quad \\frac{4xz}{y} = \\frac{4xy}{z} \\Rightarrow z^2b^2 = y^2c\n\\]\n\nBut then\n\n\\[\n\\frac{x^2}{a^2} + \\frac{y^2}{b^2} + \\frac{z^2}{c^2} - 1 = 0 \\Rightarrow 3\\frac{y^2}{b^2} = 1 \\Rightarrow y = \\frac{b}{\\sqrt{3}} \\quad \\text{and} \\quad z = \\frac{c}{\\sqrt{3}}\n\\]\n\n(they are all positive by assumption). So there is only one stationary point \\(\\left(\\frac{a}{\\sqrt{3}}, \\frac{b}{\\sqrt{3}}, \\frac{c}{\\sqrt{3}}\\right)\\).\n\nBeing the only stationary point it is the required point of maximum and consequently, the largest volume of the rectangular box inscribed in the ellipsoid is\n\n\\[\nV = \\frac{8abc}{3\\sqrt{3}}.\n\\]\n\nAuthor: Ariadna Lucia Pletea", "id": "./materials/208.pdf" }, { "contents": "MathE project\n\nContinuity for real functions of several variables\n\nLet $D \\subseteq \\mathbb{R}^k$ be a nonempty set, $a \\in D$ and let us consider a real function $f : D \\to \\mathbb{R}$.\n\n**Definition 1.1** If $a \\in D$ is a cluster point of $D$, we say that the function $f$ is continuous at $a$ if the limit of $f$ at the point $a$ exists and\n\n$$\\lim_{x \\to a} f(x) = f(a).$$\n\nIf $a \\in D$ is an isolated point, $f$ is continuous at $a$. We say that the function $f$ is continuous on the set $D$ if it is continuous at each point of $D$.\n\n**Proposition 1.1** (with sequences) Let $D \\subseteq \\mathbb{R}^k$ be a nonempty set and let $a \\in D$ a cluster point of $D$. The function $f : D \\to \\mathbb{R}$ is continuous at the point $a$ if and only if for any sequence $(x_n)_n \\subset D$ with $\\lim_{n \\to +\\infty} x_n = a$, we have that\n\n$$\\lim_{n \\to +\\infty} f(x_n) = f(a).$$\n\n**Remark 1.1** If there exists a sequence $(x_n)_n \\subset D$ with $\\lim_{n \\to +\\infty} x_n = a$ such that\n\n$$\\lim_{n \\to +\\infty} f(x_n) \\neq f(a),$$\n\nthen the function $f$ is not continuous at the point $a$.\n\n**Definition 1.2** Let $a = (a_1, a_2, \\ldots, a_k) \\in D$. Consider the function $f_i : D_i \\to \\mathbb{R}$ of variable $x_i$, $i = 1, k$, given by\n\n$$f_i(x_i) = f(a_1, a_2, \\ldots, a_{i-1}, x_i, a_{i+1}, \\ldots, a_k)$$\n\ndefined on the set $D_i = \\{x_i \\in \\mathbb{R} \\mid (a_1, a_2, \\ldots, a_{i-1}, x_i, a_{i+1}, \\ldots, a_k) \\in D\\}$. If the function $f_i$ is continuous at $a_i \\in D$, one says that the function $f$ is partially continuous with respect to variable $x_i$ at the point $a$.\n\n**Remark 1.2** If the function $f$ is continuous at the point $a \\in D$ (on $D$), then it is partially continuous with respect to each variable $x_i$, $i = 1, k$, at the point $a \\in D$ (on $D$, respectively). The partially continuous of $f$ at the point $a$ does not involve the global continuity of $f$ at $a$.\n\nFor a two-variables function $f : D \\subseteq \\mathbb{R}^2 \\to \\mathbb{R}$, $f = f(x, y)$ the above proposition is:\n\n**Proposition 1.2** (with sequences) Let $D \\subseteq \\mathbb{R}^2$ be a nonempty set and let $(a, b) \\in D$ a cluster point of $D$. The function $f : D \\to \\mathbb{R}$ is continuous at $(a, b)$ if and only if for any sequence $(x_n, y_n)_n \\subset D$ with $\\lim_{n \\to +\\infty} (x_n, y_n) = (a, b)$, we have that\n\n$$\\lim_{n \\to +\\infty} f(x_n, y_n) = f(a, b).$$\nRemark 1.3 If there exists a sequence \\((x_n, y_n)_n \\subset D\\) with \\(\\lim_{n \\to +\\infty} (x_n, y_n) = (a, b)\\) and \\(\\lim_{n \\to +\\infty} f(x_n, y_n) \\neq f(a, b)\\) then \\(f\\) is not continuous at \\((a, b)\\).\n\nExample 1.1 Study the continuity of the function \\(f : \\mathbb{R}^2 \\to \\mathbb{R}\\)\n\n\\[\nf(x, y) = \\begin{cases} \n\\sin(x^3 + y^3) \\over x^2 + y^2, & (x, y) \\neq (0, 0) \\\\\n0, & (x, y) = (0, 0).\n\\end{cases}\n\\]\n\nSolution. On the set \\(\\mathbb{R}^2 \\setminus \\{(0, 0)\\}\\) the function \\(f\\) is a composition of elementary continuous functions, so \\(f\\) is continuous. We study the continuity at \\((0, 0)\\). We use the known limit \\(\\lim_{t \\to 0} \\frac{\\sin t}{t} = 1\\) and we get\n\n\\[\n\\lim_{(x, y) \\to (0, 0)} f(x, y) = \\lim_{(x, y) \\to (0, 0)} \\frac{\\sin(x^3 + y^3)}{x^3 + y^3} \\cdot \\frac{x^3 + y^3}{x^2 + y^2} = 0 = f(0, 0).\n\\]\n\nIt follows that \\(f\\) is continuous at \\((0, 0)\\) and so it is continuous on \\(\\mathbb{R}^2\\). To prove that\n\n\\[\n\\lim_{(x, y) \\to (0, 0)} \\frac{x^3 + y^3}{x^2 + y^2} = 0\n\\]\n\nlet us observe that we can write\n\n\\[\n\\left| \\frac{x^3 + y^3}{x^2 + y^2} \\right| \\leq |x| \\cdot \\frac{x^2}{x^2 + y^2} + |y| \\cdot \\frac{y^2}{x^2 + y^2} \\leq |x| + |y| \\to 0.\n\\]\n\nExample 1.2 Study the continuity of the function \\(f : \\mathbb{R}^2 \\to \\mathbb{R}\\)\n\n\\[\nf(x, y) = \\begin{cases} \nxy \\over \\ln(1 + x^2 + y^2), & (x, y) \\neq (0, 0) \\\\\n0, & (x, y) = (0, 0).\n\\end{cases}\n\\]\n\nSolution. On the set \\(\\mathbb{R}^2 \\setminus \\{(0, 0)\\}\\) the function \\(f\\) is a composition of elementary continuous functions, so \\(f\\) is continuous. We study the continuity at \\((0, 0)\\). We use the known limit \\(\\lim_{t \\to 0} \\frac{\\ln(1 + t)}{t} = 1\\) and we get\n\n\\[\n\\lim_{(x, y) \\to (0, 0)} f(x, y) = \\lim_{(x, y) \\to (0, 0)} \\frac{x^2 + y^2}{\\ln(1 + x^2 + y^2)} \\cdot \\frac{xy}{x^2 + y^2} = \\lim_{(x, y) \\to (0, 0)} \\frac{xy}{x^2 + y^2},\n\\]\n\nand the last one is not equal with \\(f(0, 0) = 0\\) (it does not exist). Using the Remark 1.3, we can choose the sequence \\((x_n, y_n) = \\left(\\frac{1}{n}, \\frac{1}{n}\\right) \\to (0, 0)\\) and\n\n\\[\n\\lim_{n \\to +\\infty} f(x_n, y_n) = \\lim_{n \\to +\\infty} \\frac{1}{n^2} = \\frac{1}{2} \\neq f(0, 0).\n\\]\n\nIn conclusion \\(f\\) is not continuous at \\((0, 0)\\).\nExample 1.3 Prove that the function \\( f : \\mathbb{R}^2 \\to \\mathbb{R} \\)\n\n\\[\nf(x, y) = \\begin{cases} \nxy^2 + \\sin(x^3 + y^5) & , \\quad (x, y) \\neq (0, 0) \\\\\n\\frac{x^2 + y^4}{x^2 + y^4} & , \\quad (x, y) = (0, 0)\n\\end{cases}\n\\]\n\nis partially continuous with respect to both variables at the point \\((0, 0)\\), but it doesn’t continuous at this point.\n\nSolution. One of the partial function at the point \\((0, 0)\\) is \\( f_1 : \\mathbb{R} \\to \\mathbb{R} \\),\n\n\\[\nf_1(x) = f(x, 0) = \\begin{cases} \n\\sin \\frac{x^3}{x^2} & , \\quad x \\neq 0 \\\\\n0 & , \\quad x = 0\n\\end{cases}\n\\]\n\nIt is known that \\( \\lim_{t \\to 0} \\frac{\\sin t}{t} = 1 \\) and we have\n\n\\[\n\\lim_{x \\to 0} f_1(x) = \\lim_{x \\to 0} \\frac{\\sin x^3}{x^2} = \\lim_{x \\to 0} \\frac{\\sin x^3}{x^3} \\cdot x = 0 = f_1(0).\n\\]\n\nSo the function \\( f_1 \\) is continuous. The partial function \\( f_2 : \\mathbb{R} \\to \\mathbb{R} \\),\n\n\\[\nf_2(y) = f(0, y) = \\begin{cases} \n\\sin \\frac{y^4}{y^3} & , \\quad y \\neq 0 \\\\\n0 & , \\quad y = 0\n\\end{cases}\n\\]\n\nis continuous due to the relation\n\n\\[\n\\lim_{y \\to 0} f_2(y) = \\lim_{y \\to 0} \\frac{\\sin y^4}{y^3} = \\lim_{y \\to 0} \\frac{\\sin y^4}{y^4} \\cdot y = 0 = f_2(0).\n\\]\n\nBut the function \\( f \\) is not continuous at \\((0, 0)\\). To prove this, let us observe that we can write\n\n\\[\nf(x, y) = \\frac{\\sin(x^3 + y^5)}{x^2 + y^4} + \\frac{xy^2}{x^2 + y^4}\n\\]\n\nand, for the first term, we have\n\n\\[\n\\lim_{(x, y) \\to (0, 0)} \\frac{\\sin(x^3 + y^5)}{x^2 + y^4} = \\lim_{(x, y) \\to (0, 0)} \\frac{\\sin(x^3 + y^5)}{x^3 + y^5} \\cdot \\frac{x^3 + y^5}{x^2 + y^4} = 0.\n\\]\n\nIndeed, we have\n\n\\[\n\\left| \\frac{x^3 + y^5}{x^2 + y^4} \\right| \\leq |x| \\frac{x^2}{x^2 + y^4} + |y| \\frac{y^4}{x^2 + y^4} \\leq |x| + |y| \\to 0\n\\]\n\nwhen \\((x, y) \\to (0, 0)\\).\n\nFor the second term, let us observe that there exists a sequence \\((x_n, y_n) = \\left( \\frac{1}{n^2}, \\frac{1}{n} \\right) \\to (0, 0)\\) on which the limit of this function\n\n\\[\n(x, y) \\mapsto \\frac{xy^2}{x^2 + y^4}\n\\]\n\nis not equal with 0. Indeed we have\n\n\\[\n\\lim_{n \\to +\\infty} \\frac{1}{n^2} = \\frac{1}{2}.\n\\]\n\nSo \\( \\lim_{n \\to +\\infty} f \\left( \\frac{1}{n^2}, \\frac{1}{n} \\right) = \\frac{1}{2} \\neq f(0) \\). In conclusion \\( f \\) is not globally continuous at \\((0, 0)\\).", "id": "./materials/209.pdf" }, { "contents": "Circle and spherical surface\n\nCircle\n\nA circumference is a two-dimensional shape made by drawing a curve that is the same distance all around from the center.\n\nThe circle centered in \\( C = (c_1, c_2) \\) with radius \\( r \\) is the set of points \\( P = (x, y) \\) (the locus) that are distant from \\( C \\) the measure \\( r \\), that is,\n\n\\[\n||\\overrightarrow{CP}|| = r \\iff (x - c_1)^2 + (y - c_2)^2 = r^2.\n\\]\n\nThe distance between the midpoint and the circle border is called the radius.\n\nExample: Let us consider, on the Cartesian plane, the circle that contains points \\( A = (-1, 4) \\) and \\( B(3, 1) \\) and whose diameter measures \\( AB = 5 \\). Then the midpoint of \\([AB]\\), \\( M = (1, \\frac{5}{2}) \\), corresponds to the center of the circle and the radius is equal to \\( \\frac{AB}{2} = \\frac{5}{2} \\). Thus, the cartesian equation for this circle is as follows:\n\n\\[\n(x - 1)^2 + (y - \\frac{5}{2})^2 = \\frac{25}{4}.\n\\]\n\nSpherical surface\n\nA Spherical surface is a three-dimensional shape where any of its points is at the same distance from a fixed point, called the center of the spherical surface.\n\nThe Spherical surface centered in \\( C = (c_1, c_2, c_3) \\) with radius \\( r \\) is the set of points \\( P = (x, y, z) \\) (the locus) that are distant from \\( C \\) the measure \\( r \\), that is,\n\n\\[\n||\\overrightarrow{CP}|| = r \\iff (x - c_1)^2 + (y - c_2)^2 + (z - c_3)^2 = r^2.\n\\]", "id": "./materials/211.pdf" }, { "contents": "Operations on Linear Transformations\n\nLet the linear transformations:\n\n\\[ f: \\mathbb{R}^2 \\rightarrow \\mathbb{R}^2 \\quad \\text{and} \\quad g: \\mathbb{R}^2 \\rightarrow \\mathbb{R}^2 \\]\n\n\\[ (x, y) \\rightarrow (2x - y, x + 3y) \\quad \\text{and} \\quad (x, y) \\rightarrow (x + y, x - 2y) \\]\n\nwe can always add vectors from the same vector space, so\n\n\\[ (f + g)(x, y) = f(x, y) + g(x, y) = (2x - y, x + 3y) + (x + y, x - 2y) = (3x, 2x + y). \\]\n\nAlso note that \\( f(x, y) = \\begin{bmatrix} 2 & -1 \\\\ 1 & 3 \\end{bmatrix} \\begin{bmatrix} x \\\\ y \\end{bmatrix} \\) and \\( g(x, y) = \\begin{bmatrix} 1 & 1 \\\\ 1 & -2 \\end{bmatrix} \\begin{bmatrix} x \\\\ y \\end{bmatrix} \\). Then:\n\n\\[ (f+g)(x, y) = \\begin{bmatrix} 2 & -1 \\\\ 1 & 3 \\end{bmatrix} \\begin{bmatrix} x \\\\ y \\end{bmatrix} + \\begin{bmatrix} 1 & 1 \\\\ 1 & -2 \\end{bmatrix} \\begin{bmatrix} x \\\\ y \\end{bmatrix} = \\left( \\begin{bmatrix} 2 & -1 \\\\ 1 & 3 \\end{bmatrix} + \\begin{bmatrix} 1 & 1 \\\\ 1 & -2 \\end{bmatrix} \\right) \\begin{bmatrix} x \\\\ y \\end{bmatrix} = \\begin{bmatrix} 3 & 0 \\\\ 2 & 1 \\end{bmatrix} \\begin{bmatrix} x \\\\ y \\end{bmatrix}. \\]\n\n**Conclusion:** we can only add linear transformations with the same starting space and the same finishing space, and note that the sum is still a linear application.\n\n**Theorem:** Consider \\( V \\) and \\( E \\) vector spaces and \\( f: V \\rightarrow E \\) and \\( g: V \\rightarrow E \\) linear transformations. Then the following functions are also linear transformations:\n\n- \\( f + g \\) defined by \\( (f + g)(v) = f(v) + g(v) \\);\n- \\( kf \\) defined by \\( (kf)(v) = kf(v) \\);\n- \\( f \\circ g \\) defined by \\( (f \\circ g)(v) = (f[g(v)]) \\), since \\( g(V) \\subseteq D_f \\).\n\n**Example:** Let the linear transformations:\n\n\\[ f: \\mathbb{R}^3 \\rightarrow \\mathbb{R}^2 \\quad \\text{and} \\quad g: \\mathbb{R}^2 \\rightarrow \\mathbb{R}^2 \\]\n\n\\[ (x, y, z) \\rightarrow (y, x - z) \\quad \\text{and} \\quad (x, y) \\rightarrow (x + y, x - 2y) \\]\n\nNote that the application sum \\( f + g \\) does not exist, because the starting space of \\( f \\) is different from the starting space of \\( g \\).\n\nThe application \\(-5f\\) is defined by \\(-5f(x, y, z) = -5(y, x - z) = (-5y, -5x + 5z)\\).\n\nThe composite application \\( g \\circ f : \\mathbb{R}^3 \\rightarrow \\mathbb{R}^2 \\) is defined by\n\n\\[ g(f(x, y, z)) = g(f(x, y, z)) = g(y, x - z) = (x + y - z, -2x + y + 2z) \\]\nand is linear, such that are $f$ and $g$.\n\nAlso given that $f(x, y, z) = \\begin{bmatrix} 0 & 1 & 0 \\\\ 1 & 0 & -1 \\end{bmatrix} \\begin{bmatrix} x \\\\ y \\\\ z \\end{bmatrix}$ and $g(x, y) = \\begin{bmatrix} 1 & 1 \\\\ 1 & -2 \\end{bmatrix} \\begin{bmatrix} x \\\\ y \\end{bmatrix}$, it is proved that\n\n$$g \\circ f = \\begin{bmatrix} 1 & 1 \\\\ 1 & -2 \\end{bmatrix} \\begin{bmatrix} 0 & 1 & 0 \\\\ 1 & 0 & -1 \\end{bmatrix} \\begin{bmatrix} x \\\\ y \\\\ z \\end{bmatrix} = \\begin{bmatrix} 1 & 1 & -1 \\\\ -2 & 1 & 2 \\end{bmatrix} \\begin{bmatrix} x \\\\ y \\\\ z \\end{bmatrix}.$$ \n\nNotice that $f \\circ g$ does not exist.\n\n**Theorem:** Consider $V$ and $E$ vector spaces and $f : V \\to E$ and $g : E \\to W$ linear transformations, such that\n\n$$f(v) = [a_{ij}]_{n \\times m} v, \\quad v \\in V \\quad \\text{and} \\quad g(u) = [b_{ij}]_{m \\times r} u, \\quad u \\in E.$$ \n\nThen the composite function $f \\circ g : V \\to W$ is defined by\n\n$$(g \\circ f)(v) = [a_{ij}]_{n \\times m} [b_{ij}]_{m \\times r} v.$$", "id": "./materials/212.pdf" }, { "contents": "Reflections and translations\n\nComputer graphics deals with the manipulation of images, through their positioning through linear transformations such as reflections, dilations and contractions, orthogonal projections and rotations.\n\nSome reflections\n\nThe linear operator $T : \\mathbb{R}^2 \\to \\mathbb{R}^2$ defined by $T(x, y) = (y, x)$ translates a reflection around the line $y = x$, according to the figure beside. In fact:\n\n- $T(A) = T(2, 1) = (1, 2) = A'$;\n- $T(B) = T(4, 0) = (0, 4) = B'$;\n- $T(C) = T(6, 1) = (1, 6) = C'$;\n- $T(D) = T(3, 2) = (2, 3) = D'$.\n\nFollowing are some of some of the most common reflections and their matrix representations when considering the canonical basis\n\n| Reflection in $\\mathbb{R}^2$ | Operator | Matrix |\n|-----------------------------|----------|--------|\n| around of the OX axis | $T(x, y) = (x, -y)$ | $\\begin{bmatrix} 1 & 0 \\\\ 0 & -1 \\end{bmatrix}$ |\n| around of the Oy axis | $T(x, y) = (-x, y)$ | $\\begin{bmatrix} -1 & 0 \\\\ 0 & 1 \\end{bmatrix}$ |\n\n| Reflection in $\\mathbb{R}^3$ | Operator | Matrix |\n|-----------------------------|----------|--------|\n| around the xy-plane | $T(x, y, z) = (x, y, -z)$ | $\\begin{bmatrix} 1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & -1 \\end{bmatrix}$ |\n| around the yz-plane | $T(x, y, z) = (-x, y, z)$ | $\\begin{bmatrix} -1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 1 \\end{bmatrix}$ |\n| around the xz-plane | $T(x, y, z) = (x, -y, z)$ | $\\begin{bmatrix} 1 & 0 & 0 \\\\ 0 & -1 & 0 \\\\ 0 & 0 & 1 \\end{bmatrix}$ |\nDilation and contraction\n\nDilation or contraction is the operator stretching or shrinking a vector by a factor $k \\in \\mathbb{R}^+$, but keeping the direction unchanged. We call the operator a dilation if the transformed vector is at least as long as the original vector, and a contraction if the transformed vector is at most as long as the original vector.\n\nThe linear operator $f$ such that\n\n$$f(x, y) = \\begin{bmatrix} k & 0 \\\\ 0 & k \\end{bmatrix} \\begin{bmatrix} x \\\\ y \\end{bmatrix}$$\n\n- is a contraction with factor $k$ on $\\mathbb{R}^2$, if $0 < k < 1$.\n- is a dilatation with factor $k$ on $\\mathbb{R}^2$, if $k > 1$.\n\nThe linear operator $f : \\mathbb{R}^3 \\to \\mathbb{R}^3$ such that\n\n$$f(x, y, z) = \\begin{bmatrix} k & 0 & 0 \\\\ 0 & k & 0 \\\\ 0 & 0 & k \\end{bmatrix} \\begin{bmatrix} x \\\\ y \\\\ z \\end{bmatrix}$$\n\n- is a dilation on $\\mathbb{R}^3$, if $k > 1$.\n- is a contraction on $\\mathbb{R}^3$, if $0 < k < 1$.\n\nExample: The linear operator $f$ such that\n\n$$f(x, y) = \\begin{bmatrix} 1/2 & 0 \\\\ 0 & 1/2 \\end{bmatrix} \\begin{bmatrix} x \\\\ y \\end{bmatrix}$$\n\nis a contraction with factor $1/2$. ", "id": "./materials/213.pdf" }, { "contents": "Examples of linear transformations\n\nComputer graphics deals with the manipulation of images, through their positioning through linear transformations such as orthogonal projections and rotations, among others.\n\nOrthogonal projections\n\nAny vector \\( v \\) can be written as the sum of two orthogonal vectors, called components of \\( v \\).\n\nBesides that, if \\( u, v \\in \\mathbb{R}^2 \\) or \\( u, v \\in \\mathbb{R}^3 \\), then the orthogonal projection of a vector \\( v \\in E \\) in \\( u \\) is\n\n\\[\n\\text{proj}_u v = \\frac{\\|u\\| \\cos(\\theta)}{\\|v\\|} \\frac{v}{\\|v\\|} = \\frac{u \\cdot v}{\\|v\\|^2} v,\n\\]\n\naccording to the image beside.\n\nWe can say that if \\( W = \\langle v \\rangle \\) is a subspace of a vector space \\( V \\), then the orthogonal projection of a vector \\( u \\in V \\) in \\( W \\) is given by\n\n\\[\n\\text{proj}_W(u) = \\frac{u \\cdot v}{\\|v\\|^2} v.\n\\]\n\nSimilarly, if \\( B = \\{v_1, v_2, \\ldots, v_n\\} \\) is an orthogonal base of \\( W \\subseteq V \\) and \\( u \\in V \\), then\n\n\\[\n\\text{proj}_W(u) = \\frac{u \\cdot v_1}{\\|v_1\\|^2} v_1 + \\frac{u \\cdot v_2}{\\|v_2\\|^2} v_2 + \\cdots + \\frac{u \\cdot v_n}{\\|v_n\\|^2} v_n.\n\\]\n\nIn particular, \\( B = \\{(1, 0, 0), (0, 1, 0)\\} \\) is an orthogonal base of the plan \\( \\pi : z = 0 \\) which is a subspace of \\( \\mathbb{R}^3 \\). To any \\( u = (u_1, u_2, u_3) \\in V \\), we have\n\n\\[\n\\text{proj}_\\pi(u) = ((u_1, u_2, u_3) \\cdot (1, 0, 0))(1, 0, 0) + ((u_1, u_2, u_3) \\cdot (0, 1, 0))(0, 1, 0) = u_1(1, 0, 0) + u_2(0, 1, 0).\n\\]\n\nThat is, the orthogonal projection onto the \\( xy \\)-plane drops of the \\( z \\) coordinate. Formally, this can be written in matrix form as the following:\n\n\\[\n\\text{proj}_{xy}(u) = \\begin{bmatrix} 1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 0 \\end{bmatrix} \\begin{bmatrix} u_1 \\\\ u_2 \\\\ u_3 \\end{bmatrix} = \\begin{bmatrix} u_1 \\\\ u_2 \\\\ 0 \\end{bmatrix}.\n\\]\n\nNotice that this transformation preserves \\( u_1 \\) and \\( u_2 \\) but drops the last coordinate.\nAlso the orthogonal projection onto the $yz$-plane drops of the $x$ coordinate. Formally, that is:\n\n$$\\text{proj}_{yz}(u) = \\begin{bmatrix} 0 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 1 \\end{bmatrix} \\begin{bmatrix} u_1 \\\\ u_2 \\\\ u_3 \\end{bmatrix} = \\begin{bmatrix} 0 \\\\ u_2 \\\\ u_3 \\end{bmatrix}.$$ \n\n**Example:** $\\text{proj}_{yz}(-1, 2, 3) = (0, 2, 3)$\n\n**Rotation**\n\nAn operator that rotates a vector in $\\mathbb{R}^2$ through a given angle $\\theta$ is called a rotation operator in $\\mathbb{R}^2$ and is defined by\n\n$$f_R: \\mathbb{R}^2 \\rightarrow \\mathbb{R}^2 \\quad (x, y) \\rightarrow (x\\cos(\\theta) - y\\sin(\\theta), x\\sin(\\theta) + y\\cos(\\theta))$$\n\nor in the matrix form,\n\n$$f_R(x, y) = \\begin{bmatrix} \\cos(\\theta) - \\sin(\\theta) \\\\ \\sin(\\theta) + \\cos(\\theta) \\end{bmatrix} \\begin{bmatrix} x \\\\ y \\end{bmatrix},$$\n\naccording to the following:\n\nLet $(x_1, y_1) = f_R(x, y)$ and check the diagram. We can write $x_1 = r\\cos(\\theta + \\alpha)$, $y_1 = r\\sin(\\theta + \\alpha)$.\n\nAlso $x = r\\cos(\\alpha)$, $y_1 = r\\sin(\\alpha)$.\n\nUsing trigonometric identities we have\n\n$$x_1 = x\\cos(\\theta) - y\\sin(\\theta) \\quad \\text{and} \\quad y_1 = x\\sin(\\theta) + y\\cos(\\theta).$$\n\nThe operator that rotates a vector in $\\mathbb{R}^3$ about the positive $x$-axis through a given angle $\\theta$ is called a rotation operator in $\\mathbb{R}^3$ and is defined by the matrix form,\n\n$$f_R(x, y, z) = \\begin{bmatrix} 1 & 0 & 0 \\\\ 0 & \\cos(\\theta) & -\\sin(\\theta) \\\\ 0 & \\sin(\\theta) & \\cos(\\theta) \\end{bmatrix} \\begin{bmatrix} x \\\\ y \\\\ z \\end{bmatrix},$$\n\nThe operator that rotates a vector in $\\mathbb{R}^3$ about the positive $y$-axis through a given angle $\\theta$ is called a rotation operator in $\\mathbb{R}^3$ and is defined by the matrix form,\n\n$$f_R(x, y, z) = \\begin{bmatrix} \\cos(\\theta) & 0 & -\\sin(\\theta) \\\\ 0 & 1 & 0 \\\\ \\sin(\\theta) & 0 & \\cos(\\theta) \\end{bmatrix} \\begin{bmatrix} x \\\\ y \\\\ z \\end{bmatrix}.$$\n\nThe operator that rotates a vector in $\\mathbb{R}^3$ about the positive $z$-axis through a given angle $\\theta$ is called a rotation operator in $\\mathbb{R}^3$ and is defined by the matrix form,\n\n$$f_R(x, y, z) = \\begin{bmatrix} \\cos(\\theta) & -\\sin(\\theta) & 0 \\\\ \\sin(\\theta) & \\cos(\\theta) & 0 \\\\ 0 & 0 & 1 \\end{bmatrix} \\begin{bmatrix} x \\\\ y \\\\ z \\end{bmatrix}.$$", "id": "./materials/214.pdf" }, { "contents": "Orthogonal linear operator\n\nFor any linear operator $T : \\mathbb{R}^n \\to \\mathbb{R}^n$, for some $n$, you can find a matrix which implements the mapping.\n\nFor example, consider $T : \\mathbb{R}^3 \\to \\mathbb{R}^3$, such that $T(x, y, z) = (3x - y, y + 2z, x + 3z)$. The standard matrix $A = [a_{ij}]_{3 \\times 3}$ for the transformation $T$ have 3 columns. The first is the vector $T(1, 0, 0) = (3, 0, 1)$, the second vector is $T(0, 1, 0) = (-1, 1, 0)$ and the third column is $T(0, 0, 1) = (0, 2, 3)$. Therefore,\n\n$$A = \\begin{bmatrix} 3 & -1 & 0 \\\\ 0 & 1 & 2 \\\\ 1 & 0 & 3 \\end{bmatrix}.$$ \n\nNotice that $T(v) = Av$, that is, if $v = (v_1, v_2, v_3)$, then\n\n$$T(v_1, v_2, v_3) = \\begin{bmatrix} 3 & -1 & 0 \\\\ 0 & 1 & 2 \\\\ 1 & 0 & 3 \\end{bmatrix} \\begin{bmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{bmatrix}.$$ \n\nAlso remember that a square matrix is orthogonal if its inverse is equal to its transposed matrix, that is $A^{-1} = A^T$. The matrix $A$ of the linear operator $T$ above is not orthogonal. In fact,\n\n$$\\begin{bmatrix} 3 & -1 & 0 \\\\ 0 & 1 & 2 \\\\ 1 & 0 & 3 \\end{bmatrix} \\begin{bmatrix} 3 & 0 & 1 \\\\ -1 & 1 & 0 \\\\ 0 & 2 & 3 \\end{bmatrix} = \\begin{bmatrix} 10 & -1 & 3 \\\\ -1 & 5 & 6 \\\\ 3 & 6 & 10 \\end{bmatrix} \\neq \\begin{bmatrix} 1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 1 \\end{bmatrix}.$$ \n\nAn orthogonal linear operator is one which preserves not only sums and scalar multiples, but dot products and other related metrical properties such as distances, lengths and angles.\n\nSince metrical properties can all be described in terms of dot products, we use the following definition.\n\n**Definition:** A linear operator $T$ on 2-space or on 3-space is called orthogonal if it preserves inner products, that is, if $T(u) \\cdot T(v) = u \\cdot v$ for all vectors $u$ and $v$. This means that $T$ preserves the norm,\n\n$$||T(v)|| = ||v||, \\quad \\forall v \\in V.$$ \n\nSo we say that $T$ is an orthogonal operator if it is an isometry.\n\n**Example:** The linear operator $T : \\mathbb{R}^2 \\to \\mathbb{R}^2$ such that $T(x, y) = (y, -x)$ is orthogonal.\n\nIndeed, $||T(x, y)|| = ||(y, -x)|| = \\sqrt{y^2 + x^2} = ||(x, y)||$.\n\nNotice that the standard matrix\n\n$$A = \\begin{bmatrix} 0 & 1 \\\\ -1 & 0 \\end{bmatrix}.$$ \n\nof the linear transformation $T(v) = Av$ is an orthogonal matrix. In fact $A^{-1} = A^T$. \n\n\nExample: The linear operator $T : \\mathbb{R}^2 \\to \\mathbb{R}^2$ such that $T(x, y) = (x - y, 3y)$ is not orthogonal. Indeed, $||T(1, -1)|| = ||(2, -3)|| = \\sqrt{13} \\neq ||(1, -1)|| = \\sqrt{2}$.\n\nProperties:\n\n- The inverse of an orthogonal transformation is also orthogonal.\n- The composition of orthogonal transformations is orthogonal.\n\nExample: The linear operator $T : \\mathbb{R}^3 \\to \\mathbb{R}^3$, such that\n\n$$T(x, y, z) = \\left( \\frac{3}{5}x + \\frac{4}{5}y, -\\frac{4}{5}x + \\frac{3}{5}y, z \\right)$$\n\nis orthogonal. Indeed,\n\n$$||T(x, y, z)|| = \\sqrt{\\left( \\frac{3}{5}x + \\frac{4}{5}y \\right)^2 + \\left( -\\frac{4}{5}x + \\frac{3}{5}y \\right)^2 + z^2} = \\sqrt{\\frac{9}{25}x^2 + \\frac{16}{25}y^2 + \\frac{16}{25}x^2 + \\frac{9}{25}y^2 + z^2} = ||(x, y, z)||.$$\n\nConsider the standard matrix $A = [a_{ij}]_{3 \\times 3}$ of this linear operator $T$. Is $A^T$ orthogonal?\n\nYes, because if $T$ is an orthogonal linear operator, then $A^{-1} = A^T$, that is, $AA^T = A^TA = I_3$.\n\nJust now note that $(A^T)^T = A$. \n\n---\n\nEdite Martins Cordeiro \nFlora Silva \nPaula Maria Barros", "id": "./materials/215.pdf" }, { "contents": "Intersection and sum of two vector spaces and the relationship between their dimensions\n\nIntersection and sum of two vector subspaces\n\nDefinition: Let $V$ be a vector space, and let $U$ and $W$ be subspaces of $V$. Then:\n\n1. $U + W = \\{u + w : u \\in U \\land w \\in W\\}$ and is called the sum of $U$ and $W$.\n2. $U \\cap W = \\{v : v \\in U \\land v \\in W\\}$ and is called the intersection of $U$ and $W$.\n\nExample: Consider the plans $P_1 = \\{(x, y, z) \\in \\mathbb{R}^3 : z = 0\\}$ and $P_2 = \\{(x, y, z) \\in \\mathbb{R}^3 : x - y + z = 0\\}$. These are both subspaces of $\\mathbb{R}^3$, that we can define by its generic vectors as:\n\n- $P_1 = \\{(x_1, y_1, 0) : x_1, y_1 \\in \\mathbb{R}\\}$;\n- $P_2 = \\{(x_2, y_2, y_2 - x_2) : x_2, y_2 \\in \\mathbb{R}\\}$.\n\nTheir intersection is the subspace $P_1 \\cap P_2 = \\{(x, y, z) \\in \\mathbb{R}^3 : z = 0 \\land x - y = 0\\} = \\{(x, x, 0) : x \\in \\mathbb{R}\\}$.\n\nThe sum, $P_1 + P_2 = \\{(x_1 + x_2, y_1 + y_2, y_2 - x_2) : x_1, x_2, y_1, y_2 \\in \\mathbb{R}\\}$ is the vector space $\\mathbb{R}^3$.\n\nRelationship between the dimensions of the vector spaces sum and intersection of two vector spaces\n\nIn relation to the previous example, we easily determine the size of each of the spaces $P_1$, $P_2$, $P_1 \\cap P_2$ and $P_1 + P_2$.\n\nJust consider the number of free variables of the generic vector and we conclude that both subspaces $P_1 = \\{(x, y, 0) : x, y \\in \\mathbb{R}\\}$ and $P_2 = \\{(x, y, y - x) : x, y \\in \\mathbb{R}\\}$ of $\\mathbb{R}^3$ have dimension 2.\n\nTheir intersection is the subspace $P_1 \\cap P_2 = \\{(x, x, 0) : x \\in \\mathbb{R}\\}$ with dimension 1 and the subspace $P_1 + P_2$ has dimension 3, because $P_1 + P_2$ spans $\\mathbb{R}^3$. We have\n\n$$\\dim(P_1 + P_2) = \\dim(P_1) + \\dim(P_2) - \\dim(P_1 \\cap P_2).$$\n\nIt will be always like this? The answer is affirmative:\n\n**Theorem Dimension of sum:** Let $V$ be a vector space with subspaces $U$ and $W$, each one of them have finite dimension. Then $U + W$ also has finite dimension which is given by\n\n$$\\dim(U + W) = \\dim(U) + \\dim(W) - \\dim(U \\cap W).$$\n**Example:** Let the vector subspaces \\( S_1 = \\left\\{ \\begin{bmatrix} a & 0 \\\\ 3a & b \\end{bmatrix} : a, b \\in \\mathbb{R} \\right\\} \\) and \\( S_2 = \\left\\{ \\begin{bmatrix} c & d \\\\ -d & e \\end{bmatrix} : c, d, e \\in \\mathbb{R} \\right\\} \\) of the space of the square matrices of order 2.\n\n\\( S_1 \\) has dimension 2, \\( S_2 \\) has dimension 3 and their intersection, \\( S_1 \\cap S_2 = \\left\\{ \\begin{bmatrix} x & 0 \\\\ 0 & y \\end{bmatrix} : x, y \\in \\mathbb{R} \\right\\} \\), has dimension 2.\n\nThen the sum, \\( S_1 + S_2 \\) is a subspace with dimension \\( 2 + 3 - 2 = 3 \\).\n\nNotice that\n\n\\[\nB = \\left\\{ \\begin{bmatrix} 1 & 0 \\\\ 3 & 0 \\end{bmatrix}, \\begin{bmatrix} 0 & 0 \\\\ 0 & 1 \\end{bmatrix} \\right\\}\n\\]\n\nis a basis of \\( S_1 \\) and\n\n\\[\nC = \\left\\{ \\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix}, \\begin{bmatrix} 0 & 1 \\\\ -1 & 0 \\end{bmatrix}, \\begin{bmatrix} 0 & 0 \\\\ 0 & 1 \\end{bmatrix} \\right\\}\n\\]\n\nis a basis of \\( S_2 \\).\n\nThen\n\n\\[\nD = \\left\\{ \\begin{bmatrix} 1 & 0 \\\\ 3 & 0 \\end{bmatrix}, \\begin{bmatrix} 0 & 0 \\\\ 0 & 1 \\end{bmatrix}, \\begin{bmatrix} 1 & 0 \\\\ 0 & 0 \\end{bmatrix}, \\begin{bmatrix} 0 & 1 \\\\ -1 & 0 \\end{bmatrix}, \\begin{bmatrix} 0 & 0 \\\\ 0 & 1 \\end{bmatrix} \\right\\}\n\\]\n\nspans \\( S_1 + S_2 \\).\n\nAs the equality\n\n\\[\nk_1 \\begin{bmatrix} 1 & 0 \\\\ 3 & 0 \\end{bmatrix} + k_2 \\begin{bmatrix} 0 & 0 \\\\ 0 & 1 \\end{bmatrix} + k_3 \\begin{bmatrix} 1 & 0 \\\\ 0 & 0 \\end{bmatrix} + k_4 \\begin{bmatrix} 0 & 1 \\\\ -1 & 0 \\end{bmatrix} + k_5 \\begin{bmatrix} 0 & 0 \\\\ 0 & 1 \\end{bmatrix} = \\begin{bmatrix} 0 & 0 \\\\ 0 & 0 \\end{bmatrix}\n\\]\n\nrepresents a system, whose expanded matrix is\n\n\\[\n\\begin{bmatrix}\n1 & 0 & 1 & 0 & 0 & 0 & 0 \\\\\n0 & 0 & 0 & 1 & 0 & 0 & 0 \\\\\n3 & 0 & 0 & -1 & 0 & 0 & 0 \\\\\n0 & 1 & 0 & 0 & 1 & 0 & 0\n\\end{bmatrix} \\rightarrow \\begin{bmatrix}\n1 & 0 & 1 & 0 & 0 & 0 & 0 \\\\\n0 & 0 & 0 & 1 & 0 & 0 & 0 \\\\\n0 & 0 & -3 & -1 & 0 & 0 & 0 \\\\\n0 & 1 & 0 & 0 & 1 & 0 & 0\n\\end{bmatrix} \\rightarrow \\begin{bmatrix}\n1 & 0 & 1 & 0 & 0 & 0 & 0 \\\\\n0 & 1 & 0 & 0 & 1 & 0 & 0 \\\\\n0 & 0 & -3 & -1 & 0 & 0 & 0 \\\\\n0 & 0 & 0 & 1 & 0 & 0 & 0\n\\end{bmatrix}\n\\]\n\nWe can conclude that the system is doubly indeterminate, so \\( D \\) is linearly dependent. Thus, the minimum set that generates \\( S_1 + S_2 \\) has cardinality 3.", "id": "./materials/216.pdf" }, { "contents": "Orthogonal sets and basis\n\nRecall that the dot product:\n\\[ u \\cdot v = ||u|| ||v|| \\cos(\\theta), \\quad \\theta = \\hat{u} \\hat{v} \\in [0, \\pi]. \\]\n\nIf, for example, \\( u = (u_1, u_2, u_3), v = (v_1, v_2, v_3) \\in \\mathbb{R}^3 \\), we also have\n\\[ u \\cdot v = u_1 v_1 + u_2 v_2 + u_3 v_3. \\]\n\nAn inner product of 2 vectors is a generalization of the dot product, it is a way to multiply vectors, whose product being a scalar. More precisely, an inner product, \\( \\langle \\cdot, \\cdot \\rangle \\), in a real vector space \\( V \\) is any operator that satisfies the following properties:\n\n1. \\( (u + v)/w = (u/w) + (v/w), \\forall u, v, w \\in V. \\)\n2. \\( (ku)/v = k(u/v)u/w + v/w, \\forall u, v \\in V, \\forall k \\in \\mathbb{R}. \\)\n3. \\( u/v = v/u, \\forall u, v \\in V. \\)\n4. \\( u/u \\leq 0 \\) if and only if \\( u = 0. \\)\n\nThe vector space \\( V \\) together with \\( \\langle \\cdot, \\cdot \\rangle \\) is called an inner product space.\n\nAn inner product space \\( V \\) induces a norm, that is, a notion of length of a vector:\n\\[ ||v|| = \\sqrt{v \\cdot v}. \\]\n\nIn particular, the usual inner product (scalar product) in the vector space \\( \\mathbb{R}^2 \\) induces a norm\n\\[ ||v|| = \\sqrt{v \\cdot v}. \\]\n\nThat is, if \\( v = (v_1, v_2) \\), then \\( v \\cdot v = v_1^2 + v_2^2 \\) and \\( ||v|| = \\sqrt{v_1^2 + v_2^2} \\)\n\n**Definition:** The set \\( A = \\{v_1, v_2, \\ldots, v_k\\} \\in V \\setminus \\{0\\} \\) is an orthogonal set if each vector of \\( A \\) is orthogonal to each of the other vectors in the set, that is,\n\\[ v_i \\cdot v_j = 0 \\quad \\text{for} \\quad i \\neq j. \\]\n\nIf, in addition, all vectors are of unit norm, \\( ||v_i|| = 1 \\), then \\( \\{v_1, v_2, \\ldots, v_k\\} \\) is called an orthonormal set.\n\n**Examples:**\n\n1. The set \\( A = \\{(1, -2), (4, 2)\\} \\subset \\mathbb{R}^2 \\) is orthogonal but is not orthonormal.\n In fact \\( (1, -2) \\cdot (4, 2) = 1 \\times 4 - 2 \\times 2 = 0 \\), but \\( ||(1, -2)|| = \\sqrt{(1, -2) \\cdot (1, -2)} = \\sqrt{5} \\neq 1. \\)\n2. The set \\( B = \\{(1, 0, 0), (0, 1, 0)\\} \\subset \\mathbb{R}^3 \\) is orthonormal.\n**Theorem:** Any orthogonal set is linearly independent.\n\n**Definition:** An orthonormal basis for an inner product space $V$ with finite dimension is a basis for $V$ whose vectors are orthonormal to each other and are all unit vectors.\n\nThe following are examples of orthonormal bases:\n- The standard basis of $\\mathbb{R}^2$, $A = \\{(1, 0), (0, 1)\\}$;\n- The standard basis of $\\mathbb{R}^3$, $B = \\{(1, 0, 0), (0, 1, 0), (0, 0, 1)\\}$;\n- The basis $C = \\left\\{ \\left( \\frac{3}{5}, \\frac{4}{5} \\right), \\left( \\frac{4}{5}, -\\frac{3}{5} \\right) \\right\\}$ of $\\mathbb{R}^2$;\n\nLet $V$ an inner product space with an inner product, “$\\cdot$”. The following properties give us the coordinates of $V$ vectors with respect to orthogonal bases.\n\n**Theorem:** If $B = \\{w_1, w_2, \\ldots, w_n\\}$ is an orthonormal basis of $V$, then for any $v \\in V$ we have\n\n$$v = (v \\cdot w_1) w_1 + (v \\cdot w_2) w_2 + \\cdots + (v \\cdot w_n) w_n.$$ \n\n**Example:** $B = \\left\\{ \\left( \\frac{3}{5}, \\frac{4}{5} \\right), \\left( \\frac{4}{5}, -\\frac{3}{5} \\right) \\right\\}$ is an orthonormal basis of $\\mathbb{R}^2$. We can write any vector $v = (x, y)$ as\n\n$$v_B = \\left( \\frac{3x + 4y}{5}, \\frac{4x - 3y}{5} \\right).$$\n\n**Theorem:** Let $A = \\{w_1, w_2, \\ldots, w_r\\} \\subset \\mathbb{R}^n \\setminus \\{0\\}$ such that $w_i \\cdot w_j = 0$, for $i \\neq j$ and $i, j \\in \\{1, 2, \\ldots, r\\}$. Then:\n\n1. $A$ is a basis of $\\langle A \\rangle$;\n2. For any $v \\in \\langle A \\rangle$, we have $v = k_1 w_1 + k_2 w_2 + \\cdots + k_r w_r$, with\n\n$$k_i = \\frac{v \\cdot w_i}{||w_i||^2}.$$ \n\n**Example:** $A = \\{(1, -1, 0), (0, 0, 2)\\} \\subset \\mathbb{R}^3$ is an orthogonal basis, because $(1, -1, 0) \\cdot (0, 0, 2) = 0$. The norms of vectors are $||(1, -1, 0)|| = \\sqrt{2}$ and $||(0, 0, 2)|| = 2$.\n\nBesides that, $\\langle A \\rangle = \\{(x, -x, z) : x, z \\in \\mathbb{R}\\}$ and any $v = (x, -x, z) \\in \\langle A \\rangle$ is such that\n\n$$(x, -x, z) = \\frac{(x, -x, z) \\cdot (1, -1, 0)}{||(1, -1, 0)||^2} (1, -1, 0) + \\frac{(x, -x, z) \\cdot (0, 0, 2)}{||(0, 0, 2)||^2} (0, 0, 2) = \\frac{2x}{2} (1, -1, 0) + \\frac{2z}{4} (0, 0, 2).$$", "id": "./materials/217.pdf" }, { "contents": "Orthogonal projection of a vector $v$ over a vector space $S$ and distance from $v$ to $S$\n\nWe calculate the distance from $v$ to $S$, using the following theorem:\n\n**Theorem of the best approximation:** Consider the Euclidean space $E$ and a subspace $W$ of $E$. If $v \\in E$ is such that $v \\notin W$ then the vector $\\text{proj}_W(v)$ is the best approach of $v$ to $W$. That is,\n\n$$||v - \\text{proj}_W(v)|| \\leq ||v - w||, \\quad \\text{to any} \\quad w \\in W.$$ \n\nThus, $\\text{proj}_W(v)$ is the vector of $W$ that best approximates $v$. Then, the distance between $v$ and the vector space $W$ is given by $||v - \\text{proj}_W(v)||$.\n\nOrthogonal projection of one vector over another\n\nConsider two vectors $u$ and $v$. The orthogonal projection of $v$ over $w \\neq 0$ is the scalar multiple of $w$,\n\n$$\\text{proj}_w(v) = \\frac{v \\cdot w}{|w|^2} w$$\n\n**Example:** Consider, in $\\mathbb{R}^2$, $v = (-1, -1)$ and $u = (3, 4)$. The orthogonal projection of $v$ over $u$ is\n\n$$\\text{proj}_u(v) = \\frac{(-1, 1) \\cdot (2, -1)}{||(2, -1)||^2} (2, -1) = \\frac{-3}{5} (2, -1) = \\left(-\\frac{6}{5}, \\frac{3}{5}\\right).$$\n\nSo the distance from $v$ to the subspace generated by $u$, $\\langle u \\rangle$, is\n\n$$||v - \\text{proj}_u(v)|| = ||(-1, 1) - \\left(-\\frac{6}{5}, \\frac{3}{5}\\right)|| = ||\\left(\\frac{1}{5}, \\frac{2}{5}\\right)|| = \\sqrt{\\frac{5}{25}} = \\frac{\\sqrt{5}}{5}.$$ \n\nOrthogonal projection of one vector over a vector space\n\nLet $E$ be a Euclidean space, $W$ a subspace of $E$ and $B = \\{w_1, w_2, \\ldots, w_n\\}$ an orthogonal basis of $W$. Then\n\n$$\\text{proj}_W(v) = \\frac{v \\cdot w_1}{|w_1|^2} w_1 + \\frac{v \\cdot w_2}{|w_2|^2} w_2 + \\cdots + \\frac{v \\cdot w_n}{|w_n|^2} w_n.$$ \n\nScalars\n\n$$k_i = \\frac{v \\cdot w_i}{|w_i|^2}$$\n\nare said color ipb Fourier coefficients $v$ in relation to $w_i$. \n**Example:** Consider the subspace $S$ of $\\mathbb{R}^3$ generated by $A = \\{(1, -1, 2), (1, 0, 1)\\}$ and $v = (1, 2, 3) \\notin S$. The orthogonal projection of $v$ over $S$ is\n\n$$\\text{proj}_S(v) = \\frac{(1, 2, 3) \\cdot (1, -1, 2)}{||(1, -1, 2)||^2} (1, -1, 2) + \\frac{(1, 2, 3) \\cdot (1, 0, 1)}{||(1, 0, 1)||^2} (1, 0, 1) = \\frac{5}{4} (1, -1, 2) + \\frac{2}{2} (1, 0, 1).$$\n\nWe have $\\text{proj}_S(v) = \\left(\\frac{9}{4}, -\\frac{5}{4}, \\frac{7}{2}\\right)$. So the distance from $v$ to the subspace $S$ is\n\n$$||v - \\text{proj}_S(v)|| = ||(1, 2, 3) - \\left(\\frac{9}{4}, -\\frac{5}{4}, \\frac{7}{2}\\right)|| = ||(-\\frac{5}{4}, \\frac{3}{4}, -\\frac{1}{2})|| = \\sqrt{\\frac{25}{16} + \\frac{9}{16} + \\frac{1}{4}} = \\frac{19}{8}.$$", "id": "./materials/218.pdf" }, { "contents": "5.4 Directional Derivatives and the Gradient Vector\n\nObjectives\n\n- I understand the notion of a gradient vector and I know in what direction it points.\n- I know how to calculate a directional derivative in the direction of a given vector $\\vec{v}$.\n- I know how to find the unit vector $\\vec{u}$ that creates the maximum directional derivative, $D_u f$.\n\nMathematically speaking, the derivative of a function $f(x)$ at a point $x$ should be thought of as the slope of $f$ at that point $x$. If we want to be really precise, we should think of it as the slope of the tangent line of $f$ at $x$. This is the best way to conceptually understand a derivative.\n\nFor example, if you’re given the function $f(x) = -x^2 + 2$, you know the derivative of $f$ at $x = 1$ will be $-2$. This value corresponds to the slope of the line tangent to $f$ at $x = 1$.\n\nFor higher dimensions, we want to find an analogous value. That is, we want to find something that can represent a slope of a tangent. In the last section, we found partial derivatives, but as the word “partial” would suggest, we are not done! These partial derivatives are an intermediate step to the object we wish to find.\n\nRecall that slopes in three dimensions are described with vectors (see section 3.5 Lines and Planes) because vectors describe movement. So our true derivative in higher dimensions should be a vector. This vector is called the gradient vector.\n\n**Definition 5.4.1** The gradient vector of a function $f$, denoted $\\nabla f$ or $\\text{grad}(f)$, is a vector whose entries are the partial derivatives of $f$. That is,\n\n$$\\nabla f(x, y) = (f_x(x, y), f_y(x, y))$$\nThis is a generalization of a derivative of a function of several variables.\n\nA natural question you may ask is in what direction does the vector point? The answer is that it points in the direction of **steepest ascent**. If you imagine walking on a hilly area, the gradient is a vector that points you toward the steepest climb.\n\nHere is a picture of a three dimensional surface. The arrows at the bottom represent the gradient vectors. They each point in the direction where it is steepest. Longer the vector (the greater its magnitude), the steeper the surface is at that point.\n\nHere is a good thought exercise to test your understanding of gradients. What do you think the gradient vector should be for the function \\( f(x, y) = x^2 + y^2 \\) at the point \\((0, 0)\\)? Remember, this function is the paraboloid and \\((0, 0)\\) is its vertex.\n5.4.1 Examples\n\nExample 5.4.1.1 Find the gradient vector of\n\n\\[ f(x, y) = x^2 + y^2 \\]\n\nWhat are the gradient vectors at \\((1, 2), (2, 1)\\) and \\((0, 0)\\)?\n\nWe begin with the formula.\n\n\\[ \\nabla f = \\langle f_x, f_y \\rangle = \\langle 2x, 2y \\rangle \\]\n\nNow, let us find the gradient at the following points.\n\n- \\(\\nabla f(1, 2) = \\langle 2, 4 \\rangle\\)\n- \\(\\nabla f(2, 1) = \\langle 4, 2 \\rangle\\)\n- \\(\\nabla f(0, 0) = \\langle 0, 0 \\rangle\\)\n\nNotice that at \\((0, 0)\\) the gradient vector is the zero vector. Since the gradient corresponds to the notion of slope at that point, this is the same as saying the slope is zero.\n\nExample 5.4.1.2 Find the gradient vector of\n\n\\[ f(x, y) = 2xy + x^2 + y \\]\n\nWhat are the gradient vectors at \\((1, 1), (0, -1)\\) and \\((0, 0)\\)?\n\n\\[ \\nabla f = \\langle f_x, f_y \\rangle = \\langle 2y + 2x, 2x + 1 \\rangle \\]\n\nNow, let us find the gradient at the following points.\n\n- \\(\\nabla f(1, 1) = \\langle 4, 3 \\rangle\\)\n- \\(\\nabla f(0, -1) = \\langle -2, 1 \\rangle\\)\n- \\(\\nabla f(0, 0) = \\langle 0, 1 \\rangle\\)\n\nSo far, we’ve learned the definition of the gradient vector and we know that it tells us the direction of steepest ascent. What if, however, we want to know the rate of ascent in another direction? For that, we use something called a directional derivative.\nDefinition 5.4.2 The directional derivative, denoted $D_v f(x, y)$, is a derivative of a multivariable function in the direction of a vector $\\vec{v}$. It is the scalar projection of the gradient onto $\\vec{v}$.\n\n$$D_v f(x, y) = \\text{comp}_v \\nabla f(x, y) = \\frac{\\nabla f(x, y) \\cdot \\vec{v}}{|\\vec{v}|}$$\n\nThis produces a vector whose magnitude represents the rate a function ascends (how steep it is) at point $(x, y)$ in the direction of $\\vec{v}$.\n\nIf our function has three inputs, the math works out the same. Suppose $f(x, y, z) = w$. Then,\n\n$$\\nabla f(x, y, z) = \\langle f_x(x, y, z), f_y(x, y, z), f_z(x, y, z) \\rangle$$\n\nand the directional derivative in the direction of $\\vec{u}$ is\n\n$$D_v f(x, y, z) = \\frac{\\nabla f(x, y, z) \\cdot \\vec{v}}{|\\vec{v}|}$$\n\nLet’s look at some examples.\n\n5.4.2 Examples\n\nExample 5.4.2.1 Find the directional derivative of\n\n$$f(x, y) = \\frac{x}{x^2 + y^2}$$\n\nin the direction of $\\vec{v} = \\langle 3, 5 \\rangle$ at the point $(1, 2)$.\n\nFirst, we find the gradient.\n\n$$f_x(x, y) = \\frac{d}{dx} \\left( \\frac{x}{x^2 + y^2} \\right)$$\n\n$$= \\frac{(x^2 + y^2) - x(2x)}{(x^2 + y^2)^2}$$\n\n$$= \\frac{y^2 - x^2}{(x^2 + y^2)^2}$$\n\\[ f_y(x, y) = \\frac{d}{dy} \\left( \\frac{x}{x^2 + y^2} \\right) = \\frac{-2xy}{(x^2 + y^2)^2} \\]\n\nThe gradient is then\n\\[ \\nabla f(1, 2) = \\left\\langle \\frac{4 - 1}{(4 + 1)^2}, \\frac{-4}{(4 + 1)^2} \\right\\rangle = \\left\\langle \\frac{3}{25}, \\frac{-4}{25} \\right\\rangle = \\frac{1}{25} (3, -4) \\]\n\nWe now find the magnitude of \\( \\vec{v} \\). We get\n\\[ |\\vec{v}| = \\sqrt{9 + 25} = \\sqrt{34} \\]\n\nThe directional derivative is then\n\\[ D_{\\vec{v}} f(1, 2) = \\frac{\\nabla f(1, 2) \\cdot \\vec{v}}{|\\vec{v}|} = \\frac{1}{25\\sqrt{34}} (3, -4) \\cdot (3, 5) = \\frac{1}{25\\sqrt{34}} (9 - 20) = -\\frac{11}{25\\sqrt{34}} \\]\n\n**Example 5.4.2.2** Find the directional derivative of\n\\[ f(x, y, z) = \\sqrt{xyz} \\]\nin the direction of \\( \\vec{v} = \\langle -1, -2, 2 \\rangle \\) at the point \\( (3, 2, 6) \\).\n\nFirst, we find the partial derivatives to define the gradient.\n\\[ f_x(x, y, z) = \\frac{yz}{2\\sqrt{xyz}} \\]\n\\[ f_y(x, y, z) = \\frac{xz}{2\\sqrt{xyz}} \\]\n\\[ f_z(x, y, z) = \\frac{xy}{2\\sqrt{xyz}} \\]\n\nThe gradient is\n\\[ \\nabla f(3, 2, 6) = \\left\\langle \\frac{12}{2(6)}, \\frac{18}{2(6)}, \\frac{6}{2(6)} \\right\\rangle = \\left\\langle \\frac{3}{2}, \\frac{1}{2}, \\frac{1}{2} \\right\\rangle = \\frac{1}{2} (2, 3, 1) \\]\nThe magnitude of \\( \\vec{v} = (-1, -2, 2) \\) is\n\n\\[\n|\\vec{v}| = \\sqrt{1 + 4 + 4} = 3\n\\]\n\nTherefore, the directional derivative is\n\n\\[\nD_v f(3, 2, 6) = \\frac{\\nabla f(3, 2, 6) \\cdot \\vec{v}}{|\\vec{v}|} = \\frac{1}{3(2)} (2, 3, 1) \\cdot (-1, -2, 2) = \\frac{1}{6} (-2 - 6 + 2) = -1\n\\]\n\nThe next natural question is:\n\nIn what direction is the derivative maximum?\n\nAs we just saw, the directional derivative is calculated by taking the scalar projection of \\( \\nabla f \\) onto a vector \\( \\vec{v} \\). Define \\( \\theta \\) be the angle between \\( \\vec{v} \\) and \\( \\nabla f \\). Then,\n\n\\[\n\\frac{\\nabla f \\cdot \\vec{v}}{|\\vec{v}|} = \\frac{|\\nabla f||\\vec{u}|\\cos(\\theta)}{|\\vec{v}|} = |\\nabla f|\\cos(\\theta)\n\\]\n\nThis is maximized if \\( \\theta = 0 \\). From this, we know the following:\n\n- The maximum rate of change (the largest directional derivative) is \\( |\\nabla f| \\).\n- This occurs when \\( \\vec{v} \\) is parallel to \\( \\nabla f \\), i.e. when they point in the same direction.\n\nThat makes sense since \\( \\nabla f \\) is the vector pointing toward steepest ascent, so it should be the direction with the largest derivative.\n\nYou’ll typically be asked for the unit vector, \\( \\vec{u} \\), that creates the maximum directional derivative. This is because unit vectors are thought of as vectors that just contain information about direction. Based on our discussion above, this will always be\n\n\\[\n\\vec{u} = \\frac{\\nabla f}{|\\nabla f|}\n\\]\n\nLet’s look at two examples.\n\n### 5.4.3 Examples\n\n**Example 5.4.3.1** 1. Find the maximum rate of change of \\( f \\) at the given point and the direction in which it occurs.\n\n\\[\nf(s, t) = te^{st}, \\quad (0, 2)\n\\]\nThe maximum rate of change is $|\\nabla f(0, 2)|$. Let’s first find the gradient.\n\n$$\\nabla f = \\langle te^s, ste^s + e^s \\rangle$$\n\nThen\n\n$$|\\nabla f(0, 2)| = \\sqrt{(2)^2 + 1^2} = \\sqrt{5}$$\n\nThe direction is the unit vector.\n\n$$\\frac{\\nabla f(0, 2)}{|\\nabla f(0, 2)|} = \\left\\langle \\frac{2}{\\sqrt{5}}, \\frac{1}{\\sqrt{5}} \\right\\rangle$$\n\nNote: for this problem, it may not have been clear which component was the first and which was the second since $s$ and $t$ are atypical variables. For clues about the order, look out how the ordered pairs are defined in the function. It was written as “$f(s, t)$,” which tells us our vectors are $\\langle s, t \\rangle$.\n\n**Example 5.4.3.2 2.** Find the maximum rate of change of $f$ at the given point and the direction in which it occurs.\n\n$$f(x, y, z) = \\sqrt{x^2 + y^2 + z^2}, \\quad (3, 6, -2)$$\n\nAs above, the maximum rate of change is $|\\nabla f(3, 6, -2)|$.\n\n$$\\nabla f(x, y, z) = \\left\\langle \\frac{x}{\\sqrt{x^2 + y^2 + z^2}}, \\frac{y}{\\sqrt{x^2 + y^2 + z^2}}, \\frac{z}{\\sqrt{x^2 + y^2 + z^2}} \\right\\rangle$$\n\nThen\n\n$$\\nabla f(3, 6, -2) = \\left\\langle \\frac{3}{7}, \\frac{6}{7}, -\\frac{2}{7} \\right\\rangle$$\n\nThe $|\\nabla f(3, 6, -2)| = 1/7\\sqrt{9 + 36 + 4} = \\frac{1}{7}$\n\nSince it’s already a unit vector, the direction is\n\n$$\\left\\langle \\frac{3}{7}, \\frac{6}{7}, -\\frac{2}{7} \\right\\rangle$$\nThe gradient vector of a function $f$, denoted $\\nabla f$ or $\\text{grad}(f)$, is a vector whose entries are the partial derivatives of $f$.\n\n$$\\nabla f(x, y) = (f_x(x, y), f_y(x, y))$$\n\nIt is the generalization of a derivative in higher dimensions.\n\n- The gradient points in the direction of steepest ascent.\n\n- The directional derivative, denoted $D_v f(x, y)$, is a derivative of a $f(x, y)$ in the direction of a vector $\\vec{v}$. It is the scalar projection of the gradient onto $\\vec{v}$.\n\n$$D_v f(x, y) = \\text{comp}_v \\nabla f(x, y) = \\frac{\\nabla f(x, y) \\cdot \\vec{v}}{|\\vec{v}|}$$\n\nThis produces a vector whose magnitude represents the rate a function ascends (how steep it is) at point $(x, y)$ in the direction of $\\vec{v}$.\n\n- Both the gradient and the directional derivative work the same in higher variables.\n\n- The maximum directional derivative is always $|\\nabla f|$.\n\n- This happens in the direction of the unit vector\n\n$$\\vec{u} = \\frac{\\nabla f}{|\\nabla f|}$$\n\nRemember, we use the unit vector as a convention. Any vector parallel to $\\nabla f$ will work.", "id": "./materials/219.pdf" }, { "contents": "Partial Derivatives and Differentiability\n\n1. Partial Derivatives\n\n**Definition 6.1.** Let \\( f : U \\subset \\mathbb{R}^n \\to \\mathbb{R} \\) be a real-valued function. Then the \\( i^{th} \\) partial derivative is the real-valued function\n\n\\[\n\\frac{\\partial f}{\\partial x_i}(x) = \\lim_{h \\to 0} \\frac{f(x + he_i) - f(x)}{h}\n\\]\n\n**Remark 6.2.** Note that limit used in the above definition is just the limit of a function of a single variable. Indeed, fix \\( x = (x_1, \\ldots, x_{i-1}, x_i, x_{i+1}, \\ldots, x_n) \\) and define\n\n\\[\nF(h) = (x_1, \\ldots, x_{i-1}, h, x_{i+1}, \\ldots, x_n)\n\\]\n\nThen\n\n\\[\n\\frac{\\partial f}{\\partial x_i}(x) = \\lim_{h \\to 0} \\frac{F(x_i + h) - F(x_i)}{h} \\equiv \\frac{dF}{dh} \\bigg|_{h=x_i}\n\\]\n\nThis observation in fact tells us precisely how to compute the \\( i^{th} \\) partial derivative of \\( f \\). We regard all the coordinates except \\( x_i \\) to be fixed (i.e., constants) and then use ordinary calculus to differentiate the resulting function of \\( x_i \\) (regarded as a function of a single variable).\n\n**Example 6.3.** Let \\( f(x, y) = \\cos(x^2y^2) \\). Then\n\n\\[\n\\frac{\\partial f}{\\partial x} = -\\sin(x^2y^2)(2xy^2) \\\\\n\\frac{\\partial f}{\\partial y} = -\\sin(x^2y^2)(2x^2y)\n\\]\n\nThe ordinary derivative \\( \\frac{df}{dx} \\) of a function \\( f \\) of a single variable \\( x \\) tells us the rate of change of a function as \\( x \\) increases. Similarly, the partial derivative is interpretable as a rate of change: the partial derivative\n\n\\[\n\\frac{\\partial f}{\\partial x_i}(x)\n\\]\n\nis the rate at which the function \\( f \\) changes as one moves away from the point \\( x \\) in the direction \\( e_i \\).\n\n2. Differentiability of Functions of Several Variables\n\nRecall that in the case of a function of a single variable, a function \\( f(x) \\) is differentiable only if it is continuous; but that continuity does not guarantee differentiability. Intuitively, continuity of \\( f(x) \\) requires that its graph be a continuous curve; and differentiability requires also that there is always a unique tangent vector to the graph of \\( f(x) \\). In other words, a function \\( f(x) \\) is differentiable if and only if its graph is a smooth continuous curve with no sharp corners (a sharp corner would be a place where there would be two possible tangent vectors).\nIf we try to extend this graphical picture of differentiability to functions of two or more variables, it would be natural to think of a differentiable function of several variables as one whose graph is a smooth continuous surface, with no sharp peaks or folds. Because for such a surface it would always be possible to associate a unique tangent plane at a given point.\n\nHowever, “differentiability” in this sense turns out to be a much stronger condition than the mere existence of partial derivatives. For the existence of a partial derivatives at a point \\( x_0 \\) requires only a smooth approach to the point \\( f(x_0) \\) along the direction of the coordinate axes. We have seen examples of functions that are discontinuous even though\n\n\\[\n\\lim_{x \\to 0} f(x,0) = \\lim_{y \\to 0} f(0,y)\n\\]\n\nboth exist. For example the function\n\n\\[\nf(x,y) = \\frac{(x-y)^2}{x^2 + y^2}\n\\]\n\nhas this property, and in fact, both \\( \\frac{\\partial f}{\\partial x} \\) and \\( \\frac{\\partial f}{\\partial y} \\) exist and are continuous functions at the point \\((0,0)\\).\n\nWith this sort of phenomenon in mind we give the following definition of differentiability.\n\n**Definition 6.4.** We say that a function \\( f : \\mathbb{R}^2 \\to \\mathbb{R} \\) of two variables \\( x \\) and \\( y \\) is **differentiable** at \\((x_0, y_0)\\) if\n\n1. Both \\( \\frac{\\partial f}{\\partial x} \\) and \\( \\frac{\\partial f}{\\partial y} \\) exist at the point \\((x_0, y_0)\\).\n2. \\[\n\\lim_{(x,y) \\to (x_0,y_0)} \\frac{f(x,y) - f(x_0,y_0) - \\left[ \\frac{\\partial f}{\\partial x}(x_0,y_0) \\right] (x-x_0) - \\left[ \\frac{\\partial f}{\\partial y}(x_0,y_0) \\right] (y-y_0)}{|(x-x_0)^2 + (y-y_0)^2|^{1/2}} = 0\n\\]\n\n**Remark 6.5.** The limit condition simply means that\n\n\\[\nF(x,y) = f(x_0,y_0) + \\frac{\\partial f}{\\partial x}(x_0,y_0) (x-x_0) + \\frac{\\partial f}{\\partial y}(x_0,y_0) (y-y_0)\n\\]\n\nis a good approximation to \\( f(x,y) \\) near the point \\((x_0, y_0)\\). To make contact with our graphical interpretation of differentiability, we simply note that the graph of \\( F(x,y) \\) is a plane (it is linear in the variables \\( x \\) and \\( y \\)). To make this completely obvious, recall that the solution set of any equation of form\n\n\\[\nAx + By + Cz = D\n\\]\n\nis a plane in \\( \\mathbb{R}^3 \\); so taking\n\n\\[\nA = -\\left. \\frac{\\partial f}{\\partial x} \\right|_{(x_0,y_0)}\n\\]\n\n\\[\nB = -\\left. \\frac{\\partial f}{\\partial y} \\right|_{(x_0,y_0)}\n\\]\n\n\\[\nC = 1\n\\]\n\n\\[\nD = f(x_0,y_0) - \\left[ \\left. \\frac{\\partial f}{\\partial x} \\right|_{(x_0,y_0)} \\right] x_0 - \\left[ \\left. \\frac{\\partial f}{\\partial y} \\right|_{(x_0,y_0)} \\right] y_0\n\\]\n\nwe see that the equation of the graph of \\( F(x,y) \\)\n\n\\[\nz = F(x,y) = -Ax - By + D\n\\]\n\nis a plane. Thus the limit condition is saying that the graph of \\( f(x,y) \\) coincides with the graph of a plane as \\((x,y)\\) approaches the point \\((x_0, y_0)\\). This observation motivates the following definition.\nDefinition 6.6. Let \\( f : \\mathbb{R}^2 \\to \\mathbb{R} \\) be a function that is differentiable at the point \\( x_0 = (x_0, y_0) \\). Then the plane in \\( \\mathbb{R}^3 \\) defined by the equation\n\\[\nz = f(x_0, y_0) + \\frac{\\partial f}{\\partial x}(x_0, y_0)(x - x_0) + \\frac{\\partial f}{\\partial y}(x_0, y_0)(y - y_0)\n\\]\nis called the plane tangent to the graph of \\( f \\) at \\( x_0 \\).\n\nIn order to generalize these definitions to the case of a function from \\( \\mathbb{R}^n \\) to \\( \\mathbb{R}^m \\) it suffices to simply generalize our notation.\n\nDefinition 6.7. If \\( f \\) is any function from \\( \\mathbb{R}^n \\) to \\( \\mathbb{R} \\) we define the gradient \\( \\nabla f \\) of \\( f \\) to be the function from \\( \\mathbb{R}^n \\) to \\( \\mathbb{R} \\) defined by\n\\[\n\\nabla f(x) = \\left( \\frac{\\partial f}{\\partial x_1}(x), \\frac{\\partial f}{\\partial x_2}(x), \\ldots, \\frac{\\partial f}{\\partial x_n}(x) \\right)\n\\]\nOf course, for this definition to make sense all the partial derivatives \\( \\frac{\\partial f}{\\partial x_i}(x), \\ldots, \\frac{\\partial f}{\\partial x_n}(x) \\) must exist.\n\nDefinition 6.8. Let \\( U \\) be an open set in \\( \\mathbb{R}^n \\) and let \\( f \\) be a function from \\( U \\) to \\( \\mathbb{R} \\). We say that \\( f \\) is differentiable at a point \\( x_0 \\in U \\) if the partial derivatives\n\\[\n\\frac{\\partial f}{\\partial x_i}(x_0) = \\lim_{h \\to 0} \\frac{f(x_0 + he_i) - f(x_0)}{h}\n\\]\nall exist and if\n\\[\n0 = \\lim_{x \\to x_0} \\frac{f(x) - f(x_0) - \\nabla f(x_0) \\cdot (x - x_0)}{\\|x - x_0\\|}\n\\]\nWe can also extend our notion of differentiability to functions from \\( \\mathbb{R}^n \\) to \\( \\mathbb{R}^m \\).\n\nDefinition 6.9. Let \\( U \\) be an open set in \\( \\mathbb{R}^n \\) and let \\( f \\) be a function from \\( U \\) to \\( \\mathbb{R} \\). We say that \\( f \\) is differentiable at a point \\( x_0 \\in U \\) if the partial derivatives\n\\[\n\\frac{\\partial f_j}{\\partial x_i}(x_0) = \\lim_{h \\to 0} \\frac{f_j(x_0 + he_i) - f_j(x_0)}{h}\n\\]\nall exist and\n\\[\n0 = \\lim_{x \\to x_0} \\frac{\\|f(x) - f(x_0) - \\sum_{j=0}^n \\nabla f_j(x_0) \\cdot (x - x_0)e_j\\|}{\\|x - x_0\\|}\n\\]\nExample 6.10. Let \\( f(x, y) = (xy, x + y) \\). Then\n\\[\nT(x, y) = \\left( \\begin{array}{cc} \\frac{\\partial f_1}{\\partial x} & \\frac{\\partial f_1}{\\partial y} \\\\ \\frac{\\partial f_2}{\\partial x} & \\frac{\\partial f_2}{\\partial y} \\end{array} \\right) = \\left( \\begin{array}{cc} y & x \\\\ 1 & 1 \\end{array} \\right)\n\\]\nand so for \\( f \\) to be differentiable at \\( (1, 0) \\) we require\n\\[\n0 = \\lim_{x \\to x_0} \\frac{\\|f(x) - f(x_0) - \\sum_{j=0}^n \\nabla f_j(x_0) \\cdot (x - x_0)e_j\\|}{\\|x - x_0\\|}\n\\]\n\\[\n= \\lim_{x \\to x_0} \\frac{\\|f(x) - f(x_0) - (\\nabla f_1(x_0) \\cdot (x - x_0), 0) - (0, \\nabla f_2(x_0) \\cdot (x - x_0))\\|}{\\|x - x_0\\|}\n\\]\nNow\n\\[ f(x) = (xy, x + y) \\]\n\\[ f(x_0) = f(1, 0) = (0, 1) \\]\n\\[ \\nabla f_1 = \\left( \\frac{\\partial}{\\partial x}(xy), \\frac{\\partial}{\\partial y}(xy) \\right) = (y, x) \\quad \\Rightarrow \\quad \\nabla f_1(x_0) = \\nabla f_1(1, 0) = (0, 1) \\]\n\\[ \\nabla f_2 = \\left( \\frac{\\partial}{\\partial x}(x + y), \\frac{\\partial}{\\partial y}(x + y) \\right) = (1, 1) \\quad \\Rightarrow \\quad \\nabla f_2(x_0) = \\nabla f_2(1, 0) = (1, 1) \\]\n\\[ \\nabla f_1(x_0) \\cdot (x - x_0) = (0, 1) \\cdot (x - 1, y - 0) = y \\]\n\\[ \\nabla f_2(x_0) \\cdot (x - x_0) = (1, 1) \\cdot (x - 1, y - 0) = x - 1 + y \\]\n\nHence the limit condition becomes\n\\[\n0 = \\lim_{(x,y) \\to (1,0)} \\frac{\\|(xy, x + y) - (0, 1) - (y, 0) - (0, x - 1 + y)\\|}{\\|(x - 1, y)\\|}\n\\]\n\\[\n= \\lim_{(x,y) \\to (1,0)} \\frac{\\|(xy - 0 - y - 0, x + y - 1 - 0 - x + 1 - y)\\|}{\\|(x - 1, y)\\|}\n\\]\n\\[\n= \\lim_{(x,y) \\to (1,0)} \\frac{\\|(xy - y, 0)\\|}{\\|(x - 1, y)\\|}\n\\]\n\\[\n= \\lim_{(x,y) \\to (1,0)} \\frac{\\sqrt{(xy - y)^2}}{\\sqrt{(x - 1)^2 + y^2}}\n\\]\n\nUnfortunately as we have seen in the examples we discussed in the lecture on limits and continuity, it is not so easy to see whether or not this limit exists. Luckily we have a theorem at our disposal that makes it a lot easier to decide questions of differentiability.\n\n**Theorem 6.11.** Let \\( f : U \\subset \\mathbb{R}^n \\to \\mathbb{R}^m \\). Suppose that the derivatives \\( \\frac{\\partial f}{\\partial x_i} \\) all exist and are continuous on a neighborhood of a point \\( x \\) in \\( U \\). Then \\( f \\) is differentiable at \\( x \\).", "id": "./materials/220.pdf" }, { "contents": "Section 2-6 : Chain Rule\n\nWe’ve been using the standard chain rule for functions of one variable throughout the last couple of sections. It’s now time to extend the chain rule out to more complicated situations. Before we actually do that let’s first review the notation for the chain rule for functions of one variable.\n\nThe notation that’s probably familiar to most people is the following.\n\n\\[ F(x) = f(g(x)) \\quad \\Rightarrow \\quad F'(x) = f'(g(x))g'(x) \\]\n\nThere is an alternate notation however that while probably not used much in Calculus I is more convenient at this point because it will match up with the notation that we are going to be using in this section. Here it is.\n\n\\[ \\frac{dy}{dt} = \\frac{dy}{dx} \\frac{dx}{dt} \\]\n\nNotice that the derivative \\( \\frac{dy}{dt} \\) really does make sense here since if we were to plug in for \\( x \\) then \\( y \\) really would be a function of \\( t \\). One way to remember this form of the chain rule is to note that if we think of the two derivatives on the right side as fractions the \\( dx \\)'s will cancel to get the same derivative on both sides.\n\nOkay, now that we’ve got that out of the way let’s move into the more complicated chain rules that we are liable to run across in this course.\n\nAs with many topics in multivariable calculus, there are in fact many different formulas depending upon the number of variables that we’re dealing with. So, let’s start this discussion off with a function of two variables, \\( z = f(x, y) \\). From this point there are still many different possibilities that we can look at. We will be looking at two distinct cases prior to generalizing the whole idea out.\n\n**Case 1:** \\( z = f(x, y), \\ x = g(t), \\ y = h(t) \\) and compute \\( \\frac{dz}{dt} \\).\n\nThis case is analogous to the standard chain rule from Calculus I that we looked at above. In this case we are going to compute an ordinary derivative since \\( z \\) really would be a function of \\( t \\) only if we were to substitute in for \\( x \\) and \\( y \\).\n\nThe chain rule for this case is,\n\n\\[ \\frac{dz}{dt} = \\frac{\\partial f}{\\partial x} \\frac{dx}{dt} + \\frac{\\partial f}{\\partial y} \\frac{dy}{dt} \\]\n\nSo, basically what we’re doing here is differentiating \\( f \\) with respect to each variable in it and then multiplying each of these by the derivative of that variable with respect to \\( t \\). The final step is to then add all this up.\nLet's take a look at a couple of examples.\n\n**Example 1** Compute \\( \\frac{dz}{dt} \\) for each of the following.\n\n(a) \\( z = xe^{xy}, \\ x = t^2, \\ y = t^{-1} \\)\n\n(b) \\( z = x^2y^3 + y\\cos x, \\ x = \\ln(t^2), \\ y = \\sin(4t) \\)\n\n**Solution**\n\n(a) \\( z = xe^{xy}, \\ x = t^2, \\ y = t^{-1} \\)\n\nThere really isn’t all that much to do here other than using the formula.\n\n\\[\n\\frac{dz}{dt} = \\frac{\\partial f}{\\partial x} \\frac{dx}{dt} + \\frac{\\partial f}{\\partial y} \\frac{dy}{dt}\n\\]\n\n\\[\n= (e^{xy} + yxe^{xy})(2t) + x^2e^{xy}(-t^{-2})\n\\]\n\n\\[\n= 2t(e^{xy} + yxe^{xy}) - t^{-2}x^2e^{xy}\n\\]\n\nSo, technically we’ve computed the derivative. However, we should probably go ahead and substitute in for \\( x \\) and \\( y \\) as well at this point since we’ve already got \\( t \\)'s in the derivative. Doing this gives,\n\n\\[\n\\frac{dz}{dt} = 2t(e^t + te^t) - t^{-2}t^4e^t = 2te^t + t^2e^t\n\\]\n\nNote that in this case it might actually have been easier to just substitute in for \\( x \\) and \\( y \\) in the original function and just compute the derivative as we normally would. For comparison’s sake let’s do that.\n\n\\[\nz = t^2e^t \\quad \\Rightarrow \\quad \\frac{dz}{dt} = 2te^t + t^2e^t\n\\]\n\nThe same result for less work. Note however, that often it will actually be more work to do the substitution first.\n\n(b) \\( z = x^2y^3 + y\\cos x, \\ x = \\ln(t^2), \\ y = \\sin(4t) \\)\n\nOkay, in this case it would almost definitely be more work to do the substitution first so we’ll use the chain rule first and then substitute.\n\n\\[\n\\frac{dz}{dt} = (2xy^3 - y\\sin x)\\left(\\frac{2}{t}\\right) + (3x^2y^2 + \\cos x)(4\\cos(4t))\n\\]\n\n\\[\n= \\frac{4\\sin^3(4t)\\ln^2 - 2\\sin(4t)\\sin(\\ln^2)}{t} + 4\\cos(4t)(3\\sin^2(4t)[\\ln^2]^2 + \\cos(\\ln^2))\n\\]\nNote that sometimes, because of the significant mess of the final answer, we will only simplify the first step a little and leave the answer in terms of $x$, $y$, and $t$. This is dependent upon the situation, class and instructor however so be careful about not substituting in for without first talking to your instructor.\n\nNow, there is a special case that we should take a quick look at before moving on to the next case. Let’s suppose that we have the following situation,\n\n$$z = f(x, y) \\quad y = g(x)$$\n\nIn this case the chain rule for $\\frac{dz}{dx}$ becomes,\n\n$$\\frac{dz}{dx} = \\frac{\\partial f}{\\partial x} \\frac{dx}{dx} + \\frac{\\partial f}{\\partial y} \\frac{dy}{dx} = \\frac{\\partial f}{\\partial x} + \\frac{\\partial f}{\\partial y} \\frac{dy}{dx}$$\n\nIn the first term we are using the fact that,\n\n$$\\frac{dx}{dx} = \\frac{d}{dx}(x) = 1$$\n\nLet’s take a quick look at an example.\n\n**Example 2** Compute $\\frac{dz}{dx}$ for $z = x\\ln(xy) + y^2$, $y = \\cos(x^2 + 1)$\n\n**Solution**\n\nWe’ll just plug into the formula.\n\n$$\\frac{dz}{dx} = \\left( \\ln(xy) + x \\frac{y}{xy} \\right) + \\left( x \\frac{x}{xy} + 3y^2 \\right) \\left( -2x \\sin(x^2 + 1) \\right)$$\n\n$$= \\ln(x \\cos(x^2 + 1)) + 1 - 2x \\sin(x^2 + 1) \\left( \\frac{x}{\\cos(x^2 + 1)} + 3 \\cos^2(x^2 + 1) \\right)$$\n\n$$= \\ln(x \\cos(x^2 + 1)) + 1 - 2x^2 \\tan(x^2 + 1) - 6x \\sin(x^2 + 1) \\cos^2(x^2 + 1)$$\n\nNow let’s take a look at the second case.\n\n**Case 2**: $z = f(x, y)$, $x = g(s, t)$, $y = h(s, t)$ and compute $\\frac{\\partial z}{\\partial s}$ and $\\frac{\\partial z}{\\partial t}$.\n\nIn this case if we were to substitute in for $x$ and $y$ we would get that $z$ is a function of $s$ and $t$ and so it makes sense that we would be computing partial derivatives here and that there would be two of them.\n\nHere is the chain rule for both of these cases.\nSo, not surprisingly, these are very similar to the first case that we looked at. Here is a quick example of this kind of chain rule.\n\n**Example 3** Find \\( \\frac{\\partial z}{\\partial s} \\) and \\( \\frac{\\partial z}{\\partial t} \\) for \\( z = e^{2r} \\sin(3\\theta) \\), \\( r = st - t^2 \\), \\( \\theta = \\sqrt{s^2 + t^2} \\).\n\n**Solution**\n\nHere is the chain rule for \\( \\frac{\\partial z}{\\partial s} \\).\n\n\\[\n\\frac{\\partial z}{\\partial s} = \\left( 2e^{2r} \\sin(3\\theta) \\right) (t) + \\left( 3e^{2r} \\cos(3\\theta) \\right) \\frac{s}{\\sqrt{s^2 + t^2}}\n\\]\n\n\\[\n= t \\left( 2e^{2(st-t^2)} \\sin(3\\sqrt{s^2 + t^2}) \\right) + \\frac{3se^{2(st-t^2)} \\cos(3\\sqrt{s^2 + t^2})}{\\sqrt{s^2 + t^2}}\n\\]\n\nNow the chain rule for \\( \\frac{\\partial z}{\\partial t} \\).\n\n\\[\n\\frac{\\partial z}{\\partial t} = \\left( 2e^{2r} \\sin(3\\theta) \\right) (s - 2t) + \\left( 3e^{2r} \\cos(3\\theta) \\right) \\frac{t}{\\sqrt{s^2 + t^2}}\n\\]\n\n\\[\n= (s - 2t) \\left( 2e^{2(st-t^2)} \\sin(3\\sqrt{s^2 + t^2}) \\right) + \\frac{3se^{2(st-t^2)} \\cos(3\\sqrt{s^2 + t^2})}{\\sqrt{s^2 + t^2}}\n\\]\n\nOkay, now that we’ve seen a couple of cases for the chain rule let’s see the general version of the chain rule.\n\n**Chain Rule**\n\nSuppose that \\( z \\) is a function of \\( n \\) variables, \\( x_1, x_2, \\ldots, x_n \\), and that each of these variables are in turn functions of \\( m \\) variables, \\( t_1, t_2, \\ldots, t_m \\). Then for any variable \\( t_i \\), \\( i = 1, 2, \\ldots, m \\) we have the following,\n\n\\[\n\\frac{\\partial z}{\\partial t_i} = \\frac{\\partial z}{\\partial x_1} \\frac{\\partial x_1}{\\partial t_i} + \\frac{\\partial z}{\\partial x_2} \\frac{\\partial x_2}{\\partial t_i} + \\cdots + \\frac{\\partial z}{\\partial x_n} \\frac{\\partial x_n}{\\partial t_i}\n\\]\n\nWow. That’s a lot to remember. There is actually an easier way to construct all the chain rules that we’ve discussed in the section or will look at in later examples. We can build up a tree diagram that will give us the chain rule for any situation. To see how these work let’s go back and take a look at the chain rule for \\( \\frac{\\partial z}{\\partial s} \\) given that \\( z = f(x, y) \\), \\( x = g(s, t) \\), \\( y = h(s, t) \\). We already know what this is, but it may help to illustrate the tree diagram if we already know the answer. For reference here is the chain rule for this case,\n\\[\n\\frac{\\partial z}{\\partial s} = \\frac{\\partial f}{\\partial x} \\frac{\\partial x}{\\partial s} + \\frac{\\partial f}{\\partial y} \\frac{\\partial y}{\\partial s}\n\\]\n\nHere is the tree diagram for this case.\n\nWe start at the top with the function itself and the branch out from that point. The first set of branches is for the variables in the function. From each of these endpoints we put down a further set of branches that gives the variables that both \\(x\\) and \\(y\\) are a function of. We connect each letter with a line and each line represents a partial derivative as shown. Note that the letter in the numerator of the partial derivative is the upper “node” of the tree and the letter in the denominator of the partial derivative is the lower “node” of the tree.\n\nTo use this to get the chain rule we start at the bottom and for each branch that ends with the variable we want to take the derivative with respect to (\\(s\\) in this case) we move up the tree until we hit the top multiplying the derivatives that we see along that set of branches. Once we’ve done this for each branch that ends at \\(s\\), we then add the results up to get the chain rule for that given situation.\n\nNote that we don’t always put the derivatives in the tree. Some of the trees get a little large/messy and so we won’t put in the derivatives. Just remember what derivative should be on each branch and you’ll be okay without actually writing them down.\n\nLet’s write down some chain rules.\n\n**Example 4** Use a tree diagram to write down the chain rule for the given derivatives.\n\n(a) \\(\\frac{dw}{dt}\\) for \\(w = f(x, y, z)\\), \\(x = g_1(t)\\), \\(y = g_2(t)\\), and \\(z = g_3(t)\\)\n\n(b) \\(\\frac{\\partial w}{\\partial r}\\) for \\(w = f(x, y, z)\\), \\(x = g_1(s, t, r)\\), \\(y = g_2(s, t, r)\\), and \\(z = g_3(s, t, r)\\)\n\n**Solution**\n\n(a) \\(\\frac{dw}{dt}\\) for \\(w = f(x, y, z)\\), \\(x = g_1(t)\\), \\(y = g_2(t)\\), and \\(z = g_3(t)\\)\n\nSo, we’ll first need the tree diagram so let’s get that.\nFrom this it looks like the chain rule for this case should be,\n\\[\n\\frac{dw}{dt} = \\frac{\\partial f}{\\partial x} \\frac{dx}{dt} + \\frac{\\partial f}{\\partial y} \\frac{dy}{dt} + \\frac{\\partial f}{\\partial z} \\frac{dz}{dt}\n\\]\nwhich is really just a natural extension to the two variable case that we saw above.\n\n(b) \\( \\frac{\\partial w}{\\partial r} \\) for \\( w = f(x, y, z) \\), \\( x = g_1(s, t, r) \\), \\( y = g_2(s, t, r) \\), and \\( z = g_3(s, t, r) \\)\n\nHere is the tree diagram for this situation.\n\nFrom this it looks like the derivative will be,\n\\[\n\\frac{\\partial w}{\\partial r} = \\frac{\\partial f}{\\partial x} \\frac{\\partial x}{\\partial r} + \\frac{\\partial f}{\\partial y} \\frac{\\partial y}{\\partial r} + \\frac{\\partial f}{\\partial z} \\frac{\\partial z}{\\partial r}\n\\]\n\nSo, provided we can write down the tree diagram, and these aren’t usually too bad to write down, we can do the chain rule for any set up that we might run across.\n\nWe’ve now seen how to take first derivatives of these more complicated situations, but what about higher order derivatives? How do we do those? It’s probably easiest to see how to deal with these with an example.\n\n**Example 5** Compute \\( \\frac{\\partial^2 f}{\\partial \\theta^2} \\) for \\( f(x, y) \\) if \\( x = r \\cos \\theta \\) and \\( y = r \\sin \\theta \\).\n\n**Solution**\nWe will need the first derivative before we can even think about finding the second derivative so let’s get that. This situation falls into the second case that we looked at above so we don’t need a new tree diagram. Here is the first derivative.\n\\[\n\\frac{\\partial f}{\\partial \\theta} = \\frac{\\partial f}{\\partial x} \\frac{\\partial x}{\\partial \\theta} + \\frac{\\partial f}{\\partial y} \\frac{\\partial y}{\\partial \\theta}\n\\]\n\\[\n= -r \\sin(\\theta) \\frac{\\partial f}{\\partial x} + r \\cos(\\theta) \\frac{\\partial f}{\\partial y}\n\\]\n\nOkay, now we know that the second derivative is,\n\\[\n\\frac{\\partial^2 f}{\\partial \\theta^2} = \\frac{\\partial}{\\partial \\theta} \\left( \\frac{\\partial f}{\\partial \\theta} \\right) = \\frac{\\partial}{\\partial \\theta} \\left( -r \\sin(\\theta) \\frac{\\partial f}{\\partial x} + r \\cos(\\theta) \\frac{\\partial f}{\\partial y} \\right)\n\\]\n\nThe issue here is to correctly deal with this derivative. Since the two first order derivatives, \\( \\frac{\\partial f}{\\partial x} \\) and \\( \\frac{\\partial f}{\\partial y} \\), are both functions of \\( x \\) and \\( y \\) which are in turn functions of \\( r \\) and \\( \\theta \\) both of these terms are products. So, the using the product rule gives the following,\n\\[\n\\frac{\\partial^2 f}{\\partial \\theta^2} = -r \\cos(\\theta) \\frac{\\partial f}{\\partial x} - r \\sin(\\theta) \\frac{\\partial}{\\partial \\theta} \\left( \\frac{\\partial f}{\\partial x} \\right) - r \\sin(\\theta) \\frac{\\partial f}{\\partial y} + r \\cos(\\theta) \\frac{\\partial}{\\partial \\theta} \\left( \\frac{\\partial f}{\\partial y} \\right)\n\\]\n\nWe now need to determine what \\( \\frac{\\partial}{\\partial \\theta} \\left( \\frac{\\partial f}{\\partial x} \\right) \\) and \\( \\frac{\\partial}{\\partial \\theta} \\left( \\frac{\\partial f}{\\partial y} \\right) \\) will be. These are both chain rule problems again since both of the derivatives are functions of \\( x \\) and \\( y \\) and we want to take the derivative with respect to \\( \\theta \\).\n\nBefore we do these let’s rewrite the first chain rule that we did above a little.\n\\[\n\\frac{\\partial}{\\partial \\theta} (f) = -r \\sin(\\theta) \\frac{\\partial}{\\partial x} (f) + r \\cos(\\theta) \\frac{\\partial}{\\partial y} (f)\n\\]\n\nNote that all we’ve done is change the notation for the derivative a little. With the first chain rule written in this way we can think of (1) as a formula for differentiating any function of \\( x \\) and \\( y \\) with respect to \\( \\theta \\) provided we have \\( x = r \\cos \\theta \\) and \\( y = r \\sin \\theta \\).\n\nThis however is exactly what we need to do the two new derivatives we need above. Both of the first order partial derivatives, \\( \\frac{\\partial f}{\\partial x} \\) and \\( \\frac{\\partial f}{\\partial y} \\), are functions of \\( x \\) and \\( y \\) and \\( x = r \\cos \\theta \\) and \\( y = r \\sin \\theta \\) so we can use (1) to compute these derivatives.\n\nTo do this we’ll simply replace all the \\( f \\)’s in (1) with the first order partial derivative that we want to differentiate. At that point all we need to do is a little notational work and we’ll get the formula that we’re after.\nHere is the use of (1) to compute \\( \\frac{\\partial}{\\partial \\theta} \\left( \\frac{\\partial f}{\\partial x} \\right) \\).\n\n\\[\n\\frac{\\partial}{\\partial \\theta} \\left( \\frac{\\partial f}{\\partial x} \\right) = -r \\sin(\\theta) \\frac{\\partial}{\\partial x} \\left( \\frac{\\partial f}{\\partial x} \\right) + r \\cos(\\theta) \\frac{\\partial}{\\partial y} \\left( \\frac{\\partial f}{\\partial x} \\right)\n\\]\n\n\\[\n= -r \\sin(\\theta) \\frac{\\partial^2 f}{\\partial x^2} + r \\cos(\\theta) \\frac{\\partial^2 f}{\\partial y \\partial x}\n\\]\n\nHere is the computation for \\( \\frac{\\partial}{\\partial \\theta} \\left( \\frac{\\partial f}{\\partial y} \\right) \\).\n\n\\[\n\\frac{\\partial}{\\partial \\theta} \\left( \\frac{\\partial f}{\\partial y} \\right) = -r \\sin(\\theta) \\frac{\\partial}{\\partial x} \\left( \\frac{\\partial f}{\\partial y} \\right) + r \\cos(\\theta) \\frac{\\partial}{\\partial y} \\left( \\frac{\\partial f}{\\partial y} \\right)\n\\]\n\n\\[\n= -r \\sin(\\theta) \\frac{\\partial^2 f}{\\partial x \\partial y} + r \\cos(\\theta) \\frac{\\partial^2 f}{\\partial y^2}\n\\]\n\nThe final step is to plug these back into the second derivative and do some simplifying.\n\n\\[\n\\frac{\\partial^2 f}{\\partial \\theta^2} = -r \\cos(\\theta) \\frac{\\partial f}{\\partial x} - r \\sin(\\theta) \\left( -r \\sin(\\theta) \\frac{\\partial^2 f}{\\partial x^2} + r \\cos(\\theta) \\frac{\\partial^2 f}{\\partial y \\partial x} \\right) -\n\\]\n\n\\[\nr \\sin(\\theta) \\frac{\\partial f}{\\partial y} + r \\cos(\\theta) \\left( -r \\sin(\\theta) \\frac{\\partial^2 f}{\\partial x \\partial y} + r \\cos(\\theta) \\frac{\\partial^2 f}{\\partial y^2} \\right)\n\\]\n\n\\[\n= -r \\cos(\\theta) \\frac{\\partial f}{\\partial x} + r^2 \\sin^2(\\theta) \\frac{\\partial^2 f}{\\partial x^2} - r^2 \\sin(\\theta) \\cos(\\theta) \\frac{\\partial^2 f}{\\partial y \\partial x} -\n\\]\n\n\\[\nr \\sin(\\theta) \\frac{\\partial f}{\\partial y} - r^2 \\sin(\\theta) \\cos(\\theta) \\frac{\\partial^2 f}{\\partial x \\partial y} + r^2 \\cos^2(\\theta) \\frac{\\partial^2 f}{\\partial y^2}\n\\]\n\n\\[\n= -r \\cos(\\theta) \\frac{\\partial f}{\\partial x} - r \\sin(\\theta) \\frac{\\partial f}{\\partial y} + r^2 \\sin^2(\\theta) \\frac{\\partial^2 f}{\\partial x^2} -\n\\]\n\n\\[\n2r^2 \\sin(\\theta) \\cos(\\theta) \\frac{\\partial^2 f}{\\partial y \\partial x} + r^2 \\cos^2(\\theta) \\frac{\\partial^2 f}{\\partial y^2}\n\\]\n\nIt’s long and fairly messy but there it is.\n\nThe final topic in this section is a revisiting of implicit differentiation. With these forms of the chain rule implicit differentiation actually becomes a fairly simple process. Let’s start out with the implicit differentiation that we saw in a Calculus I course.\n\nWe will start with a function in the form \\( F(x, y) = 0 \\) (if it’s not in this form simply move everything to one side of the equal sign to get it into this form) where \\( y = y(x) \\). In a Calculus I course we were then\nasked to compute \\( \\frac{dy}{dx} \\) and this was often a fairly messy process. Using the chain rule from this section however we can get a nice simple formula for doing this. We’ll start by differentiating both sides with respect to \\( x \\). This will mean using the chain rule on the left side and the right side will, of course, differentiate to zero. Here are the results of that.\n\n\\[\nF_x + F_y \\frac{dy}{dx} = 0 \\quad \\Rightarrow \\quad \\frac{dy}{dx} = -\\frac{F_x}{F_y}\n\\]\n\nAs shown, all we need to do next is solve for \\( \\frac{dy}{dx} \\) and we’ve now got a very nice formula to use for implicit differentiation. Note as well that in order to simplify the formula we switched back to using the subscript notation for the derivatives.\n\nLet’s check out a quick example.\n\n**Example 6** Find \\( \\frac{dy}{dx} \\) for \\( x \\cos(3y) + x^3y^5 = 3x - e^{xy} \\).\n\n**Solution**\n\nThe first step is to get a zero on one side of the equal sign and that’s easy enough to do.\n\n\\[\nx \\cos(3y) + x^3y^5 - 3x + e^{xy} = 0\n\\]\n\nNow, the function on the left is \\( F(x, y) \\) in our formula so all we need to do is use the formula to find the derivative.\n\n\\[\n\\frac{dy}{dx} = -\\frac{\\cos(3y) + 3x^2y^5 - 3 + ye^{xy}}{-3x\\sin(3y) + 5x^3y^4 + xe^{xy}}\n\\]\n\nThere we go. It would have taken much longer to do this using the old Calculus I way of doing this.\n\nWe can also do something similar to handle the types of implicit differentiation problems involving partial derivatives like those we saw when we first introduced partial derivatives. In these cases we will start off with a function in the form \\( F(x, y, z) = 0 \\) and assume that \\( z = f(x, y) \\) and we want to find \\( \\frac{\\partial z}{\\partial x} \\) and/or \\( \\frac{\\partial z}{\\partial y} \\).\n\nLet’s start by trying to find \\( \\frac{\\partial z}{\\partial x} \\). We will differentiate both sides with respect to \\( x \\) and we’ll need to remember that we’re going to be treating \\( y \\) as a constant. Also, the left side will require the chain rule. Here is this derivative.\n\n\\[\n\\frac{\\partial F}{\\partial x} \\frac{\\partial x}{\\partial x} + \\frac{\\partial F}{\\partial y} \\frac{\\partial y}{\\partial x} + \\frac{\\partial F}{\\partial z} \\frac{\\partial z}{\\partial x} = 0\n\\]\nNow, we have the following,\n\\[\n\\frac{\\partial x}{\\partial x} = 1 \\quad \\text{and} \\quad \\frac{\\partial y}{\\partial x} = 0\n\\]\nThe first is because we are just differentiating \\( x \\) with respect to \\( x \\) and we know that is 1. The second is because we are treating the \\( y \\) as a constant and so it will differentiate to zero.\n\nPlugging these in and solving for \\( \\frac{\\partial z}{\\partial x} \\) gives,\n\\[\n\\frac{\\partial z}{\\partial x} = \\frac{F_x}{F_z}\n\\]\nA similar argument can be used to show that,\n\\[\n\\frac{\\partial z}{\\partial y} = \\frac{F_y}{F_z}\n\\]\nAs with the one variable case we switched to the subscripting notation for derivatives to simplify the formulas. Let’s take a quick look at an example of this.\n\n**Example 7** Find \\( \\frac{\\partial z}{\\partial x} \\) and \\( \\frac{\\partial z}{\\partial y} \\) for \\( x^2 \\sin(2y - 5z) = 1 + y \\cos(6zx) \\).\n\n**Solution**\nThis was one of the functions that we used the old implicit differentiation on back in the Partial Derivatives section. You might want to go back and see the difference between the two.\n\nFirst let’s get everything on one side.\n\\[\nx^2 \\sin(2y - 5z) - 1 - y \\cos(6zx) = 0\n\\]\nNow, the function on the left is \\( F(x, y, z) \\) and so all that we need to do is use the formulas developed above to find the derivatives.\n\\[\n\\frac{\\partial z}{\\partial x} = -\\frac{2x \\sin(2y - 5z) + 6yz \\sin(6zx)}{-5x^2 \\cos(2y - 5z) + 6yx \\sin(6zx)}\n\\]\n\\[\n\\frac{\\partial z}{\\partial y} = -\\frac{2x^2 \\cos(2y - 5z) - \\cos(6zx)}{-5x^2 \\cos(2y - 5z) + 6yx \\sin(6zx)}\n\\]\nIf you go back and compare these answers to those that we found the first time around you will notice that they might appear to be different. However, if you take into account the minus sign that sits in the front of our answers here you will see that they are in fact the same.\n", "id": "./materials/221.pdf" }, { "contents": "Higher order partial derivatives\n\nWhat you need to know already:\n\n- What partial derivatives are and how to compute them.\n\nWhat you can learn here:\n\n- The notation and some basic important facts about higher order partial derivatives.\n\nWe have seen that a partial derivative is just a regular derivative, but computed on a two-variable function by considering the other variable as constant. That means that we can consider higher derivatives with respect to one of the variables, just like we did for usual functions. All we need is to add is a minor change of notation to point out that we are dealing with a partial derivative.\n\n**Definition**\n\nIf \\( z = f(x, y) \\) is a two-variable function that is differentiable twice with respect to \\( x \\), then its second partial derivative for \\( x \\) is denoted by:\n\n\\[\n\\frac{\\partial^2 f(x, y)}{\\partial x^2} = f_{xx} = z_{xx}\n\\]\n\nSimilarly, the second partial derivative with respect to \\( y \\), if it exists, is denoted by:\n\n\\[\n\\frac{\\partial^2 f(x, y)}{\\partial y^2} = f_{yy} = z_{yy}\n\\]\n\n**Example:** \\( z = y^2 \\ln x \\)\n\nHere are the two partial derivatives of this function:\n\n\\[\nz_x = y^2 \\frac{1}{x} = \\frac{y^2}{x} \\quad ; \\quad z_y = 2y \\ln x\n\\]\n\nTherefore, the second partial derivatives with respect to \\( x \\) and \\( y \\) are, respectively:\n\n\\[\nz_{xx} = \\frac{\\partial}{\\partial x} \\left( \\frac{y^2}{x} \\right) = -\\frac{y^2}{x^2} \\quad ; \\quad z_{yy} = \\frac{\\partial}{\\partial y} (2y \\ln x) = 2 \\ln x\n\\]\n\nBut these partial derivatives are still functions of two variables, so do we have to keep differentiating them always with respect to the same variable?\n\nClearly there is no need for that! There is no reason why we cannot alternate variable and differentiate first with respect to one variable, then to the other. Again, all we need to do is be careful about the notation.\n**Definition**\n\nThe second mixed partial derivatives of a two-variable function \\( z = f(x, y) \\) are denoted by:\n\n\\[\n\\frac{\\partial^2 f(x, y)}{\\partial x \\partial y} = \\frac{\\partial}{\\partial x} \\left( \\frac{\\partial}{\\partial y} f(x, y) \\right) = f_{xy} = z_{xy}\n\\]\n\n\\[\n\\frac{\\partial^2 f(x, y)}{\\partial y \\partial x} = \\frac{\\partial}{\\partial y} \\left( \\frac{\\partial}{\\partial x} f(x, y) \\right) = f_{yx} = z_{yx}\n\\]\n\n**Warning bells**\n\nNotice and remember that the order of differentiation is indicated in the subscript from right to left, just as in the composition of functions, so that, for instance:\n\n\\[\nf_{xy} = \\frac{\\partial}{\\partial x} \\left( \\frac{\\partial}{\\partial y} f(x, y) \\right)\n\\]\n\n**Example:** \\( z = y^2 \\ln x \\)\n\nWe have seen earlier that here \\( z_x = \\frac{y^2}{x} \\) and \\( z_y = 2y \\ln x \\).\n\nTherefore, the second mixed partial derivatives are, respectively:\n\n\\[\nz_{xx} = \\frac{\\partial z_x}{\\partial y} = \\frac{\\partial}{\\partial y} \\left( \\frac{y^2}{x} \\right) = \\frac{2y}{x} \\quad ; \\quad z_{xy} = \\frac{\\partial z_x}{\\partial x} = \\frac{\\partial}{\\partial x} \\left( 2y \\ln x \\right) = \\frac{2y}{x}\n\\]\n\n**Technical fact**\n\nIf \\( z = f(x, y) \\) is a two-variable function, whose second mixed partial derivatives are continuous, then they are equal, regardless of the order of differentiation:\n\n\\[\nf_{xy} = f_{yx}\n\\]\n\nThis means that the second mixed partials are different only in very rare cases, in exceptions that must be constructed with some care and viciousness! For our purposes, we shall just ignore them. Moreover, I will omit the proof of this fact, since it consists of an uninspiring sequence of technical steps and checks. Instead, here is another example, after which you can try to compute these mixed higher derivatives by yourself and check that they are equal.\n\n**Example:** \\( f(x, y) = \\sin x^2 y \\)\n\nWe start from the first partials:\n\n\\[\nf_x = 2xy \\cos x^2 y \\quad ; \\quad f_y = x^2 \\cos x^2 y\n\\]\n\nThen we compute the two pure partials:\n\n\\[\nf_{xx} = (2xy)_x \\cos x^2 y + 2xy \\left( \\cos x^2 y \\right)_x = 2y \\cos x^2 y - 4x^2 y^2 \\sin x^2 y\n\\]\n\n\\[\nf_{yy} = x^2 \\left( \\cos x^2 y \\right)_y = -x^4 \\sin x^2 y\n\\]\n\nFinally we compute the two mixed partials and check that they are the same:\n\\[ f_{yx} = \\frac{\\partial}{\\partial y} (2xy) \\cos x^2 y + 2xy \\frac{\\partial}{\\partial y} (\\cos x^2 y) \\]\n\\[ = 2x \\cos x^2 y - 2x^2 y \\sin x^2 y \\]\n\\[ f_{xy} = \\frac{\\partial}{\\partial x} \\cos x^2 y + x^2 \\frac{\\partial}{\\partial x} (\\cos x^2 y) \\]\n\\[ = 2x \\cos x^2 y - 2x^2 y \\sin x^2 y \\]\n\nThey are indeed equal \\( \\odot \\).\n\n**What about even higher order derivatives?**\n\nYou are ambitious today! Well, they are done in the obvious way.\n\n**Knot on your finger**\n\n**Higher partial derivatives** can be computed to any order, provided the required derivatives exist.\n\nThe order of differentiation in the notation for a higher mixed partial is indicated from right to left, but it does not matter, provided such mixed partial is continuous.\n\n**Example:** \\( f(x, y) = \\sin x^2 y \\)\n\nLet us go one step further with this function and compute \\( f_{xy} \\):\n\n\\[ f_{xy} = \\frac{\\partial}{\\partial x} \\left( \\frac{\\partial}{\\partial y} \\sin x^2 y \\right) = \\frac{\\partial}{\\partial x} (-x^2 \\sin x^2 y) \\]\n\\[ = -4x^3 \\sin x^2 y - 2yx^2 \\cos x^2 y \\]\n\nBut we can also compute \\( f_{yx} \\) and we should get the same function, right?\n\nLet’s see:\n\n\\[ f_{yx} = \\frac{\\partial^2}{\\partial y \\partial x} (2xy \\cos x^2 y) = \\frac{\\partial}{\\partial y} (2x \\cos x^2 y - 2x^2 y \\sin x^2 y) \\]\n\\[ = -2x^2 \\sin x^2 y - 2x^3 \\sin x^2 y - 2yx^2 \\cos x^2 y \\]\n\\[ = -4x^3 \\sin x^2 y - 2yx^2 \\cos x^2 y \\]\n\nYes, as expected. I leave you the fun of checking that the third possible mixed partial \\( f_{xy} \\) is also the same.\n\n**Warning bells**\n\nThe order of differentiation for mixed partial derivatives does not matter, but only provided that:\n\n- such mixed partial is continuous AND\n- the number of times each variable is used as the variable of differentiation is the same.\n\nTherefore, while we can assume that \\( f_{xy} = f_{yx} = f_{yx} \\) in general it is NOT TRUE that \\( f_{xy} = f_{yx} \\) since we are not using each variable the same number of times in the two cases.\n\n*Well, that makes sense! If we take different derivatives, we cannot expect the same conclusion.*\n\nObviously I agree, but I have seen students make mistakes in this way, so I hope you will not join them.\nSummary\n\n- Higher order partial derivatives can be computed just as for usual derivatives.\n- Higher partial derivatives may be computed with respect to a single variable, or changing variable at each successive step, so as to obtain a mixed partial derivative.\n- If a mixed partial derivative is continuous, the order in which the variables were used in the computation is irrelevant, as long as it is done the required number of times for each variable.\n\nCommon errors to avoid\n\n- Keep track of the algebra and of which variable you are using at each step. Otherwise, you may end up with a derivative different from what you want.\n\nLearning questions for Section D 6-3\n\nReview questions:\n\n1. Describe the notation used for higher order partial derivatives.\n2. Explain how the order of differentiation in mixed partial derivatives is indicated in its notation.\n\nMemory questions:\n\n1. What is the correct Leibniz notation for all second derivatives of a two-variable function \\( z = f(x, y) \\)?\n2. What is the correct short notation for all second derivatives of a two-variable function \\( z = f(x, y) \\)?\n3. In which order are the variables indicated in a mixed partial derivative?\n4. How do we call a higher derivative of a two-variable function for which each variable is used at some stage as the variable of differentiation?\n5. When are the mixed partial derivatives of a function of two variables independent of the order of differentiation?\nComputation questions:\n\nCompute all second partial derivatives for the functions provided in questions 1-6.\n\n1. \\( f(x, y) = y^2 \\).\n2. \\( z = x^2 - xy + y^2 \\).\n3. \\( z = \\frac{x}{y} + e^x \\sin y \\).\n4. \\( z = f(x, y) = x^2 - \\frac{x}{y} + y^2 \\).\n5. \\( z = \\frac{x^4}{2y} + e^{2x} \\sin y \\).\n6. \\( u(x, y) = x^3 y^2 + 4x^2 \\cos y \\).\n\n7. Compute \\( f_{xx} \\) for \\( f(x, y) = \\frac{x^2 y}{y - \\sinh y} \\).\n8. Compute \\( f_{xy} \\) for the function defined by \\( f(x, y) = \\ln xy - 2\\cos(x - y) + xy^2 \\).\n9. Determine \\( \\frac{\\partial^3 z}{\\partial x \\partial y \\partial x} \\) for \\( z = \\frac{\\cos y}{3} - x^2 e^{\\cos y} \\). Be smart in your use of properties of partial derivatives.\n10. Given the function \\( f(x, y) = e^x \\cos y \\), determine the formula for \\( \\left( \\frac{\\partial f}{\\partial x} + \\frac{\\partial f}{\\partial y} \\right) \\left( \\frac{\\partial^2 f}{\\partial x \\partial y} \\right) \\).\n11. Given the function \\( f(x, y) = e^x \\sinh y \\), determine the formula for \\( \\left( \\frac{\\partial f}{\\partial x} + \\frac{\\partial f}{\\partial y} \\right)^2 - \\frac{\\partial^2 f}{\\partial x \\partial y} \\).\n\nTheory questions:\n\n1. Is it always correct to switch the order of differentiation for mixed partial derivatives?\n2. By visual inspection only, what is \\( \\frac{\\partial^6 \\left( x^2 y^2 \\right)}{\\partial x^4 \\partial y^2} \\)?\n\nTemplated questions:\n1. Construct a reasonably simple two-variable function and compute all its second partial derivatives.\n\nWhat questions do you have for your instructor?", "id": "./materials/222.pdf" }, { "contents": "Finding the analytical expression of linear transformation from the image vectors of a basis of the starting space\n\nConsider the basis \\( S = \\{(1, 4), (-2, 1)\\} \\) for \\( \\mathbb{R}^2 \\).\n\nLet \\( T: \\mathbb{R}^2 \\to \\mathbb{R}^3 \\) be the linear transformation for which\n\\[\nT(1, 4) = (4, -1, 1) \\quad \\text{and} \\quad T(-2, 1) = (0, -2, 3)\n\\]\n\n(a) Determine \\( T(-13, -7) \\).\n(b) Find a formula for \\( T(x, y) \\).\n\nNotice that\n\nLet \\( T: U \\to V \\) be a linear transformation, where \\( U \\) is finite-dimensional.\nIf \\( S = \\{u_1, u_2, \\ldots, u_n\\} \\) is a basis for \\( U \\) and \\( u = c_1u_1 + c_2u_2 + \\cdots + c_nu_n \\), for \\( u \\in U, c_1, c_2, \\ldots, c_n \\in \\mathbb{R} \\), then\n\\[\nT(u) = c_1T(u_1) + c_2T(u_2) + \\cdots + c_nT(u_n)\n\\]\n\n(a) Attend to\n\\[\n(-13, -7) = -3(1,4) + 5(-2,1)\n\\]\nwe have\n\\[\nT(-13, -7) = -3T(1,4) + 5T(-2,1)\n\\]\n\\[\n= -3(4, -1,1) + 5(0, -2,3)\n\\]\n\\[\n= (-12,3, -3) + (0, -10,15)\n\\]\n\\[\n= (-12, -7,12)\n\\]\nThus, \\( T(-13, -7) = (-12, -7,12) \\).\n\n(b) We must begin to find the coordinates of \\((x, y)\\) on \\( S \\) basis, this is \\( \\alpha \\) and \\( \\beta \\).\n\\[\n(x, y) = \\alpha(1,4) + \\beta(-2,1)\n\\]\n\\[\n\\begin{align*}\n\\alpha - 2\\beta &= x \\\\\n4\\alpha + \\beta &= y\n\\end{align*}\n\\]\n\\[\n\\begin{align*}\n\\alpha &= x + 2\\beta \\\\\n4\\alpha + \\beta &= y \\\\\n4(x + 2\\beta) + \\beta &= y\n\\end{align*}\n\\]\n\\[\n\\begin{align*}\n\\begin{cases}\n\\alpha = x + 2\\beta \\\\\n4x + 9\\beta = y\n\\end{cases} & \\iff \\\\\n\\begin{cases}\n\\alpha = x + 2\\beta \\\\\n\\beta = \\frac{y - 4x}{9}\n\\end{cases} & \\iff \\\\\n\\begin{cases}\n\\alpha = x + \\frac{2y - 8x}{9} \\\\\n\\beta = \\frac{y - 4x}{9}\n\\end{cases}\n\\end{align*}\n\\]\n\nSo\n\n\\[\n(x, y) = \\frac{2y + x}{9} (1,4) + \\frac{y - 4x}{9} (-2,1)\n\\]\n\nAnd\n\n\\[\nT(x, y) = \\frac{2y + x}{9} T(1,4) + \\frac{y - 4x}{9} T(-2,1)\n\\]\n\nThis is,\n\n\\[\nT(x, y) = \\frac{2y + x}{9} (4, -1,1) + \\frac{y - 4x}{9} (0, -2,3)\n\\]\n\n\\[\n= \\left( \\frac{8y + 4x}{9}, \\frac{-2y - x - 2y + 8x}{9}, \\frac{2y + x + 3y - 12x}{9} \\right)\n\\]\n\n\\[\n= \\left( \\frac{4x + 8y}{9}, \\frac{7x - 4y}{9}, \\frac{-11x + 5y}{9} \\right)\n\\]\n\nThus,\n\n\\[\nT(x, y) = \\left( \\frac{4x + 8y}{9}, \\frac{7x - 4y}{9}, \\frac{-11x + 5y}{9} \\right)\n\\]\n\nSuggestion: Determine \\( T(-13, -7) \\) using that formula.", "id": "./materials/223.pdf" }, { "contents": "Finding the analytical expression of linear transformation from the image vectors of a basis of the starting space\n\nConsider the basis $S = \\{(1, 4), (-2, 1)\\}$ for $\\mathbb{R}^2$.\n\nLet $T: \\mathbb{R}^2 \\to \\mathbb{R}^3$ be the linear transformation for which\n\n$T(1, 4) = (4, -1, 1)$ and $T(-2, 1) = (0, -2, 3)$\n\n(a) Determine $T(-13, -7)$.\n\n(b) Find a formula for $T(x, y)$.\n\nNotice that\n\nLet $T: U \\to V$ be a linear transformation, where $U$ is finite-dimensional.\n\nIf $S = \\{u_1, u_2, ..., u_n\\}$ is a basis for $U$ and $u = c_1u_1 + c_2u_2 + \\cdots + c_nu_n$, for $u \\in U$, $c_1, c_2, ..., c_n \\in \\mathbb{R}$, then\n\n$$T(u) = c_1T(u_1) + c_2T(u_2) + \\cdots + c_nT(u_n)$$\n\n(a) Attend to\n\n$$(-13, -7) = -3(1,4) + 5(-2,1)$$\n\nwe have\n\n$$T(-13, -7) = -3T(1,4) + 5T(-2,1)$$\n\n$$= -3(4, -1,1) + 5(0, -2,3)$$\n\n$$= (-12,3, -3) + (0, -10,15)$$\n\n$$= (-12, -7,12)$$\n\nThus, $T(-13, -7) = (-12, -7,12)$.\n\n(b) We must begin to find the coordinates of $(x, y)$ on $S$ basis, this is $\\alpha$ and $\\beta$.\n\n$$(x, y) = \\alpha(1,4) + \\beta(-2,1)$$\n\n$$\\begin{cases} \n\\alpha - 2\\beta = x \\\\\n4\\alpha + \\beta = y\n\\end{cases} \\iff \\begin{cases} \n\\alpha = x + 2\\beta \\\\\n4\\alpha + \\beta = y\n\\end{cases} \\iff \\begin{cases} \n\\alpha = x + 2\\beta \\\\\n4(x + 2\\beta) + \\beta = y\n\\end{cases}$$\n\\[\n\\begin{align*}\n\\begin{cases}\n\\alpha = x + 2\\beta \\\\\n4x + 9\\beta = y\n\\end{cases} & \\iff \\\\\n\\begin{cases}\n\\alpha = x + 2\\beta \\\\\n\\beta = \\frac{y - 4x}{9}\n\\end{cases} & \\iff \\\\\n\\begin{cases}\n\\alpha = x + \\frac{2y - 8x}{9} \\\\\n\\beta = \\frac{y - 4x}{9}\n\\end{cases}\n\\end{align*}\n\\]\n\nSo\n\n\\[\n(x, y) = \\frac{2y + x}{9} (1,4) + \\frac{y - 4x}{9} (-2,1)\n\\]\n\nAnd\n\n\\[\nT(x, y) = \\frac{2y + x}{9} T(1,4) + \\frac{y - 4x}{9} T(-2,1)\n\\]\n\nThis is,\n\n\\[\nT(x, y) = \\frac{2y + x}{9} (4, -1,1) + \\frac{y - 4x}{9} (0, -2,3)\n\\]\n\n\\[\n= \\left( \\frac{8y + 4x}{9}, \\frac{-2y - x - 2y + 8x}{9}, \\frac{2y + x + 3y - 12x}{9} \\right)\n\\]\n\n\\[\n= \\left( \\frac{4x + 8y}{9}, \\frac{7x - 4y}{9}, \\frac{-11x + 5y}{9} \\right)\n\\]\n\nThus,\n\n\\[\nT(x, y) = \\left( \\frac{4x + 8y}{9}, \\frac{7x - 4y}{9}, \\frac{-11x + 5y}{9} \\right)\n\\]\n\nSuggestion: Determine \\( T(-13, -7) \\) using that formula.", "id": "./materials/224.pdf" }, { "contents": "**Dimension theorem**\n\n**Theorem:** Let $T: U \\rightarrow V$ be a linear transformation between the vector spaces of finite dimension $U$ and $V$, then:\n\n$$\\dim(U) = \\dim(\\ker(T)) + \\dim(\\text{range}(T))$$\n\n1. Let $T: \\mathbb{R}^4 \\rightarrow \\mathbb{R}^6$ be a linear transformation.\n\n a) Knowing that $\\dim(\\ker(T)) = 2$, determine the dimension of the $\\text{range}(T)$.\n\n Applying the dimension theorem:\n\n $$\\dim(\\mathbb{R}^4) = \\dim(\\ker(T)) + \\dim(\\text{range}(T))$$\n\n Then,\n\n $$4 = 2 + \\dim(\\text{range}(T)) \\iff \\dim(\\text{range}(T)) = 2$$\n\n b) Knowing that $\\dim(\\text{range}(T)) = 3$, determine the dimension of the $\\ker(T)$.\n\n Applying again the dimension theorem:\n\n $$\\dim(\\mathbb{R}^4) = \\dim(\\ker(T)) + \\dim(\\text{range}(T))$$\n\n Then,\n\n $$4 = \\dim(\\ker(T)) + 3 \\iff \\dim(\\ker(T)) = 1$$\n\n2. Verify the dimension theorem for the linear transformation $T: \\mathbb{R}^2 \\rightarrow \\mathbb{R}^3$ defined by $T(x, y) = (x, x + y, y)$.\n\n Let us first determine the kernel of the transformation $T$. By definition we have:\n\n $$\\ker(T) = \\{(x, y) \\in \\mathbb{R}^2 : T(x, y) = (0, 0, 0)\\}$$\n\n Then,\n\n $$T(x, y) = (0, 0, 0) \\iff (x, x + y, y) = (0, 0, 0)$$\n\n $$\\iff \\begin{cases} x = 0 \\\\ x + y = 0 \\\\ y = 0 \\end{cases} \\iff \\begin{cases} x = 0 \\\\ 0 = 0 \\\\ y = 0 \\end{cases}$$\n\n Therefore,\n\n $$\\ker(T) = \\{(0, 0)\\}$$\n\n As the $\\ker(T) = \\{(0, 0)\\}$, then the $\\dim(\\ker(T)) = 0$. \nLet us now determine the range of the transformation $T$:\n\n$$\\text{range}(T) = \\{(a, b, c) \\in \\mathbb{R}^3 : T(x, y) = (a, b, c) \\text{ with } (x, y) \\in \\mathbb{R}^2\\}$$\n\nWe have:\n\n$$T(x, y) = (a, b, c) \\iff (x, x + y, y) = (a, b, c) \\iff \\begin{cases} x = a \\\\ x + y = b \\\\ y = c \\end{cases}$$\n\nThe matrix of the system is:\n\n$$\\begin{bmatrix} 1 & 0 & a \\\\ 1 & 1 & b \\\\ 0 & 1 & c \\end{bmatrix} \\rightarrow \\begin{bmatrix} 1 & 0 & a \\\\ 0 & 1 & -a + b \\\\ 0 & 1 & c \\end{bmatrix} \\rightarrow \\begin{bmatrix} 1 & 0 & a \\\\ 0 & 1 & -a + b \\\\ 0 & 0 & a - b + c \\end{bmatrix}$$\n\nFor the system to be possible:\n\n$$a - b + c = 0 \\iff c = b - a$$\n\nThen,\n\n$$\\text{range}(T) = \\{(a, b, c) \\in \\mathbb{R}^3 : c = b - a\\}$$\n\nLet’s determine a basis for the range($T$):\n\n$$(a, b, b - a) = (a, 0, -a) + (0, b, b) = a(1, 0, -1) + b(0, 1, 1)$$\n\nThe vectors $(1, 0, -1)$ and $(0, 1, 1)$ generate the range($T$). So, let’s verify if they are linearly independent:\n\n$$c_1(1, 0, -1) + c_2(0, 1, 1) = (0, 0, 0)$$\n\n$$\\iff \\begin{cases} c_1 = 0 \\\\ c_2 = 0 \\\\ -c_1 + c_2 = 0 \\end{cases} \\iff \\begin{cases} c_1 = 0 \\\\ c_2 = 0 \\end{cases}$$\n\nTherefore, the vectors are linearly independent.\n\nThus, the set formed by the vectors $(1, 0, -1)$ and $(0, 1, 1)$ is a basis for the range($T$) and the $\\text{dim}(\\text{range}(T)) = 2$.\n\nLet’s verify the dimension theorem for this linear transformation:\n\n$$\\text{dim}(\\mathbb{R}^2) = \\text{dim}(\\ker(T)) + \\text{dim}(\\text{range}(T))$$\n\nThus,\n\n$$2 = 0 + 2$$", "id": "./materials/225.pdf" }, { "contents": "Mediatrix line and mediating plane\n\nMediatrix line of a line segment in $\\mathbb{R}^2$\n\nGiven two points $A, B \\in \\mathbb{R}^2$, the locus of points that are equidistant from $A$ and $B$ is a line called mediatrix (or the perpendicular bissector) of $[AB]$.\n\n**Note:** The mediatrix $m$ of $[AB]$ checks the following:\n\n- is orthogonal to $[AB]$;\n- contains the midpoint of $[AB]$.\n\n**Example:** Considere the points $A = (-2, 1)$ and $B = (-1, 0)$. The set of the points $P = (x, y)$ equidistant from $A$ and $B$, are such that:\n\n$$AP = BP \\iff \\sqrt{(x + 2)^2 + (y - 1)^2} = \\sqrt{(x + 1)^2 + (y - 1)^2} \\iff 4x + 4 - 2y + 1 = 2x + 1 \\iff x - y + 2 = 0.$$ \n\nThus we have the equation of the mediatrix $m : x - y + 2 = 0$, whose director vector $\\vec{v} = (1, 1)$ is perpendicular to $\\overrightarrow{AB} = B - A = (1, -1)$. De facto, $\\overrightarrow{AB} \\cdot \\vec{v} = 0$.\n\nNotice also that $M = \\left(\\frac{-1 - 2}{2}, \\frac{1}{2}\\right) \\in m$, because $\\frac{-3}{2} - \\frac{1}{2} + 2 = 0$.\n\nMediating plane of a line segment in $\\mathbb{R}^3$\n\nGiven two points $A, B \\in \\mathbb{R}^3$, the locus of points that are equidistant from $A$ and $B$ is a plane, called mediating plane (or the perpendicular bissector) of $[AB]$.\n\n**Note:** The mediating plane $m$ of $[AB]$ checks the following:\n\n- is orthogonal to $[AB]$;\n- contains the midpoint of $[AB]$.\n\n**Example:** Considere the points $A = (2, 3, 1)$ and $B = (-1, 1, 0)$ of $\\mathbb{R}^3$. The set of the points $P = (x, y, z)$ equidistant from $A$ and $B$, are such that $\\overrightarrow{AP} = \\overrightarrow{BP} \\iff$\n\n$$\\sqrt{(x - 2)^2 + (y - 3)^2 + (z - 1)^2} = \\sqrt{(x + 1)^2 + (y - 1)^2 + z^2} \\iff -4x + 4 - 6y + 9 - 2z + 1 = 2x + 1 - 2y + 1.$$ \n\nThus, we have the equation of the mediating plane $\\pi : -6x - 4y - 2z + 12 = 0$, is orthogonal to the vector $\\vec{v} = (-6, -4, -2)$, which is collinear with $\\overrightarrow{AB} = B - A = (-3, -2, -1)$. ", "id": "./materials/228.pdf" }, { "contents": "Cross product and related properties\n\nVector Product (Cross product)\n\nThe vector product of two vectors \\( u \\) and \\( v \\) is a vector \\( u \\times v \\) that is at right angles to both and is defined by\n\n\\[\n u \\times v = ||u|| ||v|| \\sin(\\hat{uv}) n, \\quad \\text{with} \\quad ||n|| = 1 \\quad \\text{and} \\quad u, v \\perp n.\n\\]\n\nSpecifically,\n\n1. \\( u \\times v \\) is perpendicular to the vectors \\( u \\) and \\( v \\);\n2. \\( ||u \\times v|| = ||u|| \\cdot ||v|| \\sin(\\hat{uv}) \\);\n3. \\( u \\times v \\) has sense determined by the right hand (follow with the fingers of the right hand, the rotation movement of the vector \\( u \\) to approach \\( v \\) and consider the direction of the thumb).\n\nNotice that:\n\n- \\( u \\times v \\) is orthogonal to the plane containing the vectors;\n- \\( u \\times v = 0 \\) when vectors \\( u \\) and \\( v \\) point in the same, or opposite, direction.\n\nIn the 3-dimensional Cartesian system, the vector product of vectors \\( u = (u_1, u_2, u_3) \\) and \\( v = (v_1, v_2, v_3) \\) is defined as\n\n\\[\n u \\times v = (u_2v_3 - v_2u_3, v_1u_3 - u_1v_3, u_1v_2 - v_1u_2).\n\\]\n\nIt is a vector perpendicular to the vectors \\( u \\) and \\( v \\) and can more easily be represented matrix-wise as:\n\n\\[\n u \\times v = \\begin{vmatrix} i & j & k \\\\ u_1 & u_2 & u_3 \\\\ v_1 & v_2 & v_3 \\end{vmatrix} = (u_2v_3 - v_2u_3)i - (u_1v_3 - v_1u_3)j + (u_1v_2 - v_1u_2)k.\n\\]\n\nExample: \\((1, 2, -1) \\times (2, 0, 1) = \\begin{vmatrix} i & j & k \\\\ 1 & 2 & -1 \\\\ 2 & 0 & 1 \\end{vmatrix} = 2i - 3j - 4k = (2, -3, -4)\\)\nRegarding the previous example, note that \\((2, -3, -4) \\cdot (1, 2, -1) = 2 - 6 + 4 = 0\\) and \\((2, -3, -4) \\cdot (2, 0, 1) = 4 - 4 = 0\\).\n\n**Properties:** Be the vectors \\(u, v, w \\in \\mathbb{R}^3\\). We have\n\n1. \\(u \\times v \\times w = u \\times (v \\times w)\\) (associative);\n2. \\(u \\times v = -v \\times u\\) (anti-commutative);\n3. \\(u \\times v = 0 \\iff u = 0 \\lor v = 0 \\lor \\hat{uv} = 0^\\circ \\lor \\hat{uv} = 180^\\circ\\).\n\n**Example:** \\((1, -2, 3) \\times (-2, 4, -6) = \\begin{vmatrix} i & j & k \\\\ 1 & -2 & 3 \\\\ -2 & 4 & -6 \\end{vmatrix} = (0, 0, 0)\\), because the vectors \\((1, -2, 3)\\) e \\((-2, 4, -6)\\) are collinear.\n\nThe norm of the vector product \\(||u \\times v|| = ||u|| \\cdot ||v|| \\cdot \\sin(\\hat{uv})||\\) the area of the parallelogram determined by \\(u\\) and \\(v\\).\n\nIn effect, according to the figure above, the area of the parallelogram is given by \\(A = ||v|| \\cdot h\\). Besides that, \\(||u|| \\cdot \\sin(\\hat{uv}) = h\\).", "id": "./materials/229.pdf" }, { "contents": "Scalar product of two vectors and orthogonal projection of one vector over another\n\nLet’s talk about scalar product of two vectors and its application for the calculation of the orthogonal projection of one vector over another.\n\nScalar product of two vectors and related properties\n\n**Definition:** The scalar product (or dot product) $u \\cdot v$ of two vectors $u$ and $v$ is a number defined by\n\n$$||u|| ||v|| \\cos(\\theta),$$\n\nwith $\\theta = \\hat{uv} \\in [0, \\pi]$.\n\n- If $u$ and $v$ are parallel vectors, then $u \\cdot v = ||u|| ||v||$;\n- If $u$ and $v$ are antiparallel vectors, then $u \\cdot v = -||u|| ||v||$;\n- If $u$ and $v$ are two orthogonal vectors, then $u \\cdot v = 0$.\n\nNotice that $\\hat{vv} = 0$ and this implies that the dot product of a vector $a$ with itself is $a \\cdot a = ||a|| ||a||$, which gives\n\n$$||a|| = \\sqrt{a \\cdot a}.$$\n\nIn $\\mathbb{R}^n$ we have the alternative definition of scalar product:\n\n**Definition:** The dot product of two vectors $v = (v_1, v_2, \\ldots, v_n), u = (u_1, u_2, \\ldots, u_n) \\in \\mathbb{R}^n$ is\n\n$$v \\cdot u = v_1 u_1 + v_2 u_2 + v_3 u_3 + \\cdots + v_n u_n.$$\n\n**Example:** On the Cartesian plane, consider the vectors $i = (1, 0), j = (0, 1)$ and $v(-1, 1)$.\n\nOn the one hand, we have $i \\cdot j = (1, 0) \\cdot (0, 1) = 0$ and $j \\cdot v = (0, 1) \\cdot (-1, 1) = 1 + 0 = 1$.\n\nOn the other hand, we also have\n\n$$i \\cdot j = ||(1, 0)|| ||(0, 1)|| \\cos\\left(\\frac{\\pi}{2}\\right) = 0$$\n\nAlso\n\n$$j \\cdot v = ||(0, 1)|| ||(-1, 1)|| \\cos\\left(\\frac{\\pi}{2}\\right) = \\sqrt{2} \\times \\left(\\frac{\\sqrt{2}}{2}\\right) = 1.$$\n\\[ i \\cdot v = ||(1, 0)|| ||(-1, 1)|| \\cos \\left( \\frac{3\\pi}{2} \\right) = \\sqrt{2} \\times \\left( -\\frac{\\sqrt{2}}{2} \\right) = -1. \\]\n\nThe dot product fulfills the following properties if \\( u, v, \\) and \\( w \\) are vectors and \\( k \\) is a real scalar:\n\n1. \\( v \\cdot u = u \\cdot v; \\)\n2. \\( v \\cdot (u + w) = (v \\cdot u) + (v \\cdot w); \\)\n3. \\( v \\cdot (ku + w) = k(v \\cdot u) + (v \\cdot w); \\)\n4. \\( k_1 v \\cdot (k_2 u) = k_1 k_2 (v \\cdot u). \\)\n\n**Examples:** Consider \\( u = (-1, 2, 3) \\) and \\( v = (2, 0, -1). \\) We have:\n\n- \\( u \\cdot 3v = 3(-1, 2, 3) \\cdot (2, 0, -1) = 3(-2 - 3) = -10. \\)\n- \\( (u + v) \\cdot (u - v) = u \\cdot u - u \\cdot v + v \\cdot u - v \\cdot v = ||u||^2 + ||v||^2 = 14 + 5. \\)\n\nAn inner product is a generalization of the dot product. Is any operator who checks the properties above.\n\n**Orthogonal projection of one vector over another**\n\nOne important use of dot products is in projections.\n\nThe orthogonal projection of \\( u \\) onto \\( v \\) is the length of the segment \\([AD]\\) shown in the figure beside, \\( \\overrightarrow{AD}. \\)\n\nThe **vector projection** of \\( u \\) onto \\( v \\) is the vector \\( \\overrightarrow{AD} \\)\n\nNote that \\( |\\text{proj}_v u| = ||v|| \\cos(\\theta) \\) and therefore:\n\n\\[ |\\text{proj}_v u| = \\overrightarrow{AD} = \\frac{|u \\cdot v|}{||v||} \\quad \\text{and} \\quad \\text{proj}_v u = \\overrightarrow{AD} = \\frac{|u \\cdot v|}{||v||^2} v. \\]\n\n**Exemplo:** The orthogonal projection of \\( u = (-2, 1) \\) onto \\( v = (-3, -1) \\) is the vector:\n\n\\[ \\text{proj}_v u = \\frac{|(-2, 1) \\cdot (-3, -1)|}{10} (-3, -1) = \\frac{5}{10} (-3, -1). \\]\n\nWe can also decompose \\( u \\) as the sum of two vectors \\( w_1, w_2, \\) such that \\( w_1 \\parallel u \\) and \\( w_2 \\perp u. \\) In fact,\n\n\\[ w_1 = \\text{proj}_v u = \\left( -\\frac{3}{2}, -\\frac{1}{2} \\right) \\]\n\nand\n\n\\[ w_2 = u - w_1 = (-2, 1) - \\left( -\\frac{3}{2}, -\\frac{1}{2} \\right) = \\left( -\\frac{1}{2}, \\frac{3}{2} \\right). \\]", "id": "./materials/230.pdf" }, { "contents": "Distance from a point to a line and distance between two parallel lines\n\nDistance from a point to a line\n\nThe distance from a point \\( A \\) to a line \\( r \\) is equal to the distance from \\( A \\) to its orthogonal projection \\( A' \\) on the line \\( r \\), according to the figure beside.\n\nWe calculate the distance \\( d \\) by doing:\n\n1. Determine the line \\( PP' \\) that is perpendicular to \\( r \\) containing \\( P \\);\n2. Determine \\( P' = PP' \\cap r \\);\n3. Determine \\( d = PP' \\).\n\nExample:\nConsider in \\( \\mathbb{R}^3 \\), \\( P = (2, 1, 1) \\) and \\( r : (x, y, z) = (0, 0, -1) + k(1, -1, 1), \\ k \\in \\mathbb{R} \\). Let us determine the distance from \\( P \\) to \\( r \\).\n\nFor example, \\( u = (1, 2, 1) \\) is orthogonal to \\( v = (1, -1, 1) \\) because \\( u \\cdot v = 0 \\).\n\nThen, \\( PP' : (x, y, z) = (2, 1, 1) + t(1, 2, 1), t \\in \\mathbb{R} \\).\n\nBesides that \\( P' = (x, y, z) = PP' \\cap r \\) is such that\n\n\\[\n\\begin{align*}\n x &= \\frac{y}{-1} = z + 1 \\\\\n x - 2 &= \\frac{y - 1}{2} = z - 1\n\\end{align*}\n\\]\n\n\\( \\iff \\)\n\n\\[\n\\begin{align*}\n x &= 1 \\\\\n y &= -1 \\\\\n z &= 0\n\\end{align*}\n\\]\n\nThat is \\( P' = (1, -1, 0) \\).\n\nFinally \\( d = PP' = \\sqrt{(2 - 1)^2 + (1 + 1)^2 + (1 - 0)^2} = \\sqrt{6} \\).\n\nDistance between two parallel lines\n\nThe distance between two parallel lines \\( r \\) and \\( s \\) is equal to the distance between a point \\( P \\) of the line \\( r \\) and its orthogonal projection on \\( s \\).\n\n- \\( P' \\) is the orthogonal projection of \\( P \\in r \\) on \\( s \\);\n- \\( d \\) is the distance of \\( P \\) to \\( P' \\).\n\nExample: The lines \\( r : 2x - y + 2 = 0 \\) and \\( s : -4x + 2y + 1 = 0 \\) are parallel, both have the direction of \\( v = (1, 2) \\). The distance from \\( r \\) to \\( s \\) is equal to the distance from a point \\( P \\in r \\) to \\( s \\).\n\nConsider \\( P = (-1, 0) \\in r \\) and \\( u = (2, -1) \\) orthogonal to \\( v = (1, 2) \\).\nThen the line $t : y - 0 = -\\frac{1}{2}(x + 1)$ contains $P$ and is perpendicular to the lines $r$ and $s$. Thus\n\n$$P'(x, y) = t \\cap s : \\begin{cases} y = -\\frac{1}{2}(x + 1) \\\\ -4x + 2y + 1 = 0 \\end{cases} \\iff \\begin{cases} y = -\\frac{1}{2}(x + 1) \\\\ 4x - x - 1 + 1 = 0 \\end{cases} \\iff \\begin{cases} y = \\frac{1}{2} \\\\ x = 0 \\end{cases}.$$ \n\nFinally, the distance from $r$ to $s$ is equal to the distance from $P = (-1, 0)$ to $P'(0, \\frac{1}{2})$,\n\n$$d(r, s) = d(P, P') = \\sqrt{(-1 + 0)^2 + (0 - \\frac{1}{2})^2} = \\frac{\\sqrt{5}}{2}.$$", "id": "./materials/231.pdf" }, { "contents": "Distance from a point to a plane, from a line to a parallel plane or between two parallel planes\n\nDistance from a point to a plane:\n\nWe can determine the distance from point \\( P \\) to plane \\( p \\), by performing:\n\n- Calculate the line \\( r \\) that contains the point \\( P \\) and is normal to the plane \\( p \\);\n- Calculate \\( P' = r \\cap p \\);\n- Determine the distance from \\( P \\) to \\( P' \\).\n\n**Example:** To calculate the distance from \\( P = (1, 2, -1) \\) to the plane \\( p : x - y + z = 0 \\), we can take \\( n = (2, -2, 1) \\perp p \\) and the line \\( r \\) that contains \\( P \\) and is normal to the plane \\( p \\) is \\( r : \\frac{x - 1}{2} = \\frac{y - 2}{-2} = z + 1 \\).\n\n\\[\nP' = r \\cap p = \\begin{cases} x - 1 = -y + 2 \\\\ -y + 2 = z + 1 \\\\ x - y + z = 1 \\end{cases} \\iff \\begin{cases} x = 2 \\\\ y = 0 \\\\ z = 1 \\end{cases}\n\\]\n\nThen, \\( d(P, p) = d(P, P') = \\sqrt{(1 - 2)^2 + (2 - 0)^2 + (-1 - 1)^2} = 3 \\)\n\nDistance from a straight line to a parallel plane:\n\nGiven a line \\( r \\) parallel to a plane \\( p \\), the distance \\( d \\) from the line \\( r \\) is the distance from any point \\( p \\) on the line to the plane, that is,\n\n\\[\nd(r, p) = d(P, p)\n\\]\n\n**Example:** To calculate the distance from \\( r : \\frac{x - 1}{2} = -y = \\frac{z + 1}{-3} \\) to the plane \\( p : x - y + z = 0 \\) is the distance from \\( P = (1, 0, -1) \\in r \\) to the plane \\( p \\).\nDistance between two parallel planes:\n\nTo calculate the distance between two planes $\\alpha$ and $\\beta$ parallel to each other, we can perform:\n\n- Calculate the line $r$ that is normal to the planes $\\alpha$ and $\\beta$;\n- Calculate $A = r \\cap \\alpha$;\n- Calculate $B = r \\cap \\beta$;\n- Determine the distance from $A$ to $B$.\n\n**Example:** To calculate the distance from $p_1 : x - 2y + z + 6 = 0$ to the plane $p_2 : 2x - 4y + 2z + 6 = 0$, we consider:\n\n- $r : x = -\\frac{y}{2} = z$ that is normal $p_1$ and $p_2$;\n- $A = r \\cap p_1 : \\begin{cases} x = -\\frac{y}{2} \\\\ x = z \\end{cases} \\iff \\begin{cases} y = -2x \\\\ z = x \\end{cases} \\iff \\begin{cases} y = \\frac{1}{2} \\\\ z = -1 \\end{cases}$;\n- $B = r \\cap p_2 : \\begin{cases} x = -\\frac{y}{2} \\\\ x = z \\end{cases} \\iff \\begin{cases} y = -2x \\\\ z = x \\end{cases} \\iff \\begin{cases} y = \\frac{1}{2} \\\\ z = -\\frac{1}{2} \\end{cases}$.\n\nThen, $A = (-1, \\frac{1}{2}, -1), B = (-\\frac{1}{2}, 1, -\\frac{1}{2})$ and\n\n$$d(p_1, p_2) = AB = \\sqrt{(-\\frac{1}{2} + 1)^2 + (1 - \\frac{1}{2})^2 + (-\\frac{1}{2} + 1)^2} = \\frac{\\sqrt{3}}{2}.$$", "id": "./materials/232.pdf" }, { "contents": "Is \\( \\{\\emptyset\\} \\) an empty set?\n\nRoshan Poudel\n\nInstituto Politécnico de Bragança, Bragança, Portugal\nA collection of objects that somehow share a common feature - the elements - is called a set. An element can be of any nature, depending on the problem under consideration, such as numbers, functions, or lines. A set can be finite or infinite.\n\n**Example**\n\n1. \\( A = \\{1, 3, 5, 7, 9\\} \\) is an example of a finite set.\n2. \\( \\mathbb{Z} \\), the set of the integers, is an example of a finite set.\n3. The elements of a set are not only limited to numbers, the elements of a set can be anything, \\( B = \\{\\text{cow, donkey, rat, horse}\\} \\) is also a set.\n4. Sets can also be written in set builder notation: \\( A = \\{x \\in \\mathbb{N} \\mid x \\geq 4 \\text{ and } x \\leq 10\\} \\) which is same as \\( A = \\{4, 5, 6, 7, 8, 9, 10\\} \\)\nProperties of Set\n\nProperties\n\n1. The order of the elements in a set doesn’t matter.\n2. If one or more elements of a set are repeated, the set remains the same.\n For example \\{1, 2, 3, 1, 2, 3, 1, 2, 3\\} is the same as just \\{1, 2, 3\\}.\n3. Two sets are considered equal if and only if each element of one set is an element of the other.\n\nSymbol \\(\\in\\) is used to denote that an element belongs to a set. For example: \\(X = \\{a, e, i, o, u\\}\\) Then, \\(a \\in X\\) but \\(b \\notin X\\) or \\(\\{a\\} \\notin X\\).\nSome important Sets\n\n| Symbol | Name |\n|--------|-------------------------------------------|\n| $\\mathbb{Z}$ | The set of integers. |\n| $\\mathbb{N}$ | The set of natural numbers. |\n| $\\mathbb{Q}$ | The set of rational numbers. |\n| $\\mathbb{R}$ | The set of real numbers. |\n| $\\mathbb{C}$ | The set of complex numbers. |\n\nThe empty set is a set without any elements, represented by $\\{\\}$ or $\\emptyset$.\n\nA set with only one element is called a singleton set. For example $X = \\{a\\}$. \nSo, Is \\( \\{\\emptyset\\} \\) an empty set?\n\n**NO**, \\( \\{\\emptyset\\} \\) is not an empty set; it is a singleton set (it has the element \\( \\emptyset \\) in it). Empty set is indicated by \\( \\{\\} \\) or \\( \\emptyset \\).", "id": "./materials/365.pdf" }, { "contents": "Determine $|\\mathcal{P}(\\mathcal{P}(\\{\\phi, \\tau\\}))|$. \n\nRoshan Poudel\n\nInstituto Politécnico de Bragança, Bragança, Portugal\nSubset\n\nSet $A$ is a subset of set $B$ iff each element of set $A$ is also an element of set $B$. If set $A$ is a subset of set $B$ then we write as $A \\subset B$.\n\n1. If each element of set $A$ is also an element of set $B$ and $B$ may be equal to $A$, then set $A$ is an **improper subset** of set $B$.\n\n **For example:** $A = \\{1, 2, 3, 4, 5\\}$ and $B = \\{1, 2, 3, 4, 5\\}$ then $A \\subseteq B$ and $B \\subseteq A$.\n\n2. If each element of set $A$ is also element of set $B$ but set $B$ is not equal to set $A$ then Set $A$ is **proper subset** of set $B$.\n\n **For example:** $A = \\{2, 3, 4\\}$ and $B = \\{1, 2, 3, 4, 5\\}$ then $A \\subset B$ but $A \\not\\subset B$.\n## Properties of Subset\n\n| | |\n|---|---|\n| 1 | A set with \\( n \\) elements has \\( 2^n \\) subsets. |\n| 2 | Every set is subset of itself. |\n| 3 | Empty set (\\( \\emptyset \\)) is subset of every set. |\n| 4 | \\( A = B \\) if and only if \\( A \\subseteq B \\) and \\( B \\subseteq A \\). |\n| 5 | A is a subset of \\( B \\) if and only if their intersection is equal to \\( A \\), that is, \\( A \\subseteq B \\iff (A \\cap B) = A \\). |\n| 6 | Set \\( A \\) is a subset of \\( B \\) if and only if their union is equal to \\( B \\), that is, \\( A \\subseteq B \\iff (A \\cup B) = B \\). |\nWhat are the subsets of set $A = \\{x, y, z\\}$?\n\n- $\\emptyset$\n- $\\{x\\}$\n- $\\{y\\}$\n- $\\{z\\}$\n- $\\{x, y\\}$\n- $\\{x, z\\}$\n- $\\{y, z\\}$\n- $\\{x, y, z\\}$\n\nNotice, there are 8 subsets of set $A$ which is also the result of $2^{|A|} = 2^3 = 8$\nA set $A$ is a superset of another set $B$ if all elements of the set $B$ are elements of the set $A$. The notation for superset is $A \\supset B$.\n\n**Properties**\n\n- $A \\supset \\emptyset$.\n- Since every set is a subset of itself, then every set is also a superset of itself.\nThe set of all subsets of a set $A$ is called the power set of $A$. The power set of $A$ is denoted with the symbol $\\mathcal{P}(A)$.\n\n**Example**\n\nIf $A$ is the set $\\{1, 2, 3\\}$, then what is $\\mathcal{P}(A)$?\n\n$$\\mathcal{P}(A) = \\{\\emptyset, \\{1\\}, \\{2\\}, \\{3\\}, \\{1, 2\\}, \\{1, 3\\}, \\{2, 3\\}, \\{1, 2, 3\\}\\}$$\nAs we know, for any set $A$, $|\\mathcal{P}(A)| = 2^{|A|}$.\n\nIn this case,\n\n$$|\\{\\phi, \\tau\\}| = 2$$\n\nTherefore,\n\n$$|\\mathcal{P}(\\{\\phi, \\tau\\})| = 2^2 = 4$$\n\n$$|\\mathcal{P}(\\mathcal{P}(\\{\\phi, \\tau\\}))| = 2^4 = 16$$\n\nSo, $|\\mathcal{P}(\\mathcal{P}(\\{\\phi, \\tau\\}))| = 16$. ", "id": "./materials/366.pdf" }, { "contents": "Objectives\n\nAfter studying this chapter you should\n\n- understand how quadratic equations lead to complex numbers and how to plot complex numbers on an Argand diagram;\n- be able to relate graphs of polynomials to complex numbers;\n- be able to do basic arithmetic operations on complex numbers of the form \\( a + ib \\);\n- understand the polar form \\([r, \\theta]\\) of a complex number and its algebra;\n- understand Euler's relation and the exponential form of a complex number \\(re^{i\\theta}\\);\n- be able to use de Moivre's theorem;\n- be able to interpret relationships of complex numbers as loci in the complex plane.\n\n3.0 Introduction\n\nThe history of complex numbers goes back to the ancient Greeks who decided (but were perplexed) that no number existed that satisfies\n\n\\[ x^2 = -1 \\]\n\nFor example, Diophantus (about 275 AD) attempted to solve what seems a reasonable problem, namely\n\n'Find the sides of a right-angled triangle of perimeter 12 units and area 7 squared units.'\n\nLetting \\( AB = x, AC = h \\) as shown,\n\nthen \\( \\text{area} = \\frac{1}{2} x h \\)\n\nand \\( \\text{perimeter} = x + h + \\sqrt{x^2 + h^2} \\)\nChapter 3 Complex Numbers\n\nActivity 1\n\nShow that the two equations above reduce to\n\n\\[ 6x^2 - 43x + 84 = 0 \\]\n\nwhen perimeter = 12 and area = 7. Does this have real solutions?\n\nA similar problem was posed by Cardan in 1545. He tried to solve the problem of finding two numbers, a and b, whose sum is 10 and whose product is 40;\n\ni.e. \\[ a + b = 10 \\] (1)\n\\[ ab = 40 \\] (2)\n\nEliminating b gives\n\n\\[ a(10 - a) = 40 \\]\n\nor \\[ a^2 - 10a + 40 = 0. \\]\n\nSolving this quadratic gives\n\n\\[ a = \\frac{1}{2} (10 \\pm \\sqrt{-60}) = 5 \\pm \\sqrt{-15} \\]\n\nThis shows that there are no real solutions, but if it is agreed to continue using the numbers\n\n\\[ a = 5 + \\sqrt{-15}, \\quad b = 5 - \\sqrt{-15} \\]\n\nthen equations (1) and (2) are satisfied.\n\nShow that equations (1) and (2) are satisfied by these values of x and y.\n\nSo these are solutions of the original problem but they are not real numbers. Surprisingly, it was not until the nineteenth century that such solutions were fully understood.\n\nThe square root of \\(-1\\) is denoted by i, so that\n\n\\[ i = \\sqrt{-1} \\]\n\nand \\[ a = 5 + \\sqrt{15}i, \\quad b = 5 - \\sqrt{15}i \\]\n\nare examples of complex numbers.\nActivity 2 The need for complex numbers\n\nSolve if possible, the following quadratic equations by factorising or by using the quadratic formula. If a solution is not possible explain why.\n\n(a) \\( x^2 - 1 = 0 \\) \n(b) \\( x^2 - x - 6 = 0 \\) \n(c) \\( x^2 - 2x - 2 = 0 \\) \n(d) \\( x^2 - 2x + 2 = 0 \\)\n\nYou should have found (a), (b) and (c) straightforward to solve but in (d) a term appears in the solution which includes the square root of a negative number and to obtain solutions you need to use the symbol \\( i = \\sqrt{-1} \\), or\n\n\\[\ni^2 = -1\n\\]\n\nIt is then possible to obtain a solution to (d) in Activity 2.\n\nExample\n\nSolve \\( x^2 - 2x + 2 = 0 \\).\n\nSolution\n\nUsing the quadratic formula\n\n\\[\nx = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}\n\\]\n\n\\[\n\\Rightarrow x = \\frac{-(-2) \\pm \\sqrt{(-2)^2 - 4(1)(2)}}{2(1)}\n\\]\n\n\\[\n\\Rightarrow x = \\frac{2 \\pm \\sqrt{-4}}{2}\n\\]\n\nBut \\( \\sqrt{-4} = \\sqrt{4(-1)} = \\sqrt{4} \\sqrt{-1} = 2\\sqrt{-1} = 2i \\) \n(\\text{using the definition of } i).\n\nTherefore \\( x = \\frac{2 \\pm 2i}{2} \\)\n\n\\[\n\\Rightarrow x = 1 \\pm i\n\\]\n\nTherefore the two solutions are \\( x = 1 + i \\) and \\( x = 1 - i \\).\nChapter 3 Complex Numbers\n\nActivity 3\n\nSolve the following equations, leaving your answers in terms of $i$:\n\n(a) $x^2 + x + 1 = 0$\n(b) $3x^2 - 4x + 2 = 0$\n(c) $x^2 + 1 = 0$\n(d) $2x - 7 = 4x^2$\n\nThe set of solutions to a quadratic equation such as\n\n$$ax^2 + bx + c = 0$$\n\ncan be related to the intercepts on the $x$-axis when the graph of the function\n\n$$f(x) = ax^2 + bx + c$$\n\nis drawn.\n\nActivity 4 Quadratic graphs\n\nUsing a graphics calculator, a graph drawing program on a computer, a spreadsheet or otherwise, draw the graphs of the following functions and find a connection between the existence or not of real solutions to the related quadratic equations.\n\n(a) $f(x) = x^2 - 1$\n(b) $f(x) = x^2 - x - 6$\n(c) $f(x) = x^2 - 2x - 2$\n(d) $f(x) = x^2 + x + 1$\n(e) $f(x) = 3x^2 - 4x + 2$\n(f) $f(x) = x^2 + 1$\n\nYou should have noted that if the graph of the function either intercepts the $x$-axis in two places or touches it in one place then the solutions of the related quadratic equation are real, but if the graph does not intercept the $x$-axis then the solutions are complex.\n\nIf the quadratic equation is expressed as $ax^2 + bx + c = 0$, then the expression that determines the type of solution is $b^2 - 4ac$, called the discriminant.\n\nIn a quadratic equation $ax^2 + bx + c = 0$, if:\n\n- $b^2 - 4ac > 0$ then solutions are real and different\n- $b^2 - 4ac = 0$ then solutions are real and equal\n- $b^2 - 4ac < 0$ then solutions are complex\n3.1 Complex number algebra\n\nA number such as $3 + 4i$ is called a complex number. It is the sum of two terms (each of which may be zero).\n\nThe real term (not containing $i$) is called the real part and the coefficient of $i$ is the imaginary part. Therefore the real part of $3 + 4i$ is 3 and the imaginary part is 4.\n\nA number is real when the coefficient of $i$ is zero and is imaginary when the real part is zero.\n\ne.g. $3 + 0i = 3$ is real and $0 + 4i = 4i$ is imaginary.\n\nHaving introduced a complex number, the ways in which they can be combined, i.e. addition, multiplication, division etc., need to be defined. This is termed the algebra of complex numbers. You will see that, in general, you proceed as in real numbers, but using $i^2 = -1$ where appropriate.\n\nBut first equality of complex numbers must be defined.\n\nIf two complex numbers, say\n\n$$a + bi, \\ c + di$$\n\nare equal, then both their real and imaginary parts are equal;\n\n$$a + bi = c + di \\Rightarrow a = c \\text{ and } b = d$$\n\nAddition and subtraction\n\nAddition of complex numbers is defined by separately adding real and imaginary parts; so if\n\n$$z = a + bi, \\ w = c + di$$\n\nthen\n\n$$z + w = (a + c) + (b + d)i.$$ \n\nSimilarly for subtraction.\n\nExample\n\nExpress each of the following in the form $x + yi$.\n\n(a) $(3 + 5i) + (2 - 3i)$\n(b) $(3 + 5i) + 6$\n(c) $7i - (4 + 5i)$\nChapter 3 Complex Numbers\n\nSolution\n(a) \\((3 + 5i) + (2 - 3i) = 3 + 2 + (5 - 3)i = 5 + 2i\\)\n(b) \\((3 + 5i) + 6 = 9 + 5i\\)\n(c) \\(7i - (4 + 5i) = 7i - 4 - 5i = -4 + 2i\\)\n\nMultiplication\n\nMultiplication is straightforward provided you remember that \\(i^2 = -1\\).\n\nExample\nSimplify in the form \\(x + yi:\\)\n(a) \\(3(2 + 4i)\\)\n(b) \\((5 + 3i)i\\)\n(c) \\((2 - 7i)(3 + 4i)\\)\n\nSolution\n(a) \\(3(2 + 4i) = 3(2) + 3(4i) = 6 + 12i\\)\n(b) \\((5 + 3i)i = (5)i + (3i)i = 5i + 3i^2 = 5i + (-1)3 = -3 + 5i\\)\n(c) \\((2 - 7i)(3 + 4i) = (2)(3) - (7i)(3) + (2)(4i) - (7i)(4i)\\)\n\\[= 6 - 21i + 8i - (-28)\\]\n\\[= 6 - 21i + 8i + 28\\]\n\\[= 34 - 13i\\]\n\nIn general, if \\(z = a + bi, \\ w = c + di,\\)\nthen \\(zw = (a + bi)(c + di)\\)\n\\[= ac - bd + (ad + bc)i\\]\nActivity 5\n\nSimplify the following expressions:\n\n(a) \\((2 + 6i) + (9 - 2i)\\) \n(b) \\((8 - 3i) - (1 + 5i)\\)\n\n(c) \\(3(7 - 3i) + i(2 + 2i)\\) \n(d) \\((3 + 5i)(1 - 4i)\\)\n\n(e) \\((5 + 12i)(6 + 7i)\\) \n(f) \\((2 + i)^2\\)\n\n(g) \\(i^3\\) \n(h) \\(i^4\\)\n\n(i) \\((1 - i)^3\\) \n(j) \\((1 + i)^2 + (1 - i)^2\\)\n\n(k) \\((2 + i)^4 + (2 - i)^4\\) \n(l) \\((a + ib)(a - ib)\\)\n\nDivision\n\nThe complex conjugate of a complex number is obtained by changing the sign of the imaginary part. So if \\(z = a + bi\\), its complex conjugate, \\(\\bar{z}\\), is defined by\n\n\\[ \\bar{z} = a - bi \\]\n\nAny complex number \\(a + bi\\) has a complex conjugate \\(a - bi\\) and from Activity 5 it can be seen that \\((a + bi)(a - bi)\\) is a real number. This fact is used in simplifying expressions where the denominator of a quotient is complex.\n\nExample\n\nSimplify the expressions:\n\n(a) \\(\\frac{1}{i}\\) \n(b) \\(\\frac{3}{1 + i}\\) \n(c) \\(\\frac{4 + 7i}{2 + 5i}\\)\n\nSolution\n\nTo simplify these expressions you multiply the numerator and denominator of the quotient by the complex conjugate of the denominator.\n\n(a) The complex conjugate of \\(i\\) is \\(-i\\), therefore\n\n\\[ \\frac{1}{i} = \\frac{1}{i} \\times \\frac{-i}{-i} = \\frac{(1)(-i)}{(i)(-i)} = \\frac{-i}{-1} = i \\]\n\n(b) The complex conjugate of \\(1 + i\\) is \\(1 - i\\), therefore\n\n\\[ \\frac{3}{1 + i} = \\frac{3}{1 + i} \\times \\frac{1 - i}{1 - i} = \\frac{3(1 - i)}{(1 + i)(1 - i)} = \\frac{3 - 3i}{2} = \\frac{3}{2} - \\frac{3}{2}i \\]\n(c) The complex conjugate of $2 + 5i$ is $2 - 5i$ therefore\n\n$$\\frac{4 + 7i}{2 + 5i} = \\frac{4 + 7i}{2 + 5i} \\times \\frac{2 - 5i}{2 - 5i} = \\frac{43 - 6i}{29} = \\frac{43}{29} - \\frac{6}{29}i$$\n\n**Activity 6 Division**\n\nSimplify to the form $a + ib$\n\n(a) $\\frac{4}{1}$ \n(b) $\\frac{1 - i}{1 + i}$ \n(c) $\\frac{4 + 5i}{6 - 5i}$ \n(d) $\\frac{4i}{(1 + 2i)^2}$\n\n### 3.2 Solving equations\n\nJust as you can have equations with real numbers, you can have equations with complex numbers, as illustrated in the example below.\n\n**Example**\n\nSolve each of the following equations for the complex number $z$.\n\n(a) $4 + 5i = z - (1 - i)$\n(b) $(1 + 2i)z = 2 + 5i$\n\n**Solution**\n\n(a) Writing $z = x + iy$,\n\n$$4 + 5i = (x + yi) - (1 - i)$$\n\n$$4 + 5i = x - 1 + (y + 1)i$$\n\nComparing real parts $\\Rightarrow 4 = x - 1, \\ x = 5$\n\nComparing imaginary parts $\\Rightarrow 5 = y + 1, \\ y = 4$\n\nSo $z = 5 + 4i$. In fact there is no need to introduce the real and imaginary parts of $z$, since\n\n$$4 + 5i = z - (1 - i)$$\n\n$\\Rightarrow z = 4 + 5i + (1 - i)$\n\n$\\Rightarrow z = 5 + 4i$\n\n(b) $(1 + 2i)z = 2 + 5i$\n\n$$z = \\frac{2 + 5i}{1 + 2i}$$\nChapter 3 Complex Numbers\n\n\\[ z = \\frac{2 + 5i}{1 + 2i} \\times \\frac{1 - 2i}{1 - 2i} \\]\n\\[ z = \\frac{12 + i}{5} = \\frac{12}{5} + \\frac{1}{5}i \\]\n\nActivity 7\n\n(a) Solve the following equations for real \\( x \\) and \\( y \\)\n (i) \\( 3 + 5i + x - yi = 6 - 2i \\)\n (ii) \\( x + yi = (1 - i)(2 + 8i) \\).\n\n(b) Determine the complex number \\( z \\) which satisfies\n \\( z(3 + 3i) = 2 - i \\).\n\nExercise 3A\n\n1. Solve the equations:\n (a) \\( x^2 + 9 = 0 \\) \n (b) \\( 9x^2 + 25 = 0 \\)\n (c) \\( x^2 + 2x + 2 = 0 \\) \n (d) \\( x^2 + x + 1 = 0 \\)\n (e) \\( 2x^2 + 3x + 2 = 0 \\)\n\n2. Find the quadratic equation which has roots \\( 2 \\pm 3i \\).\n\n3. Write the following complex numbers in the form \\( x + yi \\):\n (a) \\( (3 + 2i) + (2 + 4i) \\) \n (b) \\( (4 + 3i) - (2 + 5i) \\)\n (c) \\( (4 + 3i) + (4 - 3i) \\) \n (d) \\( (2 + 7i) - (2 - 7i) \\)\n (e) \\( (3 + 2i)(4 - 3i) \\) \n (f) \\( (3 + 2i)^2 \\)\n (g) \\( (1 + i)(1 - i)(2 + i) \\)\n\n4. Find the value of the real number \\( y \\) such that \\( (3 + 2i)(1 + iy) \\)\n is (a) real \n (b) imaginary.\n\n5. Simplify:\n (a) \\( i^2 \\) \n (b) \\( i^4 \\) \n (c) \\( \\frac{1}{i} \\) \n (d) \\( \\frac{1}{i^2} \\) \n (e) \\( \\frac{1}{i^3} \\)\n\n6. If \\( z = 1 + 2i \\), find\n (a) \\( z^2 \\) \n (b) \\( \\frac{1}{z} \\) \n (c) \\( \\frac{1}{z^2} \\)\n\n7. Write in the form \\( x + yi \\):\n (a) \\( \\frac{2 + 3i}{1 + i} \\) \n (b) \\( \\frac{-4 + 3i}{-2 - i} \\) \n (c) \\( \\frac{4i}{2 - i} \\)\n (d) \\( \\frac{1}{2 + 3i} \\) \n (e) \\( \\frac{3 - 2i}{1} \\) \n (f) \\( \\frac{p + qi}{r + si} \\)\n\n8. Simplify:\n (a) \\( \\frac{(2 + i)(3 - 2i)}{1 + i} \\) \n (b) \\( \\frac{(1 - i)^3}{(2 + i)^2} \\)\n (c) \\( \\frac{1}{3 + i} \\) \n (d) \\( \\frac{1}{3 - i} \\)\n\n9. Solve for \\( z \\) when\n (a) \\( z(2 + i) = 3 - 2i \\) \n (b) \\( (z + i)(1 - i) = 2 + 3i \\)\n (c) \\( \\frac{1}{z} \\) \n (d) \\( \\frac{1}{2 - i} \\) \n (e) \\( \\frac{3}{1 + i} \\)\n\n10. Find the values of the real numbers \\( x \\) and \\( y \\) in each of the following:\n (a) \\( \\frac{x}{1 + i} + \\frac{y}{1 - 2i} = 1 \\)\n (b) \\( \\frac{x}{2 - i} + \\frac{yi}{1 + 3} = \\frac{2}{1 + i} \\)\n11. Given that \\( p \\) and \\( q \\) are real and that \\( 1 + 2i \\) is a root of the equation\n\\[\nz^2 + (p + 5i)z + q(2 - i) = 0\n\\]\ndetermine:\n(a) the values of \\( p \\) and \\( q \\);\n(b) the other root of the equation.\n\n12. The complex numbers \\( u, v \\) and \\( w \\) are related by\n\\[\n\\frac{1}{u} = \\frac{1}{v} + \\frac{1}{w}.\n\\]\nGiven that \\( v = 3 + 4i, w = 4 - 3i \\), find \\( u \\) in the form \\( x + iy \\).\n\n### 3.3 Argand diagram\n\nAny complex number \\( z = a + bi \\) can be represented by an ordered pair \\((a, b)\\) and hence plotted on \\( xy\\)-axes with the real part measured along the \\( x\\)-axis and the imaginary part along the \\( y\\)-axis. This graphical representation of the complex number field is called an **Argand diagram**, named after the Swiss mathematician Jean Argand (1768-1822).\n\n**Example**\n\nRepresent the following complex numbers on an Argand diagram:\n\n(a) \\( z = 3 + 2i \\) \n(b) \\( z = 4 - 5i \\) \n(c) \\( z = -2 - i \\)\n\n**Solution**\n\nThe Argand diagram is shown opposite.\n\n**Activity 8**\n\nLet \\( z_1 = 5 + 2i, \\ z_2 = 1 + 3i, \\ z_3 = 2 - 3i, \\ z_4 = -4 - 7i \\).\n\n(a) Plot the complex numbers \\( z_1, z_2, z_3, z_4 \\) on an Argand diagram and label them.\n\n(b) Plot the complex numbers \\( z_1 + z_2 \\) and \\( z_1 - z_2 \\) on the same Argand diagram. Geometrically, how do the positions of the numbers \\( z_1 + z_2 \\) and \\( z_1 - z_2 \\) relate to \\( z_1 \\) and \\( z_2 \\)?\n3.4 Polar coordinates\n\nConsider the complex number \\( z = 3 + 4i \\) as represented on an Argand diagram. The position of A can be expressed as coordinates (3, 4), the cartesian form, or in terms of the length and direction of OA.\n\nUsing Pythagoras' theorem, the length of \\( OA = \\sqrt{3^2 + 4^2} = 5 \\).\n\nThis is written as \\( |z| = r = 5 \\). \\( |z| \\) is read as the modulus or absolute value of \\( z \\).\n\nThe angle that OA makes with the positive real axis is\n\n\\[ \\theta = \\tan^{-1} \\frac{4}{3} = 53.13^\\circ \\text{ (or 0.927 radians)} \\]\n\nThis is written as \\( \\arg(z) = 53.13^\\circ \\). You say \\( \\arg(z) \\) is the argument or phase of \\( z \\).\n\nThe parameters \\( |z| \\) and \\( \\arg(z) \\) are in fact the equivalent of polar coordinates \\( r, \\theta \\) as shown opposite. There is a simple connection between the polar coordinate form and the cartesian or rectangular form \\( (a, b) \\):\n\n\\[ a = r \\cos \\theta, \\quad b = r \\sin \\theta. \\]\n\nTherefore\n\n\\[ z = a + bi = r \\cos \\theta + ri \\sin \\theta = r(\\cos \\theta + i \\sin \\theta) \\]\n\nwhere \\( |z| = r \\), and \\( \\arg(z) = \\theta \\).\n\nIt is more usual to express the angle \\( \\theta \\) in radians. Note also that it is convention to write the \\( i \\) before \\( \\sin \\theta \\), i.e. \\( i \\sin \\theta \\) is preferable to \\( \\sin \\theta i \\).\n\nIn the diagram opposite, the point A could be labelled \\( (2\\sqrt{3}, 2) \\) or as \\( 2\\sqrt{3} + 2i \\).\n\nThe angle that OA makes with the positive x-axis is given by\n\n\\[ \\theta = \\tan^{-1} \\frac{2}{2\\sqrt{3}} = \\tan^{-1} \\frac{1}{\\sqrt{3}}. \\]\n\nTherefore \\( \\theta = \\frac{\\pi}{6} \\) or \\( 2\\pi + \\frac{\\pi}{6} \\) or \\( 4\\pi + \\frac{\\pi}{6} \\) or ... etc. There is an infinite number of possible angles. The one you should normally use is in the interval \\( -\\pi < \\theta \\leq \\pi \\), and this is called the principal argument.\nUsing polar coordinates the point A could be labelled with its polar coordinates \\([r, \\theta]\\) as \\(4, \\frac{\\pi}{6}\\). Note the use of squared brackets when using polar coordinates. This is to avoid confusion with Cartesian coordinates.\n\nThus \\(2\\sqrt{3} + 2i = 4 \\cos \\frac{\\pi}{6} + i \\sin \\frac{\\pi}{6}\\).\n\n**Important note:** if you are expressing \\(a + ib\\) in its polar form, where \\(a\\) and \\(b\\) are both positive, then the formula \\(\\theta = \\tan^{-1} \\frac{b}{a}\\) is quite sufficient. But in other cases you need to think about the position of \\(a + ib\\) in the Argand diagram.\n\n**Example**\n\nWrite \\(z = -1 - i\\) in polar form.\n\n**Solution**\n\nNow \\(z = a + ib\\) where \\(a = -1\\) and \\(b = -1\\) and in polar form the modulus of \\(z = |z| = r = \\sqrt{1^2 + (-1)^2} = \\sqrt{2}\\) and the argument is \\(\\frac{5\\pi}{4}\\) (or 225°): its principal value is \\(-\\frac{3\\pi}{4}\\).\n\nHence \\(z = \\sqrt{2}, -\\frac{3\\pi}{4}\\) in polar coordinates. (The formula \\(\\tan^{-1} \\frac{b}{a}\\) would have given you \\(\\frac{\\pi}{4}\\).)\n\n**Activity 9**\n\n(a) Write the following numbers in \\([r, \\theta]\\) form:\n\n(i) \\(7 + 2i\\) \n(ii) \\(3 - i\\) \n(iii) \\(-4 + 6i\\) \n(iv) \\(-3 - i\\)\n\n(b) Write the following in \\(a + bi\\) form:\n\n(remember that the angles are in radians)\n\n(i) \\(3, \\frac{\\pi}{4}\\) \n(ii) \\([5, \\pi]\\) \n(iii) \\([6, 4.2]\\)\n\n(iv) \\(\\sqrt{2}, -\\frac{2\\pi}{3}\\)\n3.5 Complex number algebra\n\nYou will now investigate the set of complex numbers in the modulus/argument form, \\([r, \\theta]\\).\n\nSuppose you wish to combine two complex numbers of the form\n\n\\[ z_1 = [r_1, \\theta_1] \\quad z_2 = [r_2, \\theta_2] \\]\n\nNote that, in a \\(a + bi\\) form,\n\n\\[ z_1 = r_1 \\cos \\theta_1 + i r_1 \\sin \\theta_1 \\]\n\nand\n\n\\[ z_2 = r_2 \\cos \\theta_2 + i r_2 \\sin \\theta_2 \\]\n\nSo\n\n\\[ z_1 z_2 = (r_1 \\cos \\theta_1 + i r_1 \\sin \\theta_1)(r_2 \\cos \\theta_2 + i r_2 \\sin \\theta_2) \\]\n\n\\[ = r_1 r_2 (\\cos \\theta_1 + i \\sin \\theta_1)(\\cos \\theta_2 + i \\sin \\theta_2) \\]\n\n\\[ = r_1 r_2 \\left[ (\\cos \\theta_1 \\cos \\theta_2 - \\sin \\theta_1 \\sin \\theta_2) + i (\\sin \\theta_1 \\cos \\theta_2 + \\cos \\theta_1 \\sin \\theta_2) \\right]. \\]\n\nSimplify the expressions in the brackets.\n\nUsing the formulae for angles,\n\n\\[ z_1 z_2 = r_1 r_2 \\left[ \\cos (\\theta_1 + \\theta_2) + i \\sin (\\theta_1 + \\theta_2) \\right] \\]\n\nor, in polar notation\n\n\\[ z_1 z_2 = [r_1 r_2, \\theta_1 + \\theta_2]. \\]\n\nFor example, \\([3, 0.5] \\times [4, 0.3] = [12, 0.8]\\).\n\nThat is, the first elements of the ordered pairs are multiplied and the second elements are added.\n\nActivity 10\n\nGiven that \\(z_1 = [3, 0.7]\\), \\(z_2 = [2, 1.2]\\) and \\(z_3 = [4, -0.5]\\),\n\n(a) find \\(z_1 \\times z_2\\) and \\(z_1 \\times z_3\\)\n\n(b) show that \\([1, 0] \\times z_1 = z_1\\)\n(c) (i) find a complex number \\( z = [r, \\theta] \\) such that \n\\[ z \\times z_2 = [1, 0]. \\]\n(ii) find a complex number \\( z = [r, \\theta] \\) such that \n\\[ z \\times z_3 = [1, 0]. \\]\n\n(d) for any complex number \\([r, \\theta]\\) show that \n\\[ \\frac{1}{r}, -\\theta \\times [r, \\theta] = [1, 0] \\quad (r > 0). \\]\n\n---\n\n**Activity 11**\n\nUse a spreadsheet package to plot numbers on an Argand diagram by entering numbers and formulae into cells A5 to E5 as shown opposite.\n\nCells D5 and E5 calculate the \\( x \\) and \\( y \\) coordinates respectively of the complex number whose modulus and argument are in cells B5 and C5 (the argument is entered as a multiple of \\( \\pi \\)).\n\nA second number can be entered in cells B6 and C6 and its \\((x, y)\\) coordinates calculated by using appropriate formulae in cells D6 and E6.\n\nThis can be repeated for further numbers (the spreadsheet facility 'FILL DOWN' is useful here).\n\nUse the appropriate facility on your spreadsheet to plot the \\((x, y)\\) values.\n\nLabel rows and columns if it makes it easier.\n\nExperiment with different values of \\( r \\) and \\( \\theta \\).\n\nAn example is shown in the graph opposite and the related spreadsheet below.\nExercise 3B\n\n1. Mark on an Argand diagram the points representing the following numbers:\n (a) 2 (b) 3i (c) -i (d) 1+2i (e) 3-i (f) -2+3i\n\n2. The points A, B, C and D represent the numbers $z_1$, $z_2$, $z_3$ and $z_4$ and O is the origin.\n (a) If OABC is a parallelogram, and $z_1 = 1+i$, $z_2 = 4+5i$, find $z_3$.\n (b) Find $z_2$ and $z_4$ when ABCD is a square and\n (i) $z_1 = 2+i$, $z_3 = 6+7i$\n (ii) $z_1 = 6-2i$, $z_3 = 6i$\n\n3. Find the modulus and argument of\n (a) $1-i$ (b) $1+\\sqrt{3}i$ (c) $3-3i$ (d) $3+2i$\n\n4. Show that\n (a) $|z| = |\\bar{z}|$ (b) $\\arg z = -\\arg \\bar{z}$\n and illustrate these results on an Argand diagram.\n\n5. Find the modulus and argument of $z_1$, $z_2$, $z_1z_2$ and $\\frac{z_1}{z_2}$ when $z_1 = 1+i$ and $z_2 = \\sqrt{3}+i$. What do you notice?\n\n6. Write in the form $a+bi$\n (a) $4, \\frac{\\pi}{3}$ (b) $5, \\frac{\\pi}{2}$\n (c) $3\\sqrt{2}, -\\frac{3\\pi}{4}$ (d) $[4, 13\\pi]$\n\n7. Write in polar form\n (a) $1+i$ (b) $-2+i$ (c) $-5$ (d) $4i$ (e) $3+4i$\n (f) $-3-4i$ (g) $3-4i$ (h) $-3+4i$\n\n8. In this question, angles are in radians.\n (a) (i) Plot the following complex numbers on an Argand diagram and label them:\n $z_1 = [4, 0]$, $z_2 = 3, \\frac{\\pi}{2}$, $z_3 = 2, -\\frac{\\pi}{2}$\n $z_4 = 3, \\frac{\\pi}{3}$, $z_5 = 2, \\frac{5\\pi}{3}$\n (ii) Let the complex number $z = 1, \\frac{\\pi}{2}$\n Calculate $z \\times z_1$, $z \\times z_2$, etc. and plot the points on the same diagram as in (i).\n What do you notice?\n (b) Repeat (a) (ii) using $z = 1, \\frac{\\pi}{3}$\n (c) In general, what happens when a complex number is multiplied by $[1, \\theta]$? Make up some examples to illustrate your answer.\n (d) Repeat (a) (ii) using $z = 0.5, \\frac{\\pi}{2}$\n (e) In general, what happens when a complex number is multiplied by $0.5, \\frac{\\pi}{2}$? Make up some examples to illustrate your answer.\n (f) Repeat (e) for $3, \\frac{\\pi}{3}$\n (g) Describe what happens when a complex number is multiplied by $3, \\frac{\\pi}{3}$. Make up some examples to illustrate your answer.\n\n3.6 De Moivre's theorem\n\nAn important theorem in complex numbers is named after the French mathematician, Abraham de Moivre (1667-1754). Although born in France, he came to England where he made the acquaintance of Newton and Halley and became a private teacher of Mathematics. He never obtained the university position he sought but he did produce a considerable amount of research, including his work on complex numbers.\nChapter 3 Complex Numbers\n\nThe derivation of de Moivre's theorem now follows.\n\nConsider the complex number \\( z = \\cos \\frac{\\pi}{3} + i \\sin \\frac{\\pi}{3} \\).\n\nThen \\( z^2 = \\cos \\frac{\\pi}{3} + i \\sin \\frac{\\pi}{3} \\times \\cos \\frac{\\pi}{3} + i \\sin \\frac{\\pi}{3} \\)\n\n\\[ = \\cos^2 \\frac{\\pi}{3} - \\sin^2 \\frac{\\pi}{3} + 2i \\cos \\frac{\\pi}{3} \\sin \\frac{\\pi}{3} \\]\n\n\\[ = \\cos \\frac{2\\pi}{3} + i \\sin \\frac{2\\pi}{3} \\]\n\nor with the modulus/argument notation\n\n\\[ z = 1, \\frac{\\pi}{3} \\]\n\nand \\( z^2 = 1, \\frac{\\pi}{3} \\times 1, \\frac{\\pi}{3} = 1, \\frac{2\\pi}{3} \\).\n\nRemember that any complex number \\( z = x + yi \\) can be written in the form of an ordered pair \\([r, \\theta]\\) where \\( r = \\sqrt{x^2 + y^2} \\) and\n\n\\[ \\theta = \\tan^{-1} \\frac{y}{x} \\].\n\nIf the modulus of the number is 1, then \\( z = \\cos \\theta + i \\sin \\theta \\)\n\nand \\( z^2 = (\\cos \\theta + i \\sin \\theta)^2 \\)\n\n\\[ = \\cos^2 \\theta - \\sin^2 \\theta + 2i \\cos \\theta \\sin \\theta \\]\n\n\\[ = \\cos 2\\theta + i \\sin 2\\theta \\]\n\ni.e. \\( z^2 = [1, \\theta]^2 = [1, 2\\theta] \\).\n\nActivity 12\n\n(a) Use the principle that, with the usual notation,\n\n\\[ [r_1, \\theta_1] \\times [r_2, \\theta_2] = [r_1 r_2, \\theta_1 + \\theta_2] \\]\n\nto investigate \\( \\cos \\frac{\\pi}{6} + i \\sin \\frac{\\pi}{6} \\) when \\( n = 0, 1, 2, 3, ..., 12 \\).\n(b) In the same way as in (a), investigate\n\n\\[ 3\\cos \\frac{\\pi}{6} + 3i \\sin \\frac{\\pi}{6}^n \\]\n\nfor \\( n = 0, 1, 2, \\ldots, 6 \\).\n\nYou should find from the last activity that\n\n\\[ (\\cos \\theta + i \\sin \\theta)^n = \\cos(n\\theta) + i \\sin(n\\theta). \\]\n\nIn \\([r, \\theta]\\) form this is \\([r^n, n\\theta]\\) and de Moivre's theorem states that this is true for any rational number \\( n \\).\n\nA more rigorous way of deriving de Moivre's theorem follows.\n\n**Activity 13**\n\nShow that \\((\\cos \\theta + i \\sin \\theta)^n = \\cos n\\theta + i \\sin n\\theta\\) for \\( n = 3 \\) and \\( n = 4 \\).\n\n**Activity 14**\n\nShow that\n\n\\[ (\\cos k\\theta + i \\sin k\\theta)(\\cos \\theta + i \\sin \\theta) = \\cos(k + 1)\\theta + i \\sin(k + 1)\\theta. \\]\n\nHence show that if\n\n\\[ (\\cos \\theta + i \\sin \\theta)^k = \\cos k\\theta + i \\sin k\\theta \\]\n\nthen \\((\\cos \\theta + i \\sin \\theta)^{k+1} = \\cos((k + 1)\\theta) + i \\sin((k + 1)\\theta)\\).\n\nThe principle of mathematical induction will be used to prove that \\((\\cos \\theta + i \\sin \\theta)^n = \\cos(n\\theta) + i \\sin(n\\theta)\\) for all positive integers.\n\nLet \\( S(k) \\) be the statement\n\n\\[ (\\cos \\theta + i \\sin \\theta)^k = \\cos k\\theta + i \\sin k\\theta. \\]\n\nAs \\( S(1) \\) is true and you have shown in Activity 14 that \\( S(k) \\) implies \\( S(k + 1) \\) then \\( S(2) \\) is also true. But then (again by Activity 14) \\( S(3) \\) is true. But then ... Hence \\( S(n) \\) is true for \\( n = 1, 2, 3, \\ldots \\). This is the principle of mathematical induction (which you meet more fully later in the book). So for all positive integers \\( n \\),\nChapter 3 Complex Numbers\n\n\\[(\\cos \\theta + i \\sin \\theta)^n = \\cos n\\theta + i \\sin n\\theta\\]\n\nIf \\(n\\) is a **negative integer**, then let \\(m = -n\\)\n\n\\[(\\cos \\theta + i \\sin \\theta)^n = (\\cos \\theta + i \\sin \\theta)^{-m} = \\frac{1}{(\\cos \\theta + i \\sin \\theta)^m}\\]\n\nwhere \\(m\\) is positive and, from the work above,\n\n\\[(\\cos \\theta + i \\sin \\theta)^m = (\\cos m\\theta + i \\sin m\\theta).\\]\n\nTherefore \\(\\frac{1}{(\\cos \\theta + i \\sin \\theta)^n} = \\frac{1}{(\\cos m\\theta + i \\sin m\\theta)}\\)\n\n**Activity 15**\n\nShow that\n\n\\[\\frac{1}{(\\cos m\\theta + i \\sin m\\theta)} = \\cos m\\theta - i \\sin m\\theta\\]\n\nand hence that \\((\\cos \\theta + i \\sin \\theta)^n = \\cos n\\theta + i \\sin n\\theta\\) when \\(n\\) is a negative integer.\n\n**Hint**: multiply top and bottom by \\((\\cos \\theta - i \\sin \\theta)\\) and use the fact that \\(\\sin(-A) = -\\sin(A)\\).\n\nWhen \\(n\\) is a **rational number**, i.e. \\(n = \\frac{p}{q}\\) where \\(p\\) and \\(q\\) are integers, then as \\(q\\) is an integer\n\n\\[\\cos \\frac{p}{q} \\theta + i \\sin \\frac{p}{q} \\theta^q = (\\cos p\\theta + i \\sin p\\theta)\\]\n\nSince \\(p\\) is an integer\n\n\\[\\cos p\\theta + i \\sin p\\theta = (\\cos \\theta + i \\sin \\theta)^p,\\]\n\nand hence\n\n\\[\\cos \\frac{p}{q} \\theta + i \\sin \\frac{p}{q} \\theta^q = (\\cos \\theta + i \\sin \\theta)^p\\]\nThus \\[ \\cos \\frac{p}{q} \\theta + i \\sin \\frac{p}{q} \\theta = (\\cos \\theta + i \\sin \\theta)^{\\frac{p}{q}} \\]\n\nTherefore \\[ \\cos n \\theta + i \\sin n \\theta = (\\cos \\theta + i \\sin \\theta)^n \\] for any rational number \\( n \\) and clearly this leads to\n\n\\[ (r(\\cos \\theta + i \\sin \\theta))^n = r^n (\\cos n \\theta + i \\sin n \\theta) \\]\n\n### 3.7 Applications of de Moivre's theorem\n\nThere are many applications of de Moivre's theorem, including the proof of trigonometric identities.\n\n**Example**\n\nProve that \\( \\cos 3\\theta = \\cos^3 \\theta - 3\\cos \\theta \\sin^2 \\theta \\).\n\n**Solution**\n\nBy de Moivre's theorem:\n\n\\[\n\\cos 3\\theta + i \\sin 3\\theta = (\\cos \\theta + i \\sin \\theta)^3 \\\\\n= \\cos^3 \\theta + 3\\cos^2 \\theta (i \\sin \\theta) + 3\\cos \\theta (i \\sin \\theta)^2 + (i \\sin \\theta)^3 \\\\\n= \\cos^3 \\theta + 3i \\cos^2 \\theta \\sin \\theta - 3\\cos \\theta \\sin^2 \\theta - i \\sin^3 \\theta \\\\\n= \\cos^3 \\theta - 3\\cos \\theta \\sin^2 \\theta + i(3\\cos^2 \\theta \\sin \\theta - \\sin^3 \\theta)\n\\]\n\nComparing real parts of the equation above you obtain\n\n\\[ \\cos 3\\theta = \\cos^3 \\theta - 3\\cos \\theta \\sin^2 \\theta \\]\n\n**Example**\n\nSimplify the following expression:\n\n\\[ \\frac{\\cos 2\\theta + i \\sin 2\\theta}{\\cos 3\\theta + i \\sin 3\\theta} \\]\n\n**Solution**\n\n\\[\n\\frac{\\cos 2\\theta + i \\sin 2\\theta}{\\cos 3\\theta + i \\sin 3\\theta} = \\frac{(\\cos \\theta + i \\sin \\theta)^2}{(\\cos \\theta + i \\sin \\theta)^3} = \\frac{1}{(\\cos \\theta + i \\sin \\theta)^1}\n\\]\nChapter 3 Complex Numbers\n\n\\[ (\\cos \\theta + i \\sin \\theta)^{-1} = (\\cos(-\\theta) + i \\sin(-\\theta)) = \\cos \\theta - i \\sin \\theta \\]\n\n**Exercise 3C**\n\n1. Use de Moivre's theorem to prove the trig identities:\n (a) \\( \\sin 2\\theta = 2\\sin \\theta \\cos \\theta \\)\n (b) \\( \\cos 5\\theta = \\cos^5 \\theta - 10\\cos^3 \\theta \\sin^2 \\theta + 5\\cos \\theta \\sin^4 \\theta \\)\n\n2. If \\( z = \\cos \\theta + i \\sin \\theta \\) then use de Moivre's theorem to show that:\n (a) \\( z + \\frac{1}{z} = 2\\cos \\theta \\)\n (b) \\( z^2 + \\frac{1}{z^2} = 2\\cos 2\\theta \\)\n (c) \\( z^n + \\frac{1}{z^n} = 2\\cos n\\theta \\)\n\n**Activity 16**\n\nMake an educated guess at a complex solution to the equation \\( z^3 = 1 \\) and then use the facilities of the spreadsheet to raise it to the power 3 and plot it on the Argand diagram. If it is a solution of the equation then the resultant point will be plotted at distance 1 unit along the real axis. The initial spreadsheet layout from Activity 11 can be adapted. In addition, the cells shown opposite are required.\n\nWhat does the long formula in cell C7 do? Is it strictly necessary in this context?\n\nBelow are two examples of the output from a spreadsheet using these cells - the first one is not a cube root of 1 but the second is.\n3.8 Solutions of $z^3 = 1$\n\nWrite down one solution of $z^3 = 1$.\n\nDe Moivre's theorem can be used to find all the solutions of $z^3 = 1$.\n\nLet $z = [r, \\theta]$\n\nthen $z^3 = [r, \\theta]^3 = [r^3, 3\\theta]$\n\nand you can express 1 as $1 = [1, 2n\\pi]$ where $n$ is an integer.\n\nThen $[r^3, 3\\theta] = [1, 2n\\pi]$\n\nTherefore $r^3 = 1$ and $3\\theta = 2n\\pi$\n\ni.e. $r = 1$ and $\\theta = \\frac{2n\\pi}{3}$\n\nThe solutions are then given by letting $n = 0, 1, 2, ...$\n\nIf $n = 0$, $z_1 = [1, 0] = 1$\n\nIf $n = 1$, $z_2 = 1, \\frac{2\\pi}{3} = \\cos \\frac{2\\pi}{3} + i \\sin \\frac{2\\pi}{3}$\n\n$= \\frac{1}{2} + \\frac{\\sqrt{3}}{2}i$\n\nIf $n = 2$, $z_3 = 1, \\frac{4\\pi}{3} = \\cos \\frac{4\\pi}{3} + i \\sin \\frac{4\\pi}{3}$\n\n$= \\frac{1}{2} - \\frac{\\sqrt{3}}{2}i$\n\nWhat happens if $n = 3, 4, ...$?\n\nActivity 17 Cube roots of unity\n\nPlot the three distinct cube roots of unity on an Argand diagram. What do you notice?\nChapter 3 Complex Numbers\n\nActivity 18\n\nUse de Moivre's theorem to find all solutions to the following equations and plot the results on an Argand diagram.\n\n(a) \\( z^4 = 1 \\) (b) \\( z^3 = 8 \\) (c) \\( z^3 = i \\)\n\n3.9 Euler's theorem\n\nYou have probably already met the series expansion of \\( e^x \\), namely\n\n\\[\ne^x = 1 + x + \\frac{x^2}{2!} + \\frac{x^3}{3!} + \\frac{x^4}{4!} + \\ldots\n\\]\n\nAlso the series expansions for \\( \\cos \\theta \\) and \\( \\sin \\theta \\) are given by\n\n\\[\n\\cos \\theta = 1 - \\frac{\\theta^2}{2!} + \\frac{\\theta^4}{4!} - \\frac{\\theta^6}{6!} + \\ldots\n\\]\n\n\\[\n\\sin \\theta = \\theta - \\frac{\\theta^3}{3!} + \\frac{\\theta^5}{5!} - \\frac{\\theta^7}{7!} + \\ldots\n\\]\n\nActivity 19\n\n(a) For each of the following values of \\( \\theta \\), use the series for \\( e^x \\) with \\( x \\) replaced by \\( i\\theta \\) to calculate (to 4 d.p.) the value of \\( e^{i\\theta} \\). (Write your answer in the form \\( a + bi \\).)\n\n(i) \\( \\theta = 0 \\) (ii) \\( \\theta = 1 \\) (iii) \\( \\theta = 2 \\) (iv) \\( \\theta = -0.4 \\)\n\n(b) Calculate \\( \\cos \\theta \\) and \\( \\sin \\theta \\) for each of the values in (a).\n\n(c) Find a connection between the values of \\( e^{i\\theta} \\), \\( \\cos \\theta \\) and \\( \\sin \\theta \\) for each of the values of \\( \\theta \\) given in (a) and make up one other example to test your conjecture.\n\n(d) To prove this for all values of \\( \\theta \\), write down the series expansions of \\( e^{i\\theta} \\), \\( \\cos \\theta \\) and \\( \\sin \\theta \\) and show that\n\n\\[\ne^{i\\theta} = \\cos \\theta + i \\sin \\theta.\n\\]\nThe previous activity has shown that\n\n\\[ e^{i\\theta} = \\cos \\theta + i \\sin \\theta \\]\n\nwhich is sometimes known as **Euler's theorem**.\n\nIt is an important result, and can be used to derive de Moivre's theorem in a simple way. If \\( z \\) is any complex number then in polar form\n\n\\[ z = x + yi = r \\cos \\theta + ri \\sin \\theta \\]\n\n\\[ = r(\\cos \\theta + i \\sin \\theta) \\]\n\n\\[ = re^{i\\theta}, \\text{ using Euler's theorem.} \\]\n\nThus\n\n\\[ z^n = (re^{i\\theta})^n = r^n e^{ni\\theta} = r^n e^{i(n\\theta)} \\]\n\nor\n\n\\[ (r \\cos \\theta + ir \\sin \\theta)^n = r^n (\\cos(n\\theta) + i \\sin(n\\theta)) \\]\n\n\\[ \\Rightarrow (\\cos \\theta + i \\sin \\theta)^n = \\cos(n\\theta) + i \\sin(n\\theta) \\]\n\nwhich is de Moivre's theorem.\n\n**What assumptions about complex number algebra have been made in the 'proof' above?**\n\nOne interesting result can be obtained from Euler's theorem by putting \\( \\theta = \\pi \\). This gives\n\n\\[ e^{i\\pi} = \\cos \\pi + i \\sin \\pi \\]\n\n\\[ = -1 + i \\times 0. \\]\n\nSo\n\n\\[ e^{i\\pi} + 1 = 0 \\]\n\nThis is often referred to as **Euler's equation**, since it connects the five most 'famous' numbers\n\n\\[ 0, 1, \\pi, e, i \\]\n\nwith a '+' and '=' sign!\n\n**Try substituting other values of \\( \\theta \\) in Euler's theorem and see what equation is derived.**\n3.10 Exponential form of a complex number\n\nWhen a complex number \\( z \\) has modulus \\( r \\), which must be non-negative, and argument \\( \\theta \\), which is usually taken such that it satisfies \\(-\\pi < \\theta \\leq \\pi\\), you have already shown that it can be represented in the forms\n\n(i) \\( r(\\cos \\theta + i \\sin \\theta) \\)\n\n(ii) \\([r, \\theta]\\)\n\n(iii) \\(re^{i\\theta}\\)\n\nExpression (iii) is referred to as the exponential form of a complex number.\n\nActivity 20\n\nWrite each of the following complex numbers in the exponential form.\n\n(a) \\(2 \\cos \\frac{\\pi}{3} + i \\cos \\frac{\\pi}{3}\\) \n(b) \\(5, \\frac{2\\pi}{3}\\) \n(c) \\(1 - i\\sqrt{3}\\)\n\n3.11 Solving equations\n\nYou have already investigated the solutions of the equation \\(z^3 = 1\\) and similar equations using a spreadsheet and by using de Moivre’s theorem. A similar approach will now be used to solve more complicated equations.\n\nExample\n\nWrite down the modulus and argument of the complex number \\(4 - 4i\\).\n\nSolve the equation \\(z^5 = 4 - 4i\\), expressing your answers in the exponential form.\n\nSolution\n\n\\[|4 - 4i| = \\sqrt{4^2 + (-4)^2} = 4\\sqrt{2}\\]\n\nAs before it is often helpful to make a small sketch of an Argand diagram to locate the correct quadrant for the argument.\nSo \\( \\arg(4 - 4i) = -\\frac{\\pi}{4} \\)\n\nTherefore the complex number \\( 4 - 4i \\) can be expressed as\n\\[\n4\\sqrt{2}, \\quad -\\frac{\\pi}{4}\n\\]\n\nIt is quite convenient to work using the polar form of a complex number when solving \\( z^5 = 4 - 4i \\).\n\nLet \\( z = [r, \\theta] \\), then \\( z^5 = [r^5, 5\\theta] \\).\n\nSo as to obtain all five roots of the equation, the argument is considered to be \\( 2n\\pi - \\frac{\\pi}{4} \\) where \\( n \\) is an integer.\n\nEquating the results\n\\[\n[r^5, 5\\theta] = 4\\sqrt{2}, \\quad 2n\\pi - \\frac{\\pi}{4}\n\\]\n\n\\[\nr^5 = 4\\sqrt{2} \\quad \\Rightarrow \\quad r = \\sqrt{2}\n\\]\n\n\\[\n5\\theta = 2n\\pi - \\frac{\\pi}{4} \\quad \\Rightarrow \\quad \\theta = (8n - 1)\\frac{\\pi}{20}\n\\]\n\nNow choose the five appropriate values of \\( n \\) so that \\( \\theta \\) lies between \\( -\\pi \\) and \\( \\pi \\).\n\n\\[\nn = -2 \\quad \\Rightarrow \\quad \\theta = -\\frac{17\\pi}{20}\n\\]\n\n\\[\nn = -1 \\quad \\Rightarrow \\quad \\theta = -\\frac{9\\pi}{20}\n\\]\n\n\\[\nn = 0 \\quad \\Rightarrow \\quad \\theta = -\\frac{\\pi}{20}\n\\]\n\n\\[\nn = 1 \\quad \\Rightarrow \\quad \\theta = \\frac{7\\pi}{20}\n\\]\n\n\\[\nn = 2 \\quad \\Rightarrow \\quad \\theta = \\frac{15\\pi}{20} \\quad \\text{or} \\quad \\frac{3\\pi}{4}\n\\]\n\nThe solutions in exponential form are therefore\n\\[\n\\sqrt{2}e^{-\\frac{17\\pi i}{20}}, \\quad \\sqrt{2}e^{-\\frac{9\\pi i}{20}}, \\quad \\sqrt{2}e^{-\\frac{\\pi i}{20}}, \\quad \\sqrt{2}e^{\\frac{7\\pi i}{20}} \\quad \\text{and} \\quad \\sqrt{2}e^{\\frac{3\\pi i}{4}}.\n\\]\nChapter 3 Complex Numbers\n\nActivity 21\n\nShow that $1+i$ is a root of the equation $z^4 = -4$ and find each of the other roots in the form $a + bi$ where $a$ and $b$ are real.\n\nPlot the roots on an Argand diagram. By considering the diagonals, or otherwise, show that the points are at the vertices of a square. Calculate the area of the square.\n\nActivity 22\n\nGiven that $k \\neq 1$ and the roots of the equation $z^3 = k$ are $\\alpha, \\beta$ and $\\gamma$, use the substitution $z = \\frac{x - 2}{x + 1}$ to obtain the roots of the equation\n\n$$(x - 2)^3 = k(x + 1)^3$$\n\nExercise 3D\n\n1. By using de Moivre’s theorem, find all solutions to the following equations, giving your answers in polar form. Plot each set of roots on an Argand diagram and comment on the symmetry.\n (a) $z^4 = 16$\n (b) $z^3 = -27i$\n (c) $z^5 = -1$\n\n2. Find the cube roots of\n (a) $1 + i$\n (b) $2i - 2$\n giving your answers in exponential form.\n\n3. Using the answers from Question 1(a), determine the solutions of the equation\n $$(x + 1)^4 = 16(x - 1)^4$$\n giving your answers in the form $a + bi$.\n\n4. Using the results from Question 1(b), solve the equation\n $$1 + 27i(x + 1)^3 = 0$$\n giving your answers in the form $a + bi$.\n\n5. Solve the equation $z^3 = i(x - 1)^3$ giving your answers in the form $a + bi$.\n Plot the solutions on an Argand diagram and comment on your results.\n\n6. Determine the four roots of the equation\n $$(z - 2)^4 + (z + 1)^4 = 0$$\n and plot them on an Argand diagram.\n\n3.12 Loci in the complex plane\n\nSuppose $z$ is allowed to vary in such a way that $|z - 1| = 2$. You could write $z = x + iy$ and obtain\n\n$$\\sqrt{(x - 1)^2 + y^2} = 2$$\n\nor\n\n$$(x - 1)^2 + y^2 = 4$$\nYou can immediately identify this as the cartesian equation of a circle centre \\((1, 0)\\) and radius 2. In terms of the complex plane, the centre is \\(1 + 0i\\).\n\nThis approach could be adopted for most problems and the exercise is simply one in algebra, lacking any geometrical feel for the locus.\n\nInstead, if \\(\\omega\\) is a complex number, you can identify \\(|z - \\omega|\\) as the distance of \\(z\\) from the point represented by \\(\\omega\\) on the complex plane. The locus \\(|z - 1| = 2\\) can be interpreted as the set of points that are 2 units from the point \\(1 + 0i\\); in other words, a circle centre \\(1 + 0i\\) and radius 2.\n\n**Activity 23**\n\nIllustrate the locus of \\(z\\) in the complex plane if \\(z\\) satisfies\n\n(a) \\(|z - (3 + 2i)| = 5\\) \n(b) \\(|z - 2 + i| = |1 + 3i|\\)\n\n(c) \\(|z + 2i| = 2\\) \n(d) \\(|z - 4| = 0\\)\n\n**Activity 24**\n\nDescribe the path of a point which moves in a fixed plane so that it is always the same distance from two fixed points A and B.\n\nIllustrate the locus of \\(z\\) in the case when \\(z\\) satisfies\n\n\\(|z + 3| = |z - 4i|\\).\n\nYou would probably have had some difficulty in writing down a cartesian equation of the locus in Activity 24, even though you could describe the locus geometrically.\n\n**Activity 25**\n\nDescribe the locus of \\(z\\) in the case where \\(z\\) moves in such a way that\n\n\\(|z| = |z + 2 - 2i|\\).\n\nNow try to write down the cartesian equation of this locus which should be a straight line.\n\nBy writing \\(z = x + iy\\), try to obtain the same result algebraically.\nChapter 3 Complex Numbers\n\nActivity 26\n\nInvestigate the locus of P when P moves in the complex plane and represents the complex number \\( z \\) which satisfies\n\n\\[ |z + 1| = k|z - 1| \\]\n\nfor different values of the real number \\( k \\).\n\nWhy does \\( k = 1 \\) have to be treated as a special case?\n\nExample\n\nThe point P represents the complex number \\( z \\) on an Argand diagram. Describe the locus geometrically and obtain a cartesian equation for the locus in the cases\n\n(a) \\( |z| = |z - 4| \\)\n(b) \\( |z| + |z - 4| = 6 \\)\n(c) \\( |z| = 2|z - 4| \\)\n\nSolution\n\n(a) From your work in Activity 25, you should recognise this as a straight line. In fact, it is the mediator, or perpendicular bisector, of the line segment joining the origin to the point \\( 4 + 0i \\).\n\nIt should be immediately obvious that its cartesian equation is \\( x = 2 \\); however, writing\n\n\\[ z = x + iy \\]\n\n\\[ |z| = |x + iy| = |x - 4 + iy| \\]\n\nSquaring both sides gives\n\n\\[ x^2 + y^2 = (x - 4)^2 + y^2 \\]\n\nleading to\n\n\\[ 0 = -8x + 16 \\]\n\nor \\( x = 2 \\).\n\n(b) You may be aware of a curve that is traced out when the sum of the distances from two fixed points is constant. You could try using a piece of string with its ends fastened to two fixed points. The curve is called an ellipse.\n\nA sketch of the locus is shown opposite.\nYou can obtain a cartesian equation by putting \\( z = x + iy \\)\n\n\\[\n|x + iy| + |x - 4 + iy| = 6\n\\]\n\nSo\n\n\\[\n\\sqrt{x^2 + y^2} + \\sqrt{(x - 4)^2 + y^2} = 6\n\\]\n\n\\[\n(x - 4)^2 + y^2 = [6 - (x^2 + y^2)]^2\n\\]\n\n\\[\nx^2 - 8x + 16 + y^2 = 36 - 12\\sqrt{x^2 + y^2} + x^2 + y^2\n\\]\n\n\\[\n12\\sqrt{x^2 + y^2} = 20 + 8x\n\\]\n\n\\[\n3\\sqrt{x^2 + y^2} = 5 + 2x\n\\]\n\n\\[\n9(x^2 + y^2) = 25 + 20x + 4x^2\n\\]\n\n\\[\n5x^2 - 20x + 9y^2 = 25\n\\]\n\n\\[\n5(x - 2)^2 + 9y^2 = 45\n\\]\n\n\\[\n\\frac{(x - 2)^2}{9} + \\frac{y^2}{5} = 1\n\\]\n\n(c) You should have discovered in Activity 26 that the locus will be a circle when the relationship is of this form. It is called the circle of Apollonius.\n\nYou could possibly sketch the locus without finding the cartesian equation.\n\nLet \\( z = x + iy \\)\n\n\\[\n|x + iy| = 2|x - 4 + iy|\n\\]\n\n\\[\n\\sqrt{x^2 + y^2} = 2\\sqrt{(x - 4)^2 + y^2}\n\\]\n\n\\[\nx^2 + y^2 = 4(x^2 - 8x + 16 + y^2)\n\\]\n\n\\[\n0 = 3x^2 + 3y^2 - 32x + 64\n\\]\n\nIn order to find the centre and radius you can complete the square\n\n\\[\nx^2 + y^2 - \\frac{32}{3}x + \\frac{64}{3} = 0\n\\]\n\n\\[\nx - \\frac{16}{3}^2 + y^2 = \\frac{256}{9} - \\frac{64}{3} = \\frac{64}{9}\n\\]\n\nCentre of circle is at \\( \\frac{16}{3} + 0i \\) and radius is \\( \\frac{8}{3} \\).\nChapter 3 Complex Numbers\n\nActivity 27\n\nBy recognising the locus\n\n$$|z - 2| = 3|z - 10|$$\n\nas the circle of Apollonius, use the idea of simple ratios to determine the coordinates of the centre and the radius of the circle.\n\nCheck your answer by finding the cartesian equation of the circle.\n\nActivity 28\n\nBy folding a piece of paper, create an angle of $45^\\circ$ and cut it out. Now mark two fixed points on a piece of paper and explore the locus of the vertex as you keep the two sides of the cut-out in contact with the fixed points as shown.\n\nYou should find that $P$ moves on the arc of a circle.\n\nAlternatively, when you have a circle and two fixed points $A$ and $B$, if you choose a sequence of points $P_1, P_2, P_3, \\ldots$ on the circumference, what do you notice about the angles $AP_1B, AP_2B, AP_3B, \\ldots$?\n\nThis is an example of the constant angle locus.\n\nExample\n\nThe point $P$ represents $z$ in the complex plane. Find the locus of $P$ in each of the cases below when $z$ satisfies\n\n(a) $\\arg z = \\frac{5\\pi}{6}$\n\n(b) $\\arg(z - 2 + 3i) = -\\frac{\\pi}{4}$\n\n(c) $\\arg \\frac{z - 1}{z + 1} = \\frac{\\pi}{4}$\n\nSolution\n\n(a) The locus is a half-line starting at the origin making an angle $\\frac{5\\pi}{6}$ with the real axis.\n(b) The half-line to be considered here is one which starts at the point $2 - 3i$.\n\nIt makes an angle of $\\frac{\\pi}{4}$ below the real axis as shown opposite.\n\n(c) You need to make use of the fact that\n\n$$\\arg \\frac{z - 1}{z + 1} = \\arg(z - 1) - \\arg(z + 1)$$\n\nOne possible solution for $z$ is shown in the second diagram opposite.\n\nBy the results of Activity 28 you can see that the locus of $z$ is the major arc of a circle passing through $1 + 0i$ and $-1 + 0i$.\n\nSince the angle at the centre of the circle is twice that on the circumference, it can be seen that the centre of the circle is at $0 + i$ and hence the radius of the circle is $\\sqrt{2}$.\n\nThe problem can be tackled algebraically but there are difficulties that can creep in by assuming\n\n$$\\arg(x + iy) = \\tan^{-1} \\frac{y}{x}$$\n\nNevertheless, you can obtain the cartesian equation of the full circle of which the locus is only part.\n\nLet $z = x + iy$\n\n$$\\arg \\frac{z - 1}{z + 1} = \\arg \\frac{x - 1 + iy}{x + 1 + iy}$$\n\n$$= \\arg \\frac{(x - 1) + iy}{(x + 1)^2 + y^2}$$\n\n$$= \\arg \\frac{x^2 - 1 + y^2 + 2iy}{(x + 1)^2 + y^2} = \\frac{\\pi}{4}$$\n\nTaking tangents of both sides\n\n$$\\frac{2y}{x^2 - 1 + y^2} = 1$$\n\n$$\\Rightarrow x^2 + y^2 - 2y = 1$$\n\n$$x^2 + (y - 1)^2 = 2$$\n\nfrom which we see the centre is $0 + i$ and the radius is $\\sqrt{2}$.\n\nNote: this approach does not indicate whether the locus is the major or minor arc of the circle and so the first approach is recommended.\nChapter 3 Complex Numbers\n\nExercise 3E\n\n1. Sketch the locus of \\( z \\) described by\n (a) \\( |z + 3 - 4i| = 5 \\)\n (b) \\( |z + 2| = |z - 5 + i| \\)\n (c) \\( |z + 3i| = 3|z - i| \\)\n (d) \\( |z - 2i| + |z - 3 + i| = 0 \\)\n\n2. Describe geometrically and obtain a cartesian equation for the locus of \\( z \\) in each of the following cases.\n (a) \\( |z - 3|^2 = 100 \\)\n (b) \\( |z - 1|^2 + |z - 4|^2 = 9 \\)\n (c) \\( |z - 1| + |z - 4| = 5 \\)\n (d) \\( |z - 1| - |z - 4| = 1 \\)\n\n3. Describe geometrically and sketch the region on the complex plane for which\n (a) \\( 2 < |z - 3 + i| \\leq 5 \\)\n (b) \\( \\frac{\\pi}{4} \\leq \\arg(z - 2i) < \\frac{\\pi}{3} \\)\n\n4. Sketch the loci for which\n (a) \\( \\arg \\left( \\frac{z + 1}{z - 1} \\right) = \\frac{3\\pi}{2} \\)\n (b) \\( \\arg(z - 2)^3 = \\frac{\\pi}{2} \\)\n (c) \\( \\arg(z + 2) - \\arg(z - 3) = \\frac{\\pi}{3} \\)\n (d) \\( \\arg \\left( \\frac{z - 5 + 7i}{z + 4 + 4i} \\right) = \\frac{\\pi}{2} \\)\n\n3.13 Miscellaneous Exercises\n\n1. Find the modulus and argument of the complex numbers\n \\( z_1 = 1 + i \\) and \\( z_2 = 1 - \\sqrt{3}i \\).\n Hence find in the form \\( z = [r, \\theta] \\) where \\(-\\pi < \\theta \\leq \\pi\\) and \\( r > 0\\), the complex numbers\n (a) \\( z_1z_2 \\)\n (b) \\( \\frac{z_1}{z_2} \\)\n (c) \\( \\frac{z_2}{z_1} \\)\n (d) \\( z_1^2 \\)\n (e) \\( z_2^3 \\)\n (f) \\( \\frac{z_1^2}{z_2^3} \\)\n\n2. Express the numbers \\( 1, 3i, -4, z = 2 + \\sqrt{3}i \\) in the \\([r, \\theta]\\) form. Hence express in the \\([r, \\theta]\\) form\n (a) \\( \\frac{1}{z} \\)\n (b) \\( 3zi \\)\n (c) \\( \\frac{z}{3i} \\)\n (d) \\( -4z \\)\n (e) \\( \\frac{-4}{z} \\)\n\n3. Find \\( \\sqrt[3]{3 + i} \\) in the \\([r, \\theta]\\) form. Hence find\n (a) \\( \\left( \\sqrt[3]{3 + i} \\right)^3 \\)\n (b) \\( \\left( \\sqrt[3]{3 + i} \\right)^5 \\)\n in the form \\( a + bi \\).\n (c) Find the least value of the positive integer \\( n \\) for which \\( \\left( \\sqrt[3]{3 + i} \\right)^n \\) is\n (i) purely real\n (ii) purely imaginary.\n\n4. Find in the form \\( a + bi \\)\n (a) \\( \\left( 1 + \\sqrt{3}i \\right)^5 \\)\n (b) \\( \\left( \\sqrt{3} - i \\right)^{10} \\)\n (c) \\( (1 - i)^7 \\)\n by making use of de Moivre's theorem.\n\n5. Simplify \\( (1 + i)^{10} - (1 - i)^{10} \\).\n Given that \\( n \\) is a positive integer, show that\n \\( (1 + i)^{4n} - (1 - i)^{4n} = 0 \\).\n\n6. Given that \\( z = \\frac{\\sqrt{3}}{2} + \\frac{1}{2}i \\), simplify \\( z^2, z^3, z^4 \\) and illustrate each of these numbers as points on an Argand diagram.\n\n7. Show that the three roots of \\( z^3 = 1 \\) can be expressed in the form \\( 1, \\omega, \\omega^2 \\).\n Hence show that \\( 1 + \\omega + \\omega^2 = 0 \\).\n Using this relation and the fact that \\( \\omega^3 = 1 \\), simplify the following\n (a) \\( (1 + \\omega)^7 \\)\n (b) \\( (1 - \\omega)(1 - \\omega^2) \\)\n (c) \\( \\frac{\\omega^3}{1 + \\omega} \\)\n (d) \\( (1 - \\omega + \\omega^2)^4 \\)\n (e) \\( (\\omega - \\omega^2)^5 \\)\n (f) \\( \\frac{(1 + \\omega^2)(1 - \\omega)}{(1 + \\omega)} \\)\n\n8. The roots of the equation \\( z^2 + 4z + 29 = 0 \\) are \\( z_1 \\) and \\( z_2 \\). Show that \\( |z_1| = |z_2| \\) and calculate, in degrees, the argument of \\( z_1 \\) and the argument of \\( z_2 \\).\n In an Argand diagram, O is the origin and \\( z_1 \\) and \\( z_2 \\) are represented by the points P and Q.\n Calculate the radius of the circle passing through the points O, P and Q. (AEB)\n9. Sketch on an Argand diagram the loci given by\n\\[ |z - 1 - 2i| = 5 \\]\n\\[ |z - 5 + i| = |z + 3 - 5i| \\]\nShow that these loci intersect at the point \\( z_1 = -2 - 2i \\), and at a second point \\( z_2 \\). Find \\( z_2 \\) in the form \\( a + bi \\), where \\( a \\) and \\( b \\) are real.\nExpress \\( z_1 \\) in the form \\( r(\\cos \\alpha + i \\sin \\alpha) \\) where \\( r > 0 \\) and \\( -\\pi < \\alpha \\leq \\pi \\), giving the value of \\( r \\) and the value of \\( \\alpha \\). Show that \\( z_1 \\) is a root of the equation \\( z^4 + 64 = 0 \\).\nExpress \\( z^4 + 64 \\) in the form \\( (z^2 + Az + B)(z^2 + Cz + D) \\)\nwhere \\( A, B, C \\) and \\( D \\) are real, and find these numbers. (AEB)\n\n10. (a) Find the modulus and argument of the complex number \\( \\frac{\\sqrt{3} + i}{1 + i} \\), giving the argument in radians between \\( -\\pi \\) and \\( \\pi \\).\n(b) Find the value of the real number \\( \\lambda \\) in the case when \\( \\frac{\\sqrt{3} + i \\lambda}{1 + i \\sqrt{3}} \\) is real. (AEB)\n\n11. The complex number \\( u = -10 + 9i \\)\n(a) Show the complex number \\( u \\) on an Argand diagram.\n(b) Giving your answer to the nearest degree, calculate the argument of \\( u \\).\n(c) Find the complex number \\( v \\) which satisfies the equation \\( uv = -11 + 28i \\).\n(d) Verify that \\( |u + v| = 8 \\sqrt{2} \\). (AEB)\n\n12. (a) The complex number \\( z \\) satisfies the equation \\( |z + 1| = \\sqrt{2} |z - 1| \\). The point \\( P \\) represents \\( z \\) on an Argand diagram. Show that the locus of \\( P \\) is a circle with its centre on the real axis, and find its radius.\n(b) Find the four roots of the equation \\( (z + 1)^4 + 4(z - 1)^4 = 0 \\), expressing the roots \\( z_1, z_2, z_3 \\) and \\( z_4 \\) in the form \\( a + bi \\).\nShow that the points on an Argand diagram representing \\( z_1, z_2, z_3 \\) and \\( z_4 \\) are the vertices of a trapezium and calculate its area. (AEB)\n\n13. Let \\( z \\) be the complex number \\(-1 + \\sqrt{3}i\\).\n(a) Express \\( z^2 \\) in the form \\( a + bi \\).\n(b) Find the value of the real number \\( p \\) such that \\( z^2 + pz \\) is real.\n(c) Find the value of the real number \\( q \\) such that \\( \\text{Arg}(z^2 + qz) = \\frac{5\\pi}{6} \\). (AEB)\n\n14. Use the method of mathematical induction to prove that\n\\[ (\\cos \\theta + i \\sin \\theta)^n = \\cos n\\theta + i \\sin n\\theta \\]\nwhere \\( n \\) is a positive integer.\nDeduce that the result is also true when \\( n \\) is a negative integer.\nShow that\n\\[ 2 \\cos \\theta = z^n + z^{-n}, \\]\nwhere \\( z = \\cos \\theta + i \\sin \\theta \\).\nBy considering \\( (z + z^{-1})^4 \\), show that\n\\[ \\cos 4\\theta = \\frac{1}{8}(\\cos 4\\theta + 4 \\cos 2\\theta + 3). \\]\nHence evaluate \\( \\int \\cos^4 2\\theta \\, d\\theta \\). (AEB)\n\n15. You are given the complex number \\( \\omega = \\cos \\frac{2\\pi}{5} + i \\sin \\frac{2\\pi}{5} \\).\n(a) Write down the value of \\( \\omega^5 \\) and prove that \\( 1 + \\omega + \\omega^2 + \\omega^3 + \\omega^4 = 0 \\).\nSimplify \\( (\\omega + \\omega^4)(\\omega^2 + \\omega^3) \\).\nForm a quadratic equation with integer coefficients having roots \\( \\omega + \\omega^4 \\) and \\( \\omega^2 + \\omega^3 \\) and hence prove that\n\\[ \\cos \\frac{2\\pi}{5} = -\\frac{1 + \\sqrt{5}}{4}. \\]\n(b) In an Argand diagram the point \\( P \\) is represented by the complex number \\( z \\).\nSketch and describe geometrically in each case, the locus of the point \\( P \\) when\n(i) \\( |z - \\omega| = |z - 1| \\)\n(ii) \\( \\arg \\frac{z - \\omega}{z - 1} = \\frac{\\pi}{5} \\). (AEB)\n16. (a) Use de Moivre’s theorem to show that\n\\[\n(\\sqrt{3} - i)^n = 2^n \\cos \\frac{n\\pi}{6} - i\\sin \\frac{n\\pi}{6},\n\\]\nwhere \\(n\\) is an integer.\n\n(i) Find the least positive integer \\(m\\) for which \\((\\sqrt{3} - i)^m\\) is real and positive.\n\n(ii) Given that \\((\\sqrt{3} - i)\\) is a root of the equation\n\\[\nz^9 + 16(1 + i)z^3 + a + ib = 0,\n\\]\nfind the values of the real constants \\(a\\) and \\(b\\).\n\n(b) The point \\(P\\) represents a complex number \\(z\\) on an Argand diagram and\n\\[\n|z - \\omega^2| = 3|z - \\omega^3|\n\\]\nwhere \\(\\omega = \\sqrt{3} - i\\).\n\nShow that the locus of \\(P\\) is a circle and find its radius and the complex number represented by its centre.\n\n(AEB)", "id": "./materials/367.pdf" }, { "contents": "Basic Mathematics\n\nIntroduction to Complex Numbers\n\nMartin Lavelle\n\nThe aim of this package is to provide a short study and self assessment programme for students who wish to become more familiar with complex numbers.\nTable of Contents\n\n1. The Square Root of Minus One!\n2. Real, Imaginary and Complex Numbers\n3. Adding and Subtracting Complex Numbers\n4. Multiplying Complex Numbers\n5. Complex Conjugation\n6. Dividing Complex Numbers\n7. Quiz on Complex Numbers\n Solutions to Exercises\n Solutions to Quizzes\n\nThe full range of these packages and some instructions, should they be required, can be obtained from our web page Mathematics Support Materials.\n1. The Square Root of Minus One!\n\nIf we want to calculate the square root of a negative number, it rapidly becomes clear that neither a positive or a negative number can do it.\n\nE.g., \\( \\sqrt{-1} \\neq \\pm 1 \\), since \\( 1^2 = (-1)^2 = +1 \\).\n\nTo find \\( \\sqrt{-1} \\) we introduce a new quantity, \\( i \\), defined to be such that \\( i^2 = -1 \\). (Note that engineers often use the notation \\( j \\).)\n\n**Example 1**\n\n(a) \\( \\sqrt{-25} = 5i \\)\n\nSince \\( (5i)^2 = 5^2 \\times i^2 \\)\n\n\\[ = 25 \\times (-1) \\]\n\n\\[ = -25. \\]\n(b) \\[ \\sqrt{-\\frac{16}{9}} = \\frac{4}{3}i \\]\n\nSince \\( \\left(\\frac{4}{3}i\\right)^2 = \\frac{16}{9} \\times (i^2) \\)\n\n\\[ = -\\frac{16}{9}. \\]\n\n2. Real, Imaginary and Complex Numbers\n\nReal numbers are the usual positive and negative numbers.\n\nIf we multiply a real number by \\( i \\), we call the result an imaginary number. Examples of imaginary numbers are: \\( i \\), \\( 3i \\) and \\( -i/2 \\).\n\nIf we add or subtract a real number and an imaginary number, the result is a complex number. We write a complex number as\n\n\\[ z = a + ib \\]\n\nwhere \\( a \\) and \\( b \\) are real numbers.\n3. Adding and Subtracting Complex Numbers\n\nIf we want to add or subtract two complex numbers, \\( z_1 = a + ib \\) and \\( z_2 = c + id \\), the rule is to add the real and imaginary parts separately:\n\n\\[\n\\begin{align*}\n z_1 + z_2 &= a + ib + c + id = a + c + i(b + d) \\\\\n z_1 - z_2 &= a + ib - c - id = a - c + i(b - d)\n\\end{align*}\n\\]\n\nExample 2\n\n(a) \\((1 + i) + (3 + i) = 1 + 3 + i(1 + 1) = 4 + 2i\\)\n\n(b) \\((2 + 5i) - (1 - 4i) = 2 + 5i - 1 + 4i = 1 + 9i\\)\n\nExercise 1. Add or subtract the following complex numbers. (Click on the green letters for the solutions.)\n\n(a) \\((3 + 2i) + (3 + i)\\) \n(b) \\((4 - 2i) - (3 - 2i)\\)\n\n(c) \\((-1 + 3i) + \\frac{1}{2}(2 + 2i)\\) \n(d) \\(\\frac{1}{3}(2 - 5i) - \\frac{1}{6}(8 - 2i)\\)\nQuiz To which of the following does the expression\n\\[(4 - 3i) + (2 + 5i)\\]\nsimplify?\n\n(a) 6 − 8i \n(b) 6 + 2i \n(c) 1 + 7i \n(d) 9 − i\n\nQuiz To which of the following does the expression\n\\[(3 - i) - (2 - 6i)\\]\nsimplify?\n\n(a) 3 − 9i \n(b) 2 + 4i \n(c) 1 − 5i \n(d) 1 + 5i\n4. Multiplying Complex Numbers\n\nWe multiply two complex numbers just as we would multiply expressions of the form \\((x + y)\\) together (see the package on Brackets)\n\n\\[\n(a + ib)(c + id) = ac + a(id) + (ib)c + (ib)(id) \\\\\n= ac + iad + ibc - bd \\\\\n= ac - bd + i(ad + bc)\n\\]\n\nExample 3\n\n\\[\n(2 + 3i)(3 + 2i) = 2 \\times 3 + 2 \\times 2i + 3i \\times 3 + 3i \\times 2i \\\\\n= 6 + 4i + 9i - 6 \\\\\n= 13i\n\\]\nExercise 2. Multiply the following complex numbers. (Click on the green letters for the solutions.)\n\n(a) $(3 + 2i)(3 + i)$ \n(b) $(4 - 2i)(3 - 2i)$ \n(c) $(-1 + 3i)(2 + 2i)$ \n(d) $(2 - 5i)(8 - 3i)$\n\nQuiz To which of the following does the expression $(2 - i)(3 + 4i)$ simplify?\n\n(a) $5 + 4i$ \n(b) $6 + 11i$ \n(c) $10 + 5i$ \n(d) $6 + i$\n5. Complex Conjugation\n\nFor any complex number, \\( z = a + ib \\), we define the complex conjugate to be: \\( z^* = a - ib \\). It is very useful since the following are real:\n\n\\[\n\\begin{align*}\n z + z^* &= a + ib + (a - ib) = 2a \\\\\n zz^* &= (a + ib)(a - ib) = a^2 + iab - iab - (ib)^2 = a^2 + b^2\n\\end{align*}\n\\]\n\nThe modulus of a complex number is defined as: \\( |z| = \\sqrt{zz^*} \\)\n\n**Exercise 3.** Combine the following complex numbers and their conjugates. (Click on the green letters for the solutions.)\n\n(a) If \\( z = (3 + 2i) \\), find \\( z + z^* \\) \n(b) If \\( z = (3 - 2i) \\), find \\( zz^* \\) \n(c) If \\( z = (-1 + 3i) \\), find \\( zz^* \\) \n(d) If \\( z = (4 - 3i) \\), find \\( |z| \\)\n\n**Quiz** Which of the following is the modulus of \\( 4 - 2i \\)?\n\n(a) \\( \\sqrt{20} \\) \n(b) 2 \n(c) 20 \n(d) \\( \\sqrt{12} \\)\n6. Dividing Complex Numbers\n\nThe trick for dividing two complex numbers is to multiply top and bottom by the complex conjugate of the denominator:\n\n\\[\n\\frac{z_1}{z_2} = \\frac{z_1}{z_2} \\times \\frac{z_2^*}{z_2^*} = \\frac{z_1 z_2^*}{z_2 z_2^*}\n\\]\n\nThe denominator, \\(z_2 z_2^*\\), is now a real number.\n\nExample 4\n\n\\[\n\\frac{1}{i} = \\frac{1}{i} \\times \\frac{-i}{-i} = \\frac{-i}{i \\times (-i)} = \\frac{-i}{1} = -i\n\\]\nExample 5\n\n\\[\n\\frac{2 + 3i}{1 + 2i} = \\frac{(2 + 3i)(1 - 2i)}{(1 + 2i)(1 - 2i)} = \\frac{(2 + 3i)(1 - 2i)}{1 + 4} = \\frac{1}{5}(2 + 3i)(1 - 2i) = \\frac{1}{5}(2 - 4i + 3i + 6) = \\frac{1}{5}(8 - i)\n\\]\n\nExercise 4. Perform the following divisions: (Click on the green letters for the solutions.)\n\n(a) \\(\\frac{2 + 4i}{i}\\) \n(b) \\(\\frac{-2 + 6i}{1 + 2i}\\) \n(c) \\(\\frac{1 + 3i}{2 + i}\\) \n(d) \\(\\frac{3 + 2i}{3 + i}\\)\nQuiz To which of the following does the expression \\( \\frac{8 - i}{2 + i} \\) simplify?\n\n(a) \\( 3 - 2i \\) \n(b) \\( 2 + 3i \\) \n(c) \\( 4 - \\frac{1}{2}i \\) \n(d) \\( 4 \\)\n\nQuiz To which of the following does the expression \\( \\frac{-2 + i}{2 + i} \\) simplify?\n\n(a) \\( -1 \\) \n(b) \\( \\frac{1}{5} (-5 + 7i) \\) \n(c) \\( -1 + \\frac{1}{2}i \\) \n(d) \\( \\frac{1}{5} (-3 + 4i) \\)\n7. Quiz on Complex Numbers\n\nBegin Quiz In each of the following, simplify the expression and choose the solution from the options given.\n\n1. \\((3 + 4i) - (2 - 3i)\\)\n (a) 3 - i (b) 5 + 7i (c) 1 + 7i (d) 1 - i\n\n2. \\((3 + 3i)(2 - 3i)\\)\n (a) 6 - 8i (b) 6 + 8i (c) -3 + 3i (d) 15 - 3i\n\n3. \\(|12 - 5i|\\)\n (a) 13 (b) \\(\\sqrt{7}\\) (c) \\(\\sqrt{119}\\) (d) -12.5\n\n4. \\((7 - 17i)/(5 - i)\\)\n (a) \\(\\frac{7}{2} + 17i\\) (b) 3 + i (c) -2 + 2i (d) 2 - 3i\n\nEnd Quiz Score: Correct\nSolutions to Exercises\n\nExercise 1(a)\n\n\\[(3 + 2i) + (3 + i) = 3 + 2i + 3 + i\\]\n\\[= 3 + 3 + 2i + 2i\\]\n\\[= 6 + 3i\\]\n\nClick on the green square to return\nExercise 1(b) Here we need to be careful with the signs!\n\n\\[\n4 - 2i - (3 - 2i) = 4 - 2i - 3 + 2i \\\\\n= 4 - 3 - 2i + 2i \\\\\n= 1\n\\]\n\nA purely real result\n\nClick on the green square to return\nExercise 1(c) The factor of $\\frac{1}{2}$ multiplies both terms in the complex number.\n\n\\[-1 + 3i + \\frac{1}{2}(2 + 2i) = -1 + 3i + 1 + i = 4i\\]\n\nA purely imaginary result.\n\nClick on the green square to return\nExercise 1(d)\n\n\\[\n\\frac{1}{3}(2 - 5i) - \\frac{1}{6}(8 - 2i) = \\frac{2}{3} - \\frac{5}{3}i - \\frac{8}{6} + \\frac{2}{6}i\n\\]\n\n\\[\n= \\frac{2}{3} - \\frac{5}{3}i - \\frac{4}{3} + \\frac{1}{3}i\n\\]\n\n\\[\n= \\frac{2}{3} - \\frac{4}{3}i + \\frac{1}{3}i\n\\]\n\n\\[\n= -\\frac{2}{3} - \\frac{4}{3}i\n\\]\n\nwhich we could also write as \\(-\\frac{2}{3}(1 + 2i)\\).\n\nClick on the green square to return\nExercise 2(a)\n\n\\[(3 + 2i)(3 + i) = 3 \\times 3 + 3 \\times i + 2i \\times 3 + 2i \\times i\\]\n\\[= 9 + 3i + 6i - 2\\]\n\\[= 9 - 2 + 3i + 6i\\]\n\\[= 7 + 9i\\]\nExercise 2(b)\n\n\\[(4 - 2i)(3 - 2i) = 4 \\times 3 + 4 \\times (-2i) - 2i \\times 3 - 2i \\times -2i\\]\n\\[= 12 - 8i - 6i - 4\\]\n\\[= 12 - 4 - 8i - 6i\\]\n\\[= 8 - 14i\\]\n\nClick on the green square to return\nExercise 2(c)\n\n\\((-1 + 3i)(2 + 2i) = -1 \\times 2 - 1 \\times 2i + 3i \\times 2 + 3i \\times 2i\\)\n\n\\[= -2 - 2i + 6i - 6\\]\n\n\\[= -2 - 6 - 2i + 6i\\]\n\n\\[= -8 + 4i\\]\nExercise 2(d)\n\n\\[(2 - 5i)(8 - 3i) = 2 \\times 8 + 2 \\times (-3i) - 5i \\times 8 - 5i \\times (-3i)\\]\n\\[= 16 - 6i - 40i - 15\\]\n\\[= 16 - 15 - 6i - 40i\\]\n\\[= 1 - 46i\\]\nExercise 3(a)\n\n\\[(3 + 2i) + (3 + 2i)^* = (3 + 2i) + (3 - 2i)\\]\n\\[= 3 + 2i + 3 - 2i\\]\n\\[= 3 + 3 + 2i - 2i\\]\n\\[= 6\\]\nExercise 3(b)\n\n\\[(3 - 2i)(3 - 2i)^* = (3 - 2i)(3 + 2i)\\]\n\\[= 9 + 6i - 6i - 2i \\times (2i)\\]\n\\[= 9 - 4i^2\\]\n\\[= 9 + 4 = 13\\]\nExercise 3(c)\n\n\\((-1 + 3i)(-1 + 3i)^*\\) \n\\[= (-1 + 3i)(-1 - 3i)\\] \n\\[= (-1) \\times (-1) + (-1)(-3i) + 3i(-1) + 3i(-3i)\\] \n\\[= 1 + 3i - 3i - 9i^2\\] \n\\[= 1 + 9 = 10\\]\nExercise 3(d)\n\n\\[\n\\sqrt{(4 - 3i)(4 + 3i)} = \\sqrt{4^2 + 4 \\times 3i - 3i \\times 4 - 3i \\times 3i} \\\\\n= \\sqrt{16 + 12i - 12i - 9i^2} \\\\\n= \\sqrt{16 + 9} \\\\\n= \\sqrt{25} = 5\n\\]\n\nClick on the green square to return\nExercise 4(a)\n\n\\[\n\\frac{2 + 4i}{i} = \\frac{(2 + 4i) \\times -i}{i \\times -i} = \\frac{(2 + 4i) \\times (-i)}{-1} = (2 + 4i)(-i) = -2i - 4i^2 = 4 - 2i\n\\]\nExercise 4(b)\n\n\\[\n\\frac{-2 + 6i}{1 + 2i} = \\frac{(-2 + 6i)}{(1 + 2i)} \\times \\frac{(1 - 2i)}{(1 - 2i)}\n\\]\n\n\\[\n= \\frac{(-2 + 6i)(1 - 2i)}{1 + 4}\n\\]\n\n\\[\n= \\frac{1}{5}(-2 + 6i)(1 - 2i)\n\\]\n\n\\[\n= \\frac{1}{5}(-2 + 4i + 6i - 12i^2)\n\\]\n\n\\[\n= \\frac{1}{5}(-2 + 10i + 12)\n\\]\n\n\\[\n= \\frac{1}{5}(10 + 10i) = 2 + 2i\n\\]\nExercise 4(c)\n\n\\[\n\\frac{1 + 3i}{2 + i} = \\frac{(1 + 3i)(2 - i)}{(2 + i)(2 - i)} = \\frac{(1 + 3i)(2 - i)}{4 + 1} = \\frac{1}{5}(2 - i + 6i - 3i^2) = \\frac{1}{5}(2 + 3 + 5i) = \\frac{1}{5}(5 + 5i) = 1 + i\n\\]\nExercise 4(d)\n\n\\[\n\\frac{(3 + 2i)}{(3 + i)} = \\frac{(3 + 2i)}{(3 + i)} \\times \\frac{(3 - i)}{(3 - i)}\n\\]\n\n\\[\n= \\frac{(3 + 2i)(3 - i)}{9 + 1}\n\\]\n\n\\[\n= \\frac{1}{10} (3 + 2i)(3 - i)\n\\]\n\n\\[\n= \\frac{1}{10} (9 - 3i + 6i - 2i^2)\n\\]\n\n\\[\n= \\frac{1}{10} (9 + 2 + 3i)\n\\]\n\n\\[\n= \\frac{1}{10} (11 + 3i)\n\\]\nSolutions to Quizzes\n\nSolution to Quiz:\n\n\\[(4 - 3i) + (2 + 5i) = 4 - 3i + 2 + 5i = 4 + 2 - 3i + 5i = 6 + 2i\\]\n\nEnd Quiz\nSolution to Quiz:\n\nBe careful with the signs!\n\n\\[(3 - i) - (2 - 6i) = 3 - i - 2 + 6i\\]\n\\[= 3 - 2 - i + 6i\\]\n\\[= 1 + 5i\\]\n\nEnd Quiz\nSolution to Quiz:\n\n\\[(2 - i)(3 + 4i) = 2 \\times 3 + 2 \\times (4i) - i \\times 3 - i \\times (4i)\\]\n\\[= 6 + 8i - 3i - 4i^2\\]\n\\[= 6 + 5i + 4\\]\n\\[= 10 + 5i\\]\n\nEnd Quiz\nSolution to Quiz:\n\n\\[ |4 - 2i| = \\sqrt{(4 - 2i)(4 + 2i)} \\]\n\\[ = \\sqrt{4^2 + 2^2} \\]\n\\[ = \\sqrt{16 + 4} \\]\n\\[ = \\sqrt{20} \\]\n\nEnd Quiz\nSolution to Quiz:\n\n\\[\n\\frac{8 - i}{2 + i} = \\frac{8 - i}{2 + i} \\times \\frac{2 - i}{2 - i} \\\\\n= \\frac{(8 - i)(2 - i)}{2^2 + 1^2} \\\\\n= \\frac{(8 \\times 2 + 8 \\times (-i) - i \\times 2 - i \\times (-i))}{5} \\\\\n= \\frac{1}{5} (16 - 8i - 2i - 1) \\\\\n= \\frac{1}{5} (15 - 10i) = 3 - 2i\n\\]\n\nEnd Quiz\nSolution to Quiz:\n\n\\[\n\\frac{-2 + i}{2 + i} = \\frac{-2 + i}{2 + i} \\cdot \\frac{2 - i}{2 - i} = \\frac{(-2 + i)(2 - i)}{2^2 + 1^2} = \\frac{1}{5} (-2 \\times 2 - 2 \\times (-i) + i \\times 2 + i \\times (-i)) = \\frac{1}{5} (-4 + 2i + 2i + 1) = \\frac{1}{5} (-3 + 4i)\n\\]\n\nEnd Quiz", "id": "./materials/368.pdf" }, { "contents": "Complex numbers - Exercises with detailed solutions\n\n1. Compute real and imaginary part of \\( z = \\frac{i - 4}{2i - 3} \\).\n\n2. Compute the absolute value and the conjugate of\n\n\\[\nz = (1 + i)^6, \\quad w = i^{17}.\n\\]\n\n3. Write in the “algebraic” form \\((a + ib)\\) the following complex numbers\n\n\\[\nz = i^5 + i + 1, \\quad w = (3 + 3i)^8.\n\\]\n\n4. Write in the “trigonometric” form \\((\\rho \\cos \\theta + i \\sin \\theta)\\) the following complex numbers\n\n\\[\na) 8 \\quad b) 6i \\quad c) \\left(\\cos \\frac{\\pi}{3} - i \\sin \\frac{\\pi}{3}\\right)^7.\n\\]\n\n5. Simplify\n\n\\[\n(a) \\quad \\frac{1 + i}{1 - i} - (1 + 2i)(2 + 2i) + \\frac{3 - i}{1 + i};\n\\]\n\n\\[\n(b) \\quad 2i(i - 1) + \\left(\\sqrt{3} + i\\right)^3 + (1 + i)(1 + i).\n\\]\n\n6. Compute the square roots of \\( z = -1 - i \\).\n\n7. Compute the cube roots of \\( z = -8 \\).\n\n8. Prove that there is no complex number such that \\(|z| - z = i\\).\n\n9. Find \\( z \\in \\mathbb{C} \\) such that\n\n\\[\na) \\pi = i(z - 1) \\quad b) z^2 \\cdot \\pi = z \\quad c) |z + 3i| = 3|z|.\n\\]\n\n10. Find \\( z \\in \\mathbb{C} \\) such that \\( z^2 \\in \\mathbb{R} \\).\n\n11. Find \\( z \\in \\mathbb{C} \\) such that\n\n\\[\n(a) \\quad \\text{Re} \\left( z(1 + i) \\right) + z\\pi = 0; \\\\\n(b) \\quad \\text{Re} \\left( z^2 \\right) + i \\text{Im} \\left( \\pi(1 + 2i) \\right) = -3; \\\\\n(c) \\quad \\text{Im} \\left( (2 - i)z \\right) = 1.\n\\]\n\n12. Find \\( a \\in \\mathbb{R} \\) such that \\( z = -i \\) is a root for the polynomial \\( P(z) = z^3 - z^2 + z + 1 + a \\). Furthermore, for such value of \\( a \\) find the factors of \\( P(z) \\) in \\( \\mathbb{R} \\) and in \\( \\mathbb{C} \\).\nSolutions\n\n1. \\( z = \\frac{i - 4}{2i - 3} = \\frac{i - 4 \\cdot 2i + 3}{2i - 3 \\cdot 2i + 3} = \\frac{-2 + 3i - 8i - 12}{-4 - 9} = \\frac{14}{13} + \\frac{5}{13}i \\) hence \\( \\text{Re}(z) = \\frac{14}{13} \\) and \\( \\text{Im}(z) = \\frac{5}{13} \\).\n\n2. \\( z = (1 + i)^6 = \\left( \\sqrt{2}(\\cos \\frac{\\pi}{4} + i \\sin \\frac{\\pi}{4}) \\right)^6 = 8 \\left( \\cos \\frac{3\\pi}{2} + i \\sin \\frac{3\\pi}{2} \\right) = -8i. \\) Hence \\( |z| = 8 \\) and \\( \\bar{z} = 8i. \\)\n\n3. \\( w = i^{17} = i \\cdot i^{16} = i \\cdot (i^4)^4 = i \\cdot (1)^4 = i. \\) Hence \\( |w| = 1 \\) and \\( \\bar{w} = -i. \\)\n\n4. If \\( z = a + ib, a, b \\in \\mathbb{R}, \\) its trigonometric form is\n\n\\[ z = \\rho (\\cos \\theta + i \\sin \\theta), \\quad \\text{where } \\rho := \\sqrt{a^2 + b^2} \\text{ and } \\theta \\text{ is such that } \\cos \\theta = \\frac{a}{\\rho}, \\sin \\theta = \\frac{b}{\\rho}. \\]\n\na) \\( a = 8, b = 0, \\cos \\theta = 1 \\) e \\( \\sin \\theta = 0. \\) Hence \\( 8 = 8 (\\cos 0 + i \\sin 0). \\)\n\nb) \\( 6i = 6 (0 + i) = 6 (\\cos \\frac{\\pi}{2} + i \\sin \\frac{\\pi}{2}). \\)\n\nc) We use the de Moivre’s Formula:\n\n\\[ \\left( \\cos \\left( \\frac{\\pi}{3} \\right) - i \\sin \\left( \\frac{\\pi}{3} \\right) \\right)^7 = \\cos \\frac{7\\pi}{3} - i \\sin \\frac{7\\pi}{3} = \\cos 2\\pi + \\frac{\\pi}{3} - i \\sin 2\\pi + \\frac{\\pi}{3} = \\cos \\frac{\\pi}{3} - i \\sin \\frac{\\pi}{3}. \\]\n\n5. (a) We compute\n\n\\[ \\frac{1 + i}{1 - i} - (1 + 2i)(2 + 2i) + \\frac{3 - i}{1 + i} = \\frac{1 + i}{1 - i} \\cdot \\frac{1 + i}{1 + i} - (1 + 2i)(2 + 2i) + \\frac{3 - i}{1 + i} \\cdot \\frac{1 - i}{1 - i} \\]\n\n\\[ = i - 2 - 2i - 4i + 4 + \\frac{3 - 1 - 3i - i}{2} = i + 2 - 6i + \\frac{2 - 4i}{2} = 2 - 5i + 1 - 2i = 3 - 7i. \\]\n\n(b) Since\n\n\\[ \\left( \\sqrt{3} + i \\right)^3 = \\left( \\sqrt{3} - i \\right)^3 = \\left( \\sqrt{3} - i \\right)^2 \\left( \\sqrt{3} - i \\right) = \\left( 3 - 1 - 2i \\sqrt{3} \\right) \\left( \\sqrt{3} - i \\right) \\]\n\n\\[ = \\left( 2 - 2i \\sqrt{3} \\right) \\left( \\sqrt{3} - i \\right) = 2 \\sqrt{3} - 2i - 6i - 2 \\sqrt{3} = -8i, \\]\n\nwe obtain\n\n\\[ 2i(i - 1) + \\left( \\sqrt{3} + i \\right)^3 + (1 + i)(1 + i) = -2 - 2i - 8i + 2 = -10i. \\]\n\n6. Every \\( z \\in \\mathbb{C} \\) has \\( n \\) distinct roots of order \\( n \\), which correspond (in the complex plane) to the vertices of a regular \\( n \\)-agon inscribed in the circle of radius \\( \\sqrt{|z|} \\) centered at the origin.\n\nWhen \\( z = \\rho (\\cos \\theta + i \\sin \\theta) = \\rho e^{i \\theta}, \\) then the roots of order \\( n \\) of \\( z \\) are\n\n\\[ \\sqrt[n]{\\rho} \\left( \\cos \\left( \\frac{\\theta + 2k\\pi}{n} \\right) + i \\sin \\left( \\frac{\\theta + 2k\\pi}{n} \\right) \\right) = \\sqrt[n]{\\rho} e^{\\frac{\\theta + 2k\\pi}{n}}. \\quad k = 0, 1, 2, \\ldots, n - 1. \\]\n\nThe square roots of \\( z = -1 - i = \\sqrt{2} (\\cos \\frac{5\\pi}{4} + i \\sin \\frac{5\\pi}{4}) \\) are\n\n\\[ z_1 = \\sqrt[2]{2} \\left( \\cos \\left( \\frac{5\\pi}{2} \\right) + i \\sin \\left( \\frac{5\\pi}{2} \\right) \\right) = \\sqrt{2} \\left( \\cos \\frac{5\\pi}{8} + i \\sin \\frac{5\\pi}{8} \\right) \\text{ and } \\]\nThe trigonometric form of\n\nSuppose that some\n\nWe will use the notation\n\nIf\n\nRe (\\(z\\)) = \\(\\sqrt{2}\\) \\(\\cos \\frac{13\\pi}{8} + i \\sin \\frac{13\\pi}{8}\\).\n\nWe could also argue as follows: the equation\n\n\\((x + iy)^2 = -1 - i\\)\n\nis equivalent to the system\n\n\\[\n\\begin{align*}\n x^2 - y^2 &= -1, \\\\\n 2xy &= -1,\n\\end{align*}\n\\]\n\nwhich admits solutions\n\n\\[z = \\pm \\left(\\sqrt{\\frac{\\sqrt{2} - 1}{2}} - i \\frac{\\sqrt{2}}{\\sqrt{2} - 1}\\right)\\]\n\nwhich coincide with \\(z_1\\) and \\(z_2\\).\n\n7. The trigonometric form of \\(z = -8\\), is \\(z = 8(\\cos \\pi + i \\sin \\pi)\\). Then\n\n\\[z_1 = \\sqrt{8} \\left(\\cos \\left(\\frac{\\pi}{3}\\right) + i \\sin \\left(\\frac{\\pi}{3}\\right)\\right) = 2 \\left(\\cos \\left(\\frac{\\pi}{3}\\right) + i \\sin \\left(\\frac{\\pi}{3}\\right)\\right) = 1 + i\\sqrt{3},\\]\n\n\\[z_2 = \\sqrt{8} (\\cos \\pi + i \\sin \\pi) = 2 (\\cos \\pi + i \\sin \\pi) = -2,\\]\n\nand\n\n\\[z_3 = \\sqrt{8} \\left(\\cos \\left(\\frac{5\\pi}{3}\\right) + i \\sin \\left(\\frac{5\\pi}{3}\\right)\\right) = 2 \\left(\\cos \\left(\\frac{5\\pi}{3}\\right) + i \\sin \\left(\\frac{5\\pi}{3}\\right)\\right) = 1 - i\\sqrt{3}.\n\\]\n\n8. Suppose that some \\(z \\in \\mathbb{C}\\) satisfies the equation. Then \\(|z| = \\text{Re}(z) + i \\text{Im}(z) + 1\\). Hence, since \\(|z| \\in \\mathbb{R}\\), necessarily \\(\\text{Im}(z) = -1\\). The equation then is \\(\\sqrt{(\\text{Re}(z))^2 + 1} = \\text{Re}(z)\\), and, squaring, we obtain \\(1 = 0\\).\n\n9. We will use the notation \\(z = a + ib, a, b \\in \\mathbb{R}\\).\n\na) The equation becomes \\(a - ib = i(a + ib - 1)\\), that is \\(a - ib = -b + i(a - 1)\\). Then \\(a = -b\\) and \\(-b = a - 1\\), which has no solution; We conclude that the equation has no solution.\n\nb) The equation becomes \\(z \\cdot (z \\bar{z} - 1) = 0\\). Hence a first solution is \\(z = 0\\), while the others satisfy \\(z \\bar{z} = |z|^2 = 1\\). Then also all the points of the circle of radius 1 centered at the origin satisfies the equation.\n\nc) We square both terms and we obtain\n\n\\[|z + 3i|^2 = |a + i(b + 3)|^2 = a^2 + (b + 3)^2, \\quad (3|z|)^2 = 9(a^2 + b^2).\n\\]\n\nHence we have to solve the equation\n\n\\[a^2 + (b + 3)^2 = 9(a^2 + b^2) \\iff 8(a^2 + b^2) = 6b + 9 \\iff a^2 + b^2 - \\frac{3}{4}b = \\frac{9}{8} \\iff a^2 + \\left(b - \\frac{3}{8}\\right)^2 = \\left(\\frac{9}{8}\\right)^2.\n\\]\n\nThen the solution are all the points of the circle of radius \\(9/8\\) centered at \\((0, 3/8)\\).\n\n10. If \\(z = a + ib, a, b \\in \\mathbb{R}\\) then \\(z^2 \\in \\mathbb{R}\\) if and only if \\(a^2 - b^2 + 2iab \\in \\mathbb{R}\\), that is if and only if \\(ab = 0\\). Hence \\(z^2 \\in \\mathbb{R}\\) if and only if \\(z \\in \\mathbb{R}\\) \\((b = 0)\\) or if \\(z\\) is a pure imaginary number \\((a = 0)\\).\n\n11. Let \\(z = a + ib, a, b \\in \\mathbb{R}\\).\n\na) \\(\\text{Re}(z(1 + i)) = \\text{Re}((a + ib)(1 + i)) = \\text{Re}(a - b + i(a + b)) = a - b\\). The equation is then equivalent to\n\n\\[a - b + a^2 + b^2 = 0 \\iff \\left(a + \\frac{1}{2}\\right)^2 + \\left(b - \\frac{1}{2}\\right)^2 = \\frac{1}{2},\n\\]\n\nwhose solutions are the points of the circle with center in \\((-1/2, 1/2)\\) and radius \\(\\sqrt{2}/2\\).\n(b) Since \\( z^2 = a^2 - b^2 + 2iab \\) and \\( \\bar{z}(1 + 2i) = (a - ib)(1 + 2i) = a + 2b + i(2a - b) \\), the equation can be written as\n\n\\[\na^2 - b^2 + i(2a - b) = -3,\n\\]\n\nand we deduce \\( 2a = b \\) and \\( a^2 - b^2 = -3 \\). The solution of this system are \\( z_1 = 1 + 2i \\) and \\( z_2 = -1 - 2i \\), the unique solutions of the starting equation.\n\n(c) Since \\( (2 - i)(a + ib) = 2a + b + i(2b - a) \\), the equation can be written as\n\n\\[\n2b - a = 1.\n\\]\n\nwhose solutions are the points of the line \\( x - 2y + 1 = 0 \\).\n\n12. If \\( z = -i \\), then \\( z^2 = -1 \\), \\( z^3 = i \\), and \\( P(-i) = i + 1 - i + 1 + a = 2 + a \\). Then \\(-i\\) is a root for \\( P \\) if and only if \\( a = -2 \\). Since \\( P(z) = z^3 - z^2 + z - 1 \\) contains \\( z - 1 \\), we have \\( P(z) = (z - 1)(z - i)(z + i) \\).", "id": "./materials/369.pdf" }, { "contents": "The Argand Diagram\n\nIt is very useful to have a graphical or pictorial representation of complex numbers.\n\nFor example, the complex number \\( z = 3 + 4i \\) is represented as a point in the \\( xy \\) plane with coordinates \\((3, 4)\\) as shown in Figure 1. That is, the real part, 3, is plotted on the \\( x \\) axis, and the imaginary part, 4, is plotted on the \\( y \\) axis.\n\n![Figure 1. Argand diagram which represents the complex number 3+4i by the point P(3,4).](image)\n\nMore generally, the complex number \\( z = a + bi \\) is plotted as a point with coordinates \\((a, b)\\) as shown in Figure 2.\n\n![Figure 2. Argand diagram which represents the complex number \\( a + bi \\) by the point P\\((a, b)\\).](image)\n\nBecause the real part of \\( z \\) is plotted on the horizontal axis we often refer to this as the **real axis**. The imaginary part of \\( z \\) is plotted on the vertical axis and so we refer to this as the **imaginary axis**. Such a diagram is called an **Argand diagram**. Engineers often refer to this diagram as the **complex plane**.\nExamples\n\nPlot the complex numbers $z_1 = 2 + 3i$, $z_2 = -3 + 2i$, $z_3 = -3 - 2i$, $z_4 = 2 - 5i$, $z_5 = 6$, $z_6 = i$ on an Argand diagram.\n\nSolution\n\nThe Argand diagram is shown in Figure 3.\n\n![Argand diagram showing several complex numbers.](image)\n\nFigure 3. Argand diagram showing several complex numbers.\n\nNote that purely real numbers lie on the real axis. Purely imaginary numbers lie on the imaginary axis.\n\nAnother observation is that complex conjugate pairs (such as $-3 + 2i$ and $-3 - 2i$) lie symmetrically about the $x$ axis.\n\nFinally, because every real number, $a$ say, can be written as a complex number, $a + 0i$, that is as a complex number with a zero imaginary part, it follows that all real numbers are also complex numbers. As such we see that complex numbers form an extension of the sets of numbers with which we were already familiar.", "id": "./materials/370.pdf" }, { "contents": "Operations with complex numbers\n\nMathE\n\n2 March 2023\n\nGiven two complex numbers, how can we add, subtract, multiply and divide them?\n\nLet\n\n\\[ z_1 = a + bi \\]\n\nand\n\n\\[ z_2 = c + di \\]\n\nbe two complex numbers, where \\( a, b, c, d \\in \\mathbb{R} \\).\n\n- **Addition**\n\n\\[ z_1 + z_2 = \\]\n\n\\[ (a + bi) + (c + di) \\]\n\n\\[ = (a + c) + (b + d)i \\]\n\n(1) Group terms with \\( i \\) and without \\( i \\) separately\n\n(2) Factor \\( i \\)\n\n(1) You can do this step because addition in \\( \\mathbb{R} \\) is commutative.\n\nYou can see addition of complex numbers as a sum of vectors in \\( \\mathbb{R}^2 \\)\n\n(if you want to know a bit more about this, see the section “For the curious...” below).\n• **Subtraction**\n\n\\[ z_1 - z_2 = (a + bi) - (c + di) = \\]\n\n\\[ a + bi - c - di = \\]\n\n(1) Group terms with \\( i \\) and without \\( i \\) separately\n\n\\[ (a - c) + (bi - di) = \\]\n\n(2) Factor \\( i \\)\n\n\\[ (a - c) + (b - d)i \\]\n\n(1) You can do this step because addition in \\( \\mathbb{R} \\) is commutative\n\n(subtracting numbers can be seen as adding a positive numbers with the symmetric of another positive number, this is, if \\( r, s \\in \\mathbb{R} \\), then \\( r - s = r + (-s) \\)).\n\nYou can see subtraction of complex numbers as a difference of vectors in \\( \\mathbb{R}^2 \\)\n\n![Diagram of complex numbers subtraction](image)\n\n• **Product**\n\n\\[ z_1 \\cdot z_2 = \\]\n\n\\[ (a + bi) \\cdot (c + di) = \\]\n\n(1) Distributive property of \\( \\times \\) over +\n\n\\[ a \\cdot c + a \\cdot di + bi \\cdot c + (bi) \\cdot (di) = \\]\n\n\\[ ac + adi + bci + bdi^2 = \\]\n\n(2) \\( i^2 = -1 \\)\n\n\\[ ac + adi + bci - bd = \\]\n\n(3) Group terms with \\( i \\) and without \\( i \\) separately\n\n\\[ (ac - bd) + (adi + bci) = \\]\n\n(4) Factor \\( i \\)\n\n\\[ (ac - bd) + (ad + bc)i \\]\n\nFor the product, it is a bit harder to get the intuition for the geometric interpretation, so it will be added to the “For the curious...” section.\n• Division\n\n\\[\n\\frac{z_1}{z_2} = \\frac{a + bi}{c + di} \\quad (1)\n\\]\n\n\\[\n(a + bi)(c - di) = (c + di)(c - di) \\quad (2)\n\\]\n\n\\[\nac + a \\times (-di) + bi \\times c + bi \\times (-di) \\quad (3)\n\\]\n\n\\[\n\\frac{ac - adi + bci - bdi^2}{c^2 - (di)^2} \\quad (4)\n\\]\n\n\\[\n\\frac{ac - adi + bci + bd}{c^2 + d^2} \\quad (5)\n\\]\n\n\\[\n\\frac{(ac + bd) + (bci - adi)}{c^2 + d^2} \\quad (6)\n\\]\n\n\\[\n\\frac{ac + bd + (bc - ad)i}{c^2 + d^2 + c^2 + d^2i} \\quad (7)\n\\]\n\n(1) Multiply the numerator and the denominator by conjugate of the denominator \\((c + di = c - di)\\).\n\n(2) The numerator is now a product of complex numbers (the previous case).\n\nThe denominator, despite also being a product of complex numbers (and you can proceed as the previous case, as well), can be simplified more quickly if you remember the difference of two perfect squares: \\((x - y)(x + y) = x^2 - y^2\\).\n\n(3) Simplify both the numerator and the denominator. Note that the steps we take are possible because usual multiplication in \\(\\mathbb{R}\\) is commutative.\n\n(4) \\(i^2 = -1\\).\n\n(5) Group terms with \\(i\\) and without \\(i\\) separately.\n\n(6) Factor \\(i\\).\n\n(7) Separate in two fractions.\nExample 1. Let $z_1 = 2 + 4i$ and $z_2 = -3 - 7i$. Then we have:\n\n- **Addition**\n\n $z_1 + z_2 =$\n\n $(2 + 4i) + (-3 - 7i) \\overset{(1)}{=} (2 + (-3)) + (4i + (-7i)) \\overset{(2)}{=} -1 + (4 + (-7))i = -1 + (-3)i = -1 - 3i$\n\n (1) Group terms with $i$ and without $i$ separately\n (2) Factor $i$\nSubtraction\n\n\\[ z_1 - z_2 = \\]\n\\[ (2 + 4i) - (-3 - 7i) \\]\n\\[ = (2 - (-3)) + (4i - (-7i)) \\]\n\\[ = (2 + 3) + (4 + 7)i \\]\n\\[ = 5 + 11i \\]\n\n(1) Group terms with \\( i \\) and without \\( i \\) separately\n\n(2) Factor \\( i \\)\nProduct\n\n\\[ z_1 \\cdot z_2 = \\]\n\\[ (2 + 4i) \\cdot (-3 - 7i) \\]\n\\[ = (2 + 4i) \\cdot (-3 - 7i) \\quad (1) \\]\n\\[ = 2 \\times (-3) + 2 \\times (-7i) + 4i \\times (-3) + 4i \\times (-7i) \\quad (2) \\]\n\\[ = -6 - 14i - 12i - 28i^2 \\quad (3) \\]\n\\[ = -6 - 14i - 12i - 28 \\times (-1) \\]\n\\[ = -6 - 14i - 12i + 28 \\quad (4) \\]\n\\[ = (-6 + 28) + (-14i - 12i) \\quad (5) \\]\n\\[ = 22 + (-14 - 12)i \\]\n\\[ = 22 - 26i \\]\n\n(1) Distributive property of \\( \\times \\) over +\n(2) Simplify expression\n(3) \\( i^2 = -1 \\)\n(4) Group terms with \\( i \\) and without \\( i \\) separately\n(5) Factor \\( i \\)\nDivision\n\n\\[\n\\frac{z_1}{z_2} = \\frac{2 + 4i}{-3 - 7i} = \\frac{(2 + 4i)(-3 + 7i)}{(-3 - 7i)(-3 + 7i)}\n\\]\n\n\\[\n2 \\times (-3) + 2 \\times (7i) + (4i) \\times (-3) + (4i) \\times (7i) = (-3)^2 - (7i)^2\n\\]\n\n\\[\n\\frac{-6 + 14i - 12i + 28i^2}{9 - 49i^2} = \\frac{-6 + 14i - 12i - 28}{9 - 49}\n\\]\n\n\\[\n\\frac{(-6 - 28) + (14i - 12i)}{-40} = \\frac{-34 + (14 - 12)i}{-40}\n\\]\n\n\\[\n\\frac{-34 + 2i}{-40} = \\frac{-34 + 2i}{-40} = \\frac{17}{20} - \\frac{1}{20}i\n\\]\n\n(1) Multiply the numerator and the denominator by the conjugate of the denominator \\((-3 - 7i = -3 + 7i)\\).\n\n(2) The numerator and the denominator are now a product of complex numbers.\n\nFor the numerator, proceed as in the product case (above).\n\nFor the denominator, notice that you have the difference of two perfect squares - remember that\n\\[(a - b)(a + b) = a^2 - b^2\\]. In this specific example, \\(a = -3\\) and \\(b = 7i\\).\n\n(3) \\(i^2 = -1\\).\n\n(4) Group terms with \\(i\\) and without \\(i\\) separately.\n\n(5) Factor \\(i\\).\n\n(6) Separate into two fractions.\nFor the curious...\n\nSometimes, instead of thinking of complex numbers, it may be useful to think of ordered pairs in $\\mathbb{R}^2$. We can do this because there exists an isomorphism between $\\mathbb{C}$ and $\\mathbb{R}^2$. This means it is possible to define a bijective function whose domain is $\\mathbb{C}$ and whose codomain is $\\mathbb{R}^2$ (or a bijective function whose domain is $\\mathbb{R}^2$ and whose codomain is $\\mathbb{C}$).\n\nDefine\n\n$$f : \\mathbb{C} \\rightarrow \\mathbb{R}^2$$\n\n$$a + bi \\mapsto (a, b)$$\n\n$f$ is injective: To show that $f$ is injective, we need to see that\n\n$$(\\forall a + bi, c + di \\in \\mathbb{C}) : f(a + bi) = f(c + di) \\Rightarrow a + bi = c + di$$\n\nLet $a + bi, c + di \\in \\mathbb{C}$ such that $f(a + bi) = f(c + di)$.\n\n$$f(a + bi) = f(c + di) \\iff (a, b) = (c, d) \\iff \\begin{cases} a = c \\\\ b = d \\end{cases} \\iff a + bi = c + di$$\n\n$f$ is surjective: To see that $f$ is surjective, we have to show that\n\n$$(\\forall (a, b) \\in \\mathbb{R}^2)(\\exists z_1 \\in \\mathbb{C}) : f(z_1) = (a, b)$$\n\nGiven $a, b \\in \\mathbb{R}$, it follows that $z_1 = a + bi \\in \\mathbb{C}$.\n\nHence, $f(z_1) = f(a + bi) = (a, b)$.\n\nNote: The existence of this bijection is what makes it possible to talk about the Argand’s plane. If you notice, when you use the Argand’s plane what you are doing is representing a complex number in $\\mathbb{R}^2$ (in the xOy plane).\n\nA geometric interpretation of the product and division of two complex numbers\n\nHave you ever heard of Euler’s formula? When specified to the complex plane, it states that\n\n$$|z| \\cdot e^{i\\theta} = |z| \\cdot (\\cos(\\theta) + i \\sin(\\theta))$$\n\nwhere $z = a + bi \\in \\mathbb{C}$, $|z| = \\sqrt{a^2 + b^2}$ and $\\theta \\in [0, 2\\pi]$ is the angle between the x-axis and the ray whose endpoint is the origin and that passes through the complex number $z \\sim (a, b)$:\nSo, given two complex numbers $z_1 = a + bi$ and $z_2 = c + di$, with $a, b, c, d \\in \\mathbb{R}$, we have that\n\n$$z_1 = |z_1| \\cdot e^{i\\theta_1} \\quad \\text{and} \\quad z_2 = |z_2| \\cdot e^{i\\theta_2}$$\n\nfor some $\\theta_1, \\theta_2 \\in [0, 2\\pi]$\n\nHence,\n\n$$z_1 \\cdot z_2 = |z_1| \\cdot e^{i\\theta_1} \\cdot |z_1| \\cdot e^{i\\theta_2} = |z_1| \\cdot |z_2| \\cdot e^{i(\\theta_1 + \\theta_2)}$$\n\nthis is, the product gives origin to a new complex number whose:\n\n- angle between the $x$-axis and the ray whose endpoint is the origin and that passes through it is $\\theta_1 + \\theta_2 \\pmod{2\\pi}$\n- length is $|z_1| \\cdot |z_2|$\nIt is now easy to see how to proceed for the division:\n\n\\[\n\\frac{z_1}{z_2} = \\frac{|z_1|}{|z_2|} \\cdot e^{i(\\theta_1 - \\theta_2)}\n\\]\n\nThe division gives origin to a new complex number whose:\n\n- angle between the x-axis and the ray whose endpoint is the origin and that passes through it is \\(\\theta_1 - \\theta_2 \\pmod{2\\pi}\\)\n- length is \\(\\frac{|z_1|}{|z_2|}\\)\n\n**Note:** What does \\(\\theta_1 + \\theta_2 \\pmod{2\\pi}\\) mean?\n\nIn mathematics, this is a common notation for referring to the remainder of the division of \\(\\theta_1 + \\theta_2\\) by \\(2\\pi\\).\n\nYou learnt that after dividing \\(\\theta_1 + \\theta_2\\) by \\(2\\pi\\), it is possible to write\n\n\\[\n\\theta_1 + \\theta_2 = 2\\pi \\cdot q + r\n\\]\n\nwhere \\(q \\in \\mathbb{Z}\\) and \\(0 \\leq r < 2\\pi\\).\n\nSo we have \\(\\theta_1 + \\theta_2 \\pmod{2\\pi} = r\\).", "id": "./materials/372.pdf" }, { "contents": "A complex number \\( z \\in \\mathbb{C} \\) can be represented in its:\n\n- **Algebraic form:** \\( z = a + bi \\) for some \\( a, b \\in \\mathbb{R} \\);\n- **Trigonometric form:** \\( z = |z| \\cdot e^{i \\theta} = |z| \\cdot (\\cos(\\theta) + i \\cdot \\sin(\\theta)) \\) with \\( |z| > 0 \\) and \\( \\theta \\in [0, 2\\pi] \\);\n\n### From algebraic to trigonometric form\n\n**Example 1.** Represent the complex number \\( z = -2 - \\frac{2\\sqrt{3}}{3}i \\) in trigonometric form.\n\nOur objective is to write \\( z = -2 - \\frac{2\\sqrt{3}}{3}i \\) in the form\n\n\\[\n|z| \\cdot (\\cos(\\theta) + i \\sin(\\theta))\n\\]\n\nwith \\( |z| > 0 \\) and \\( \\theta \\in [0, 2\\pi] \\).\n\nBy the Pythagorean theorem, we have that:\n\n\\[\n|z|^2 = 2^2 + \\left( \\frac{2\\sqrt{3}}{3} \\right)^2 \\iff |z|^2 = 4 + \\frac{4 \\times 3}{9} \\iff |z|^2 = 12 \\iff |z| = \\pm \\sqrt{12} = \\pm 2\\sqrt{3}\n\\]\n\nSince \\( |z| \\) is a length, we have that \\( |z| > 0 \\), so \\( |z| = 2\\sqrt{3} \\).\nUsing trigonometry, we see that:\n\n\\[\n\\cos(\\theta) = \\frac{-2}{|z|} = \\frac{-2}{2\\sqrt{3}} = -\\frac{1}{\\sqrt{3}} = -\\frac{\\sqrt{3}}{3} \\quad \\sin(\\theta) = \\frac{-2\\sqrt{3}}{|z|} = \\frac{-2\\sqrt{3}}{2\\sqrt{3}} = -\\frac{\\sqrt{3}}{3} = \\frac{-2\\sqrt{3}}{3 \\times 2\\sqrt{3}} = -\\frac{1}{3}\n\\]\n\nAs a result,\n\n\\[\n\\tan(\\theta) = \\frac{\\sin(\\theta)}{\\cos(\\theta)} = \\frac{-\\frac{1}{3}}{-\\frac{\\sqrt{3}}{3}} = \\frac{3}{\\sqrt{3}} = \\frac{\\sqrt{3}}{3} \\quad \\iff \\quad \\theta = \\arctan \\left( \\frac{\\sqrt{3}}{3} \\right)\n\\]\n\nNow, we have to be careful. We have two options:\n\nOption 1: We look at the usual trigonometric table:\n\n| \\(\\theta^\\circ\\) | 0° | 30° | 45° | 60° | 90° |\n|------------------|----|-----|-----|-----|-----|\n| \\(\\sin \\theta\\) | 0 | \\(\\frac{1}{2}\\) | \\(\\frac{\\sqrt{2}}{2}\\) | \\(\\frac{\\sqrt{3}}{2}\\) | 1 |\n| \\(\\cos \\theta\\) | 1 | \\(\\frac{\\sqrt{3}}{2}\\) | \\(\\frac{\\sqrt{2}}{2}\\) | \\(\\frac{1}{2}\\) | 0 |\n| \\(\\tan \\theta\\) | 0 | \\(\\frac{\\sqrt{3}}{3}\\) | 1 | undefined | |\n\nOption 2: We input on a calculator \\(\\arctan \\left( \\frac{\\sqrt{3}}{3} \\right)\\):\n\n\\[\n\\arctan(\\sqrt{3}) = 30^\\circ = \\frac{30\\pi}{180} \\text{ radians} = \\frac{\\pi}{6} \\text{ radians}\n\\]\n\nHowever, the answer is none of the results... Can you see why?\nThe first thing we should note is that our angle $\\theta$ does not look like an angle whose measure is $30^\\circ$:\n\nSo, is the table and the calculator lying to us? Not really.\nTo understand why, we need to remember what tangent of an angle is geometrically.\n\nIn the general case, given an angle $\\alpha \\in [0, 2\\pi \\setminus \\{\\pm \\frac{\\pi}{2}\\}$ (the tangent function is not defined for $\\frac{\\pi}{2}$ and $-\\frac{\\pi}{2}$):\n\n$$\\tan(\\alpha) = \\tan((\\alpha + \\pi) \\mod 2\\pi)$$\n\nAs a result, we conclude that\n\n$$\\theta = 30^\\circ + 180^\\circ = 210^\\circ$$\n\nwhich is equivalent to say that\n\n$$\\theta = \\frac{\\pi}{6} + \\pi = \\frac{7\\pi}{6} \\text{ radians}$$\n\nNote here that $30^\\circ$ is the measure of the angle $\\alpha$ in the image.\n\nAs a result, we have that\n\n$$z = -2 - \\frac{2\\sqrt{3}}{3}i = 2\\sqrt{3} \\cdot e^{\\frac{7\\pi}{6}i}$$\n\nthis is, $|z| = 2\\sqrt{3}$ and $\\theta = \\frac{7\\pi}{6}$.\n\n(usually, radians rather than degrees are used for the final answer)\nFrom trigonometric to algebraic form\n\nExample 2. Represent the complex number $z = 6 \\cdot e^{\\frac{\\pi}{3}i}$ in algebraic form.\n\nOur objective is to write $z = 6 \\cdot e^{\\frac{\\pi}{3}i}$ in the form\n\n$$a + bi$$\n\nwith $a, b \\in \\mathbb{R}$.\n\nWe can do the following calculations:\n\n$$z = 6 \\cdot e^{\\frac{\\pi}{3}i} = 6 \\cdot (\\cos \\left( \\frac{\\pi}{3} \\right) + i \\sin \\left( \\frac{\\pi}{3} \\right)) \\quad (1)$$\n\n$$= 6 \\cdot \\left( \\frac{1}{2} + \\frac{\\sqrt{3}}{2}i \\right) = \\frac{6}{2} + \\frac{6\\sqrt{3}}{2}i = 3 + 3\\sqrt{3}i$$\n\n(1) use the trigonometric table above, knowing that $\\frac{\\pi}{3}$ radians $= 60^\\circ$\n\nAs a result, we immediately conclude that\n\n$$z = 6 \\cdot e^{\\frac{\\pi}{3}i} = 3 + 3\\sqrt{3}i$$\n\nthis is, $a = 3$ and $b = 3\\sqrt{3}$. ", "id": "./materials/373.pdf" }, { "contents": "Solved exercises about powers of $i$\n\nMathE\n\n22nd of March of 2023\n\nIn order to solve this kind of exercises, we first need to experiment some powers of $i$ to see how it works.\n\nBy definition, we know that:\n\n- $i = \\sqrt{-1}$\n- Any non-zero number raised to the power of 0 is 1, this is,\n $$a^0 = 1, \\forall a \\in \\mathbb{C} \\setminus \\{0\\}$$\n- Any number raised to the power of 1 is the number itself, this is,\n $$a^1 = a, \\forall a \\in \\mathbb{C}$$\n\nHence,\n\n- $i^0 = 1$\n- $i^1 = i$\n- $i^2 = (\\sqrt{-1})^2 = -1$\n- $i^3 = i^2 \\cdot i = -1 \\cdot i = -i$\n- $i^4 = i^2 \\cdot i^2 = (-1) \\times (-1) = 1$\n\nNote that we have $i^4 = i^0$, so we conclude that the powers are the same in steps of 4, this is\n\n$$i^0 = i^{0+4} = i^{0+4+4} = i^{0+4+4+4} = \\cdots = 1$$\n$$i^1 = i^{1+4} = i^{1+4+4} = i^{1+4+4+4} = \\cdots = i$$\n$$i^2 = i^{2+4} = i^{2+4+4} = i^{2+4+4+4} = \\cdots = -1$$\n$$i^3 = i^{3+4} = i^{3+4+4} = i^{3+4+4+4} = \\cdots = -i$$\n\nWe can compress the above information by saying that\n\n$$i^{4k} = (i^4)^k = 1, \\forall k \\in \\mathbb{N}_0$$\n$$i^{4k+1} = (i^4)^k \\cdot i^1 = 1 \\cdot i = i, \\forall k \\in \\mathbb{N}_0$$\n$$i^{4k+2} = (i^4)^k \\cdot i^2 = 1 \\times (-1) = -1, \\forall k \\in \\mathbb{N}_0$$\n$$i^{4k+3} = (i^4)^k \\cdot i^3 = 1 \\cdot (-i) = -i, \\forall k \\in \\mathbb{N}_0$$\nExercise 1. Simplify $i^{148}$.\n\nWe have that\n\n$$i^{148} \\overset{(1)}{=} i^{4 \\times 37} = (i^4)^{37} \\overset{(2)}{=} 1^{37} = 1$$\n\n(1) We have $\\frac{148}{4} = 37$, so $148 = 37 \\times 4 = 4 \\times 37$\n\n(2) We saw here that $i^4 = 1$\n\nAnswer: $i^{148} = 1$\n\nExercise 2. Simplify $i^3 + i^6 + i^9$.\n\nWe have that\n\n$$i^3 + i^6 + i^9 = i^3 + (i^3)^2 + (i^3)^3 \\overset{(1)}{=} -i + (-i)^2 + (-i)^3 \\overset{(2)}{=}$$\n\n$$\\overset{(2)}{=} -i + i^2 + (-i)^2 \\cdot (-i) \\overset{(3)}{=} -i + (-1) + i^2 \\cdot (-i) =$$\n\n$$= -i - 1 + (-1) \\cdot (-i) = -i - 1 + i \\overset{(4)}{=} (-1 + 1)i - 1 = 0 \\cdot i - 1 = -1$$\n\n(1) We saw here that $i^3 = -i$\n\n(2) We know $a^{2k} = (-a)^{2k}$, $\\forall k \\in \\mathbb{N}_0$, $\\forall a \\in \\mathbb{R}$\n\n(3) We saw here that $i^2 = -1$\n\n(4) Factor $i$\n\nAnswer: $-1$\n\nExercise 3. Simplify $i^2 + 8i^3 + i^5 - 2i^7$.\n\nWe have that\n\n$$i^2 + 8i^3 + i^5 - 2i^7 = i^2 + 8i^3 + i^3 \\cdot i^2 - 2 \\cdot (i^2)^3 \\cdot i \\overset{(1)}{=}$$\n\n$$\\overset{(1)}{=} (-1) + 8 \\cdot (-i) + (-i) \\cdot (-1) - 2 \\cdot (-1)^3 \\cdot i =$$\n\n$$= -1 - 8i + i + 2 \\cdot (-1) \\cdot i =$$\n\n$$= -1 - 8i + i + 2i =$$\n\n$$= -1 - 5i$$\n\n(1) We saw here that $i^2 = -1$ and that $i^3 = -i$\n\nAnswer: $-1 - 5i$", "id": "./materials/374.pdf" }, { "contents": "Solved exercises about operations with complex numbers\n\nMathE\n\n22nd of March of 2023\n\nExercise 1. Simplify \\( \\frac{2 + 4i}{1 + i} (2 - i) \\).\n\nWe have that\n\n\\[\n\\frac{2 + 4i}{1 + i} (2 - i) = \\frac{(2 + 4i) \\cdot (2 - i)}{1 + i} = \\frac{2 \\times 2 + 2 \\cdot (-i) + 2 \\times 4i + (4i) \\cdot (-i)}{1 + i} = \\frac{4 - 2i + 8i - 4i^2}{1 + i} = \\frac{4 + 6i - 4 \\times (-1)}{1 + i} = \\frac{4 + 6i + 4}{1 + i} = \\frac{8 + 6i}{1 + i} = \\frac{(8 + 6i) \\cdot (1 - i)}{(1 + i) \\cdot (1 - i)} = \\frac{8 - 8i + 6i - 6i^2}{1^2 - i^2} = \\frac{8 - 2i - 6 \\times (-1)}{1 - (-1)} = \\frac{8 - 2i + 6}{1 + 1} = \\frac{14 - 2i}{2} = \\frac{14}{2} - \\frac{2}{2}i = 7 - i\n\\]\n\n(1) By the distributive property of multiplication over addition\n\n(2) \\( i^2 = -1 \\)\n\n(3) Multiply both the numerator and the denominator by the conjugate of the denominator:\n\n\\( \\frac{1}{1 + i} = 1 - i \\)\n\n(4) Apply the difference of two squares to the denominator:\n\n\\( (a + b)(a - b) = a^2 - b^2, \\forall a, b \\in \\mathbb{R} \\)\n\n(5) Separate into two fractions\n\nAnswer: \\( 7 - i \\)\n\nExercise 2. Simplify \\( \\frac{1 + i}{1 - i} + \\frac{1 - i}{1 + i} \\).\n\nWe have that\n\n\\[\n\\frac{1 + i}{1 - i} + \\frac{1 - i}{1 + i} = \\frac{(1 + i) \\cdot (1 + i)}{(1 - i) \\cdot (1 + i)} + \\frac{(1 - i) \\cdot (1 - i)}{(1 + i) \\cdot (1 - i)} = \\frac{(1 + i)^2 + (1 - i)^2}{(1 + i) \\cdot (1 - i)} = \\frac{1 + 2i + i^2 + 1 - 2i + i^2}{1 - i^2} = \\frac{2 + 2i^2}{2} = \\frac{2 - 2}{2} = 1\n\\]\n\\[\n\\begin{align*}\n(2) \\quad & \\frac{(1^2 + 2 \\cdot i + i^2) + (1^2 + 2 \\cdot (-i) + i^2)}{1^2 - i^2} \\\\\n& = \\frac{1 + 2i - 1 + 2i - 1}{1 - (-1)} \\\\\n& = \\frac{0}{1+1} = \\frac{0}{2} = 0\n\\end{align*}\n\\]\n\n(1) Give the same denominator to both fractions\n\n(2) In the denominator, apply the difference of two squares:\n\n\\[(a + b)(a - b) = a^2 - b^2, \\ \\forall a, b \\in \\mathbb{R}\\]\n\nIn the numerator, apply the square of a sum:\n\n\\[(a + b)^2 = a^2 + 2 \\cdot a \\cdot b + b^2, \\ \\forall a, b \\in \\mathbb{R}\\]\n\n(3) \\(i^2 = -1\\)\n\nAnswer: 0", "id": "./materials/375.pdf" }, { "contents": "Solved exercises about passing from algebraic to trigonometric form and vice-versa\n\nMathE\n\n23th of March of 2023\n\nExercise 1. Find the polar form of $z = 1 + i$.\n\nStep 1: Find the absolute value of $z$, $|z|$.\n\nThe first thing to note is that $Re(z) = 1$ and $Im(z) = 1$. Hence,\n\n$$|z| = \\sqrt{(Re(z))^2 + (Im(z))^2} = \\sqrt{1^2 + 1^2} = \\sqrt{1 + 1} = \\sqrt{2}$$\n\nStep 2: Find the argument of $z$, $Arg(z) = \\theta$.\n\nWe start by finding the cosine and sine of the angle $\\theta \\in [0, 2\\pi]$:\n\n$$\\cos(\\theta) = \\frac{Re(z)}{|z|} = \\frac{1}{\\sqrt{2}} = \\frac{\\sqrt{2}}{2} \\quad \\sin(\\theta) = \\frac{Im(z)}{|z|} = \\frac{1}{\\sqrt{2}} = \\frac{\\sqrt{2}}{2}$$\n\nNow, we have to analyze the results we got:\n\n- We have that $\\cos(\\theta) > 0$ and $\\sin(\\theta) > 0$, so $\\theta$ is an angle in the first quadrant, meaning that $\\theta \\in \\left(0, \\frac{\\pi}{2}\\right)$.\n\n- Using a calculator or the trigonometric table, we conclude that $\\theta = Arg(z) = \\frac{\\pi}{4}$.\n\nAs a result,\n\n$$z = \\sqrt{2} \\cdot e^{i \\frac{\\pi}{4}}$$\n\nAnswer: $\\sqrt{2} \\cdot e^{i \\frac{\\pi}{4}}$\nExercise 2. Express $z = e^{i \\frac{2\\pi}{3}}$ in the form $z = a + bi$.\n\nWe have that\n\n$$e^{i \\frac{2\\pi}{3}} = \\cos \\left( \\frac{2\\pi}{3} \\right) + i \\cdot \\sin \\left( \\frac{2\\pi}{3} \\right) \\quad (1) = -\\cos \\left( \\frac{\\pi}{3} \\right) + i \\cdot \\sin \\left( \\frac{\\pi}{3} \\right) \\quad (2)$$\n\n$$= -\\frac{1}{2} + \\frac{\\sqrt{3}}{2}i$$\n\n(2) Look at the table\n\n(1)\n\nAnswer: $-\\frac{1}{2} + \\frac{\\sqrt{3}}{2}i$\n\nExercise 3. Represent $z = (1 + i)^3$ in the form $a + bi$.\n\nThis exercise has, at least, two different approaches.\n\n1st approach: We already saw, in Exercise 1, that the trigonometric form of the complex number $w = 1 + i$ is\n\n$$\\sqrt{2} \\cdot e^{i \\frac{\\pi}{4}}$$\n\nHence, we can proceed with the following substitution:\n\n$$z = (1 + i)^3 = \\left( \\sqrt{2} \\cdot e^{i \\frac{\\pi}{4}} \\right)^3 = \\left( \\sqrt{2} \\right)^3 \\cdot \\left( e^{i \\frac{\\pi}{4}} \\right)^3 = \\left( \\sqrt{2} \\right)^2 \\cdot \\sqrt{2} \\cdot \\left( e^{i \\frac{\\pi}{4}} \\right)^3 = 2\\sqrt{2} \\cdot e^{i \\left( \\frac{3\\pi}{4} \\right)}$$\n\nNow, we just have to go back to algebraic form:\n\n$$2\\sqrt{2} \\cdot e^{i \\left( \\frac{3\\pi}{4} \\right)} = 2\\sqrt{2} \\cdot \\left( \\cos \\left( \\frac{3\\pi}{4} \\right) + i \\cdot \\sin \\left( \\frac{3\\pi}{4} \\right) \\right) =$$\n\n$$= 2\\sqrt{2} \\cdot \\left( -\\cos \\left( \\frac{\\pi}{4} \\right) + i \\sin \\left( \\frac{3\\pi}{4} \\right) \\right) =$$\n\n$$= 2\\sqrt{2} \\cdot \\left( -\\frac{\\sqrt{2}}{2} + \\frac{\\sqrt{2}}{2}i \\right) =$$\n\n$$= -2 + 2i$$\n2nd approach: We use the Binomial theorem, which states that, given \\( n \\geq 0 \\) and \\( x, y \\in \\mathbb{C} \\),\n\n\\[\n(x + y)^n = \\sum_{k=0}^{n} \\binom{n}{k} x^k y^{n-k}\n\\]\n\nIn our specific exercise, \\( x = 1 \\), \\( y = i \\) and \\( n = 3 \\). Hence,\n\n\\[\n(1 + i)^3 = \\sum_{k=0}^{3} \\binom{3}{k} 1^k i^{3-k} = \\binom{3}{0} 1^0 i^3 + \\binom{3}{1} 1^1 i^2 + \\binom{3}{2} 1^2 i^1 + \\binom{3}{3} 1^3 i^0 =\n\\]\n\n\\[\n= i^3 + 3i^2 + 3i + 1 = -i + 3 \\cdot (-1) + 3i + 1 = -i - 3 + 3i + 1 =\n\\]\n\n\\[\n= -2 + 2i\n\\]\n\nAnswer: \\(-2 + 2i\\)", "id": "./materials/376.pdf" }, { "contents": "Find \\( \\int_0^1 \\int_0^{2y} (4 + 2x - y^2) \\, dx \\, dy \\)\n\n- As we can see, we should first integrate in order to \\( x \\), and only then in order to \\( y \\).\n\n\\[\n\\int_0^1 \\int_0^{2y} (4 + 2x - y^2) \\, dx \\, dy \\\\\n= \\int_0^1 \\left[ 4x + x^2 - y^2 x \\right]_{x=0}^{x=2y} \\, dy \\\\\n= \\int_0^1 (8y + 4y^2 - 2y^3 - 0 - 0 + 0) \\, dy \\\\\n= \\left[ 4y^2 + \\frac{4y^3}{3} - \\frac{y^4}{2} \\right]_{y=0}^{y=1} \\\\\n= 4 + \\frac{4}{3} - \\frac{1}{2} - 0 - 0 + 0 \\\\\n= \\frac{24}{6} + \\frac{8}{6} - \\frac{3}{6} \\\\\n= \\frac{29}{6}\n\\]", "id": "./materials/378.pdf" }, { "contents": "Find \\( \\int_0^2 \\int_1^3 (xy^2) \\, dy \\, dx \\)\n\n- As we can see, we should first integrate in order to \\( y \\), and only then in order to \\( x \\).\n\n\\[\n\\int_0^2 \\int_1^3 (xy^2) \\, dy \\, dx \\\\\n= \\int_0^2 \\left[ \\frac{xy^3}{3} \\right]_{y=1}^{y=3} \\, dx \\\\\n= \\int_0^2 \\left( 9x - \\frac{x}{3} \\right) \\, dx \\\\\n= \\left[ \\frac{9x^2}{2} - \\frac{x^2}{6} \\right]_{x=0}^{x=2} \\\\\n= 18 - \\frac{2}{3} - 0 + 0 \\\\\n= \\frac{54}{3} - \\frac{2}{3} \\\\\n= \\frac{52}{3}\n\\]", "id": "./materials/379.pdf" }, { "contents": "Integrate $\\int_0^1 \\int_0^x \\cos(x^2) \\, dy \\, dx$\n\n- As we can see, we should first integrate in order to $y$, and only then in order to $x$.\n\n\\[\n\\int_0^1 \\int_0^x \\cos(x^2) \\, dy \\, dx = \\int_0^1 \\left[ y \\cos(x^2) \\right]_{y=0}^{y=x} \\, dx = \\int_0^1 (x \\cos(x^2) - 0) \\, dx = \\int_0^1 x \\cos(x^2) \\, dx\n\\]\n\n- In this case it is worth making a substitution, were\n\n\\[\nx^2 = u \\\\\ndu = 2x \\, dx \\\\\n\\Rightarrow x \\cos(x^2) \\, dx = \\cos(u) \\times \\frac{du}{2}\n\\]\nThis means that we can proceed with the substitution\n\n\\[\n\\int_0^1 x \\cos(x^2) \\, dx = \\int_0^1 \\left( \\frac{\\cos(u)}{2} \\right) \\, du\n\\]\n\n\\[\n= \\frac{1}{2} \\left[ \\sin(u) \\right]_u^1\n\\]\n\n\\[\n= \\frac{1}{2} \\left( \\sin(1) - \\sin(0) \\right)\n\\]\n\n\\[\n= \\frac{\\sin(1)}{2}\n\\]", "id": "./materials/380.pdf" }, { "contents": "Find the area of one loop of the rose $\\rho = \\cos(3\\theta)$, using a double integral.\n\n- Let’s first sketch the region we are working with.\n\n![Figure 1: 2D sketch of a three leaf rose (left); Detail of the leaf we are analysing and evaluating its area – D (right).](image)\n\n- We can define $\\rho$ as\n\n$$0 \\leq \\rho \\leq \\cos(3\\theta)$$\n\nwhich allows us to define also $\\theta$ as\n\n$$\\rho = 0 \\Rightarrow \\left( \\frac{\\pi}{2} = 3\\theta \\lor -\\frac{\\pi}{2} = 3\\theta \\right) \\iff \\left( \\theta = \\frac{\\pi}{6} \\lor \\theta = -\\frac{\\pi}{6} \\right)$$\n\n- Having everything set, we can start defining the integral that will allow\nus to evaluate the area of the region D.\n\n\\[\n\\iint_D 1 \\, dA = \\int_{-\\pi/6}^{\\pi/6} \\int_0^{\\cos(3\\theta)} \\rho \\, d\\rho \\, d\\theta\n\\]\n\n\\[\n= \\int_{-\\pi/6}^{\\pi/6} \\left[ \\frac{\\rho^2}{2} \\right]_{\\rho=0}^{\\rho=\\cos(3\\theta)} \\, d\\theta\n\\]\n\n\\[\n= \\int_{-\\pi/6}^{\\pi/6} \\frac{\\cos^2(3\\theta)}{2} \\, d\\theta\n\\]\n\n\\[\n= \\int_{-\\pi/6}^{\\pi/6} \\frac{1 - \\cos(6\\theta)}{4} \\, d\\theta\n\\]\n\n\\[\n= \\left[ \\frac{\\theta}{4} - \\frac{\\sin(6\\theta)}{24} \\right]_{\\theta=-\\pi/6}^{\\theta=\\pi/6}\n\\]\n\n\\[\n= \\frac{\\pi}{24} - 0 + \\frac{\\pi}{24} + 0\n\\]\n\n\\[\n= \\frac{\\pi}{12}\n\\]", "id": "./materials/381.pdf" }, { "contents": "Find \\( \\int_0^1 \\int_{-\\sqrt{1-y^2}}^{1-y} (x^2 + y^2) \\, dx \\, dy \\)\n\n- As we can see, we should first integrate in order to \\( x \\), and only then in order to \\( y \\).\n\n\\[\n\\int_0^1 \\int_{-\\sqrt{1-y^2}}^{1-y} (x^2 + y^2) \\, dx \\, dy \\\\\n= \\int_0^1 \\left[ \\frac{x^3}{3} + xy^2 \\right]_{x=-\\sqrt{1-y^2}}^{x=1-y} \\, dy \\\\\n= \\int_0^1 \\left( \\frac{(1-y)^3}{3} + (1-y)y^2 - \\frac{(-\\sqrt{1-y^2})^3}{3} - \\left( -\\sqrt{1-y^2} \\right)y^2 \\right) \\, dy \\\\\n= \\int_0^1 \\left( \\frac{(1-y)^3}{3} \\right) \\, dy + \\int_0^1 (1-y)y^2 \\, dy + \\int_0^1 \\frac{(-\\sqrt{1-y^2})^3}{3} \\, dy + \\int_0^1 \\left( \\sqrt{1-y^2} \\right)y^2 \\, dy\n\\]\n\n- Now we should solve each integral separately.\n• First integral:\n\n\\[\n\\int_0^1 \\left( \\frac{(1-y)^3}{3} \\right) dy\n\\]\n\n\\[\n= \\frac{1}{3} \\int_0^1 (1-y)^3 dy\n\\]\n\n\\[\n= \\frac{1}{3} \\int_0^1 (1-y)(1-2y+y^2) dy\n\\]\n\n\\[\n= \\frac{1}{3} \\int_0^1 (1-2y+y^2-y+2y^2-y^3) dy\n\\]\n\n\\[\n= \\frac{1}{3} \\int_0^1 (1-3y+3y^2-y^3) dy\n\\]\n\n\\[\n= \\frac{1}{3} \\left[ y - \\frac{3y^2}{2} + y^3 - \\frac{y^4}{4} \\right]_{y=0}^{y=1}\n\\]\n\n\\[\n= \\frac{1}{3} \\left( 1 - \\frac{3}{2} + 1 - \\frac{1}{4} \\right)\n\\]\n\n\\[\n= \\frac{1}{3} \\left( \\frac{1}{4} \\right)\n\\]\n\n\\[\n= \\frac{1}{12}\n\\]\n\n• The idea now is to solve the other three integrals in order to evaluate the value of the double integral asked at the top of the first page.\n\n• At the end of the second integral, you should get: \\( \\frac{1}{12} \\)\n\n• At the end of the third integral, you should get: \\( \\frac{\\pi}{16} \\)\n\n• At the end of the fourth integral, you should get: \\( \\frac{\\pi}{16} \\)\n\n• At the end of everything, should get:\n\n\\[\n\\frac{1}{12} + \\frac{1}{12} + \\frac{\\pi}{16} + \\frac{\\pi}{16}\n\\]\n\n\\[\n= \\frac{8}{48} + \\frac{6\\pi}{48}\n\\]\n\n\\[\n= \\frac{4 + 3\\pi}{24}\n\\]", "id": "./materials/382.pdf" }, { "contents": "Determine \\( \\int \\int_R 2x - 7y^2 \\, dA \\) where \\( R = \\{(x, y) \\mid 0 \\leq y \\leq 2, -1 \\leq x \\leq 1\\} \\).\n\n- Let’s first sketch the region \\( R \\).\n\n![Figure 1: 2D sketch of the region R.](image)\n\n- At first we have to define the double integral over the region \\( R \\).\n- In this case, since \\( x \\) and \\( y \\) are independent from one-another, this means that we can choose the order of integration as we please.\n- Let’s choose to integrate first in order to \\( x \\) and after to \\( y \\), so we can\ndefine\n\n\\[\n\\iint_{R} 2x - 7y^2 \\, dA\n= \\int_{0}^{2} \\int_{-1}^{1} 2x - 7y^2 \\, dx \\, dy\n= \\int_{0}^{2} \\left[ x^2 - 7y^2 \\right]_{x=-1}^{x=1} \\, dy\n= \\int_{0}^{2} ((1 - 7y^2) - (1 + 7y^2)) \\, dy\n= \\int_{0}^{2} (-14y^2) \\, dy\n= \\left[ -\\frac{14y^3}{3} \\right]_{y=0}^{y=2}\n= -\\frac{14 \\times 8}{3}\n= -\\frac{112}{3}\n\\]\n\n- You could also choose to integrate first in order to \\( y \\) and only after in order to \\( x \\).\n• The result is going to be the same in this case. Let’s explore further.\n\n\\[\n\\iint_R 2x - 7y^2 \\, dA\n\\]\n\n\\[\n= \\int_{-1}^{1} \\int_{0}^{2} 2x - 7y^2 \\, dy \\, dx\n\\]\n\n\\[\n= \\int_{-1}^{1} \\left[ 2xy - \\frac{7y^3}{3} \\right]_{y=0}^{y=2} \\, dx\n\\]\n\n\\[\n= \\int_{-1}^{1} \\left( 4x - \\frac{56}{3} \\right) \\, dx\n\\]\n\n\\[\n= \\left[ 2x^2 - \\frac{56x}{3} \\right]_{x=-1}^{x=1}\n\\]\n\n\\[\n= \\left( 2 - \\frac{56}{3} - 2x - \\frac{56}{3} \\right)\n\\]\n\n\\[\n= -\\frac{112}{3}\n\\]", "id": "./materials/383.pdf" }, { "contents": "Find the area of the region defined as\n\\[ D = \\{(x, y) \\in \\mathbb{R}^2 : 1 \\leq x \\leq 2, \\ y \\geq 0, \\ xy \\leq 1\\} \\]\n\n- Let’s first sketch the region D.\n\n![Figure 1: 2d sketch of the region D.](image)\n\n- First, we can assess that \\( x \\) is independent and \\( y \\) depends on the previous one, meaning that the order of integration is not indifferent – We should integrate first in order to \\( y \\) and after in order to \\( x \\).\n\n- We can also define\n\n\\[ 1 \\leq x \\leq 2 \\ , \\ 0 \\leq y \\leq \\frac{1}{x} \\]\nHaving everything set, we can start defining the integral that will allow us to evaluate the area of the region D.\n\n\\[\n\\iint_D 1 \\, dA = \\int_1^2 \\int_0^{\\frac{1}{x}} 1 \\, dy \\, dx\n\\]\n\n\\[\n= \\int_1^2 \\left[ y \\right]_{y=0}^{y=\\frac{1}{x}} \\, dx\n\\]\n\n\\[\n= \\int_1^2 \\frac{1}{x} - 0 \\, dx\n\\]\n\n\\[\n= \\left[ \\ln(x) \\right]_{x=1}^{x=2}\n\\]\n\n\\[\n= \\ln(2) - \\ln(1)\n\\]\n\n\\[\n= \\ln(2)\n\\]", "id": "./materials/384.pdf" }, { "contents": "Change the order of integration of \\( \\int_0^1 \\int_y^1 f(x, y) \\, dx \\, dy \\).\n\n- To change the order of integration, we have to be careful. There are some cases where the order of integration is indifferent (when all variables are independent), but most cases are more complex.\n\n- As we can see, the integral in order to \\( x \\) depends on the value of the other variable.\n\n- We can define\n \\[\n 0 \\leq y \\leq 1 \\quad , \\quad y \\leq x \\leq 1\n \\]\n\n- Let’s sketch the region defined by the previous inequations.\n\nFigure 1: 2D sketch of the region D.\n• This means that we can easily define D, but having $x$ as the independent variable.\n\n$$0 \\leq x \\leq 1 \\ , \\ 0 \\leq y \\leq x$$\n\n• This means that we can rewrite\n\n$$\\int_0^1 \\int_y^1 f(x, y) \\, dx \\, dy = \\int_0^1 \\int_0^x f(x, y) \\, dy \\, dx$$", "id": "./materials/385.pdf" }, { "contents": "Determine \\( \\int_0^1 \\int_{x^2}^x x + 3 \\, dy \\, dx \\)\n\n- As we can see in the double integral we were asked to evaluate, we are going to integrate first in respect to \\( y \\) and only after to \\( x \\).\n\n\\[\n\\int_0^1 \\int_{x^2}^x x + 3 \\, dy \\, dx \\\\\n= \\int_0^1 \\left[ xy + 3y \\right]_{y=x^2}^{y=x} \\, dx \\\\\n= \\int_0^1 (x^2 + 3x - x^3 - 3x^2) \\, dx \\\\\n= \\int_0^1 (3x - 2x^2 - x^3) \\, dx \\\\\n= \\left[ \\frac{3x^2}{2} - \\frac{2x^3}{3} - \\frac{x^4}{4} \\right]_{x=0}^{x=1} \\\\\n= \\frac{3}{2} - \\frac{2}{3} - \\frac{1}{4} - 0 - 0 - 0 \\\\\n= \\frac{12}{7} - \\frac{12}{12} - \\frac{12}{12} \\\\\n= \\frac{7}{12}\n\\]", "id": "./materials/386.pdf" }, { "contents": "Determine \\( \\int_0^1 \\int_{x^2}^x x + 3 \\, dy \\, dx \\)\n\n- As we can see in the double integral we were asked to evaluate, we are going to integrate first in respect to \\( y \\) and only after to \\( x \\).\n\n\\[\n\\int_0^1 \\int_{x^2}^x x + 3 \\, dy \\, dx \\\\\n= \\int_0^1 \\left[ xy + 3y \\right]_{y=x^2}^{y=x} \\, dx \\\\\n= \\int_0^1 (x^2 + 3x - x^3 - 3x^2) \\, dx \\\\\n= \\int_0^1 (3x - 2x^2 - x^3) \\, dx \\\\\n= \\left[ \\frac{3x^2}{2} - \\frac{2x^3}{3} - \\frac{x^4}{4} \\right]_{x=0}^{x=1} \\\\\n= \\frac{3}{2} - \\frac{2}{3} - \\frac{1}{4} - 0 - 0 - 0 \\\\\n= \\frac{12}{7} - \\frac{12}{12} - \\frac{12}{12} \\\\\n= \\frac{7}{12}\n\\]", "id": "./materials/387.pdf" }, { "contents": "Determine \\( \\int \\int_D 2yx^2 + 9y^3 \\, dA \\), where \\( D \\) is the region bounded by \\( y = \\frac{2x}{3} \\) and by \\( y = 2\\sqrt{x} \\).\n\n- Let’s first sketch the region in question (D)\n\n![Figure 1: 2D sketch of the region D.](image)\n\n- Let’s start with the definition of \\( D \\), where the intersection of the two function must be evaluated.\n\n\\[\n\\frac{2x}{3} = 2\\sqrt{x} \\iff \\left( \\frac{x}{3} \\right)^2 = x \\iff x = 0 \\lor x = 9\n\\]\nmeaning we can define D as\n\n\\[ 0 \\leq x \\leq 9, \\quad \\frac{2x}{3} \\leq y \\leq 2\\sqrt{x} \\]\n\n- With that being said, we are now able to evaluate the double integral we were asked.\n\n\\[\n\\iint_D 2yx^2 + 9y^3 \\, dA\n= \\int_0^9 \\int_{\\frac{2x}{3}}^{2\\sqrt{x}} 2yx^2 + 9y^3 \\, dy \\, dx\n= \\int_0^9 \\left[ y^2x^2 + \\frac{9y^4}{4} \\right]_{y=\\frac{2x}{3}}^{y=2\\sqrt{x}} \\, dx\n= \\int_0^9 \\left( 4x^3 + 36x^2 - \\frac{4x^4}{9} - \\frac{4x^4}{9} \\right) \\, dx\n= \\int_0^9 \\left( 4x^3 + 36x^2 - \\frac{8x^4}{9} \\right) \\, dx\n= \\left[ x^4 + 12x^3 - \\frac{8x^5}{45} \\right]_{x=0}^{x=9}\n= 6561 + 8748 - \\frac{472392}{45}\n= 15309 \\times 45 - 472392\n= \\frac{24057}{5}\n\\]", "id": "./materials/388.pdf" }, { "contents": "Absolute value inequalities\n\nMathE\n\n19th of April of 2023\n\nThe absolute value of a number $x \\in \\mathbb{R}$ is defined as\n\n$$|x| = \\begin{cases} x & \\text{if } x \\geq 0 \\\\ -x & \\text{if } x < 0 \\end{cases}$$\n\nGeometrically, you can think of in the following way:\n\n- Given $x \\in \\mathbb{R}$, you can represent $x$ on the real line.\n 1. If $x = 0$, it coincides with the origin of the real line.\n\n ![Graph showing x = 0]\n\n 2. If $x > 0$, it is to the right of the origin (we are considering the numbers are increasing from left to right).\n\n ![Graph showing x > 0]\n\n 3. If $x < 0$, it is to the left of the origin.\n\n ![Graph showing x < 0]\n\n- The absolute value of $x$ is the distance from $x$ to the origin.\n 1. If $x = 0$, the distance from 0 to 0 is 0, so $|0| = 0$.\n\n ![Graph showing |0| = 0]\n\n 2. If $x > 0$, then the distance from $x$ to 0 (the origin) is $x$.\n\n ![Graph showing |x| = x]\n\n 3. If $x < 0$, then the distance from $x$ to 0 (the origin) is $-x$ (note that $x < 0$ implies that $-x > 0$).\n\n ![Graph showing |x| = -x]\nExercise 1. Solve the inequality $|x| < 5$.\n\nAs we saw above, this means that the distance from $x$ to the origin is less than 5.\n\nIn other words, $|x| < 5$ means that we have to have simultaneously $x < 5$ and $x > -5$:\n\n$$|x| < 5 \\iff x < 5 \\land x > -5 \\iff -5 < x < 5$$\n\nAnswer: $-5 < x < 5$ or $x \\in (-5, 5)$\n\nExercise 2. Solve the inequality $|x| > 5$.\n\nAs we saw above, this means that the distance from $x$ to the origin is greater than 5.\n\nIn other words, $|x| < 5$ means that we have to have $x < -5$ or $x > 5$:\n\n$$|x| > 5 \\iff x > 5 \\land x < -5$$\n\nAnswer: $x > 5 \\land x < -5$ or $x \\in (-\\infty, -5) \\cup (5, +\\infty)$\n\nExercise 3. Solve the inequality $|3x - 7| \\leq 2$.\n\nAs we saw above, this means that the distance from $3x - 7$ to the origin is less than or equal to 2.\n\nIn other words, $|3x - 7| \\leq 2$ means that we have to have simultaneously $3x - 7 \\leq 2$ and $3x - 7 \\geq -2$:\n\n$$|3x - 7| \\leq 2 \\iff 3x - 7 \\leq 2 \\land 3x - 7 \\geq -2 \\iff 3x - 7 + 7 \\leq 2 + 7 \\land 3x - 7 + 7 \\geq -2 + 7 \\iff 3x \\leq 9 \\land 3x \\geq 5 \\iff$$\n\\[\n\\frac{3x}{\\beta} \\leq \\frac{9}{3} \\land \\frac{3x}{\\beta} \\geq \\frac{5}{3} \\iff \\\\\nx \\leq 3 \\land x \\geq \\frac{5}{3}\n\\]\n\n**Answer:** \\( x \\leq 3 \\land x \\geq \\frac{5}{3} \\) or \\( x \\in \\left[ \\frac{5}{3}, 3 \\right] \\)\n\n**Exercise 4.** Solve the inequality \\( 5|4 - 3x| - 10 \\leq 0 \\).\n\nIn this type of inequalities, the first step is to isolate the modulus on one of the sides of the inequality:\n\n\\[\n5|4 - 3x| - 10 \\leq 0 \\iff \\\\\n5|4 - 3x| \\geq 10 \\iff \\\\\n\\frac{5|4 - 3x|}{5} \\geq \\frac{10}{5} \\iff \\\\\n|4 - 3x| \\geq 2\n\\]\n\nAs we saw above, this means that the distance from \\( 4 - 3x \\) to the origin is less than or equal to 2.\n\nIn other words, \\( |4 - 3x| \\leq 2 \\) means that we have to have simultaneously \\( 4 - 3x \\leq 2 \\) and \\( 4 - 3x \\geq -2 \\):\n\n\\[\n|4 - 3x| \\leq 2 \\iff \\\\\n4 - 3x \\leq 2 \\land 4 - 3x \\geq -2 \\iff \\\\\n-3x \\leq -2 \\land -3x \\geq -6 \\iff \\\\\n-3x \\geq -2 \\land -3x \\leq -6 \\iff \\\\\nx \\geq \\frac{2}{3} \\land x \\leq 2\n\\]\nOur goal is to isolate $x$ on one of the sides of the inequalities. In order to do so, in this step we must divide both terms of the inequalities by $-3$, because this allow us to “get rid” of $-3$.\n\nHowever, when we divide the terms of an inequality by a negative number,\n\n- $\\leq$ changes to $\\geq$\n- $\\geq$ changes to $\\leq$\n- $<$ changes to $>$\n- $>$ changes to $<$\n\nbecause the minus sign implies a reflection on the origin. For example, if we have\n\n$$-3 \\leq x \\leq 2$$\n\nand if we divide all the terms by $-1$, we get\n\n$$-2 \\leq -x \\leq 3$$\n\nso $-2 \\leq -x \\leq 3$.\n\n**Answer:** $x \\leq 2 \\land x \\geq \\frac{2}{3}$ or $x \\in \\left[\\frac{2}{3}, 2\\right]$.\n\n**Exercise 5.** Solve the inequality $\\frac{5}{4} - \\left|\\frac{1}{2} - \\frac{1}{4}x\\right| < \\frac{3}{8}$.\n\nIn this type of inequalities, the first step is to isolate the modulus on one of the sides of the inequality:\n\n$$\\frac{5}{4} - \\left|\\frac{1}{2} - \\frac{1}{4}x\\right| < \\frac{3}{8} \\iff$$\n\n$$\\frac{5}{4} + \\frac{5}{8} - \\left|\\frac{1}{2} - \\frac{1}{4}x\\right| < \\frac{5}{4} + \\frac{3}{8} \\iff$$\n\n$$\\left(\\times 2\\right)$$\n\n$$-\\frac{1}{2} - \\frac{1}{4}x < -\\frac{10}{8} + \\frac{3}{8} \\iff$$\n\n$$-\\frac{1}{2} - \\frac{1}{4}x < -\\frac{7}{8} \\iff$$\nAs we saw above, this means that the distance from \\( \\frac{1}{2} - \\frac{1}{4}x \\) to the origin is greater than \\( \\frac{7}{8} \\).\n\nIn other words, \\( \\left| \\frac{1}{2} - \\frac{1}{4}x \\right| > \\frac{7}{8} \\) means that we have to have\n\n\\[\n\\frac{1}{2} - \\frac{1}{4}x < \\frac{7}{8} \\quad \\text{or} \\quad \\frac{1}{2} - \\frac{1}{4}x > \\frac{7}{8}\n\\]\n\nso\n\n\\[\n\\left| \\frac{1}{2} - \\frac{1}{4}x \\right| > \\frac{7}{8} \\iff \\\\\n\\frac{1}{2} - \\frac{1}{4}x < -\\frac{7}{8} \\quad \\text{or} \\quad \\frac{1}{2} - \\frac{1}{4}x > \\frac{7}{8} \\iff \\\\\n\\frac{\\sqrt{2}}{2} + \\frac{\\sqrt{2}}{2} - \\frac{1}{4}x < -\\frac{1}{2} + \\frac{7}{8} \\quad \\text{or} \\quad \\frac{\\sqrt{2}}{2} + \\frac{\\sqrt{2}}{2} - \\frac{1}{4}x > -\\frac{1}{2} + \\frac{7}{8} \\iff \\\\\n-\\frac{1}{4}x < -\\frac{4}{8} - \\frac{7}{8} \\quad \\text{or} \\quad -\\frac{1}{4}x > -\\frac{4}{8} + \\frac{7}{8} \\iff \\\\\n-\\frac{1}{4}x < -\\frac{11}{8} \\quad \\text{or} \\quad -\\frac{1}{4}x > \\frac{3}{8} \\iff \\\\\n-\\frac{1}{4}x \\times x < -\\frac{11}{8} \\times 4 \\quad \\text{or} \\quad -\\frac{1}{4}x \\times x > \\frac{3}{8} \\times 4 \\iff \\\\\n-x < -\\frac{11}{2} \\quad \\text{or} \\quad -x > \\frac{3}{2} \\iff \\\\\n-x < -\\frac{11}{2} \\quad \\text{or} \\quad -x > \\frac{3}{2} \\iff \\\\\nx > \\frac{11}{2} \\quad \\text{or} \\quad x < -\\frac{3}{2}\n\\]\n\nAnswer: \\( x < -\\frac{3}{2} \\quad \\text{or} \\quad x > \\frac{11}{2} \\quad \\text{or} \\quad x \\in (-\\infty, -\\frac{3}{2}) \\cup (\\frac{11}{2}, +\\infty) \\)\nExercise 6. Sketch the graph of \\( y = |x| + 3 \\) and then explain why \\( |x| + 3 \\geq |x + 3| \\) for all \\( x \\in \\mathbb{R} \\).\n\nIt helps to write the function \\( f(x) = |x| + 3 \\) in the form of a piecewise function.\n\nWe know that\n\n\\[\n|x| = \\begin{cases} \n x & \\text{if } x \\geq 0 \\\\\n -x & \\text{if } x < 0\n\\end{cases}\n\\]\n\nso\n\n\\[\nf(x) = |x| + 3 = \\begin{cases} \n x + 3 & \\text{if } x \\geq 0 \\\\\n -x + 3 & \\text{if } x < 0\n\\end{cases}\n\\]\n\nThe first sub-function is\n\n\\[\nf_1 : \\mathbb{R}_0^+ \\rightarrow \\mathbb{R} \\\\\nx \\mapsto x + 3\n\\]\n\nThe second sub-function is\n\n\\[\nf_2 : \\mathbb{R}^- \\rightarrow \\mathbb{R} \\\\\nx \\mapsto -x + 3\n\\]\nPutting the functions $f_1$ and $f_2$ together, we get the graph of the function\n\n$$f: \\mathbb{R} \\rightarrow \\mathbb{R}$$\n\n$$x \\mapsto |x| + 3$$\n\nRegarding the function $g(x) = |x + 3|$, we have that\n\n$$g(x) = |x + 3| = \\begin{cases} \n x + 3 & \\text{if } x + 3 \\geq 0 \\\\\n -(x + 3) & \\text{if } x + 3 < 0 \n\\end{cases} = \\begin{cases} \n x + 3 & \\text{if } x + 3 \\geq 0 \\\\\n -x - 3 & \\text{if } x + 3 < 0 \n\\end{cases} = \\begin{cases} \n x + 3 & \\text{if } x \\geq -3 \\\\\n -x - 3 & \\text{if } x < -3 \n\\end{cases}$$\n\nThe first sub-function is\n\n$$g_1: [-3, +\\infty] \\rightarrow \\mathbb{R}$$\n\n$$x \\mapsto x + 3$$\n\nThe second sub-function is\n\n$$g_2: (-\\infty, -3) \\rightarrow \\mathbb{R}$$\n\n$$x \\mapsto -x - 3$$\nPutting the functions $g_1$ and $g_2$ together, we get the graph of the function\n\n$$g: \\mathbb{R} \\rightarrow \\mathbb{R}$$\n\n$$x \\mapsto |x + 3|$$\n\nDrawing the graphs of $f$ and $g$ on the same Cartesian plane\n\nit is possible to see that $f(x) \\geq g(x)$, $\\forall x \\in \\mathbb{R}$. \nSince \\( f(x) = |x| + 3 \\) and \\( g(x) = |x + 3| \\), this is equivalent as saying that\n\n\\[\n|x| + 3 \\geq |x + 3|, \\quad \\forall x \\in \\mathbb{R}\n\\]\n\n**Exercise 7.** Solve the inequality \\(|3x - 2| \\geq |x + 4|\\).\n\nNote the following: Let \\( a, b \\in \\mathbb{R}_0^+ \\) such that \\( a \\geq b \\).\n\nThen \\( a \\times a \\geq b \\times b \\), which is equivalent to say that \\( a^2 \\geq b^2 \\).\n\nNote that here it is essential that both \\( a \\) and \\( b \\) are non-negative real numbers.\n\nOtherwise, the statement can be false. Suppose, for example, that \\( a = 1 \\) and \\( b = -2 \\).\n\nThen we have \\( 1 \\geq -2 \\), but \\( 1 \\times 1 \\not\\geq -2 \\times (-2) \\).\n\nSince \\(|3x - 2| \\geq 0 \\) and \\(|x + 4| \\geq 0 \\), we have that\n\n\\[\n|3x - 2| \\geq |x + 4| \\Rightarrow\n|3x - 2| \\times |3x - 2| \\geq |x + 4| \\times |x + 4| \\iff\n(|3x - 2|)^2 \\geq (|x + 4|)^2\n\\]\n\nIn the next step, we can make the modulus disappear because if \\( 3x - 2 \\) is negative, then when we multiply \\( 3x - 2 \\) with itself, we get a positive number. So we can say that\n\n\\[\n(|3x - 2|)^2 \\geq (|x + 4|)^2 \\iff (3x - 2)^2 \\geq (x + 4)^2 \\quad (1)\n\\]\n\n\\[\n(3x)^2 + 2 \\times (3x) \\times (-2) + (-2)^2 \\geq x^2 + 2 \\times x \\times 4 + 4^2 \\iff\n9x^2 - 12x + 4 \\geq x^2 + 8x + 16 \\iff\n9x^2 - x^2 - 12x - 8x + 4 - 16 \\geq 0 \\iff\n8x^2 - 20x - 12 \\geq 0 \\quad (2)\n\\]\n\n\\[\n2x^2 - 5x - 3 \\geq 0\n\\]\n\n(1) Remember that \\((a + b)^2 = a^2 + 2ab + b^2\\), \\( \\forall a, b \\in \\mathbb{R} \\).\n\n(2) Divide both sides of the inequality by 4.\n\nNow we have to solve \\( 2x^2 - 5x - 3 \\geq 0 \\).\n\nTo do so, we first have to find the roots of the polynomial \\( 2x^2 - 5x - 3 \\):\n\n\\[\n2x^2 - 5x - 3 = 0 \\iff\nx = \\frac{-(-5) \\pm \\sqrt{(-5)^2 - 4 \\times 2 \\times (-3)}}{2 \\times 2} \\iff\n\\]\n\\[ x = \\frac{5 \\pm \\sqrt{25 + 24}}{4} \\iff \\]\n\\[ x = \\frac{5 \\pm \\sqrt{49}}{4} \\iff \\]\n\\[ x = \\frac{5 \\pm 7}{4} \\iff \\]\n\\[ x = \\frac{5 - 7}{4} \\lor \\frac{5 + 7}{4} \\iff \\]\n\\[ x = \\frac{-2}{4} \\lor x = \\frac{12}{4} \\iff \\]\n\\[ x = -\\frac{1}{2} \\lor x = 3 \\]\n\n| \\( x \\) | \\(-\\infty\\) | \\(-\\frac{1}{2}\\) | 3 | \\(+\\infty\\) |\n|---|---|---|---|---|\n| \\( 2x^2 - 5x - 3 \\) | + | 0 | - | 0 | + |\n\nAs we can see, the solution set of the inequality \\( 2x^2 - 5x - 3 \\geq 0 \\) is\n\n\\[ \\left( -\\infty, -\\frac{1}{2} \\right] \\cup \\left[ 3, +\\infty \\right) \\]", "id": "./materials/389.pdf" }, { "contents": "Manipulation of Algebraic Expressions\n\nIndices\nIndices\n\n- For the number $2^3$, 2 is the base and 3 is the index.\n- The index tells us how many times the number is multiplied by itself.\n- In the above case $2^3$ means 2 is multiplied by itself 3 times.\n- $2 \\times 2 \\times 2 = 8$ i.e. $2^3 = 8$\n- Every number has an index, for example, $5 = 5^1$; $16 = 16^1$\n- There are Laws of Indices which can be applied but only where the bases are the same.\n- If index is an integer it is called a power.\nIndices\n\nLaw 1: When multiplying numbers with the same base add the indices:\n\n\\[ a^m \\times a^n = a^{m+n} \\]\n\n\\[ 3^2 \\times 3^4 = 3^{2+4} = 3^6 \\]\nLaw 2: When dividing numbers with the same base subtract the indices.\n\n\\[\n\\frac{a^m}{a^n} = a^{m-n}\n\\]\n\n\\[\n\\frac{3^5}{3^2} = 3^{5-2} = 3^3\n\\]\nIndices\n\nLaw 3: When a number which is raised to a power is raised to a further power, the indices are multiplied. Thus:\n\n\\[(a^m)^n = a^{mxn}\\]\n\n\\[(3^5)^2 = 3^{5\\times2} = 3^{10}\\]\nIndices\n\n**Law 4:** When a number has an index of 0 its value is 1 thus:\n\n\\[ a^0 = 1 \\]\n\n\\[ 3^0 = 1 \\]\nIndices\n\n**Law 5:** A number raised to a **negative power** is the **reciprocal** of that number raised to a positive power. Thus:\n\n\\[ a^{-n} = \\frac{1}{a^n} \\]\n\n\\[ 3^{-4} = \\frac{1}{3^4} \\]\nIndices\n\n**Law 6:** When a number is raised to a fractional power the denominator of the fraction is the root of the number and the numerator is the power. Thus:\n\n\\[ a^{\\frac{m}{n}} = \\sqrt[n]{a^m} \\]\n\n\\[ 4^{\\frac{2}{3}} = \\sqrt[3]{4^2} = 2.52 \\]\nIndices\n\nOften used applications of Laws of Indices:\n\n\\[ \\sqrt{x} = x^{\\frac{1}{2}} \\]\n\nAlso \\[ \\frac{1}{\\sqrt{x}} = x^{-\\frac{1}{2}} \\]", "id": "./materials/39.pdf" }, { "contents": "Find \\( \\int \\int_{R} \\frac{e^{x-y}}{2} \\, dA \\), with \\( R = \\{(x, y) : -1 \\leq y \\leq 1, -1 \\leq x \\leq 1\\} \\).\n\n- As we can see in the double integral we were asked to evaluate, the region \\( R \\) is defined with two independent variables. This means that the order of integration is indifferent.\n\n- Let’s consider that we want to integrate first in respect to \\( x \\) and only then in respect to \\( y \\).\n\n- This means that we can write the double integral as\n\n\\[\n\\int \\int_{R} \\frac{e^{x-y}}{2} \\, dA = \\int_{-1}^{1} \\int_{-1}^{1} \\frac{e^{x-y}}{2} \\, dx \\, dy\n\\]\n\n\\[\n= \\frac{1}{2} \\int_{-1}^{1} e^{x} \\, dx \\, dy\n\\]\n\n\\[\n= \\frac{1}{2} \\int_{-1}^{1} \\left[ e^{x} \\right]_{x=-1}^{x=1} \\, dy\n\\]\n\n\\[\n= \\frac{1}{2} \\int_{-1}^{1} \\left( e - \\frac{1}{e} \\right) \\, dy\n\\]\n\n- Now is just to evaluate a definite integral.\n\n- At the end of the exercise you should get: \\( \\frac{e^{4} - 2e^{2} + 1}{2e^{2}} \\)", "id": "./materials/390.pdf" }, { "contents": "Find the value of the region’s area bounded by $y = x^2 - 4$ and $y = -x^2 + 4$.\n\n- We can say that $y$ is a dependent variable, whereas $x$ is independent. This means that we need to integrate first in order to $y$ and only after to $x$.\n\n- Let’s first sketch the region in question (D)\n\n![Figure 1: 2D sketch of the region D.](image)\n\n- Evaluating the intersection of both equations, even tough we can see through the graph the solution\n\n$$x^2 - 4 = -x^2 + 4 \\iff x^2 = 4 \\iff x = \\pm 2$$\n\n- So we can define the region D as\n\n$$-2 \\leq x \\leq 2 \\quad , \\quad x^2 - 4 \\leq y \\leq -x^2 + 4$$\nThis means that we can evaluate D’s area with\n\n\\[\n\\iint_D 1 \\, dA = \\int_{-2}^{2} \\int_{x^2-4}^{-x^2+4} 1 \\, dy \\, dx\n\\]\n\n\\[\n= \\int_{-2}^{2} \\left[ \\int_{y=x^2-4}^{y=-x^2+4} 1 \\, dy \\right] \\, dx\n\\]\n\n\\[\n= \\int_{-2}^{2} (-x^2 + 4 - x^2 + 4) \\, dx\n\\]\n\n\\[\n= \\int_{-2}^{2} (-2x^2 + 8) \\, dx\n\\]\n\n\\[\n= \\left[ -\\frac{2x^3}{3} + 8x \\right]_{x=-2}^{x=2}\n\\]\n\n\\[\n= -\\frac{16}{3} + 16 - \\frac{16}{3} + 16\n\\]\n\n\\[\n= -16 \\times 2 + 16 \\times 2 \\times 3\n\\]\n\n\\[\n= \\frac{64}{3}\n\\]", "id": "./materials/391.pdf" }, { "contents": "Evaluate \\( \\int \\int_D 4x + 2 \\, dA \\), where \\( D \\) is the region enclosed by \\( y = x^2 \\) and \\( y = 2x \\).\n\n- We can say that \\( y \\) is a dependent variable, whereas \\( x \\) is independent. This means that we need to integrate first in order to \\( y \\) and only after to \\( x \\).\n- Let’s first sketch the region in question (\\( D \\))\n\n![Figure 1: 2D sketch of the region D.](image)\n\n- Evaluating the intersection of both equations, even tough we can see through the graph the solution\n \\[ x^2 = 2x \\iff x = 2 \\lor x = 0 \\]\n- So we can define the region \\( D \\) as\n \\[ 0 \\leq x \\leq 2 \\quad , \\quad 2x \\leq y \\leq x^2 \\]\nThis means that we can evaluate the integral we were asked to\n\n\\[\n\\iint_D (4x + 2) \\, dA\n\\]\n\n\\[\n= \\int_0^2 \\int_{x^2}^{2x} 4x + 2 \\, dy \\, dx\n\\]\n\n\\[\n= \\int_0^2 \\left[ 4xy + 2y \\right]_{y=x^2}^{y=2x} \\, dx\n\\]\n\n\\[\n= \\int_0^2 (8x^2 + 4x - 4x^3 - 2x^2) \\, dx\n\\]\n\n\\[\n= \\int_0^2 (6x^2 + 4x - 4x^3) \\, dx\n\\]\n\n\\[\n= \\left[ 2x^3 + 2x^2 - x^4 \\right]_{x=0}^{x=2}\n\\]\n\n\\[\n= 16 + 8 - 16\n\\]\n\n\\[\n= 8\n\\]", "id": "./materials/392.pdf" }, { "contents": "Find \\( \\int \\int_D 3x + 4y^2 \\, dA \\) in the upper half of the plane \\((y \\geq 0)\\) and between the curves \\(x^2 + y^2 = 1\\) and \\(x^2 + y^2 = 4\\).\n\n- Let’s first sketch the region in question \\((D)\\)\n\n![Figure 1: 2D sketch of the region D.](image)\n\n- According to the geometry of \\(D\\), it is convenient to change to polar coordinates, where\n\n\\[\nx = r \\cos (\\theta) \\quad , \\quad y = r \\sin (\\theta)\n\\]\n\n- Let’s start with the definition of \\(D\\)\n\n\\[\n1 \\leq r \\leq 2 \\quad , \\quad 0 \\leq \\theta \\leq \\pi\n\\]\n• With that being said, we are now able to evaluate the double integral we were asked.\n\n\\[\n\\int \\int_D 3x + 4y^2 \\, dA\n\\]\n\n\\[\n= \\int_1^2 \\int_0^\\pi (3r \\cos (\\theta) + 4(r \\sin (\\theta))^2) r \\, d\\theta dr\n\\]\n\n\\[\n= \\int_1^2 \\int_0^\\pi 3r^2 \\cos (\\theta) + 4r^3 \\sin^2 (\\theta) \\, d\\theta dr\n\\]\n\n\\[\n= \\int_1^2 \\int_0^\\pi 3r^2 \\cos (\\theta) + 4r^3 (1 - \\cos^2 (\\theta)) \\, d\\theta dr\n\\]\n\n\\[\n= \\int_1^2 \\int_0^\\pi 3r^2 \\cos (\\theta) + 4r^3 \\left(1 - \\frac{1 + \\cos (2\\theta)}{2}\\right) \\, d\\theta dr\n\\]\n\n\\[\n= \\int_1^2 \\left[ 3r^2 \\sin (\\theta) + 2r^3 \\theta + 4r^3 \\sin (2\\theta) \\right]_{\\theta=0}^{\\theta=\\pi} dr\n\\]\n\n\\[\n= \\int_1^2 2r^3 \\pi \\, dr\n\\]\n\n\\[\n= \\left[ \\frac{r^4}{2} \\pi \\right]_{r=1}^{r=2}\n\\]\n\n\\[\n= 8\\pi - \\frac{\\pi}{2}\n\\]\n\n\\[\n= \\frac{15\\pi}{2}\n\\]", "id": "./materials/393.pdf" }, { "contents": "Find \\( \\int \\int_D x(y - 1) \\, dA \\), where \\( D \\) is bounded by \\( y = x^2 - 4 \\) and \\( y = -x^2 + 4 \\).\n\n- We can say that \\( y \\) is a dependent variable, whereas \\( x \\) is independent. This means that we need to integrate first in order to \\( y \\) and only after to \\( x \\).\n- Let’s first sketch the region in question (\\( D \\))\n\n![Figure 1: 2D sketch of the region D.](image)\n\n- Evaluating the intersection of both equations, even tough we can see through the graph the solution\n\n\\[\nx^2 - 4 = -x^2 + 4 \\iff x^2 = 4 \\iff x = \\pm 2\n\\]\n• So we can define the region D as\n\n\\[-2 \\leq x \\leq 2, \\quad x^2 - 4 \\leq y \\leq -x^2 + 4\\]\n\n• This means that we can evaluate the double integral\n\n\\[\n\\int \\int_D x(y - 1) \\, dA = \\int_{-2}^{2} \\int_{x^2 - 4}^{-x^2 + 4} x(y - 1) \\, dy \\, dx\n\\]\n\n• At the end you should get: 0", "id": "./materials/394.pdf" }, { "contents": "Convert to polar form \\( \\int \\int_D 3 - y \\, dA \\), where the region D is bounded by the circle with radius 1.\n\n- We can define D using polar coordinates, this is\n \\[\n x = \\rho \\cos (\\theta) \\quad , \\quad y = \\rho \\sin (\\theta)\n \\]\n\n- Since we also know that D is the circle with radius 1, we can define D as\n \\[\n 0 \\leq \\rho \\leq 1 \\quad , \\quad 0 \\leq \\theta \\leq 2\\pi\n \\]\n\n- So, we may rewrite the initial double integral as\n \\[\n \\int \\int_D 3 - y \\, dA = \\int_0^{2\\pi} \\int_0^1 (3 - \\rho \\sin (\\theta)) \\rho \\, d\\rho d\\theta\n \\]\n \\[\n = \\int_0^{2\\pi} \\int_0^1 3\\rho - \\rho^2 \\sin (\\theta) \\, d\\rho d\\theta\n \\]", "id": "./materials/395.pdf" }, { "contents": "Convert to polar coordinates \\( \\int \\int_D xy \\, dA \\) where\n\n\\[ D = \\{(x, y) \\in \\mathbb{R}^2 : 0 \\leq x \\leq 1 \\land 0 \\leq y \\leq x\\}. \\]\n\n- Let’s first sketch the region in question (D)\n\n![Figure 1: 2D sketch of the region D.](image)\n\n- We are asked to change to polar coordinates, so, proceeding with that transformation we can define\n\n\\[ x = r \\cos(\\theta) \\quad , \\quad y = r \\sin(\\theta) \\]\n\n- Let’s start with the definition of D\n\n\\[ 0 \\leq x \\leq 1 \\quad , \\quad 0 \\leq y \\leq x \\]\n• We can easily evaluate the points that define D as a polygon: (0, 0), (1, 1) and (1, 0).\n\n• This means that we can define\n\n\\[ 0 \\leq \\theta \\leq \\tan \\left( \\frac{1}{1} \\right) \\iff 0 \\leq \\theta \\leq \\frac{\\pi}{4} \\]\n\n• Now we are just missing to define the interval of integration for r. Keep in mind that r is a dependent variable, this is, its value depends of \\( \\theta \\). Since r has a monotonic increase, we may say\n\n\\[ \\frac{0}{\\cos (0)} \\leq r \\leq \\frac{1}{\\cos (\\theta)} \\iff 0 \\leq r \\leq \\frac{1}{\\cos (\\theta)} \\]\n\n• With that being said, we are now able to make the conversion we were asked\n\n\\[\n\\iint_D xy \\, dA = \\int_0^{\\frac{\\pi}{4}} \\int_0^{\\frac{1}{\\cos (\\theta)}} (r \\cos (\\theta))(r \\sin (\\theta))r \\, dr \\, d\\theta \\\\\n= \\int_0^{\\frac{\\pi}{4}} \\int_0^{\\frac{1}{\\cos (\\theta)}} r^3 \\cos (\\theta) \\sin (\\theta) \\, dr \\, d\\theta\n\\]", "id": "./materials/396.pdf" }, { "contents": "Find \\( \\int_0^{\\pi/4} \\int_0^{1/\\cos(\\theta)} \\rho^3 \\sin(\\theta) \\cos(\\theta) \\, d\\rho \\, d\\theta \\).\n\n- We can evaluate said integral, through\n\n\\[\n\\int_0^{\\pi/4} \\int_0^{1/\\cos(\\theta)} \\rho^3 \\sin(\\theta) \\cos(\\theta) \\, d\\rho \\, d\\theta \\\\\n= \\int_0^{\\pi/4} \\sin(\\theta) \\cos(\\theta) \\left[ \\frac{\\rho^4}{4} \\right]_{\\rho=0}^{\\rho=1/\\cos(\\theta)} \\, d\\theta \\\\\n= \\int_0^{\\pi/4} \\sin(\\theta) \\cos(\\theta) \\frac{1}{4 \\cos^4(\\theta)} \\, d\\theta \\\\\n= \\int_0^{\\pi/4} \\sin(\\theta) \\frac{1}{4 \\cos^3(\\theta)} \\, d\\theta \\\\\n= \\frac{1}{4} \\int_0^{\\pi/4} \\tan(\\theta) \\frac{1}{\\cos^2(\\theta)} \\, d\\theta \\\\\n= \\frac{1}{4} \\left[ \\tan^2(\\theta) \\right]_{\\theta=0}^{\\theta=\\pi/4} \\\\\n= \\frac{1}{4} \\left( \\frac{1}{2} - 0 \\right) \\\\\n= \\frac{1}{8}\n\\]", "id": "./materials/397.pdf" }, { "contents": "Find $\\int \\int_D y^2 + 3x \\, dA$ where $D$ is the region in the third quadrant between $x^2 + y^2 = 1$ and $x^2 + y^2 = 9$.\n\n- Let’s first sketch the region in question ($D$)\n\n![Figure 1: 2D sketch of the region D.](image)\n\n- According to the geometry of $D$, it is convenient to change to polar coordinates, where\n \n $$x = r \\cos(\\theta) \\quad , \\quad y = r \\sin(\\theta)$$\n\n- Let’s start with the definition of $D$\n \n $$1 \\leq r \\leq 3 \\quad , \\quad \\pi \\leq \\theta \\leq \\frac{3\\pi}{2}$$\n• With that being said, we are now able evaluate the double integral we were asked.\n\n\\[\n\\iint_D y^2 + 3x \\, dA\n\\]\n\n\\[\n= \\int_1^3 \\int_{\\pi/2}^{3\\pi/2} \\left( (r \\sin (\\theta))^2 + 3r \\cos (\\theta) \\right) r \\, d\\theta \\, dr\n\\]\n\n\\[\n= \\int_1^3 \\int_{\\pi/2}^{3\\pi/2} r^3 \\sin^2 (\\theta) + 3r^2 \\cos (\\theta) \\, d\\theta \\, dr\n\\]\n\n• Now is just to solve the double integral.\n\n• At the end you should get: \\(5\\pi - 26\\)", "id": "./materials/398.pdf" }, { "contents": "Convert to polar coordinates and find \\[ \\int_0^3 \\int_{-\\sqrt{9-x^2}}^{\\sqrt{9-x^2}} x^3 + xy^2 \\, dy \\, dx. \\]\n\n- Let’s first sketch the region defined by the interval of integration written on the double integral (D)\n\nFigure 1: 2D sketch of the region D.\n• As we can assess from Figure 1, D is half of a circle of radius 3.\n\n• We are asked to change to polar coordinates, so, proceeding with that transformation we can define\n\n\\[ x = r \\cos(\\theta) \\quad , \\quad y = r \\sin(\\theta) \\]\n\n• We can easily define D as\n\n\\[ -\\frac{\\pi}{2} \\leq \\theta \\leq \\frac{\\pi}{2} \\quad , \\quad 0 \\leq r \\leq 3 \\]\n\n• With that being said, we are now able to make the conversion we were asked\n\n\\[\n\\int_{0}^{3} \\int_{-\\sqrt{9-x^2}}^{\\sqrt{9-x^2}} x^3 + xy^2 \\, dy \\, dx \\\\\n= \\int_{-\\frac{\\pi}{2}}^{\\frac{\\pi}{2}} \\int_{0}^{3} \\left( (r \\cos(\\theta))^3 + (r \\cos(\\theta))(r \\sin(\\theta))^2 \\right) r \\, dr \\, d\\theta \\\\\n= \\int_{-\\frac{\\pi}{2}}^{\\frac{\\pi}{2}} \\int_{0}^{3} r^4 \\cos^3(\\theta) + r^4 \\cos(\\theta) \\sin^2(\\theta) \\, dr \\, d\\theta \\\\\n= \\int_{-\\frac{\\pi}{2}}^{\\frac{\\pi}{2}} \\left[ \\frac{r^5}{5} (\\cos^3(\\theta) + \\cos(\\theta) \\sin^2(\\theta)) \\right]_{r=0}^{r=3} \\, d\\theta \\\\\n= \\frac{243}{5} \\int_{-\\frac{\\pi}{2}}^{\\frac{\\pi}{2}} \\cos^3(\\theta) + \\cos(\\theta) \\sin^2(\\theta) \\, d\\theta \\\\\n= \\frac{243}{5} \\int_{-\\frac{\\pi}{2}}^{\\frac{\\pi}{2}} \\cos(\\theta)(\\cos^2(\\theta) + \\sin^2(\\theta)) \\, d\\theta \\\\\n= \\frac{243}{5} \\int_{-\\frac{\\pi}{2}}^{\\frac{\\pi}{2}} \\cos(\\theta) \\, d\\theta \\\\\n= \\frac{243}{5} \\left[ \\sin(\\theta) \\right]_{\\theta=-\\frac{\\pi}{2}}^{\\theta=\\frac{\\pi}{2}} \\\\\n= \\frac{243}{5} \\left( 1 - (-1) \\right) \\\\\n= \\frac{486}{5}\n\\]", "id": "./materials/399.pdf" }, { "contents": "Determine the value of the area of the region bounded by the parabolas $y = 2x^2$ and $y = 1 + x^2$.\n\n- Let’s first begin sketching said area.\n\n![Figure 1: 2D sketch of the area we were asked to evaluate.](image)\n\n- When we are asked to evaluate an area, we can translate that as\n\n$$\\int \\int_D 1 \\, dA$$\n\nwhere D is said area.\n\n- D can be defined as\n\n$$2x^2 = 1 + x^2 \\iff x^2 = 1 \\iff x = \\pm 1$$\n\n$$\\Rightarrow -1 \\leq x \\leq 1 \\quad , \\quad 2x^2 \\leq y \\leq 1 + x^2$$\n• This means that we only have to define the double integral\n\n\\[\n\\iint_D 1 \\, dA = \\int_{-1}^{1} \\int_{2x^2}^{1+x^2} 1 \\, dy \\, dx\n\\]\n\n• At the end of everything you should get: \\( \\frac{4}{3} \\)", "id": "./materials/400.pdf" }, { "contents": "Inequalities of degree greater than or equal to 3\n\nMathE\n\n27th of April of 2023\n\nExercise 1. Find the values of $x$ that satisfy the inequality $x^3 + 6x^2 - 6 > 2x^2 - x$.\n\nStep 1: Put all terms on the same side of the inequality.\n\n$$x^3 + 6x^2 - 6 > 2x^2 - x \\iff x^3 + 6x^2 - 6 - 2x^2 + x > 2x^2 - x - 2x^2 + x \\iff x^3 + 4x^2 + x - 6 > 0$$\n\nStep 2: Find a root of the polynomial $p(x) = x^3 + 4x^2 + x - 6$ (usually, it is a good idea to try 0, 1, -1, 2 or -2).\n\nFor example, you can ask yourself:\n\n- Is 0 a root of $p(x)$?\n - The answer is no, because $p(0) = 0^3 + 4 \\cdot 0^2 + 0 - 6 = -6 \\neq 0$.\n\n- Is -1 a root of $p(x)$?\n - The answer is no, because $p(-1) = (-1)^3 + 4 \\cdot (-1)^2 + (-1) - 6 = -1 + 4 - 1 - 6 = -4 \\neq 0$.\n\n- Is 1 a root of $p(x)$?\n - The answer is yes, because $p(1) = 1^3 + 4 \\cdot 1^2 + 1 - 6 = 1 + 4 + 1 - 6 = 0$.\n\nSo we found a root of the polynomial $p(x)$, which is $x = 1$ (there can be other roots, but we only need one).\nStep 3: Draw a table in the following way:\n\nAs we can see in the image above:\n\n1. The coefficients of the polynomial are written in the first line respecting the degree of $x$; that is, we start by writing the coefficients of the powers of $x$ with the highest degree from left to right.\n2. The root stays above the horizontal line segment and to the left of the vertical line segment.\n3. The coefficient of $x$ with the highest degree comes down.\n (in this case it is 1 because $x^3$ is the term of the polynomial with the highest degree).\n\nStep 4:\n\nAs a result, we get\nStep 5:\n\nAs a result, we get\n\nStep 6:\nAs a result, we get\n\nStep 7:\n\nAs a result, we get\nStep 8:\n\nAs a result, we get\n\nStep 9:\nAs a result, we get\n\n\\[\n\\begin{array}{cccc}\n1 & 4 & 1 & -6 \\\\\n1 & 1 & 5 & 6 \\\\\n1 & 5 & 6 & 0 \\\\\n\\end{array}\n\\]\n\nThe numbers in orange give us the polynomial \\( 6 + 5x + x^2 \\) (we \"ignore\" the last number, which is always 0).\n\nAs a result, we get that\n\n\\[\n\\frac{x^3 + 4x^2 + x - 6}{p(x)} = (x - 1) \\cdot (x^2 + 5x + 6)\n\\]\n\n\\( x - 1 \\) is the binomial obtained by subtracting the root found on Step 2 to \\( x \\).\n\\(x^2 + 5x + 6\\) is the trinomial we got from Ruffini’s Rule.\n\nThis is helpful because now solving \\(p(x) = 0\\) is equivalent to solve \\((x - 1) \\cdot (x^2 + 5x + 6) = 0\\):\n\n\\[\n(x - 1) \\cdot (x^2 + 6x + 6) = 0 \\iff x - 1 = 0 \\lor x^2 + 5x + 6 = 0 \\iff \\\\\nx = 1 \\lor x = \\frac{-5 \\pm \\sqrt{25 - 24}}{2} \\iff \\\\\nx = 1 \\lor x = \\frac{-5 \\pm 1}{2} \\iff \\\\\nx = 1 \\lor x = -3 \\lor x = -2\n\\]\n\nWe just found out that the roots of \\(p(x)\\) are \\(-3, -2\\) and 1.\n\nBecause \\(p(x)\\) is a continuous function, it follows from the Bolzano’s theorem that the sign of the function on the interval \\((-\\infty, -3)\\) will not change (it will either be positive or negative).\n\nThe same happens on the intervals \\((-3, -2), (-2, 1)\\) and \\((1, +\\infty)\\).\n\nTo see if the sign is either positive or negative, we can evaluate \\(p(x)\\) at any point of the interval:\n\n- On the interval \\((-\\infty, -3)\\), \\(p(x)\\) is negative because \\(-4 \\in (-\\infty, -3)\\) and \\(p(-4) = -10 < 0\\).\n- On the interval \\((-3, -2)\\), \\(p(x)\\) is positive because \\(-2.5 \\in (-3, -2)\\) and \\(p(-2.5) = 0.875 > 0\\).\n- On the interval \\((-2, 1)\\), \\(p(x)\\) is negative because \\(0 \\in (-2, 1)\\) and \\(p(0) = -6 < 0\\).\n- On the interval \\((1, +\\infty)\\), \\(p(x)\\) is positive because \\(2 \\in (1, +\\infty)\\) and \\(p(2) = 20 > 0\\).\n\nAs a result, we can construct the following table:\n\n| \\(x\\) | \\(-\\infty\\) | \\(-3\\) | \\(-2\\) | 1 | \\(+\\infty\\) |\n|-------|-------------|--------|--------|---|-------------|\n| \\(p(x)\\) | - | - | 0 | + | 0 | - | 0 | + | + |\n\nWe saw that the original inequality, which is\n\n\\[x^3 + 6x^2 - 6 > 2x^2 - x\\]\n\nis equivalent to the inequality\n\n\\[x^3 + 4x^2 + x - 6 > 0\\]\nThe table above tells us that the values of $x$ that satisfy this last inequality are\n\n$$x \\in (-3, -2) \\text{ and } x \\in (1, +\\infty)$$\n\nAs a result, the solution is $x \\in (-3, -2) \\cup (1, +\\infty)$.\n\n**Answer:** $x \\in (-3, -2) \\cup (1, +\\infty)$\nExercise 2. Solve the inequality $4x^4 < 3 - 11x^2$.\n\nStep 1: Put all terms on the same side of the inequality.\n\n$$4x^4 < 3 - 11x^2 \\iff 4x^4 - 3 + 11x^2 < 3 - 11x^2 - 3 + 11x^2 \\iff 4x^4 + 11x^2 - 3 < 0$$\n\nStep 2: Find a root of the polynomial $q(x) = 4x^4 + 11x^2 - 3$.\n\nIn order to do that, we first do the change of variable $y = x^2$:\n\n$$q(x) = 4x^4 + 11x^2 - 3 = 4(x^2)^2 + 11x^2 - 3 = 4y^2 + 11y - 3$$\n\nWe did that because we know how to solve a quadratic equation (we use the quadratic formula):\n\n$$q(x) = 0 \\iff 4y^2 + 11y - 3 = 0 \\iff y = \\frac{-11 \\pm \\sqrt{121 + 4 \\times 4 \\times (-3)}}{2 \\times 4} \\iff$$\n\n$$y = \\frac{-11 \\pm \\sqrt{121 + 48}}{8} \\iff y = \\frac{-11 \\pm 13}{8} \\iff$$\n\n$$y = \\frac{-11 - 13}{8} \\lor y = \\frac{-11 + 13}{8} \\iff$$\n\n$$y = \\frac{-24}{8} \\lor y = \\frac{2}{8} \\iff$$\n\n$$y = -3 \\lor y = \\frac{1}{4}$$\n\nWe now take into account the change of variable we did before:\n\n$$y = x^2 \\iff$$\n\n$$-3 = x^2 \\lor \\frac{1}{4} = x^2 \\iff$$\n\n(1) impossible condition in $\\mathbb{R}$\n\n$$\\frac{1}{4} = x^2 \\iff$$\n\n$$x = \\pm \\sqrt{\\frac{1}{4}} \\iff$$\n\n$$x = \\pm \\frac{1}{2}$$\n\n(1) We have an impossible condition because, in the real numbers, a number squared is always non-negative, so it cannot be equal to $-3$. \nThis tells us the real roots of the polynomial \\( q(x) = 4x^4 + 11x^2 - 3 \\) are \\(-\\frac{1}{2}\\) and \\(\\frac{1}{2}\\).\n\nAs in the previous exercise, because \\( q(x) \\) is a continuous function, it follows from the Bolzano’s theorem that the sign of the function on the interval \\((-\\infty, -\\frac{1}{2})\\) will not change (it will either be positive or negative).\n\nThe same happens on the intervals \\((-\\frac{1}{2}, \\frac{1}{2})\\) and \\((\\frac{1}{2}, +\\infty)\\).\n\nTo see if the sign is either positive or negative, we can evaluate \\( q(x) \\) at any point of the interval:\n\n- On the interval \\((-\\infty, -\\frac{1}{2})\\), \\( q(x) \\) is positive because \\(-1 \\in (-\\infty, -\\frac{1}{2}) \\) and \\( q(-1) = 11 > 0 \\).\n- On the interval \\((-\\frac{1}{2}, \\frac{1}{2})\\), \\( q(x) \\) is negative because \\(0 \\in (-\\frac{1}{2}, \\frac{1}{2}) \\) and \\( q(0) = -3 < 0 \\).\n- On the interval \\((\\frac{1}{2}, +\\infty)\\), \\( q(x) \\) is positive because \\(1 \\in (\\frac{1}{2}, +\\infty) \\) and \\( q(1) = 11 > 0 \\).\n\nAs a result, we can construct the following table:\n\n| \\( x \\) | \\(-\\infty\\) | \\(-\\frac{1}{2}\\) | \\(\\frac{1}{2}\\) | \\(+\\infty\\) |\n|---------|-------------|----------------|-------------|-------------|\n| \\( q(x) \\) | + | + | 0 | 0 |\n\nWe saw that the original inequality, which is\n\n\\[\n4x^4 < 3 - 11x^2\n\\]\n\nis equivalent to the inequality\n\n\\[\n4x^4 + 11x^2 - 3 < 0\n\\]\n\nThe table above tells us that the values of \\( x \\) that satisfy this last inequality are\n\n\\[\nx \\in \\left(-\\frac{1}{2}, \\frac{1}{2}\\right)\n\\]\n\nAs a result, the solution is \\( x \\in \\left(-\\frac{1}{2}, \\frac{1}{2}\\right) \\).\n\n**Answer:** \\( x \\in \\left(-\\frac{1}{2}, \\frac{1}{2}\\right) \\)\nExercise 3. Solve the inequality $x^4 - 25x^2 < -2x^3 + 26x - 120$.\n\nStep 1: Put all terms on the same side of the inequality.\n\n\\[\nx^4 - 25x^2 < -2x^3 + 26x - 120 \\iff x^4 - 25x^2 + 2x^3 - 26x + 120 < -2x^3 + 26x - 120 + 2x^3 - 26x + 120 \\iff x^4 + 2x^3 - 25x^2 - 26x + 120 < 0\n\\]\n\nStep 2: Find a root of the polynomial $r(x) = x^4 + 2x^3 - 25x^2 - 26x + 120$ (usually, it is a good idea to try 0, 1, -1, 2 or -2).\n\nFor example, you can ask yourself:\n\n- Is 0 a root of $r(x)$?\n - The answer is no, because $r(0) = 0^4 + 2 \\times 0^3 - 25 \\times 0^2 - 26 \\times 0 + 120 = 120 \\neq 0$.\n\n- Is 1 a root of $r(x)$?\n - The answer is no, because $r(1) = 1^4 + 2 \\times 1^3 - 25 \\times 1^2 - 26 \\times 1 + 120 = 72 \\neq 0$.\n\n- Is -1 a root of $r(x)$?\n - The answer is no, because\n \\[\n r(-1) = (-1)^4 + 2 \\times (-1)^3 - 25 \\times (-1)^2 - 26 \\times (-1) + 120 = 120 \\neq 0\n \\]\n\n- Is 2 a root of $r(x)$?\n - The answer is yes, because $r(2) = 2^4 + 2 \\times 2^3 - 25 \\times 2^2 - 26 \\times 2 + 120 = 0$\n\nSo we found a root of the polynomial $r(x)$, which is $x = 2$ (there can be other roots, but we only need one).\nStep 3: Draw a table in the following way:\n\n| | 1 | 2 | -25 | -26 | 120 |\n|---|-----|-----|-----|-----|-----|\n| 2 | | | | | |\n| 1 | | | | | |\n\nProceeding as we did in Exercise 1, we obtain the following final table:\n\n| | 1 | 2 | -25 | -26 | 120 |\n|---|-----|-----|-----|-----|-----|\n| 2 | | 8 | -34 | -120| |\n| 1 | 4 | -17 | -60 | | x |\n\nThe numbers in orange give us the polynomial $-60 - 17x + 4x^2 + 1x^3$ (we “ignore” the last number, which is always 0).\n\nAs a result, we get that\n\n$$x^4 + 2x^3 - 25x^2 - 26x + 120 = (x - 2) \\cdot (x^3 + 4x^2 - 17x - 60)$$\n\nThis is helpful because now solving $r(x) = 0$ is equivalent to solve $(x - 2) \\cdot (x^3 + 4x^2 - 17x - 60) = 0$:\n\n$$(x - 2) \\cdot (x^3 + 4x^2 - 17x - 60) = 0 \\iff x - 2 = 0 \\lor x^3 + 4x^2 - 17x - 60 = 0$$\n\n$$\\iff x = 2 \\lor x^3 + 4x^2 - 17x - 60 = 0$$\n\nThe next step is to find the roots of $x^3 + 4x^2 - 17x - 60 = 0$. \nStep 4: Find a root of the polynomial \\( s(x) = x^3 + 4x^2 - 17x - 60 \\) (usually, it is a good idea to try 0, 1, -1, 2 or -2).\n\nFor example, you can ask yourself:\n\n- Is 0 a root of \\( s(x) \\)?\n - The answer is no, because \\( s(0) = 0^3 + 4 \\times 0^2 - 17 \\times 0 - 60 = -60 \\neq 0 \\).\n\n- Is 1 a root of \\( s(x) \\)?\n - The answer is no, because \\( s(1) = 1^3 + 4 \\times 1^2 - 17 \\times 1 - 60 = -72 \\neq 0 \\).\n\n- Is -1 a root of \\( s(x) \\)?\n - The answer is no, because \\( s(-1) = (-1)^3 + 4 \\times (-1)^2 - 17 \\times (-1) - 60 = -40 \\neq 0 \\).\n\n- (...) \n\n- Is -3 a root of \\( s(x) \\)?\n - The answer is yes, because \\( s(-3) = (-3)^3 + 4 \\times (-3)^2 - 17 \\times (-3) - 60 = 0 \\).\n\nSo we found a root of the polynomial \\( s(x) \\), which is \\( x = -3 \\) (there can be other roots, but we only need one).\n\nStep 5: Draw a table in the following way:\n\n\\[\n\\begin{array}{cccc}\n\\text{Polynomial} & s(x) = x^3 + 4x^2 - 17x - 60 = \\\\\n& = 1x^3 + 4x^2 - 17x - 60 \\\\\n\\hline\n\\text{Root of } s(x) & x = -3 \\\\\n\\hline\n1 & 4 & -17 & -60 \\\\\n\\hline\n-3 & 1 \\\\\n\\end{array}\n\\]\n\nProceeding as we did before, we obtain the following final table:\nAs a result, we get that\n\n\\[ x^3 + 4x^2 - 17x - 60 = (x - (-3)) \\cdot (x^2 + x - 20) = (x + 3) \\cdot (x^2 + x - 20) \\]\n\nThis is helpful because now solving \\( s(x) = 0 \\) is equivalent to solve \\( (x + 3) \\cdot (x^2 + x - 20) = 0 \\):\n\n\\[\n\\begin{align*}\n(x + 3) \\cdot (x^2 + x - 20) &= 0 \\\\\nx + 3 &= 0 \\lor x^2 + x - 20 &= 0 \\\\\nx &= -3 \\lor x &= \\frac{-1 \\pm \\sqrt{1^2 - 4 \\times 1 \\times (-20)}}{2 \\times 1} \\\\\nx &= -3 \\lor x &= \\frac{-1 \\pm \\sqrt{1 + 80}}{2} \\\\\nx &= -3 \\lor x &= \\frac{-1 \\pm 9}{2} \\\\\nx &= -3 \\lor x &= \\frac{-10}{2} \\lor x &= \\frac{8}{2} \\\\\nx &= -3 \\lor x &= -5 \\lor x &= 4\n\\end{align*}\n\\]\n\nWe just found out that the roots of \\( s(x) \\) are \\(-5, -3\\) and 4.\n\nWe saw the roots of \\( r(x) \\) are 2 and the roots of \\( s(x) \\), so we conclude the roots of \\( r(x) \\) are \\(-5, -3, 2\\) and 4.\n\nBecause \\( r(x) \\) is a continuous function, it follows from the Bolzano’s theorem that the sign of the function on the interval \\((-\\infty, -5)\\) will not change (it will either be positive or negative).\n\nThe same happens on the intervals \\((-5, -3), (-3, 2), (2, 4)\\) and \\((4, +\\infty)\\).\n\nTo see if the sign is either positive or negative, we can evaluate \\( r(x) \\) at any point of the interval:\n• On the interval \\((-\\infty, -5)\\), \\(r(x)\\) is positive because \\(-6 \\in (-\\infty, -5)\\) and \\(r(-6) = 240 > 0\\).\n• On the interval \\((-5, -3)\\), \\(r(x)\\) is negative because \\(-4 \\in (-5, -3)\\) and \\(r(-4) = -48 < 0\\).\n• On the interval \\((-3, 2)\\), \\(r(x)\\) is positive because \\(0 \\in (-3, 2)\\) and \\(r(0) = 120 > 0\\).\n• On the interval \\((2, 4)\\), \\(r(x)\\) is negative because \\(3 \\in (2, 4)\\) and \\(r(3) = -48 < 0\\).\n• On the interval \\((4, +\\infty)\\), \\(r(x)\\) is positive because \\(5 \\in (4, +\\infty)\\) and \\(r(5) = 240 > 0\\).\n\nAs a result, we can construct the following table:\n\n| \\(x\\) | \\(-\\infty\\) | \\(-5\\) | \\(-3\\) | \\(2\\) | \\(4\\) | \\(+\\infty\\) |\n|-------|-------------|--------|--------|------|------|-----------|\n| \\(r(x)\\) | + | + | 0 | - | 0 | + | - | 0 | + | + |\n\nWe saw that the original inequality, which is\n\n\\[ x^4 - 25x^2 < -2x^3 + 26x - 120 \\]\n\nis equivalent to the inequality\n\n\\[ x^4 + 2x^3 - 25x^2 - 26x + 120 < 0 \\]\n\nThe table above tells us that the values of \\(x\\) that satisfy this last inequality are\n\n\\[ x \\in (-5, -3) \\text{ and } x \\in (2, 4) \\]\n\nAs a result, the solution is \\(x \\in (-5, -3) \\cup (2, 4)\\).\n\n**Answer:** \\(x \\in (-5, -3) \\cup (2, 4)\\)", "id": "./materials/401.pdf" }, { "contents": "Reverse the order of integration of: \\( \\int_0^3 \\int_{x^2}^9 f(x, y) \\, dy \\, dx \\)\n\n- As we can assess, right now we are integrating first in respect to \\( y \\) and only then to \\( x \\). The idea is to reverse it, so we should begin the integration first in respect to \\( x \\) and only the to \\( y \\).\n\n- For that, let’s first begin analysing the intervals of integration.\n\n![Figure 1: 2D sketch.](image)\n\n- At this moment we have\n\n\\[\n0 \\leq x \\leq 3 \\quad , \\quad x^2 \\leq y \\leq 9\n\\]\nwhere\n\\[ g(x) = x^2 \\quad \\Rightarrow \\quad g(x) = 9 \\iff x = 3 \\]\naccording to our domain.\n\n- This means that we can rewrite the integration intervals as\n \\[ 0 \\leq y \\leq 9, \\quad 0 \\leq x \\leq \\sqrt{y} \\]\n\n- At last, we can translate the first integral as\n \\[\n \\int_0^3 \\int_{x^2}^9 f(x, y) \\, dy \\, dx \\\\\n = \\int_0^9 \\int_0^{\\sqrt{y}} f(x, y) \\, dx \\, dy\n \\]", "id": "./materials/402.pdf" }, { "contents": "Reverse the order of integration of: \\[ \\int_0^3 \\int_{x^2}^9 f(x, y) \\, dy \\, dx \\]\n\n- As we can assess, right now we are integrating first in respect to \\( y \\) and only then to \\( x \\). The idea is to reverse it, so we should begin the integration first in respect to \\( x \\) and only the to \\( y \\).\n\n- For that, let’s first begin analysing the intervals of integration.\n\n![Figure 1: 2D sketch.](image)\n\n- At this moment we have\n\n\\[ 0 \\leq x \\leq 3 \\quad , \\quad x^2 \\leq y \\leq 9 \\]\nwhere\n\\[ g(x) = x^2 \\quad \\Rightarrow \\quad g(x) = 9 \\quad \\iff \\quad x = 3 \\]\naccording to our domain.\n\n- This means that we can rewrite the integration intervals as\n \\[ 0 \\leq y \\leq 9, \\quad 0 \\leq x \\leq \\sqrt{y} \\]\n\n- At last, we can translate the first integral as\n \\[\n \\int_0^3 \\int_{x^2}^9 f(x, y) \\, dy \\, dx \\\\\n = \\int_0^9 \\int_0^{\\sqrt{y}} f(x, y) \\, dx \\, dy\n \\]", "id": "./materials/403.pdf" }, { "contents": "Find the area of the region bounded by $y = 1 - x^2$ and $y = x^2 - 3$.\n\n- Let’s first begin sketching said area.\n\n![Figure 1: 2D sketch of the area we were asked to evaluate.](image)\n\n- When we are asked to evaluate an area, we can translate that as\n\n$$\\int \\int_D 1 \\, dA$$\n\nwhere D is said area.\n• D can be defined as\n\n\\[ 1 - x^2 = x^2 - 3 \\iff x^2 = 2 \\iff x = \\pm \\sqrt{2} \\]\n\n\\[ \\Rightarrow -\\sqrt{2} \\leq x \\leq \\sqrt{2}, \\quad x^2 - 3 \\leq y \\leq 1 - x^2 \\]\n\n• This means that we only have to define the double integral\n\n\\[\n\\int \\int_D 1 \\, dA = \\int_{-\\sqrt{2}}^{\\sqrt{2}} \\int_{x^2 - 3}^{1 - x^2} 1 \\, dy \\, dx\n\\]\n\n• At the end of everything you should get: \\( \\frac{16\\sqrt{2}}{3} \\)", "id": "./materials/404.pdf" }, { "contents": "Change the order of integration of: \\[ \\int_0^1 \\int_{-\\sqrt{1-y^2}}^{1-y} (x^2 + y^2) \\, dx \\, dy. \\]\n\n- As we can assess, right now we are integrating first in respect to \\( x \\) and only then to \\( y \\). The idea is to reverse it, so we should begin the integration first in respect to \\( y \\) and only the to \\( x \\). The function we are integrating doesn’t change.\n\n- For that, let’s first begin analysing the intervals of integration (attention: the horizontal axis is \\( y \\) and the vertical axis is \\( x \\)).\n\n![Figure 1: 2D sketch.](image)\n\n- At this moment we have\n \\[ 0 \\leq y \\leq 1 \\quad , \\quad -\\sqrt{1-y^2} \\leq x \\leq 1-y \\]\nwhere\n\\[-\\sqrt{1 - y^2} = 1 - y \\quad \\Rightarrow y = 1\\]\n\n- As we can see from the graph, there is need to split the new interval of integration in two.\n\n- This means that we can rewrite the integration intervals as\n\n\\[\n\\left( -1 \\leq x \\leq 0 \\ , \\ 0 \\leq y \\leq \\sqrt{1 - x^2} \\right) \\land \\left( 0 \\leq x \\leq 1 \\ , \\ 0 \\leq y \\leq 1-x \\right)\n\\]\n\n- At last, we can translate the first integral as\n\n\\[\n\\int_0^1 \\int_{-\\sqrt{1-y^2}}^{1-y} (x^2 + y^2) \\, dx \\, dy\n\\]\n\n\\[\n= \\int_{-1}^0 \\int_0^{\\sqrt{1-x^2}} (x^2 + y^2) \\, dy \\, dx + \\int_0^1 \\int_0^{1-x} (x^2 + y^2) \\, dy \\, dx\n\\]", "id": "./materials/406.pdf" }, { "contents": "Change the order of integration of:\n\\[ \\int_0^1 \\int_{\\sqrt{y}}^{y^2} f(x, y) \\, dx \\, dy + \\int_2^3 \\int_1^y f(x, y) \\, dx \\, dy. \\]\n\n- As we can assess, right now we are integrating first in respect to \\( x \\) and only then to \\( y \\). The idea is to reverse it, so we should begin the integration first in respect to \\( y \\) and only then to \\( x \\). The function we are integrating doesn’t change.\n\n- For that, let’s first begin analysing the intervals of integration (Attention: the horizontal axis is \\( y \\) and the vertical axis is \\( x \\)).\n\n- For the integral on the left, we have\n\n![Figure 1: 2D sketch for the integral on the left.](image)\n\n- For the integral on the right, we have\n• At this moment we have\n\\[\n\\left( 0 \\leq y \\leq 1 \\ , \\ y^2 \\leq x \\leq \\sqrt{y} \\right) \\land \\left( 2 \\leq y \\leq 3 \\ , \\ 1 \\leq x \\leq y \\right)\n\\]\n\n• As we can see from the graph, there is need to split the new interval of integration in three (one for the integral on the left and two for the integral on the right).\n\n• This means that we can rewrite the integration intervals as\n\\[\n\\left( 0 \\leq x \\leq 1 \\ , \\ x^2 \\leq y \\leq \\sqrt{x} \\right) \\land \\\\\n\\land \\left( 1 \\leq x \\leq 2 \\ , \\ 2 \\leq y \\leq 3 \\right) \\land \\\\\n\\land \\left( 2 \\leq x \\leq 3 \\ , \\ x \\leq y \\leq 3 \\right)\n\\]\n\n• At last, we can translate the first integral as\n\\[\n\\int_0^1 \\int_{y^2}^{\\sqrt{y}} f(x, y) \\, dx \\, dy + \\int_2^3 \\int_1^y f(x, y) \\, dx \\, dy \\\\\n= \\int_0^1 \\int_{x^2}^{\\sqrt{x}} f(x, y) \\, dy \\, dx + \\int_1^2 \\int_2^3 f(x, y) \\, dy \\, dx + \\int_2^3 \\int_x^3 f(x, y) \\, dy \\, dx\n\\]", "id": "./materials/407.pdf" }, { "contents": "Find \\( \\int_0^3 \\int_1^{(4-y)^{\\frac{1}{2}}} (x + y) \\, dx \\, dy \\).\n\n- We can evaluate said integral, through\n\n\\[\n\\int_0^3 \\int_1^{(4-y)^{\\frac{1}{2}}} (x + y) \\, dx \\, dy = \\int_0^3 \\left[ \\frac{x^2}{2} + xy \\right]_{x=1}^{x=(4-y)^{\\frac{1}{2}}} \\, dy\n\\]\n\n\\[\n= \\int_0^3 \\left( \\frac{4 - y}{2} + y(4 - y)^{\\frac{1}{2}} - \\frac{1}{2} - y \\right) \\, dy\n\\]\n\n\\[\n= \\int_0^3 \\left( \\frac{3}{2} - \\frac{3y}{2} + y(4 - y)^{\\frac{1}{2}} \\right) \\, dy\n\\]\n\n\\[\n= \\int_0^3 \\frac{3}{2} \\, dy + \\int_0^3 -\\frac{3y}{2} \\, dy + \\int_0^3 y(4 - y)^{\\frac{1}{2}} \\, dy\n\\]\n\n\\[\n= \\left[ \\frac{3y}{2} \\right]_{y=0}^{y=3} - \\left[ \\frac{3y^2}{4} \\right]_{y=0}^{y=3} + \\int_0^3 y(4 - y)^{\\frac{1}{2}} \\, dy\n\\]\n\n\\[\n= \\frac{9}{2} - \\frac{27}{4} + \\int_0^3 y(4 - y)^{\\frac{1}{2}} \\, dy\n\\]\n\n\\[\n= -\\frac{9}{4} + \\int_0^3 y(4 - y)^{\\frac{1}{2}} \\, dy\n\\]\nTo solve the last integral, we should resort to integration by parts\n\n\\[- \\frac{9}{4} + \\int_{0}^{3} y(4 - y)^{\\frac{1}{2}} \\, dy\\]\n\n\\[= - \\frac{9}{4} + \\left[ - \\frac{2y}{3} (4 - y)^{\\frac{3}{2}} \\right]_{y=0}^{y=3} - \\int_{0}^{3} - \\frac{2}{3} (4 - y)^{\\frac{3}{2}} \\, dy\\]\n\n\\[= - \\frac{9}{4} - 2 - \\int_{0}^{3} - \\frac{2}{3} (4 - y)^{\\frac{3}{2}} \\, dy\\]\n\n\\[= - \\frac{17}{4} - \\frac{2}{3} \\left[ \\frac{2}{5} (4 - y)^{\\frac{5}{2}} \\right]_{y=0}^{y=3}\\]\n\n\\[= - \\frac{17}{4} - \\frac{4}{15} \\left( 1 - (4)^{\\frac{5}{2}} \\right)\\]\n\n\\[= - \\frac{17}{4} - \\frac{4}{15} \\left( 1 - 32 \\right)\\]\n\n\\[= - \\frac{17}{4} + \\frac{124}{15}\\]\n\n\\[= - \\frac{17 \\times 15 + 124 \\times 4}{60}\\]\n\n\\[= \\frac{241}{60}\\]", "id": "./materials/408.pdf" }, { "contents": "Find the value of the region’s area bounded by the curves $y = 2x$ and $y = x^2$.\n\n- Let’s first begin sketching said area.\n\n![Figure 1: 2D sketch of the area we were asked to evaluate.](image)\n\n- When we are asked to evaluate an area, we can translate that as\n\n$$\\int \\int_D 1 \\, dA$$\n\nwhere $D$ is said area.\n\n- $D$ can be defined as\n\n$$2x = x^2 \\iff x = 2 \\lor x = 0$$\n\n$$\\Rightarrow 0 \\leq x \\leq 2 \\quad , \\quad x^2 \\leq y \\leq 2x$$\n• This means that we only have to define the double integral\n\n\\[\n\\iint_D 1 \\, dA = \\int_0^2 \\int_{x^2}^{2x} 1 \\, dy \\, dx\n\\]\n\n• At the end of everything you should get: \\( \\frac{4}{3} \\)", "id": "./materials/409.pdf" }, { "contents": "Find $\\int \\int_D xy^2 \\, dA$ knowing that the region $D$ is bounded by $y = x^2$ and $y = \\sqrt{x}$.\n\n- Let’s first begin sketching $D$.\n\n![Figure 1: 2D sketch of D.](image)\n\n- $D$ can be defined as\n\n \\[ x^2 = \\sqrt{x} \\iff x = 1 \\lor x = 0 \\]\n\n \\[ \\Rightarrow 0 \\leq x \\leq 1 \\quad , \\quad x^2 \\leq y \\leq \\sqrt{x} \\]\n• This means that we only have to define the double integral\n\n\\[\n\\iint_D xy^2 \\, dA\n\\]\n\n\\[\n= \\int_0^1 \\int_{x^2}^{\\sqrt{x}} xy^2 \\, dy \\, dx\n\\]\n\n• At the end of everything you should get: \\( \\frac{3}{56} \\)", "id": "./materials/410.pdf" }, { "contents": "Choose which double integral represents the area of the region bounded by $y = 4 - x^2$ and $y = x^2 - 4$.\n\n- Let’s first begin sketching said area.\n\n![Figure 1: 2D sketch of the area we were asked to evaluate.](image)\n\n- When we are asked to evaluate an area, we can translate that as\n\n$$\\int \\int_D 1 \\, dA$$\n\nwhere D is said area.\n• D can be defined as\n\n\\[ 4 - x^2 = x^2 - 4 \\iff x^2 = 4 \\iff x = \\pm 2 \\]\n\n\\[ \\Rightarrow -2 \\leq x \\leq 2 , \\quad x^2 - 4 \\leq y \\leq 4 - x^2 \\]\n\n• This means that we only have to define the double integral\n\n\\[\n\\iint_D 1 \\, dA = \\int_{-2}^{2} \\int_{x^2-4}^{4-x^2} 1 \\, dy \\, dx\n\\]", "id": "./materials/411.pdf" }, { "contents": "Reverse the order of integration of \\( \\int_{-2}^{2} \\int_{x^2-4}^{4-x^2} 1 \\, dy \\, dx \\).\n\n- To change the order of integration, we have to be careful. There are some cases where the order of integration is indifferent (when all variables are independent), but most cases are more complex.\n\n- As we can see, the integral in order to \\( y \\) depends on the value of the other variable.\n\n- We can define\n\n\\[\n-2 \\leq x \\leq 2, \\quad x^2 - 4 \\leq y \\leq 4 - x^2\n\\]\n\n- Let’s sketch the region defined by the previous inequations.\n\nFigure 1: 2D sketch of the region D.\n• This means that we can easily define D, but having \\( y \\) as the independent variable. However, we are going to need to divide D into two sections (separate the region from the first and second quadrants from the third and fourth quadrants).\n\n\\[\n(3, 4Q) \\land (1, 2Q) \\\\\nx^2 - 4 \\leq y \\land y \\leq 4 - x^2 \\\\\n\\iff x^2 \\leq y + 4 \\land x^2 \\leq 4 - y \\\\\n\\iff \\left( x \\leq \\sqrt{y + 4} \\land x \\geq -\\sqrt{y + 4} \\right) \\land \\left( x \\leq \\sqrt{4 - y} \\land x \\geq -\\sqrt{4 - y} \\right)\n\\]\n\n• This means that we can rewrite\n\n\\[\n\\int_{-2}^{2} \\int_{x^2 - 4}^{4 - x^2} 1 \\, dy \\, dx \\\\\n= \\int_{-4}^{0} \\int_{-\\sqrt{y + 4}}^{\\sqrt{y + 4}} 1 \\, dx \\, dy + \\int_{0}^{4} \\int_{-\\sqrt{4 - y}}^{\\sqrt{4 - y}} 1 \\, dx \\, dy\n\\]", "id": "./materials/412.pdf" }, { "contents": "Determine \\( \\int_{1}^{4} \\int_{0}^{\\sqrt{y}} e^{\\frac{x}{\\sqrt{y}}} \\, dx \\, dy \\).\n\n- As we can see, we should first integrate in order to \\( x \\), and only then in order to \\( y \\).\n\n\\[\n\\int_{1}^{4} \\int_{0}^{\\sqrt{y}} e^{\\frac{x}{\\sqrt{y}}} \\, dx \\, dy \\\\\n= \\int_{1}^{4} \\left[ \\sqrt{y} e^{\\frac{x}{\\sqrt{y}}} \\right]_{x=0}^{x=\\sqrt{y}} \\, dy \\\\\n= \\int_{1}^{4} \\left( \\sqrt{y}(e-1) \\right) \\, dy \\\\\n=(e-1) \\int_{1}^{4} y^{\\frac{1}{2}} \\, dy \\\\\n=(e-1) \\left[ \\frac{2}{3} y^{\\frac{3}{2}} \\right]_{y=1}^{y=4} \\\\\n=(e-1) \\left( \\frac{16}{3} - \\frac{2}{3} \\right) \\\\\n=\\frac{14}{3} (e-1)\n\\]", "id": "./materials/413.pdf" }, { "contents": "Find \\( \\int_0^2 \\int_x^{x^2} y^2 x \\, dy \\, dx \\).\n\n- We can evaluate said integral, through\n\n\\[\n\\int_0^2 \\int_x^{x^2} y^2 x \\, dy \\, dx = \\int_0^2 \\left[ \\frac{xy^3}{3} \\right]_{y=x}^{y=x^2} \\, dx = \\int_0^2 \\left( \\frac{x^7 - x^4}{3} \\right) \\, dx = \\frac{1}{3} \\int_0^2 (x^7 - x^4) \\, dx = \\frac{1}{3} \\left[ \\frac{x^8}{8} - \\frac{x^5}{5} \\right]_{x=0}^{x=2} = \\frac{1}{3} \\left( \\frac{2^8}{8} - \\frac{2^5}{5} \\right) = \\frac{1}{3} \\left( \\frac{256}{8} - \\frac{32}{5} \\right) = \\frac{1}{3} \\left( \\frac{256 \\times 5 - 32 \\times 8}{40} \\right) = \\frac{1024}{120} = \\frac{128}{15}\n\\]", "id": "./materials/414.pdf" }, { "contents": "Find \\( \\int \\int_D (3 - x - y) \\, dA \\), where \\( D = [1, 3] \\times [1, 2] \\).\n\n- Let’s first begin sketching \\( D \\).\n\n![Figure 1: 2D sketch of D.](image)\n\n- This means that we can define \\( D \\) as the rectangle, where\n\n\\[\n1 \\leq x \\leq 3 \\quad , \\quad 1 \\leq y \\leq 2\n\\]\nThus, we can define the double integral as\n\n\\[\n\\iint_D (3 - x - y) \\, dA\n\\]\n\n\\[\n= \\int_1^3 \\int_1^2 (3 - x - y) \\, dy \\, dx\n\\]\n\n\\[\n= \\int_1^3 \\left[ 3y - xy - \\frac{y^2}{2} \\right]_{y=1}^{y=2} \\, dx\n\\]\n\n\\[\n= \\int_1^3 \\left( 6 - 2x - 2 - 3 + x + \\frac{1}{2} \\right) \\, dx\n\\]\n\n\\[\n= \\int_1^3 \\left( -x + \\frac{3}{2} \\right) \\, dx\n\\]\n\n\\[\n= \\left[ -\\frac{x^2}{2} + \\frac{3x}{2} \\right]_{x=1}^{x=3}\n\\]\n\n\\[\n= -\\frac{9}{2} + \\frac{9}{2} + \\frac{1}{2} - \\frac{3}{2}\n\\]\n\n\\[\n= -1\n\\]", "id": "./materials/415.pdf" }, { "contents": "Find $\\int \\int_D 3 - x - y \\, dA$, where $D = \\{(x, y) \\in \\mathbb{R}^2 : 2 \\leq y \\leq 3, 1 \\leq x \\leq y\\}$.\n\n- Let’s first begin sketching $D$.\n\n![Figure 1: 2D sketch of D.](image)\n\n- As we can assess from the definition of $D$, we can define $D$ with $y$ as an independent variable and $x$ as a dependent variable.\nThus, we can define the double integral as\n\n\\[\n\\iint_D (3 - x - y) \\, dA\n\\]\n\n\\[\n= \\int_2^3 \\int_1^y (3 - x - y) \\, dx \\, dy\n\\]\n\n\\[\n= \\int_2^3 \\left[ 3x - \\frac{x^2}{2} - xy \\right]_{x=1}^{x=y} \\, dy\n\\]\n\n\\[\n= \\int_2^3 \\left( 3y - \\frac{y^2}{2} - y^2 - 3 + \\frac{1}{2} + y \\right) \\, dy\n\\]\n\n\\[\n= \\int_2^3 \\left( 4y - \\frac{3y^2}{2} - \\frac{5}{2} \\right) \\, dy\n\\]\n\n\\[\n= \\left[ 2y^2 - \\frac{y^3}{2} - \\frac{5y}{2} \\right]_{y=2}^{y=3}\n\\]\n\n\\[\n= 18 - \\frac{27}{2} - \\frac{15}{2} - 8 + 4 + 5\n\\]\n\n\\[\n= -2\n\\]", "id": "./materials/416.pdf" }, { "contents": "Change the order of integration in \\( \\int_0^6 \\int_{\\frac{x}{3}}^2 f(x, y) \\, dy \\, dx \\).\n\n- As we can assess, right now we are integrating first in respect to \\( y \\) and only then to \\( x \\). The idea is to reverse it, so we should begin the integration first in respect to \\( x \\) and only then to \\( y \\). The function we are integrating doesn’t change.\n\n- For that, let’s first begin analysing the intervals of integration.\n\n![Figure 1: 2D sketch of the region described in the integration interval.](image)\n\n- At this moment we have\n\n\\[\n0 \\leq x \\leq 6 \\quad , \\quad \\frac{x}{3} \\leq y \\leq 2\n\\]\n• This means that we can rewrite the integration intervals, using Figure 1, as\n\n\\[ 0 \\leq y \\leq 2, \\quad 0 \\leq x \\leq 3y \\]\n\n• At last, we can rewrite the double integral as\n\n\\[\n\\int_0^6 \\int_{\\frac{x}{3}}^2 f(x, y) \\, dy \\, dx = \\int_0^2 \\int_0^{3y} f(x, y) \\, dx \\, dy\n\\]", "id": "./materials/417.pdf" }, { "contents": "Find \\( \\int_{2}^{4} \\int_{-1}^{1} x^2 + y^2 \\, dx \\, dy \\).\n\n- Let’s first begin sketching D.\n\n![Figure 1: 2D sketch of D.](image)\n\n- As we can assess from Figure 1, we can define D with \\( y \\) and \\( x \\) as independent variables, which complies with the definition of a rectangle.\n• Proceeding with some calculations, we get that\n\n\\[\n\\int_{-1}^{1} \\int_{2}^{4} x^2 + y^2 \\, dx \\, dy\n\\]\n\n\\[\n= \\int_{2}^{4} \\left[ \\frac{x^3}{3} + y^2 x \\right]_{x=-1}^{x=1} \\, dy\n\\]\n\n\\[\n= \\int_{2}^{4} \\left( \\frac{1^3}{3} + y^2 + \\frac{1^3}{3} + y^2 \\right) \\, dy\n\\]\n\n\\[\n= \\int_{2}^{4} \\left( \\frac{2}{3} + 2y^2 \\right) \\, dy\n\\]\n\n\\[\n= \\left[ \\frac{2y}{3} + \\frac{2y^3}{3} \\right]_{y=2}^{y=4}\n\\]\n\n\\[\n= \\frac{8}{3} + \\frac{128}{3} - \\frac{4}{3} - \\frac{16}{3}\n\\]\n\n\\[\n= \\frac{116}{3}\n\\]", "id": "./materials/418.pdf" }, { "contents": "Find \\( \\int_0^1 \\int_0^{x^2} 1 \\, dy \\, dx \\).\n\n- Let’s first begin sketching D.\n\n![Figure 1: 2D sketch of D.](image)\n\n- Since\n\n\\[\nf(x, y) = 1\n\\]\n\nthis means that we are being asked to evaluate the area of the region D.\n\n- As we can assess from Figure 1, we can define D with \\( x \\) as an independent variable and \\( y \\) as the dependent one.\n• Proceeding with some calculations, we get that\n\n\\[\n\\int_0^1 \\int_0^{x^2} 1 \\, dy \\, dx\n\\]\n\n\\[\n= \\int_0^1 \\left[ y \\right]_{y=0}^{y=x^2} \\, dx\n\\]\n\n\\[\n= \\int_0^1 x^2 \\, dx\n\\]\n\n\\[\n= \\left[ \\frac{x^3}{3} \\right]_{x=0}^{x=1}\n\\]\n\n\\[\n= \\frac{1}{3}\n\\]", "id": "./materials/419.pdf" }, { "contents": "Manipulation of Algebraic Expressions\n\nManipulating Formula\nManipulating Formula\n\n• A formula is a statement that two quantities are equal. E.g.\n\n\\[ S = \\frac{a}{1-r} \\]\n\n• **Transposing formula** involves the manipulation of the formula when a value other than the subject is required. For example if the value ‘\\(a\\)’ is required in the above formula it would read as follows:\n\n\\[ a = S(1-r) \\]\nManipulating Formula\n\nBasic Rule of Manipulating of Formula:\n\n1. That the equality of an equation must be maintained\n2. Whatever is done on the left hand side must be done on the right hand side\nManipulating Formula\n\nProblem 1:\nManipulate $k=x+y+z$ to make $y$ the subject.\nFirstly, change the equation around so that $y$ is on the LHS:\n\n$$x+y+z=k$$\n\nSubtract $x+z$ from both sides to get the $y$ isolated\n\n$$x-x+y+z-z=k-x-z$$\n\n$$y=k-x-z$$\n\nThis proves that a quantity can be moved from one side of an equation to the other with a simple change of sign.\nManipulating Formula\n\nProblem 2:\nIf \\( a+b = p-q-s \\) express \\( q \\) as the subject.\nRearrange: \\( p-q-s = a+b \\)\nMultiply both sides by \\(-1\\):\n\\[\n(-1)p-q-s=(-1) a+b\n\\]\n\\[-p+q+s=-a-b\n\\]\n\\[q=p-s-a-b\n\\]\nMultiplying across the equation by \\(-1\\) resulted in all signs changing.\nThe reason for multiplying by \\(-1\\) was to change the \\(-q\\) to \\(+q\\) as we generally express answers with a positive quantity first i.e. in this case \\( q \\).\nManipulating Formula\n\nProblem 3:\nMake \\( d \\) the subject matter of the formula: \\( p = \\frac{\\pi d}{2} \\)\nRewrite as \\( \\frac{\\pi d}{2} = p \\)\n\nMultiply both sides by 2: \\( \\pi d = 2p \\)\n\nDivide both sides by \\( \\pi \\) to obtain \\( d = \\frac{2p}{\\pi} \\)\n\nMultiplication is used to change a formula, which includes a fraction, to whole numbers (also called integers). To remove \\( \\pi \\) from the \\( d \\) in the above example, we divide both sides by \\( \\pi \\).\nManipulating Formula\n\n**Problem 4:**\nThe formula for calculating the surface area of a sphere is: \\( A = 4 \\pi r^2 \\)\nMake \\( r \\) (radius) the subject matter of this formula.\n\n\\[\n4 \\pi r^2 = A\n\\]\n\n\\[\nr^2 = \\frac{A}{4\\pi}\n\\]\n\nDivide by \\( 4 \\pi \\) both sides to isolate \\( r^2 \\)\n\n\\[\nr = \\sqrt{\\frac{A}{4\\pi}}\n\\]\n\nTo obtain \\( r \\), square root both sides\n\nIf you have an element which is cubed, you would cube root it and so on.", "id": "./materials/42.pdf" }, { "contents": "Find \\( \\int_1^3 \\int_0^1 1 + 4xy \\, dx \\, dy \\).\n\n- Let’s first begin sketching D.\n\n![Figure 1: 2D sketch of D.](image)\n\n- As we can assess from Figure 1, we can define D with \\( y \\) and \\( x \\) as an independent variables, which complies with the definition of a rectangle.\n• Proceeding with some calculations, we get that\n\n\\[\n\\int_1^3 \\int_0^1 1 + 4xy \\, dx \\, dy\n\\]\n\n\\[\n= \\int_1^3 \\left[ x + 2x^2y \\right]_{x=0}^{x=1} \\, dy\n\\]\n\n\\[\n= \\int_1^3 \\left( 1 + 2y \\right) \\, dy\n\\]\n\n\\[\n= \\left[ y + y^2 \\right]_{y=1}^{y=3}\n\\]\n\n\\[\n= 3 + 9 - 1 - 1\n\\]\n\n\\[\n= 10\n\\]", "id": "./materials/420.pdf" }, { "contents": "Find \\( \\int_0^1 \\int_{y^2}^y x^m + y^n \\, dx \\, dy \\), where \\( m, n > 0 \\).\n\n- As we can assess from the interval of integration, we can define \\( D \\) with \\( y \\) as an independent variable and \\( x \\) as a dependent variable.\n\n- Proceeding with some calculations, we get that\n\n\\[\n\\int_0^1 \\int_{y^2}^y x^m + y^n \\, dx \\, dy = \\int_0^1 \\left[ \\frac{x^{m+1}}{m+1} + xy^n \\right]_{x=y^2}^{x=y} \\, dy\n\\]\n\n\\[\n= \\int_0^1 \\left( \\frac{y^{m+1}}{m+1} + y^{n+1} - \\frac{y^{2m+2}}{m+1} - y^{n+2} \\right) \\, dy\n\\]\n\n\\[\n= \\left[ \\frac{y^{m+2}}{(m+1)(m+2)} + \\frac{y^{n+2}}{(n+2)} - \\frac{y^{2m+3}}{(m+1)(2m+3)} - \\frac{y^{n+3}}{(n+3)} \\right]_{y=0}^{y=1}\n\\]\n\n\\[\n= \\frac{1}{(m+1)(m+2)} + \\frac{1}{(n+2)} - \\frac{1}{(m+1)(2m+3)} - \\frac{1}{(n+3)}\n\\]\n\n\\[\n= \\frac{1}{(m+1)(m+2)(2m+3)} + \\frac{1}{(n+2)(n+3)}\n\\]\n\n\\[\n= \\frac{1}{(m+2)(2m+3)} + \\frac{1}{(n+2)(n+3)}\n\\]", "id": "./materials/421.pdf" }, { "contents": "Find \\( \\int_0^1 \\int_{-\\sqrt{1-y^2}}^{1-y} 1 \\, dx \\, dy \\).\n\n- Since \n \\[ f(x, y) = 1 \\]\n this means that we are being asked to evaluate the area of the region D.\n\n- Proceeding with some calculations, we get that\n \\[\n \\int_0^1 \\int_{-\\sqrt{1-y^2}}^{1-y} 1 \\, dx \\, dy = \\int_0^1 \\left[ x \\right]_{x=-\\sqrt{1-y^2}}^{x=1-y} \\, dy\n = \\int_0^1 \\left( 1 - y + \\sqrt{1 - y^2} \\right) \\, dy\n = \\left[ y - \\frac{y^2}{2} \\right]_{y=0}^{y=1} + \\int_0^1 \\sqrt{1 - y^2} \\, dy\n = \\frac{1}{2} + \\int_0^1 \\sqrt{1 - y^2} \\, dy\n \\]\n\n- Let’s make a substitution using trigonometric properties, this is\n \\[\n \\sqrt{1 - y^2} = \\cos^2(u)\n \\Rightarrow \\frac{du}{dy} = 1\n \\]\n Since the function we are integrating is monotonically non-decreasing, this allows us to evaluate the new interval of integration\n \\[\n 0 \\leq y \\leq 1 \\quad \\Rightarrow \\quad 0 \\leq u \\leq \\frac{\\pi}{2}\n \\]\nThis means that we can continue our evaluation, where\n\n\\[\n\\frac{1}{2} + \\int_0^1 \\sqrt{1 - y^2} \\, dy\n\\]\n\n\\[\n= \\frac{1}{2} + \\int_0^{\\frac{\\pi}{2}} \\cos^2 (u) \\, du\n\\]\n\n\\[\n= \\frac{1}{2} + \\int_0^{\\frac{\\pi}{2}} \\frac{1 + \\cos (2u)}{2} \\, du\n\\]\n\n\\[\n= \\frac{1}{2} + \\left[ \\frac{u}{2} + \\sin (2u) \\right]_{u=0}^{u=\\frac{\\pi}{2}}\n\\]\n\n\\[\n= \\frac{1}{2} + \\frac{\\pi}{4}\n\\]", "id": "./materials/422.pdf" }, { "contents": "Reverse the order of integration of \\( \\int_0^4 \\int_{\\frac{y}{4}}^y f(x, y) \\, dx \\, dy \\).\n\n- To change the order of integration, we have to be careful. There are some cases where the order of integration is indifferent (when all variables are independent), but most cases are more complex.\n\n- As we can see, the integral in order to \\( x \\) depends on the value of the other variable.\n\n- We can define\n \\[\n 0 \\leq y \\leq 4, \\quad \\frac{y}{4} \\leq x \\leq y\n \\]\n\n- Let’s sketch the region defined by the previous inequations.\n\n![Figure 1: 2D sketch of the region D.](image)\n• This means that we can easily define D, but having $x$ as the independent variable. However, we are going to need to divide D into two sections. For that we need to find the intersection of $x = \\frac{y}{4}$ and $y = 4$, which is: $x = 1$\n\n• So, we can redefine D as\n\n$$(0 \\leq x \\leq 1) \\land (1 \\leq x \\leq 4)$$\n\n$\\Rightarrow (x \\leq y \\leq 4x) \\land (x \\leq y \\leq 4)$\n\n• This means that we can rewrite\n\n$$\\int_0^4 \\int_{\\frac{x}{4}}^y f(x, y) \\, dx \\, dy$$\n\n$$= \\int_0^1 \\int_x^{4x} f(x, y) \\, dy \\, dx + \\int_1^4 \\int_x^4 f(x, y) \\, dy \\, dx$$", "id": "./materials/423.pdf" }, { "contents": "Find the volume of \\( B = \\{(x, y, z) \\in \\mathbb{R}^3 : x^2 + y^2 \\leq 1, \\ z \\geq 0, \\ y + z \\leq 3\\} \\), using double integrals.\n\n- Let’s first sketch the solid B.\n\n![3D sketch of the solid B.](image)\n\n- To be able to evaluate a volume with a double integral, we are going to need the projection over a plane. In this case, it is more suitable to use the x-y plane. Let’s call it the region D\n\n- We can define D using cylindrical coordinates as\n\n\\[\nD = \\{(r, \\theta) \\mid 0 \\leq r \\leq 1, \\ 0 \\leq \\theta \\leq 2\\pi\\}\n\\]\n\n- By definition, we can evaluate a volume of a solid using double integrals as\n\n\\[\nV = \\iint_D (f(x, y) - g(x, y)) \\, dA\n\\]\nwhere, in this case,\n\n\\[ z + r \\sin(\\theta) = 3 \\quad \\Rightarrow \\quad f(r, \\theta) = 3 - r \\sin(\\theta) \\quad \\Rightarrow \\]\n\\[ z = 0 \\quad \\Rightarrow \\quad g(r, \\theta) = 0 \\]\n\n- This means that we can write the double integral asked as\n\n\\[\nV_B = \\iint_D (f(r, \\theta) - g(r, \\theta)) \\, dA\n\\]\n\\[\n= \\int_0^{2\\pi} \\int_0^1 (3 - r \\sin(\\theta) - 0) \\, r \\, dr \\, d\\theta\n\\]\n\\[\n= \\int_0^{2\\pi} \\left[ \\frac{3r^2}{2} - \\frac{r^3 \\sin(\\theta)}{3} \\right]_{r=0}^{r=1} \\, d\\theta\n\\]\n\\[\n= \\int_0^{2\\pi} \\left( \\frac{3}{2} - \\frac{\\sin(\\theta)}{3} \\right) \\, d\\theta\n\\]\n\\[\n= \\left[ \\frac{3}{2} \\theta + \\frac{\\cos(\\theta)}{3} \\right]_{\\theta=0}^{\\theta=2\\pi}\n\\]\n\\[\n= \\left( 3\\pi + \\frac{1}{3} - 0 - \\frac{1}{3} \\right)\n\\]\n\\[\n= 3\\pi\n\\]", "id": "./materials/424.pdf" }, { "contents": "Evaluate \\( \\int \\int_S x^2 \\, dS \\), where \\( S \\) is defined by \\( x^2 + y^2 + z^2 = 1 \\)\n\n- First, we have to acknowledge that the surface in question is the unit sphere.\n\n![Figure 1: 3D sketch of the surface S](image)\n\n- Using parametric representation, we can define:\n \\[\n x = \\sin(\\phi) \\cos(\\theta), \\quad y = \\sin(\\phi) \\sin(\\theta), \\quad z = \\cos(\\phi), \\quad 0 \\leq \\phi \\leq \\pi, \\quad 0 \\leq \\theta \\leq 2\\pi\n \\]\n\n- This means that we can also define\n \\[\n \\mathbf{r}(\\phi, \\theta) = \\sin(\\phi) \\cos(\\theta)\\mathbf{i} + \\sin(\\phi) \\sin(\\theta)\\mathbf{j} + \\cos(\\phi)\\mathbf{k}\n \\]\n• Evaluating $\\mathbf{r}_\\phi$ and $\\mathbf{r}_\\theta$\n\n$$\\mathbf{r}_\\phi = \\cos(\\phi) \\cos(\\theta) \\mathbf{i} + \\cos(\\phi) \\sin(\\theta) \\mathbf{j} - \\sin(\\phi) \\mathbf{k}$$\n\n$$\\mathbf{r}_\\theta = -\\sin(\\phi) \\sin(\\theta) \\mathbf{i} + \\sin(\\phi) \\cos(\\theta) \\mathbf{j}$$\n\n$$\\Rightarrow |\\mathbf{r}_\\phi \\times \\mathbf{r}_\\theta| = \\sin(\\phi)$$\n\n• By definition, we can transform a surface integral in a double integral by:\n\n$$\\iint_S f(x, y, z) \\, dS = \\iint_D f(\\mathbf{r}(u, v)) \\cdot |\\mathbf{r}_u \\times \\mathbf{r}_v| \\, dA$$\n\n• This means that we can rewrite the initial surface integral as\n\n$$\\iint_S x^2 \\, dS = \\iint_D (\\sin(\\phi) \\cos(\\theta))^2 |\\mathbf{r}_\\phi \\times \\mathbf{r}_\\theta| \\, dA$$\n\n$$= \\int_0^{2\\pi} \\int_0^\\pi \\sin^2(\\phi) \\cos^2(\\theta) \\sin(\\phi) \\, d\\phi d\\theta$$\n\n• Now is just to solve the double integral.\n\n• At the end you should get: $\\frac{4\\pi}{3}$", "id": "./materials/425.pdf" }, { "contents": "Evaluate \\( \\iint_S y \\, dS \\), where \\( S \\) is defined by \\( z = x + y^2 \\), \\( 0 \\leq x \\leq 1 \\), \\( 0 \\leq y \\leq 2 \\).\n\nFigure 1: 3D sketch of the surface \\( S \\)\n\n- By definition, we can transform a surface integral in a double integral by:\n \\[\n \\iint_S f(x, y, z) \\, dS = \\iint_D f(r(u, v)) \\cdot |r_u \\times r_v| \\, dA\n \\]\n\n- We can regard this surface as a parametric surface with the following parametric equations:\n \\[\n x = x \\quad y = y \\quad z = g(x, y)\n \\]\nwhere\n\\[ \\mathbf{r}_x = \\mathbf{i} + \\left( \\frac{\\partial g}{\\partial x} \\right) \\mathbf{k} \\quad \\mathbf{r}_y = \\mathbf{j} + \\left( \\frac{\\partial g}{\\partial y} \\right) \\mathbf{k} \\]\n\n- This means that\n\\[ \\mathbf{r}_x \\times \\mathbf{r}_y = -\\frac{\\partial g}{\\partial x} \\mathbf{i} - \\frac{\\partial g}{\\partial y} \\mathbf{j} + \\mathbf{k} \\]\n\\[ \\Rightarrow |\\mathbf{r}_x \\times \\mathbf{r}_y| = \\sqrt{\\left( \\frac{\\partial z}{\\partial x} \\right)^2 + \\left( \\frac{\\partial z}{\\partial y} \\right)^2 + 1} \\]\n\n- So, we can rewrite the first definition as\n\\[\n\\iint_S f(x, y, z) \\, dS = \\iint_D f(\\mathbf{r}(u, v)) \\cdot |\\mathbf{r}_u \\times \\mathbf{r}_v| \\, dA \\\\\n= \\iint_D f(x, y, g(x, y)) \\sqrt{\\left( \\frac{\\partial z}{\\partial x} \\right)^2 + \\left( \\frac{\\partial z}{\\partial y} \\right)^2 + 1} \\, dA\n\\]\n\n- Solving now the exercise we were given, we can assess that\n\\[ \\frac{\\partial z}{\\partial x} = 1 \\quad , \\quad \\frac{\\partial z}{\\partial y} = 2y \\]\n• Proceeding with the transformation, we get that\n\n\\[\n\\iint_S y \\, dS = \\iint_D y \\sqrt{1 + \\left( \\frac{\\partial z}{\\partial x} \\right)^2 + \\left( \\frac{\\partial z}{\\partial y} \\right)^2} \\, dA\n\\]\n\n\\[\n= \\int_0^1 \\int_0^2 y \\sqrt{1 + 1 + 4y^2} \\, dy \\, dx\n\\]\n\n\\[\n= \\sqrt{2} \\int_0^1 \\int_0^2 y(1 + 2y^2)^{\\frac{1}{2}} \\, dy \\, dx\n\\]\n\n\\[\n= \\frac{2\\sqrt{2}}{12} \\int_0^1 \\int_0^2 \\frac{12y}{2} (1 + 2y^2)^{\\frac{1}{2}} \\, dy \\, dx\n\\]\n\n\\[\n= \\frac{2\\sqrt{2}}{12} \\int_0^1 \\left[ (1 + 2y^2)^{\\frac{3}{2}} \\right]_{y=0}^{y=2} \\, dx\n\\]\n\n\\[\n= \\frac{2\\sqrt{2}}{12} \\int_0^1 (27 - 1) \\, dx\n\\]\n\n\\[\n= \\frac{52\\sqrt{2}}{12} \\left[ x \\right]_{x=0}^{x=1}\n\\]\n\n\\[\n= \\frac{13\\sqrt{2}}{3} (1 - 0)\n\\]\n\n\\[\n= \\frac{13\\sqrt{2}}{3}\n\\]", "id": "./materials/426.pdf" }, { "contents": "Evaluate the surface integral \\( \\iint_S z \\, dS \\), where \\( S \\) is the surface whose sides \\( S_1 \\) are given by \\( x^2 + y^2 = 1 \\), whose bottom \\( S_2 \\) is given by \\( x^2 + y^2 \\leq 1 \\), and whose top \\( S_3 \\) is the part of \\( z = 1 + x \\) that lies on top of \\( S_2 \\).\n\nFigure 1: 3D sketch of the surface \\( S \\)\n\n- Using parametric representation, we can define:\n\n\\[\nx = \\cos(\\theta), \\quad y = \\sin(\\theta), \\quad z = z\n\\]\n\nwhere\n\n\\[\n0 \\leq \\theta \\leq 2\\pi, \\quad 0 \\leq z \\leq 1 + \\cos(\\theta)\n\\]\n• This means that\n\\[\n\\mathbf{r}_\\theta \\times \\mathbf{r}_z = \\begin{vmatrix}\n\\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\\n-\\sin(\\theta) & \\cos(\\theta) & 0 \\\\\n0 & 0 & 1\n\\end{vmatrix} = \\cos(\\theta)\\mathbf{i} + \\sin(\\theta)\\mathbf{j}\n\\]\nwhich leads us to\n\\[\n|\\mathbf{r}_\\theta \\times \\mathbf{r}_z| = \\sqrt{\\cos^2(\\theta) + \\sin^2(\\theta)} = 1\n\\]\n• Now, we have to evaluate the integrals separately for each surface \\((S_1, S_2\\) and \\(S_3\\)).\n• By definition, we can transform a surface integral in a double integral by:\n\\[\n\\iint_S f(x, y, z) \\, d\\mathbf{S} = \\iint_D f(\\mathbf{r}(u, v)) \\cdot |\\mathbf{r}_u \\times \\mathbf{r}_v| \\, dA\n\\]\n• Let’s check, for instance, the surface integral over \\(S_1\\).\n\\[\n\\iint_{S_1} z \\, dS = \\iint_D z|\\mathbf{r}_\\theta \\times \\mathbf{r}_z| \\, dA = \\int_0^{2\\pi} \\int_0^{1+\\cos(\\theta)} z \\, dz \\, d\\theta\n\\]\n• Now is just to solve the double integral.\n• The idea is to do the same for the other two surfaces.\n• At last, you can assess the original integral you were asked.\n\\[\n\\iint_S z \\, dS = \\iint_{S_1} z \\, dS + \\iint_{S_2} z \\, dS + \\iint_{S_3} z \\, dS\n\\]\n• At the end you should get: \\(\\pi \\left(\\sqrt{2} + \\frac{3}{2}\\right)\\)", "id": "./materials/427.pdf" }, { "contents": "Evaluate the surface integral \\( \\iint_S (x + y + z) \\, dS \\), where \\( S \\) is the parallelogram with parametric equations \\( x = u + v, \\ y = u - v, \\ z = 1 + 2u + v, \\ 0 \\leq u \\leq 2, \\ 0 \\leq v \\leq 1 \\).\n\n- Since\n\n\\[\nx = u + v, \\quad y = u - v, \\quad z = 1 + 2u + v\n\\]\n\nwhere\n\n\\[\n0 \\leq u \\leq 2, \\quad 0 \\leq v \\leq 1\n\\]\n\nthis means that\n\n\\[\n\\mathbf{r}(u, v) = (u + v)\\mathbf{i} + (u - v)\\mathbf{j} + (1 + 2u + v)\\mathbf{k}\n\\]\n\n\\[\n\\Rightarrow \\mathbf{r}_u \\times \\mathbf{r}_v = \\begin{vmatrix}\n\\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\\n1 & 1 & 2 \\\\\n1 & -1 & 1\n\\end{vmatrix} = 3\\mathbf{i} + \\mathbf{j} - 2\\mathbf{k}\n\\]\n\nwhich leads us to\n\n\\[\n|\\mathbf{r}_u \\times \\mathbf{r}_v| = \\sqrt{3^2 + 1^2 + (-2)^2} = \\sqrt{14}\n\\]\n\n- By definition, we can transform a surface integral in a double integral by:\n\n\\[\n\\iint_S f(x, y, z) \\, dS = \\iint_D f(\\mathbf{r}(u, v)) \\cdot |\\mathbf{r}_u \\times \\mathbf{r}_v| \\, dA\n\\]\n• Proceeding to the transformation, we get that\n\n\\[\n\\iint_S (x + y + z) \\, dS = \\iint_D (u + v + u - v + 1 + 2u + v) |\\mathbf{r}_u \\times \\mathbf{r}_v| \\, dA\n\\]\n\n\\[\n= \\int_0^1 \\int_0^2 (4u + v + 1) \\cdot \\sqrt{14} \\, dudv\n\\]\n\n\\[\n= \\sqrt{14} \\int_0^1 \\left[ 2u^2 + uv + u \\right]_{u=0}^{u=2} \\, dv\n\\]\n\n\\[\n= \\sqrt{14} \\int_0^1 10 + 2v \\, dv\n\\]\n\n\\[\n= \\sqrt{14} \\left[ 10v + v^2 \\right]_{v=0}^{v=1}\n\\]\n\n\\[\n= \\sqrt{14} \\cdot (10 + 1)\n\\]\n\n\\[\n= 11\\sqrt{14}\n\\]", "id": "./materials/428.pdf" }, { "contents": "Evaluate the surface integral \\( \\iint_S xyz \\, dS \\), where \\( S \\) is the cone with parametric equations \\( x = u \\cos(v), \\ y = u \\sin(v), \\) with \\( 0 \\leq u \\leq 1, \\ 0 \\leq v \\leq \\frac{\\pi}{2} \\).\n\n- Since\n\n\\[\nx = u \\cos(v), \\quad y = u \\sin(v), \\quad z = u\n\\]\n\nwhere\n\n\\[\n0 \\leq u \\leq 1, \\quad 0 \\leq v \\leq \\frac{\\pi}{2}\n\\]\n\nthis means that\n\n\\[\n\\mathbf{r}(u, v) = u \\cos(v) \\mathbf{i} + u \\sin(v) \\mathbf{j} + u \\mathbf{k}\n\\]\n\n\\[\n\\Rightarrow \\mathbf{r}_u \\times \\mathbf{r}_v = \\begin{vmatrix}\n\\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\\n\\cos(v) & \\sin(v) & 1 \\\\\n-u \\sin(v) & u \\cos(v) & 0\n\\end{vmatrix} = -u \\cos(v) \\mathbf{i} - u \\sin(v) \\mathbf{j} + u \\mathbf{k}\n\\]\n\nwhich leads us to\n\n\\[\n|\\mathbf{r}_u \\times \\mathbf{r}_v| = \\sqrt{(-u \\cos(v))^2 + (-u \\sin(v))^2 + u^2} = \\sqrt{2u^2} = \\sqrt{2}u, \\quad u \\geq 0\n\\]\n\n- By definition, we can transform a surface integral in a double integral by:\n\n\\[\n\\iint_S f(x, y, z) \\, dS = \\iint_D f(\\mathbf{r}(u, v)) \\cdot |\\mathbf{r}_u \\times \\mathbf{r}_v| \\, dA\n\\]\n• Proceeding to the transformation, we get that\n\n\\[\n\\int \\int_S xyz \\, dS = \\int \\int_D [(u \\cos(v))(u \\sin(v))(u)]|r_u \\times r_v| \\, dA\n\\]\n\n\\[\n= \\int_0^1 \\int_0^{\\frac{\\pi}{2}} (u^3 \\sin(v) \\cos(v)) \\sqrt{2} u \\, dv \\, du\n\\]\n\n• Now is just to solve the double integral.\n\n• At the end you should get: \\( \\frac{\\sqrt{2}}{10} \\)", "id": "./materials/429.pdf" }, { "contents": "Evaluate the surface integral \\( \\iint_S y \\, dS \\), where \\( S \\) is the helicoid surface with vector equation \\( \\mathbf{r}(u, v) = (u \\cos(v), u \\sin(v), v) \\), with \\( 0 \\leq u \\leq 1 \\), \\( 0 \\leq v \\leq \\pi \\).\n\n- Since\n \\[\n \\mathbf{r}(u, v) = u \\cos(v) \\mathbf{i} + u \\sin(v) \\mathbf{j} + v \\mathbf{k}\n \\]\n \\[\n \\Rightarrow \\mathbf{r}_u \\times \\mathbf{r}_v = \\begin{vmatrix}\n \\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\\n \\cos(v) & \\sin(v) & 0 \\\\\n -u \\sin(v) & u \\cos(v) & 1\n \\end{vmatrix} = \\sin(v) \\mathbf{i} - u \\cos(v) \\mathbf{j} + u \\mathbf{k}\n \\]\n which leads us to\n \\[\n |\\mathbf{r}_u \\times \\mathbf{r}_v| = \\sqrt{\\sin^2(v) + \\cos^2(v) + u^2} = \\sqrt{u^2 + 1}\n \\]\n\n- By definition, we can transform a surface integral in a double integral by:\n \\[\n \\iint_S f(x, y, z) \\, dS = \\iint_D f(\\mathbf{r}(u, v)) \\cdot |\\mathbf{r}_u \\times \\mathbf{r}_v| \\, dA\n \\]\n\n- Proceeding to the transformation, we get that\n \\[\n \\iint_S y \\, dS = \\iint_D (u \\sin(v)) |\\mathbf{r}_u \\times \\mathbf{r}_v| \\, dA\n \\]\n \\[\n = \\int_0^1 \\int_0^\\pi (u \\sin(v)) \\sqrt{u^2 + 1} \\, dv \\, du\n \\]\n\n- Now is just to solve the double integral.\n\n- At the end you should get: \\( \\frac{2}{3} \\left( 2\\sqrt{2} - 1 \\right) \\)", "id": "./materials/430.pdf" }, { "contents": "Evaluate the surface integral \\( \\iint_S (x^2 + y^2) \\, dS \\), where \\( S \\) is the surface with vector equation \\( \\mathbf{r}(u, v) = (2uv, u^2 - v^2, u^2 + v^2) \\), with \\( u^2 + v^2 \\leq 1 \\).\n\n- Since\n \\[\n \\mathbf{r}(u, v) = 2uv \\mathbf{i} + (u^2 - v^2) \\mathbf{j} + (u^2 + v^2) \\mathbf{k}\n \\]\n \\[\n \\Rightarrow \\mathbf{r}_u \\times \\mathbf{r}_v = \\begin{vmatrix}\n \\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\\n 2v & 2u & 2u \\\\\n 2u & -2v & 2v\n \\end{vmatrix} = 8uv \\mathbf{i} + (4u^2 - 4v^2) \\mathbf{j} + (-4u^2 - 4v^2) \\mathbf{k}\n \\]\n which leads us to\n \\[\n |\\mathbf{r}_u \\times \\mathbf{r}_v| = \\sqrt{(8uv)^2 + (4u^2 - 4v^2)^2 + (-4u^2 - 4v^2)^2}\n \\]\n \\[\n = \\sqrt{64u^2v^2 + 32u^4 + 32v^4}\n \\]\n \\[\n = \\sqrt{32(u^2 + v^2)^2}\n \\]\n \\[\n = 4\\sqrt{2}(u^2 + v^2)\n \\]\n\n- By definition, we can transform a surface integral in a double integral by:\n \\[\n \\iint_S f(x, y, z) \\, dS = \\iint_D f(\\mathbf{r}(u, v)) \\cdot |\\mathbf{r}_u \\times \\mathbf{r}_v| \\, dA\n \\]\n• Proceeding to the transformation, we get that\n\n\\[\n\\iint_S (x^2 + y^2) \\, dS = \\iint_D \\left( (2uv)^2 + (u^2 - v^2)^2 \\right) |\\mathbf{r}_u \\times \\mathbf{r}_v| \\, dA\n\\]\n\n\\[\n= \\iint_D (4u^2v^2 + u^4 - 2u^2v^2 + v^4) \\cdot 4\\sqrt{2}(u^2 + v^2) \\, dA\n\\]\n\n\\[\n= 4\\sqrt{2} \\iint_D (u^2 + v^2)^3 \\, dA\n\\]\n\n\\[\n= 4\\sqrt{2} \\int_0^{2\\pi} \\int_0^1 (r^2)^3 r \\, dr \\, d\\theta\n\\]\n\n• Now is just to solve the double integral.\n\n• At the end you should get: \\( \\sqrt{2}\\pi \\)", "id": "./materials/431.pdf" }, { "contents": "Evaluate the surface integral \\( \\iint_S x^2yz \\, dS \\), where \\( S \\) is the part of \n\\( z = 1 + 2x + 3y \\) that lies above \\([0, 3] \\times [0, 2]\\).\n\n- By definition, we can transform a surface integral in a double integral by:\n \\[\n \\iint_S f(x, y, z) \\, dS = \\iint_D f(r(u, v)) \\cdot |r_u \\times r_v| \\, dA\n \\]\n\n- We can regard this surface as a parametric surface with the following parametric equations:\n \\[\n x = x \\quad y = y \\quad z = g(x, y)\n \\]\n where\n \\[\n r_x = i + \\left( \\frac{\\partial g}{\\partial x} \\right) k \\quad r_y = j + \\left( \\frac{\\partial g}{\\partial y} \\right) k\n \\]\n\n- This means that\n \\[\n r_x \\times r_y = -\\frac{\\partial g}{\\partial x} i - \\frac{\\partial g}{\\partial y} j + k\n \\]\n \\[\n \\Rightarrow |r_x \\times r_y| = \\sqrt{\\left( \\frac{\\partial z}{\\partial x} \\right)^2 + \\left( \\frac{\\partial z}{\\partial y} \\right)^2 + 1}\n \\]\n\n- So, we can rewrite the first definition as\n \\[\n \\iint_S f(x, y, z) \\, dS = \\iint_D f(r(u, v)) \\cdot |r_u \\times r_v| \\, dA\n \\]\n \\[\n = \\iint_D f(x, y, g(x, y)) \\sqrt{\\left( \\frac{\\partial z}{\\partial x} \\right)^2 + \\left( \\frac{\\partial z}{\\partial y} \\right)^2 + 1} \\, dA\n \\]\n• We can assess\n\n\\[ z = 1 + 2x + 3y \\quad \\Rightarrow \\quad \\left( \\frac{\\partial z}{\\partial x} = 2, \\quad \\frac{\\partial z}{\\partial y} = 3 \\right) \\]\n\n• Proceeding to the transformation, we get that\n\n\\[\n\\int \\int_S x^2yz \\, dS = \\int \\int_D x^2yz \\sqrt{1 + \\left( \\frac{\\partial z}{\\partial x} \\right)^2 + \\left( \\frac{\\partial z}{\\partial y} \\right)^2} \\, dA\n\\]\n\n\\[\n= \\int_0^3 \\int_0^2 x^2y(1 + 2x + 3y)\\sqrt{1 + 4 + 9} \\, dy \\, dx\n\\]\n\n\\[\n= \\sqrt{14} \\int_0^3 \\int_0^2 x^2y + 2x^3y + 3x^2y^2 \\, dy \\, dx\n\\]\n\n\\[\n= \\sqrt{14} \\int_0^3 \\left[ \\frac{x^2y^2}{2} + x^3y^2 + x^2y^3 \\right]_{y=0}^{y=2} \\, dx\n\\]\n\n\\[\n= \\sqrt{14} \\int_0^3 2x^2 + 4x^3 + 8x^2 \\, dx\n\\]\n\n\\[\n= \\sqrt{14} \\left[ \\frac{10x^3}{3} + x^4 \\right]_{x=0}^{x=3}\n\\]\n\n\\[\n= \\sqrt{14} \\cdot (90 + 81)\n\\]\n\n\\[\n= 171\\sqrt{14}\n\\]", "id": "./materials/432.pdf" }, { "contents": "Evaluate the surface integral \\( \\iint_S xz \\, dS \\), where \\( S \\) is part of \\( 2x + 2y + z = 4 \\) that lies in the first octant.\n\n![3D sketch of the surface S](image)\n\nFigure 1: 3D sketch of the surface \\( S \\)\n\n- By definition, we can transform a surface integral in a double integral by:\n \\[\n \\iint_S f(x, y, z) \\, dS = \\iint_D f(r(u, v)) \\cdot |r_u \\times r_v| \\, dA\n \\]\n\n- We can regard this surface as a parametric surface with the following parametric equations:\n \\[\n x = x \\quad y = y \\quad z = g(x, y)\n \\]\nwhere\n\\[ \\mathbf{r}_x = \\mathbf{i} + \\left( \\frac{\\partial g}{\\partial x} \\right) \\mathbf{k} \\quad \\mathbf{r}_y = \\mathbf{j} + \\left( \\frac{\\partial g}{\\partial y} \\right) \\mathbf{k} \\]\n\n- This means that\n\\[ \\mathbf{r}_x \\times \\mathbf{r}_y = -\\frac{\\partial g}{\\partial x} \\mathbf{i} - \\frac{\\partial g}{\\partial y} \\mathbf{j} + \\mathbf{k} \\]\n\\[ \\Rightarrow |\\mathbf{r}_x \\times \\mathbf{r}_y| = \\sqrt{\\left( \\frac{\\partial z}{\\partial x} \\right)^2 + \\left( \\frac{\\partial z}{\\partial y} \\right)^2 + 1} \\]\n\n- So, we can rewrite the first definition as\n\\[ \\iint_S f(x, y, z) \\, dS = \\iint_D f(\\mathbf{r}(u, v)) \\cdot |\\mathbf{r}_u \\times \\mathbf{r}_v| \\, dA \\]\n\\[ = \\iint_D f(x, y, g(x, y)) \\sqrt{\\left( \\frac{\\partial z}{\\partial x} \\right)^2 + \\left( \\frac{\\partial z}{\\partial y} \\right)^2 + 1} \\, dA \\]\n\n- Since\n\\[ z = 4 - 2x - 2y \\]\n\nwhere\n\\[ D = \\{(x, y) \\mid 0 \\leq x \\leq 2, 0 \\leq y \\leq 2 - x\\} \\]\n\nwe get that\n\\[ \\iint_S xz \\, dS \\]\n\\[ = \\iint_D x(4 - 2x - 2y) \\sqrt{(-2)^2 + (-2)^2 + 1} \\, dA \\]\n\\[ = 3 \\int_0^2 \\int_0^{2-x} (4x - 2x^2 - 2xy) \\, dy \\, dx \\]\n\n- Now is just to solve the double integral.\n\n- At the end you should get: 4", "id": "./materials/433.pdf" }, { "contents": "Evaluate the surface integral \\( \\iint_S x \\, dS \\), where \\( S \\) is the region with vertices \\((1, 0, 0), (0, -2, 0), \\) and \\((0, 0, 4)\\).\n\n- The plane through the points \\((1, 0, 0), (0, -2, 0), \\) and \\((0, 0, 4)\\) is defined as\n \\[\n z = 4 - 4x + 2y\n \\]\n where\n \\[\n D = \\{(x, y) \\mid 0 \\leq x \\leq 1, \\ 2x - 2 \\leq y \\leq 0\\}\n \\]\n- By definition, we can write\n \\[\n \\iint_S f(x, y, z) \\, dS = \\iint_D f(x, y, g(x, y)) \\sqrt{\\left(\\frac{\\partial z}{\\partial x}\\right)^2 + \\left(\\frac{\\partial z}{\\partial y}\\right)^2 + 1} \\, dA\n \\]\n- We can assess\n \\[\n z = 4 - 4x + 2y \\quad \\Rightarrow \\quad \\frac{\\partial z}{\\partial x} = -4, \\quad \\frac{\\partial z}{\\partial y} = 2\n \\]\n- This means that we can proceed to the transformation\n \\[\n \\iint_S x \\, dS = \\iint_D x \\sqrt{(-4)^2 + (2)^2 + 1} \\, dA\n \\]\n \\[\n = \\sqrt{21} \\int_0^1 \\int_{2x-2}^0 x \\, dy \\, dx\n \\]\n- Now is just to solve the double integral.\n- At the end you should get: \\( \\frac{\\sqrt{21}}{3} \\)", "id": "./materials/434.pdf" }, { "contents": "Evaluate the surface integral \\( \\iint_S y \\, dS \\), where \\( S \\) is the surface given by\n\\[\nz = \\frac{2}{3} \\left( x^2 + y^2 \\right), \\text{ with } 0 \\leq x \\leq 1, \\ 0 \\leq y \\leq 1.\n\\]\n\n- By definition, we can write\n\\[\n\\iint_S f(x, y, z) \\, dS = \\iint_D f(x, y, g(x, y)) \\sqrt{\\left( \\frac{\\partial z}{\\partial x} \\right)^2 + \\left( \\frac{\\partial z}{\\partial y} \\right)^2 + 1} \\, dA\n\\]\n\n- Since\n\\[\nz = \\frac{2}{3} \\left( x^2 + y^2 \\right)\n\\]\nwhere\n\\[\nD = \\{(x, y) \\mid 0 \\leq x \\leq 1, \\ 0 \\leq y \\leq 1\\}\n\\]\nthis means that we can proceed to the transformation\n\\[\n\\iint_S y \\, dS = \\iint_D y \\sqrt{(\\sqrt{x})^2 + (\\sqrt{y})^2 + 1} \\, dA\n\\]\n\\[\n= \\int_0^1 \\int_0^1 y \\sqrt{x + y + 1} \\, dx \\, dy\n\\]\n\n- Now is just to solve the double integral. Suggestion: substitute \\( u = y + 2 \\) and \\( t = y + 1 \\), after solving the first integration.\n\n- At the end you should get: \\( \\frac{4(9\\sqrt{3} + 4\\sqrt{2} - 2)}{105} \\)", "id": "./materials/435.pdf" }, { "contents": "Evaluate the surface integral \\( \\iint_S x^2 z^2 \\, dS \\), where \\( S \\) is part of the surface given by \\( z^2 = x^2 + y^2 \\) that is bounded between \\( z = 1 \\) and \\( z = 3 \\).\n\nFigure 1: 3D sketch of the surface \\( S \\)\n• We can define\n\n\\[ z = \\sqrt{x^2 + y^2} \\]\n\nwhere\n\n\\[ D = \\{(x, y) \\mid 1 \\leq x^2 + y^2 \\leq 9\\} \\]\n\n• By definition, we can write\n\n\\[\n\\iint_S f(x, y, z) \\, dS = \\iint_D f(x, y, g(x, y)) \\sqrt{\\left(\\frac{\\partial z}{\\partial x}\\right)^2 + \\left(\\frac{\\partial z}{\\partial y}\\right)^2 + 1} \\, dA\n\\]\n\n• This means that we can proceed to the transformation\n\n\\[\n\\iint_S x^2 z^2 \\, dS = \\iint_D x^2(x^2 + y^2) \\sqrt{\\left(\\frac{x}{\\sqrt{x^2 + y^2}}\\right)^2 + \\left(\\frac{y}{\\sqrt{x^2 + y^2}}\\right)^2 + 1} \\, dA\n\\]\n\n\\[\n= \\iint_D x^2(x^2 + y^2) \\sqrt{\\frac{x^2 + y^2}{x^2 + y^2} + 1} \\, dA\n\\]\n\n\\[\n= \\iint_D \\sqrt{2}x^2(x^2 + y^2) \\, dA\n\\]\n\n\\[\n= \\sqrt{2} \\int_0^{2\\pi} \\int_1^3 (r \\cos(\\theta))^2(r)^2 r \\, dr \\, d\\theta\n\\]\n\n• Now is just to solve the double integral.\n\n• At the end you should get: \\( \\frac{364\\pi \\sqrt{2}}{3} \\)", "id": "./materials/436.pdf" }, { "contents": "Evaluate the surface integral \\( \\iint_S z \\, dS \\), where \\( S \\) is the surface \\( x = y + 2z^2 \\) with \\( 0 \\leq y \\leq 1, \\ 0 \\leq z \\leq 1 \\).\n\n- Using \\( y \\) and \\( z \\) as parameters, we have\n \\[\n \\mathbf{r}(y, z) = (y + 2z^2)\\mathbf{i} + y\\mathbf{j} + z\\mathbf{k}\n \\]\n \\[\n \\Rightarrow \\mathbf{r}_y \\times \\mathbf{r}_z = \\begin{vmatrix}\n \\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\\n 1 & 1 & 0 \\\\\n 4z & 0 & 1\n \\end{vmatrix} = \\mathbf{i} - \\mathbf{j} - 4z\\mathbf{k}\n \\]\n which leads us to\n \\[\n |\\mathbf{r}_y \\times \\mathbf{r}_z| = \\sqrt{2 + 16z^2}\n \\]\n\n- By definition, we can transform a surface integral in a double integral by:\n \\[\n \\iint_S f(x, y, z) \\, dS = \\iint_D f(\\mathbf{r}(u, v)) \\cdot |\\mathbf{r}_u \\times \\mathbf{r}_v| \\, dA\n \\]\n• Proceeding to the transformation, we get that\n\n\\[\n\\int \\int_S z \\, dS = \\int_0^1 \\int_0^1 z \\sqrt{2 + 16z^2} \\, dy \\, dz\n\\]\n\n\\[\n= \\int_0^1 \\left[ zy \\sqrt{2 + 16z^2} \\right]_{y=0}^{y=1} \\, dz\n\\]\n\n\\[\n= \\int_0^1 z(2 + 16z^2)^{\\frac{1}{2}} \\, dz\n\\]\n\n\\[\n= \\frac{2}{3} \\cdot \\frac{1}{32} \\int_0^1 \\frac{3}{2} \\cdot 32 \\cdot z(2 + 16z^2)^{\\frac{1}{2}} \\, dz\n\\]\n\n\\[\n= \\frac{2}{3} \\cdot \\frac{1}{32} \\left[ (2 + 16z^2)^{\\frac{3}{2}} \\right]_{z=0}^{z=1}\n\\]\n\n\\[\n= \\frac{1}{48} (\\sqrt{18^3} - \\sqrt{2^3})\n\\]\n\n\\[\n= \\frac{1}{48} (54\\sqrt{2} - 2\\sqrt{2})\n\\]\n\n\\[\n= \\frac{13\\sqrt{2}}{12}\n\\]", "id": "./materials/437.pdf" }, { "contents": "Evaluate the surface integral \\( \\iint_S y \\, dS \\), where \\( S \\) is part of \\( y = x^2 + z^2 \\) that lies inside \\( x^2 + z^2 = 4 \\).\n\n- Using \\( x \\) and \\( z \\) as parameters, we have\n \\[\n \\mathbf{r}(x, z) = xi + (x^2 + z^2)j + zk\n \\]\n \\[\n \\Rightarrow \\mathbf{r}_x \\times \\mathbf{r}_z = \\begin{vmatrix}\n i & j & k \\\\\n 1 & 2x & 0 \\\\\n 0 & 2z & 1\n \\end{vmatrix} = 2xi - j + 2zk\n \\]\n which leads us to\n \\[\n |\\mathbf{r}_x \\times \\mathbf{r}_z| = \\sqrt{4(x^2 + z^2) + 1}\n \\]\n\n- By definition, we can transform a surface integral in a double integral by:\n \\[\n \\iint_S f(x, y, z) \\, dS = \\iint_D f(\\mathbf{r}(u, v)) \\cdot |\\mathbf{r}_u \\times \\mathbf{r}_v| \\, dA\n \\]\n\n- Proceeding to the transformation, we get that\n \\[\n \\iint_S y \\, dS = \\iint_D (x^2 + z^2)\\sqrt{4(x^2 + z^2) + 1} \\, dA\n \\]\n \\[\n = \\int_0^{2\\pi} \\int_0^2 r^2 \\sqrt{1 + 4r^2} \\cdot r \\, dr \\, d\\theta\n \\]\n \\[\n = \\pi(391\\sqrt{17} + 1)\n \\]\n \\[\n \\frac{60}{60}\n \\]\n At the end you should get: \\( \\frac{\\pi(391\\sqrt{17} + 1)}{60} \\)", "id": "./materials/438.pdf" }, { "contents": "Evaluate the surface integral \\( \\iint_S y^2 \\, dS \\), where \\( S \\) is the part of \\( x^2 + y^2 + z^2 = 4 \\) that lies inside \\( x^2 + y^2 = 1 \\) and above the xy-plane.\n\nFigure 1: 3D sketch of the surface \\( S \\)\n\n- The sphere intersects the cylinder in the circle\n \\[ x^2 + y^2 = 1 \\quad , \\quad z = \\sqrt{3} \\]\n meaning that \\( S \\) is the portion of the sphere defined as\n \\[ z \\geq \\sqrt{3} \\]\n• Using spherical coordinates to parameterize the sphere we have, where \\( \\phi \\) and \\( \\theta \\) are the parameters, we have\n\n\\[\n\\mathbf{r}(\\phi, \\theta) = 2 \\sin(\\phi) \\cos(\\theta) \\mathbf{i} + 2 \\sin(\\phi) \\sin(\\theta) \\mathbf{j} + 2 \\cos(\\phi) \\mathbf{k}\n\\]\n\nwhich leads us to\n\n\\[\n|\\mathbf{r}_\\phi \\times \\mathbf{r}_\\theta| = 4 \\sin(\\phi)\n\\]\n\n• By definition, we can transform a surface integral in a double integral by:\n\n\\[\n\\iint_S f(x, y, z) \\, dS = \\iint_D f(\\mathbf{r}(u, v)) \\cdot |\\mathbf{r}_u \\times \\mathbf{r}_v| \\, dA\n\\]\n\n• Proceeding to the transformation, and keeping in mind that for\n\n\\[\nz \\geq \\sqrt{3} \\quad \\Rightarrow \\quad \\left( 0 \\leq \\phi \\leq \\frac{\\pi}{6} , \\ 0 \\leq \\theta \\leq 2\\pi \\right)\n\\]\n\nwe get that\n\n\\[\n\\iint_S y^2 \\, dS = \\int_0^{2\\pi} \\int_0^{\\pi/6} (2 \\sin(\\phi) \\sin(\\theta))^2 (4 \\sin(\\phi)) \\, d\\phi d\\theta\n\\]\n\n• Now is just to solve the double integral.\n\n• At the end you should get: \\( \\left( \\frac{32}{3} - 6\\sqrt{3} \\right) \\pi \\)", "id": "./materials/439.pdf" }, { "contents": "Evaluate the surface integral \\( \\iint_S (x^2z + y^2z) \\, dS \\), where \\( S \\) is defined as \\( x^2 + y^2 + z^2 = 4 \\), with \\( z \\geq 0 \\).\n\n- Using spherical coordinates to parameterize the sphere we have, where \\( \\phi \\) and \\( \\theta \\) are the parameters, we have\n \\[\n \\mathbf{r}(\\phi, \\theta) = 2 \\sin(\\phi) \\cos(\\theta) \\mathbf{i} + 2 \\sin(\\phi) \\sin(\\theta) \\mathbf{j} + 2 \\cos(\\phi) \\mathbf{k}\n \\]\n which leads us to\n \\[\n |\\mathbf{r}_\\phi \\times \\mathbf{r}_\\theta| = 4 \\sin(\\phi)\n \\]\n\n- By definition, we can transform a surface integral in a double integral by:\n \\[\n \\iint_S f(x, y, z) \\, dS = \\iint_D f(\\mathbf{r}(u, v)) \\cdot |\\mathbf{r}_u \\times \\mathbf{r}_v| \\, dA\n \\]\n\n- Proceeding to the transformation, we get that\n \\[\n \\iint_S (x^2z + y^2z) \\, dS = \\int_0^{2\\pi} \\int_0^{\\frac{\\pi}{2}} (4 \\sin^2(\\phi))(2 \\cos(\\phi))(4 \\sin(\\phi)) \\, d\\phi d\\theta\n \\]\n \\[\n = 16\\pi\n \\]\n\n- Now is just to solve the double integral.\n\n- At the end you should get: \\( 16\\pi \\)", "id": "./materials/440.pdf" }, { "contents": "Evaluate the surface integral \\( \\iint_S xz \\, dS \\), where \\( S \\) is the boundary of the region enclosed by \\( y^2 + z^2 = 9 \\), \\( x = 0 \\) and \\( x + y = 5 \\).\n\nFigure 1: 3D sketch of the surface \\( S \\)\n\n- \\( S \\) consists of three surfaces: \\( S_1 \\), the lateral surface of the cylinder; \\( S_2 \\), the front formed by the plane \\( x + y = 5 \\) and the back, \\( S_3 \\), in the plane \\( x = 0 \\).\n• This means that to evaluate the surface integral you were asked, you will need to divide it\n\n\\[\n\\iint_S xz \\, dS = \\iint_{S_1} xz \\, dS + \\iint_{S_2} xz \\, dS + \\iint_{S_3} xz \\, dS\n\\]\n\n• Let’s assess the situation for \\( S_1 \\), for instance.\n\n• Using \\( u \\) and \\( v \\) as parameters, we have\n\n\\[\n\\mathbf{r}(u, v) = u\\mathbf{i} + 3\\cos(v)\\mathbf{j} + 3\\sin(v)\\mathbf{k}, \\quad (0 \\leq v \\leq 2\\pi, \\quad 0 \\leq x \\leq 5 - y)\n\\]\n\n\\[\n\\Rightarrow \\mathbf{r}_u \\times \\mathbf{r}_v = -3\\cos(v)\\mathbf{j} - 3\\sin(v)\\mathbf{k}\n\\]\n\nwhich leads us to\n\n\\[\n|\\mathbf{r}_u \\times \\mathbf{r}_v| = \\sqrt{9\\cos^2(v) + 9\\sin^2(v)} = 3\n\\]\n\n• By definition, we can transform a surface integral in a double integral by:\n\n\\[\n\\iint_S f(x, y, z) \\, dS = \\iint_D f(\\mathbf{r}(u, v)) \\cdot |\\mathbf{r}_u \\times \\mathbf{r}_v| \\, dA\n\\]\n\n• Proceeding to the transformation, we get that\n\n\\[\n\\iint_{S_1} xz \\, dS = \\int_0^{2\\pi} \\int_0^{5-3\\cos(v)} u(3\\sin(v)) \\cdot 3 \\, du \\, dv\n\\]\n\n• At last, it is just to repeat the same process for \\( S_2 \\) and \\( S_3 \\).\n\n• At the end you should get: 0", "id": "./materials/441.pdf" }, { "contents": "Evaluate the surface integral \\( \\iint_S (z + x^2y) \\, dS \\), where \\( S \\) is the part of \\( y^2 + z^2 = 1 \\) that lies between \\( x = 0 \\) and \\( x = 3 \\) in the first octant.\n\n- Using \\( u \\) and \\( v \\) as parameters, we have\n \\[\n \\mathbf{r}(u, v) = u \\mathbf{i} + \\cos(v) \\mathbf{j} + \\sin(v) \\mathbf{k} \\quad \\left( 0 \\leq u \\leq 3 \\quad , \\quad 0 \\leq v \\leq \\frac{\\pi}{2} \\right)\n \\]\n \\[\n \\Rightarrow \\mathbf{r}_u \\times \\mathbf{r}_v = -\\cos(v) \\mathbf{j} - \\sin(v) \\mathbf{k}\n \\]\n which leads us to\n \\[\n |\\mathbf{r}_u \\times \\mathbf{r}_v| = \\sqrt{\\cos^2(v) + \\sin^2(v)} = 1\n \\]\n\n- By definition, we can transform a surface integral in a double integral by:\n \\[\n \\iint_S f(x, y, z) \\, dS = \\iint_D f(\\mathbf{r}(u, v)) \\cdot |\\mathbf{r}_u \\times \\mathbf{r}_v| \\, dA\n \\]\n\n- Proceeding to the transformation, we get that\n \\[\n \\iint_S (z + x^2y) \\, dS\n \\]\n \\[\n = \\int_0^{\\frac{\\pi}{2}} \\int_0^3 (\\sin(v) + u^2 \\cos(v)) \\cdot 1 \\, du \\, dv\n \\]\n \\[\n = \\int_0^{\\frac{\\pi}{2}} \\left[ u \\sin(v) + \\frac{u^3}{3} \\cos(v) \\right]_{u=0}^{u=3} \\, dv\n \\]\n \\[\n = \\int_0^{\\frac{\\pi}{2}} 3 \\sin(v) + 9 \\cos(v) \\, dv\n \\]\n \\[\n = \\left[ -3 \\cos(v) + 9 \\sin(v) \\right]_{v=0}^{v=\\frac{\\pi}{2}}\n \\]\n \\[\n = 0 + 9 + 3 - 0\n \\]\n \\[\n = 12\n \\]", "id": "./materials/442.pdf" }, { "contents": "Evaluate the surface integral \\( \\iint_S (x^2 + y^2 + z^2) \\, dS \\), where \\( S \\) is part of \\( x^2 + y^2 = 9 \\), between \\( z = 0 \\) and \\( z = 2 \\), together with both its top and bottom disks.\n\n- \\( S \\) consists of three surfaces: \\( S_1 \\), the lateral surface; \\( S_2 \\), the top disk and the bottom disk, \\( S_3 \\).\n- This means that to evaluate the surface integral you were asked, you will need to divide it\n\n\\[\n\\iint_S (x^2 + y^2 + z^2) \\, dS = \\iint_{S_1} (x^2 + y^2 + z^2) \\, dS + \\iint_{S_2} (x^2 + y^2 + z^2) \\, dS + \\iint_{S_3} (x^2 + y^2 + z^2) \\, dS\n\\]\n\n- Let’s assess the situation for \\( S_2 \\), for instance (top disk).\n- Using cylindrical coordinates, with \\( \\theta \\) and \\( r \\) as parameters, we have\n\n\\[\nr(\\theta, r) = r \\cos(\\theta)i + r \\sin(\\theta)j + 2k\n\\]\n\nwith\n\n\\[\n0 \\leq r \\leq 3 \\quad , \\quad 0 \\leq \\theta \\leq 2\\pi\n\\]\n\nwhich leads us to\n\n\\[\n|r_\\theta \\times r_r| = r\n\\]\n\n- By definition, we can transform a surface integral in a double integral by:\n\n\\[\n\\iint_S f(x, y, z) \\, dS = \\iint_D f(r(u, v)) \\cdot |r_u \\times r_v| \\, dA\n\\]\n• Proceeding to the transformation, we get that\n\n\\[\n\\int \\int_{S_2} (x^2 + y^2 + z^2) \\, dS\n\\]\n\n\\[\n= \\int_0^{2\\pi} \\int_0^3 (r^2 + 4)(r) \\, dr \\, d\\theta\n\\]\n\n\\[\n= \\frac{153\\pi}{2}\n\\]\n\n• At last, it is just to repeat the same process for \\( S_1 \\) and \\( S_3 \\).\n\n• At the end you should get: \\( 241\\pi \\)", "id": "./materials/443.pdf" }, { "contents": "Find the flux of the vector field \\( \\mathbf{F}(x, y, z) = z\\mathbf{i} + y\\mathbf{j} + x\\mathbf{k} \\) across \\( x^2 + y^2 + z^2 = 1 \\).\n\n- First, we have to acknowledge that the surface in question is the unit sphere.\n\n![3D sketch of the surface S](image)\n\n**Figure 1: 3D sketch of the surface S**\n\n- Using parametric representation, we can define:\n \\[\n x = \\sin(\\phi) \\cos(\\theta), \\quad y = \\sin(\\phi) \\sin(\\theta), \\quad z = \\cos(\\phi), \\quad 0 \\leq \\phi \\leq \\pi, \\quad 0 \\leq \\theta \\leq 2\\pi\n \\]\n\n- This means that we can also define\n \\[\n \\mathbf{r}(\\phi, \\theta) = \\sin(\\phi) \\cos(\\theta)\\mathbf{i} + \\sin(\\phi) \\sin(\\theta)\\mathbf{j} + \\cos(\\phi)\\mathbf{k}\n \\]\nThen\n\n\\[ F(r(\\phi, \\theta)) = \\cos(\\theta)i + \\sin(\\phi)\\sin(\\theta)j + \\sin(\\phi)\\cos(\\theta)k \\]\n\n- Evaluating \\( r_\\phi \\times r_\\theta \\):\n\n\\[\n\\begin{align*}\n r_\\phi &= \\cos(\\phi)\\cos(\\theta)i + \\cos(\\phi)\\sin(\\theta)j - \\sin(\\phi)k \\\\\n r_\\theta &= -\\sin(\\phi)\\sin(\\theta)i + \\sin(\\phi)\\cos(\\theta)j \\\\\n \\Rightarrow r_\\phi \\times r_\\theta &= \\sin^2(\\phi)\\cos(\\theta)i + \\sin^2(\\phi)\\sin(\\theta)j + \\sin(\\phi)\\cos(\\phi)k\n\\end{align*}\n\\]\n\n- Therefore\n\n\\[\nF(r(\\phi, \\theta)) \\cdot (r_\\phi \\times r_\\theta) = \\sin^2(\\phi)\\cos(\\theta)\\cos(\\phi) + \\sin^3(\\phi)\\sin^2(\\theta) + \\sin^2(\\phi)\\cos(\\phi)\\cos(\\theta)\n\\]\n\n\\[= 2\\sin^2(\\phi)\\cos(\\theta)\\cos(\\phi) + \\sin^3(\\phi)\\sin^2(\\theta)\\]\n\n- By definition we know that\n\n\\[\n\\iint_S F \\cdot dS = \\iint_D F \\cdot (r_\\phi \\times r_\\theta) \\, dA\n\\]\n\nwe can proceed to the substitution\n\n\\[\n\\iint_S F \\cdot dS = \\int_0^{2\\pi} \\int_0^\\pi (2\\sin^2(\\phi)\\cos(\\theta)\\cos(\\phi) + \\sin^3(\\phi)\\sin^2(\\theta)) \\, d\\phi d\\theta\n\\]\n\n- Now is just to solve the double integral.\n\n- At the end you should get: \\( \\frac{4\\pi}{3} \\)", "id": "./materials/444.pdf" }, { "contents": "Evaluate $\\int \\int_S \\mathbf{F} \\cdot d\\mathbf{S}$ where $\\mathbf{F}(x, y, z) = yi + xj + zk$ and $S$ is the boundary of the solid region $E$ enclosed by $z = 1 - x^2 - y^2$ and $z = 0$.\n\nFigure 1: 3D sketch of the surface $S$\n\n- $S$ consists of a parabolic top surface $S_1$ and a circular bottom surface $S_2$. Since $S$ is a closed surface, we can use the convention of positive (outward) orientation. $D$ is the projection of $S_1$ onto the $xy$-plane, namely, the disk $x^2 + y^2 \\leq 1$.\n\n- We can assess, for $S_1$\n\n$$P(x, y, z) = y \\quad , \\quad Q(x, y, z) = x \\quad , \\quad R(x, y, z) = z = 1 - x^2 - y^2$$\nand\n\n\\[ z = g(x, y) = 1 - x^2 - y^2 \\quad \\Rightarrow \\quad \\left( \\frac{\\partial g}{\\partial x} = -2x, \\quad \\frac{\\partial g}{\\partial y} = -2y \\right) \\]\n\n- Knowing that a flux, in the case where \\( z = g(x, y) \\), is defined as\n\n\\[\n\\iint_{S_1} \\mathbf{F} \\cdot d\\mathbf{S} = \\iint_{D} \\left( -P \\frac{\\partial g}{\\partial x} - Q \\frac{\\partial g}{\\partial y} + R \\right) dA\n\\]\n\nwe can proceed to the substitution\n\n\\[\n\\iint_{S_1} \\mathbf{F} \\cdot d\\mathbf{S} \\\\\n= \\iint_{D} (-y(-2x) - x(-2y) + 1 - x^2 - y^2) dA \\\\\n= \\iint_{D} (1 + 4xy - x^2 - y^2) dA \\\\\n= \\int_{0}^{2\\pi} \\int_{0}^{1} (1 + 4r^2 \\cos(\\theta) \\sin(\\theta) - r^2) r \\, dr \\, d\\theta\n\\]\n\n- Now is just to solve the double integral.\n\n- We should evaluate now the surface integral for \\( S_2 \\). The disk \\( S_2 \\) is oriented downward, so its unit normal vector is \\( \\mathbf{n} = -\\mathbf{k} \\), so we have:\n\n\\[\n\\iint_{S_2} \\mathbf{F} \\cdot d\\mathbf{S} = \\iint_{S_2} \\mathbf{F} \\cdot (-\\mathbf{k}) \\, dS = \\iint_{D} (-z) \\, dA = \\iint_{D} 0 \\, dA = 0 \\quad , \\quad z = 0\n\\]\n\n- Since \\( S = S_1 + S_2 \\), we have\n\n\\[\n\\iint_{S} \\mathbf{F} \\cdot d\\mathbf{S} = \\iint_{S_1} \\mathbf{F} \\cdot d\\mathbf{S} + \\iint_{S_2} \\mathbf{F} \\cdot d\\mathbf{S}\n\\]\n\n- At the end you should get: \\( \\frac{\\pi}{2} \\)", "id": "./materials/445.pdf" }, { "contents": "Find the flux of $\\mathbf{F}$ across $S$, where $\\mathbf{F}(x, y, z) = z e^{xy} \\mathbf{i} - 3z e^{xy} \\mathbf{j} + xy \\mathbf{k}$, and $S$ is the parallelogram (with upward orientation) with parametric equations $x = u + v$, $y = u - v$, $z = 1 + 2u + v$, $0 \\leq u \\leq 2$, $0 \\leq v \\leq 1$.\n\n- Since\n\n$$x = u + v, \\quad y = u - v, \\quad z = 1 + 2u + v$$\n\nwhere\n\n$$0 \\leq u \\leq 2, \\quad 0 \\leq v \\leq 1$$\n\nthis means that\n\n$$\\mathbf{r}(u, v) = (u + v)\\mathbf{i} + (u - v)\\mathbf{j} + (1 + 2u + v)\\mathbf{k}$$\n\n$$\\Rightarrow \\mathbf{r}_u \\times \\mathbf{r}_v = \\begin{vmatrix} \\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\ 1 & 1 & 2 \\\\ 1 & -1 & 1 \\end{vmatrix} = 3\\mathbf{i} + \\mathbf{j} - 2\\mathbf{k}$$\n\nThen\n\n$$\\mathbf{F}(\\mathbf{r}(u, v)) = (1 + 2u + v) e^{(u+v)(u-v)} \\mathbf{i} - 3(1 + 2u + v) e^{(u+v)(u-v)} \\mathbf{j} + (u + v)(u - v)\\mathbf{k}$$\n\n$$= (1 + 2u + v) e^{u^2-v^2} \\mathbf{i} - 3(1 + 2u + v) e^{u^2-v^2} \\mathbf{j} + (u^2 - v^2)\\mathbf{k}$$\n\n- By definition, we know that\n\n$$\\iint_S \\mathbf{F} \\cdot d\\mathbf{S} = \\iint_D \\mathbf{F} \\cdot (\\mathbf{r}_\\phi \\times \\mathbf{r}_\\theta) \\, dA$$\n\n- However, since the $z$-component of $\\mathbf{r}_u \\times \\mathbf{r}_v$ is negative and to achieve the upward orientation we were asked, we accommodate that in the previous definition as:\n\n$$\\iint_S \\mathbf{F} \\cdot d\\mathbf{S} = \\iint_D \\mathbf{F} \\cdot (-\\mathbf{r}_u \\times \\mathbf{r}_v) \\, dA$$\nSo we can proceed to the substitution\n\n\\[\n\\int \\int_S \\mathbf{F} \\cdot d\\mathbf{S} = \\int_0^1 \\int_0^2 \\left[ -3(1 + 2u + v) e^{u^2 - v^2} + 3(1 + 2u + v) e^{u^2 - v^2} + 2(u^2 - v^2) \\right] dudv\n\\]\n\n\\[\n= \\int_0^1 \\int_0^2 2(u^2 - v^2) dudv\n\\]\n\n\\[\n= 2 \\int_0^1 \\left[ \\frac{u^3}{3} - v^2 u \\right]_{u=0}^{u=2} dv\n\\]\n\n\\[\n= 2 \\int_0^1 \\frac{8}{3} - 2v^2 dv\n\\]\n\n\\[\n= 2 \\left[ \\frac{8v}{3} - \\frac{2v^3}{3} \\right]_{v=0}^{v=1}\n\\]\n\n\\[\n= \\frac{2}{3} (8 - 2)\n\\]\n\n\\[\n= 4\n\\]", "id": "./materials/446.pdf" }, { "contents": "Find the flux of $\\mathbf{F}$ across $S$, where $\\mathbf{F}(x, y, z) = z\\mathbf{i} + y\\mathbf{j} + x\\mathbf{k}$, where $S$ is the helicoid surface with vector equation $\\mathbf{r}(u, v) = (u \\cos(v), u \\sin(v), v)$, with $0 \\leq u \\leq 1$, $0 \\leq v \\leq \\pi$ with upward orientation.\n\n- Since\n\n$$\\mathbf{r}(u, v) = u \\cos(v)\\mathbf{i} + u \\sin(v)\\mathbf{j} + v\\mathbf{k}$$\n\nand\n\n$$0 \\leq u \\leq 1, \\quad 0 \\leq v \\leq \\pi$$\n\nthis means that\n\n$$\\Rightarrow \\mathbf{r}_u \\times \\mathbf{r}_v = \\begin{vmatrix} \\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\ \\cos(v) & \\sin(v) & 0 \\\\ -u \\sin(v) & u \\cos(v) & 1 \\end{vmatrix} = \\sin(v)\\mathbf{i} - \\cos(v)\\mathbf{j} + u\\mathbf{k}$$\n\nThen\n\n$$\\mathbf{F}(\\mathbf{r}(u, v)) = v\\mathbf{i} + u \\sin(v)\\mathbf{j} + u \\cos(v)\\mathbf{k}$$\n\n- By definition, we know that\n\n$$\\iint_S \\mathbf{F} \\cdot d\\mathbf{S} = \\iint_D \\mathbf{F} \\cdot (\\mathbf{r}_u \\times \\mathbf{r}_v) \\, dA$$\n• Proceeding with the substitution\n\n\\[\n\\int \\int_S \\mathbf{F} \\cdot d\\mathbf{S} = \\int_0^1 \\int_0^\\pi (v \\sin(v) - u \\sin(v) \\cos(v) + u^2 \\cos(v)) \\, dv \\, du\n\\]\n\n\\[\n= \\int_0^1 \\left[ \\sin(v) - v \\cos(v) - \\frac{u}{2} \\sin^2(v) + u^2 \\sin(v) \\right]_{v=0}^{v=\\pi} \\, du\n\\]\n\n\\[\n= \\int_0^1 \\pi \\, du\n\\]\n\n\\[\n= \\left[ \\pi u \\right]_{u=0}^{u=1}\n\\]\n\n\\[\n= \\pi\n\\]", "id": "./materials/447.pdf" }, { "contents": "Find the flux of $\\mathbf{F}$ across $S$, where $\\mathbf{F}(x, y, z) = xy\\mathbf{i} + yz\\mathbf{j} + zx\\mathbf{k}$, where $S$ is part of the paraboloid $z = 4 - x^2 - y^2$ that lies above the square $0 \\leq x \\leq 1$, $0 \\leq y \\leq 1$, and has upward orientation.\n\n- Since\n \n $$\\mathbf{F}(x, y, z) = xy\\mathbf{i} + yz\\mathbf{j} + zx\\mathbf{k}$$\n\n and\n \n $$z = g(x, y) = 4 - x^2 - y^2$$\nand D is the square \\([0, 1] \\times [0, 1]\\), we can proceed with the substitution\n\n\\[\n\\iint_S \\mathbf{F} \\cdot d\\mathbf{S} = \\iint_D \\left( -P \\frac{\\partial g}{\\partial x} - Q \\frac{\\partial g}{\\partial y} + R \\right) dA\n\\]\n\n\\[\n= \\iint_D \\left[ -xy(-2x) - yz(-2y) + zx \\right] dA\n\\]\n\n\\[\n= \\int_0^1 \\int_0^1 \\left[ 2x^2y + 2y^2(4 - x^2 - y^2) + x(4 - x^2 - y^2) \\right] dydx\n\\]\n\n\\[\n= \\int_0^1 \\int_0^1 \\left[ 2x^2y + 8y^2 - 2y^2x^2 - 2y^4 + 4x - x^3 - xy^2 \\right] dydx\n\\]\n\n\\[\n= \\int_0^1 \\int_0^1 \\left[ 2x^2y + 8y^2 - 2y^2x^2 - 2y^4 + 4x - x^3 - xy^2 \\right] dydx\n\\]\n\n\\[\n= \\int_0^1 \\left[ x^2y^2 + \\frac{8y^3}{3} - \\frac{2y^3x^2}{3} - \\frac{2y^5}{5} + 4xy - x^3y - \\frac{xy^3}{3} \\right]_{y=0}^{y=1} dx\n\\]\n\n\\[\n= \\int_0^1 x^2 + \\frac{8}{3} - \\frac{2x^2}{3} - \\frac{2}{5} + 4x - x^3 - \\frac{x}{3} dx\n\\]\n\n\\[\n= \\int_0^1 -x^3 + \\frac{x^2}{3} + \\frac{11x}{3} + \\frac{34}{15} dx\n\\]\n\n\\[\n= \\left[ -\\frac{x^4}{4} + \\frac{x^3}{9} + \\frac{11x^2}{6} + \\frac{34x}{15} \\right]_{x=0}^{x=1}\n\\]\n\n\\[\n= -\\frac{1}{4} + \\frac{1}{9} + \\frac{11}{6} + \\frac{34}{15}\n\\]\n\n\\[\n= -\\frac{1 \\times 45 + 1 \\times 20 + 11 \\times 30 + 34 \\times 12}{180}\n\\]\n\n\\[\n= -\\frac{45 + 20 + 330 + 408}{180}\n\\]\n\n\\[\n= \\frac{713}{180}\n\\]", "id": "./materials/448.pdf" }, { "contents": "Find the flux of $\\mathbf{F}$ across $S$, where $\\mathbf{F}(x, y, z) = -xi - yj + z^3k$, where $S$ is part of the cone $z = \\sqrt{x^2 + y^2}$ between $z = 1$ and $z = 3$, with downward orientation.\n\n- We know that\n \n $$\\mathbf{F}(x, y, z) = -xi - yj + z^3k$$\n\n as well as\n \n $$z = g(x, y) = \\sqrt{x^2 + y^2}$$\n\n and\n \n $$D = \\{(x, y) \\mid 1 \\leq x^2 + y^2 \\leq 9\\}$$\n\n- Since $S$ has a downward orientation, we have\n \n $$\\int \\int_S \\mathbf{F} \\cdot d\\mathbf{S}$$\n\n $$= - \\int \\int_D \\left( - P \\frac{\\partial g}{\\partial x} - Q \\frac{\\partial g}{\\partial y} + R \\right) dA$$\n\n $$= - \\int \\int_D \\left[ - (-x) \\left( \\frac{x}{\\sqrt{x^2 + y^2}} \\right) - (-y) \\left( \\frac{y}{\\sqrt{x^2 + y^2}} \\right) + z^3 \\right] dA$$\n\n $$= - \\int \\int_D \\left[ \\frac{x^2 + y^2}{\\sqrt{x^2 + y^2}} + \\left( \\sqrt{x^2 + y^2} \\right)^3 \\right] dA$$\n\n $$= - \\int_0^{2\\pi} \\int_1^3 \\left( \\frac{r^2}{r} + r^3 \\right) r \\, dr \\, d\\theta$$\n\n- Now is just to solve the double integral.\n\n- At the end you should get: $-\\frac{1712\\pi}{15}$", "id": "./materials/449.pdf" }, { "contents": "Find the flux of $\\mathbf{F}$ across $S$, where $\\mathbf{F}(x, y, z) = xi - zj + yk$, where $S$ is part of the sphere $x^2 + y^2 + z^2 = 4$ in the first octant, with orientation toward the origin.\n\n- We know that\n \n $$\\mathbf{F}(x, y, z) = xi - zj + yk$$\n\n as well as\n \n $$z = g(x, y) = \\sqrt{4 - x^2 - y^2}$$\n\n and $D$ is the quarter disk\n \n $$D = \\{(x, y) \\mid 0 \\leq x \\leq 2 \\land 0 \\leq y \\leq \\sqrt{4 - x^2}\\}$$\n\n- Since $S$ has a downward orientation, we have\n\n $$\\iint_S \\mathbf{F} \\cdot d\\mathbf{S}$$\n\n $$= - \\iint_D \\left( - P \\frac{\\partial g}{\\partial x} - Q \\frac{\\partial g}{\\partial y} + R \\right) dA$$\n\n $$= - \\iint_D \\left[ - x \\left( \\frac{4 - x^2 - y^2}{2} \\right)^{-\\frac{1}{2}} \\times (-2x) - (-z) \\left( \\frac{4 - x^2 - y^2}{2} \\right)^{-\\frac{1}{2}} \\times (-2y) + y \\right] dA$$\n\n- Now is just to proceed to the substitution of $D$ and solve the double integral.\n\n- At the end you should get: $-\\frac{4\\pi}{3}$", "id": "./materials/450.pdf" }, { "contents": "Find the flux of $\\mathbf{F}$ across $S$, where $\\mathbf{F}(x, y, z) = xz \\mathbf{i} + x \\mathbf{j} + y \\mathbf{k}$, where $S$ is the hemisphere $x^2 + y^2 + z^2 = 25$, $y \\geq 0$, oriented in the direction of the positive $y$-axis.\n\n- We know that $\\mathbf{F}(x, y, z) = xz \\mathbf{i} - x \\mathbf{j} + y \\mathbf{k}$\n- Using spherical coordinates, $S$ is given by\n \n \\[\n \\begin{align*}\n x &= 5 \\sin(\\phi) \\cos(\\theta) \\\\\n y &= 5 \\sin(\\phi) \\sin(\\theta) \\\\\n z &= 5 \\cos(\\phi)\n \\end{align*}\n \\]\n\n where $0 \\leq x \\leq \\pi$, $0 \\leq \\phi \\leq \\pi$\n\n- This means that we can rewrite $\\mathbf{F}$ as\n \n \\[\n \\mathbf{F}(\\mathbf{r}(\\phi, \\theta)) = (5 \\sin(\\phi) \\cos(\\theta))(5 \\cos(\\phi))\\mathbf{i} + (5 \\sin(\\phi) \\cos(\\theta))\\mathbf{j} + (5 \\sin(\\phi) \\sin(\\theta))\\mathbf{k}\n \\]\n\n- Evaluating\n \n \\[\n \\mathbf{r}_\\phi \\times \\mathbf{r}_\\theta = 25 \\sin^2(\\phi) \\cos(\\theta)\\mathbf{i} + 25 \\sin^2(\\phi) \\sin(\\theta)\\mathbf{j} + 25 \\cos(\\phi) \\sin(\\phi)\\mathbf{k}\n \\]\n\n- This means that\n \n \\[\n \\mathbf{F}(\\mathbf{r}(\\phi, \\theta)) \\cdot (\\mathbf{r}_\\phi \\times \\mathbf{r}_\\theta) = 625 \\sin^3(\\phi) \\cos(\\phi) \\cos^2(\\theta) + 125 \\sin^3(\\phi) \\cos(\\theta) \\sin(\\theta) + 125 \\sin^2(\\phi) \\cos(\\phi) \\sin(\\theta)\n \\]\n\n- Proceeding with the substitution\n \n \\[\n \\int \\int_S \\mathbf{F} \\cdot d\\mathbf{S} = \\int \\int_D \\mathbf{F}(\\mathbf{r}(\\phi, \\theta)) \\cdot (\\mathbf{r}_\\phi \\times \\mathbf{r}_\\theta) dA\n \\]\n\n \\[\n = \\int_0^\\pi \\int_0^\\pi (625 \\sin^3(\\phi) \\cos(\\phi) \\cos^2(\\theta) + 125 \\sin^3(\\phi) \\cos(\\theta) \\sin(\\theta) + 125 \\sin^2(\\phi) \\cos(\\phi) \\sin(\\theta)) d\\theta d\\phi\n \\]\n\n- Now is just to solve the double integral.\n- At the end you should get: 0", "id": "./materials/451.pdf" }, { "contents": "Find the flux of $\\mathbf{F}$ across $S$, where $\\mathbf{F}(x, y, z) = y\\mathbf{j} - z\\mathbf{k}$, where $S$ consists of the paraboloid $y = x^2 + z^2$, $0 \\leq y \\leq 1$, and the disk $x^2 + z^2 \\leq 1$, $y = 1$.\n\n- Let $S_1$ be the paraboloid and $S_2$ the disk. Since $S$ is a closed surface, we use the outward orientation.\n\n- This means that to evaluate the surface integral you were asked, you will need to divide it\n\n$$\\iint_S \\mathbf{F} \\cdot d\\mathbf{S} = \\iint_{S_1} \\mathbf{F} \\cdot d\\mathbf{S} + \\iint_{S_2} \\mathbf{F} \\cdot d\\mathbf{S}$$\n\n- Let’s assess the situation for $S_1$, for instance.\n\n- We can write\n\n$$\\mathbf{F}(\\mathbf{r}(x, z)) = (x^2 + z^2)\\mathbf{j} - z\\mathbf{k}$$\n\nand, keeping in mind that the $\\mathbf{j}$-component must be negative on $S_1$,\n\n$$\\mathbf{r}_x \\times \\mathbf{r}_z = 2x\\mathbf{i} - \\mathbf{j} + 2z\\mathbf{k}$$\n\n- Then, we can proceed to the substitution\n\n$$\\iint_{S_1} \\mathbf{F} \\cdot d\\mathbf{S} = \\iint_{x^2 + z^2 \\leq 1} \\left[ -(x^2 + y^2) - 2z^2 \\right] dA$$\n\n$$= -\\int_0^{2\\pi} \\int_0^1 (r^2 + 2r^2 \\sin^2(\\theta))r \\, dr \\, d\\theta$$\n\n- When concluding this double integral, you should get: $-\\pi$\n• At last, it is just to repeat the same process for $S_2$.\n\n• At the end you should get: 0", "id": "./materials/452.pdf" }, { "contents": "Find the flux of $\\mathbf{F}$ across $S$, where $\\mathbf{F}(x, y, z) = xy\\mathbf{i} + 4x^2\\mathbf{j} + yz\\mathbf{k}$.\n\n$S$ is the surface $z = xe^y$, where $0 \\leq x \\leq 1$, $0 \\leq y \\leq 1$, with upward orientation.\n\n- We know that\n \n $$\\mathbf{F}(x, y, z) = xy\\mathbf{i} + 4x^2\\mathbf{j} + yz\\mathbf{k}$$\n\n as well as\n \n $$z = g(x, y) = xe^y$$\n\n and $D$ is the square $[0, 1] \\times [0, 1]$.\n\n- Proceeding with the substitution\n\n $$\\iint_S \\mathbf{F} \\cdot d\\mathbf{S} = \\iint_D \\left( -P \\frac{\\partial g}{\\partial x} - Q \\frac{\\partial g}{\\partial y} + R \\right) dA$$\n\n $$= \\iint_D \\left[ -xy(e^y) - 4x^2(xe^y) + yz \\right] dA$$\n\n $$= \\int_0^1 \\int_0^1 (-xye^y - 4x^3e^y + xy e^y) dy dx$$\n\n- Now is just to solve the double integral.\n\n- At the end you should get: $1 - e$", "id": "./materials/453.pdf" }, { "contents": "Find the flux of $\\mathbf{F}$ across $S$, where $\\mathbf{F}(x, y, z) = xi + 2yj + 3zk$, where $S$ is the cube with vertices $(\\pm 1, \\pm 1, \\pm 1)$.\n\n- Here $S$ consists of the six faces of the cube: $S_1, S_2, S_3, S_4, S_5, S_6$.\n- This means that to evaluate the surface integral you were asked, you will need to divide it\n\n$$\\iint_S \\mathbf{F} \\cdot d\\mathbf{S} = \\iint_{S_1} \\mathbf{F} \\cdot d\\mathbf{S} + \\iint_{S_2} \\mathbf{F} \\cdot d\\mathbf{S} + \\iint_{S_3} \\mathbf{F} \\cdot d\\mathbf{S} + \\iint_{S_4} \\mathbf{F} \\cdot d\\mathbf{S} + \\iint_{S_5} \\mathbf{F} \\cdot d\\mathbf{S} + \\iint_{S_6} \\mathbf{F} \\cdot d\\mathbf{S}$$\n\n- Let’s assess the situation for $S_1$, considered co-planar with the plane $x = 1$, for instance.\n- We can write\n\n$$\\mathbf{F} = i + 2yj + 3zk$$\n\nand\n\n$$\\mathbf{r}_y \\times \\mathbf{r}_z = i$$\n\n- Then, we can proceed to the substitution\n\n$$\\iint_{S_1} \\mathbf{F} \\cdot d\\mathbf{S} = \\int_{-1}^{1} \\int_{-1}^{1} 1 \\, dy \\, dz$$\n\n- When concluding this double integral, you should get: 4\n- At last, it is just to repeat the same process for $S_2, S_3, S_4, S_5, S_6$.\n- At the end you should get: 48", "id": "./materials/454.pdf" }, { "contents": "Find the flux of $\\mathbf{F}$ across $S$, where $\\mathbf{F}(x, y, z) = xi + yj + 5k$, where $S$ is the boundary of the region enclosed by the cylinder $x^2 + z^2 = 1$ and the planes $y = 0$ and $x + y = 2$.\n\n- Here $S$ consists of the three surfaces: $S_1$, the lateral surface of the cylinder; $S_2$, the front formed by the plane $x + y = 2$; and the back, $S_3$, in the plane $y = 0$.\n\n- This means that to evaluate the surface integral you were asked, you will need to divide it\n\n$$\\iint_S \\mathbf{F} \\cdot d\\mathbf{S} = \\iint_{S_1} \\mathbf{F} \\cdot d\\mathbf{S} + \\iint_{S_2} \\mathbf{F} \\cdot d\\mathbf{S} + \\iint_{S_3} \\mathbf{F} \\cdot d\\mathbf{S}$$\n\n- Let’s assess the situation for $S_1$, for instance.\n\n- We can write\n\n$$\\mathbf{F}(\\mathbf{r}(\\theta, y)) = \\sin(\\theta)i + yj + 5k$$\n\nand\n\n$$\\mathbf{r}_\\theta \\times \\mathbf{r}_y = \\sin(\\theta)i + \\cos(\\theta)k$$\n\n- Then, we can proceed to the substitution\n\n$$\\iint_{S_1} \\mathbf{F} \\cdot d\\mathbf{S} = \\iint_D \\mathbf{F} \\cdot (\\mathbf{r}_\\theta \\times \\mathbf{r}_y) \\, dA$$\n\n$$= \\int_0^{2\\pi} \\int_0^{2-\\sin(\\theta)} (\\sin^2(\\theta) + 5 \\cos(\\theta)) \\, dy \\, d\\theta$$\n• When concluding this double integral, you should get: $2\\pi$\n\n• At last, it is just to repeat the same process for $S_2, S_3$.\n\n• At the end you should get: $4\\pi$", "id": "./materials/455.pdf" }, { "contents": "Find the flux of $\\mathbf{F}$ across $S$, where $\\mathbf{F}(x, y, z) = x^2 \\mathbf{i} + y^2 \\mathbf{j} + z^2 \\mathbf{k}$, where $S$ is the boundary of the solid half-cylinder $0 \\leq z \\leq \\sqrt{1 - y^2}$, $0 \\leq x \\leq 2$.\n\n- Here $S$ consists of the four surfaces: $S_1$, the top surface (a portion of the circular cylinder $y^2 + z^2 = 1$); $S_2$, the bottom surface (a portion of the xy-plane); $S_3$, the front half-disk in the plane $x = 2$; and $S_4$, the back half-disk in the plane $x = 0$.\n\n- This means that to evaluate the surface integral you were asked, you will need to divide it\n\n$$\\iint_S \\mathbf{F} \\cdot d\\mathbf{S} = \\iint_{S_1} \\mathbf{F} \\cdot d\\mathbf{S} + \\iint_{S_2} \\mathbf{F} \\cdot d\\mathbf{S} + \\iint_{S_3} \\mathbf{F} \\cdot d\\mathbf{S} + \\iint_{S_4} \\mathbf{F} \\cdot d\\mathbf{S}$$\n\n- Let’s assess the situation for $S_1$, for instance.\n\n- We know that $S_1$ is the surface with upward orientation where\n\n$$\\mathbf{F}(x, y, z) = x^2 \\mathbf{i} + y^2 \\mathbf{j} + z^2 \\mathbf{k}$$\n\nas well as\n\n$$z = g(x, y) = \\sqrt{1 - y^2} \\quad , \\quad 0 \\leq x \\leq 2 \\quad , \\quad -1 \\leq y \\leq 1$$\n\n- Proceeding with the substitution\n\n$$\\iint_S \\mathbf{F} \\cdot d\\mathbf{S} = \\iint_D \\left( - P \\frac{\\partial g}{\\partial x} - Q \\frac{\\partial g}{\\partial y} + R \\right) dA$$\n\n$$= \\int_0^2 \\int_{-1}^1 \\left[ - x^2(0) - y^2 \\left( - \\frac{y}{\\sqrt{1 - y^2}} \\right) + z^2 \\right] dydx$$\n• When concluding this double integral, you should get: \\( \\frac{8}{3} \\)\n\n• At last, it is just to repeat the same process for \\( S_2, S_3, S_4 \\).\n\n• At the end you should get: \\( 2\\pi + \\frac{8}{3} \\)", "id": "./materials/456.pdf" }, { "contents": "Find the flux of $\\mathbf{F}$ across $S$, where $\\mathbf{F}(x, y, z) = yi + (z - y)j + xk$, where $S$ is the surface of the tetrahedron with vertices $(0, 0, 0), (1, 0, 0), (0, 1, 0), (0, 0, 1)$.\n\n- Here $S$ consists of the four surfaces: $S_1$, the triangular face with vertices $(1, 0, 0), (0, 1, 0), (0, 0, 1)$; $S_2$, the face of the tetrahedron in the $xy$-plane; $S_3$, the face in the $xz$-plane; and $S_4$, the face in the $yz$-plane.\n\n- This means that to evaluate the surface integral you were asked, you will need to divide it\n\n$$\\iint_S \\mathbf{F} \\cdot d\\mathbf{S} = \\iint_{S_1} \\mathbf{F} \\cdot d\\mathbf{S} + \\iint_{S_2} \\mathbf{F} \\cdot d\\mathbf{S} + \\iint_{S_3} \\mathbf{F} \\cdot d\\mathbf{S} + \\iint_{S_4} \\mathbf{F} \\cdot d\\mathbf{S}$$\n\n- Let’s assess the situation for $S_1$, for instance.\n\n- We know $S_1$ is the surface with upward orientation\n\n$$\\mathbf{F}(x, y, z) = yi + (z - y)j + xk$$\n\nas well as\n\n$$z = g(x, y) = 1 - x - y \\quad , \\quad 0 \\leq x \\leq 1 \\quad , \\quad 0 \\leq y \\leq 1$$\n\n- Proceeding with the substitution\n\n$$\\iint_S \\mathbf{F} \\cdot d\\mathbf{S} = \\iint_D \\left( - P \\frac{\\partial g}{\\partial x} - Q \\frac{\\partial g}{\\partial y} + R \\right) dA$$\n\n$$= \\int_0^1 \\int_0^{1-x} \\left[ - y(-1) - (z - y)(-1) + x \\right] dydx$$\n• When concluding this double integral, you should get: \\( \\frac{1}{3} \\)\n\n• At last, it is just to repeat the same process for \\( S_2, S_3, S_4 \\).\n\n• At the end you should get: \\( -\\frac{1}{6} \\)", "id": "./materials/457.pdf" }, { "contents": "Evaluate the surface integral \\( \\iint_S yz \\, dS \\), where \\( S \\) is part of the plane \\( x + y + z = 1 \\) that lies in the first octant.\n\n- \\( S \\) is part of the plane \\( z = 1 - x - y \\) over the region\n\n\\[\nD = \\{(x, y) \\mid 0 \\leq x \\leq 1, \\ 0 \\leq y \\leq 1 - x\\}\n\\]\nThis means that\n\n\\[\n\\iint_S yz \\, dS = \\iint_D y(1 - x - y)\\sqrt{(-1)^2 + (-1)^2 + 1} \\, dA\n\\]\n\n\\[\n= \\sqrt{3} \\int_0^1 \\int_0^{1-x} (y - xy - y^2) \\, dy \\, dx\n\\]\n\n\\[\n= \\sqrt{3} \\int_0^1 \\left[ \\frac{(1 - x)y^2}{2} - \\frac{y^3}{3} \\right]_{y=0}^{y=1-x} \\, dx\n\\]\n\n\\[\n= \\sqrt{3} \\int_0^1 \\frac{(1 - x)^3}{2} - \\frac{(1 - x)^3}{3} \\, dx\n\\]\n\n\\[\n= \\sqrt{3} \\int_0^1 \\frac{(1 - x)^3}{6} \\, dx\n\\]\n\n\\[\n= -\\frac{\\sqrt{3}}{24} \\int_0^1 -4(1 - x)^3 \\, dx\n\\]\n\n\\[\n= -\\frac{\\sqrt{3}}{24} \\left[ (1 - x)^4 \\right]_{x=0}^{x=1}\n\\]\n\n\\[\n= -\\frac{\\sqrt{3}}{24} (0 - 1)\n\\]\n\n\\[\n= \\frac{\\sqrt{3}}{24}\n\\]", "id": "./materials/458.pdf" }, { "contents": "Evaluate \\( \\iint_S yz \\, dS \\), where \\( S \\) is the surface with parametric equations\n\\[\nx = u^2, \\quad y = u \\sin(v), \\quad z = u \\cos(v), \\quad \\text{where} \\quad 0 \\leq u \\leq 1, \\quad 0 \\leq v \\leq \\frac{\\pi}{2}.\n\\]\n\n- We can define\n\\[\n\\mathbf{r}(u, v) = u^2 \\mathbf{i} + u \\sin(v) \\mathbf{j} + u \\cos(v) \\mathbf{k}\n\\]\nwhere\n\\[\n0 \\leq u \\leq 1, \\quad 0 \\leq v \\leq \\frac{\\pi}{2}\n\\]\n- This means that\n\\[\n\\mathbf{r}_u \\times \\mathbf{r}_v = \\begin{vmatrix}\n\\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\\n2u \\sin(v) & \\cos(v) & 0 \\\\\n0 & u \\cos(v) & -u \\sin(v)\n\\end{vmatrix} = -u \\mathbf{i} + 2u^2 \\sin(v) \\mathbf{j} + 2u^2 \\cos(v) \\mathbf{k}\n\\]\nwhich leads us to\n\\[\n|\\mathbf{r}_u \\times \\mathbf{r}_v| = \\sqrt{u^2 + 4u^4 \\sin^2(v) + 4u^4 \\cos^2(v)} = u \\sqrt{1 + 4u^4}, \\quad u \\geq 0\n\\]\n- This means that\n\\[\n\\iint_S yz \\, dS = \\iint_D (u \\sin(v))(u \\cos(v)) |\\mathbf{r}_u \\times \\mathbf{r}_v| \\, dA\n\\]\n\\[\n= \\int_0^{\\frac{\\pi}{2}} \\int_0^1 (u \\sin(v))(u \\cos(v)) \\cdot u \\sqrt{1 + 4u^4} \\, du \\, dv\n\\]\n- Now is just to solve the double integral.\n- At the end you should get: \\( \\frac{5\\sqrt{5}}{48} + \\frac{1}{240} \\)", "id": "./materials/459.pdf" }, { "contents": "Evaluate $\\int \\int_S x \\, dS$, where $S$ is the surface defined by $y = x^2$, $0 \\leq x \\leq 2$, $0 \\leq z \\leq 3$.\n\n- Let the parametrization be\n \\[ r(x, z) = xi + x^2j + zk \\]\n which means that we can define\n \\[ r_x = i + 2xj \\]\n \\[ r_z = k \\]\n \\[ \\Rightarrow r_x \\times r_z = \\begin{vmatrix} i & j & k \\\\ 1 & 2x & 0 \\\\ 0 & 0 & 1 \\end{vmatrix} = 2xi + j \\]\n \\[ \\Rightarrow |r_x \\times r_z| = \\sqrt{4x^2 + 1} \\]\n\n- Proceeding to the transformation, we get that\n \\[ \\int \\int_S x \\, dS = \\int \\int_D x \\cdot |r_x \\times r_z| \\, dA \\]\n \\[ = \\int_0^3 \\int_0^2 x \\sqrt{4x^2 + 1} \\, dx \\, dz \\]\n\n- Now is just to solve the double integral.\n\n- At the end you should get: $\\frac{17\\sqrt{17} - 1}{4}$", "id": "./materials/460.pdf" }, { "contents": "Evaluate $\\int \\int_S z \\, dS$, where $S$ is the surface defined by $y^2 + z^2 = 4$, $0 \\leq z$, $1 \\leq x \\leq 4$.\n\n- Let the parametrization be\n \\[ r(x, y) = xi + yj + \\sqrt{4 - y^2}k, \\quad -2 \\leq y \\leq 2 \\]\n which means that we can define\n \\[ r_x = i \\]\n \\[ r_y = j - \\frac{y}{\\sqrt{4 - y^2}}k \\]\n \\[ \\Rightarrow r_x \\times r_y = \\begin{vmatrix} i & j & k \\\\ 1 & 0 & 0 \\\\ 0 & 1 & -\\frac{y}{\\sqrt{4 - y^2}} \\end{vmatrix} = \\frac{y}{\\sqrt{4 - y^2}}j + k \\]\n \\[ \\Rightarrow |r_x \\times r_y| = \\sqrt{\\left(\\frac{y}{\\sqrt{4 - y^2}}\\right)^2 + 1^2} = \\frac{2}{\\sqrt{4 - y^2}} \\]\n\n- Proceeding to the transformation, we get that\n \\[ \\int \\int_S z \\, dS = \\int \\int_D z \\cdot |r_x \\times r_y| \\, dA \\]\n \\[ = \\int_1^4 \\int_{-2}^2 \\sqrt{4 - y^2} \\cdot \\frac{2}{\\sqrt{4 - y^2}} \\, dy \\, dx \\]\n• Now is just to solve the double integral.\n\n• At the end you should get: 24", "id": "./materials/461.pdf" }, { "contents": "Evaluate \\( \\int_S z^2 \\, dS \\), where \\( S \\) is the hemisphere defined by \\( x^2 + y^2 + z^2 = a^2 \\), \\( z \\geq 0 \\).\n\n- Using parametric representation, we can define:\n \\[\n x = a \\sin(\\phi) \\cos(\\theta), \\quad y = a \\sin(\\phi) \\sin(\\theta), \\quad z = a \\cos(\\phi), \\quad 0 \\leq \\phi \\leq \\frac{\\pi}{2}, \\quad 0 \\leq \\theta \\leq 2\\pi\n \\]\n\n- This means that we can also define\n \\[\n \\mathbf{r}(\\phi, \\theta) = a \\sin(\\phi) \\cos(\\theta) \\mathbf{i} + a \\sin(\\phi) \\sin(\\theta) \\mathbf{j} + a \\cos(\\phi) \\mathbf{k}\n \\]\n\n- Evaluating \\( \\mathbf{r}_\\phi \\) and \\( \\mathbf{r}_\\theta \\)\n \\[\n \\mathbf{r}_\\phi = a \\cos(\\phi) \\cos(\\theta) \\mathbf{i} + a \\cos(\\phi) \\sin(\\theta) \\mathbf{j} - a \\sin(\\phi) \\mathbf{k}\n \\]\n \\[\n \\mathbf{r}_\\theta = -a \\sin(\\phi) \\sin(\\theta) \\mathbf{i} + a \\sin(\\phi) \\cos(\\theta) \\mathbf{j}\n \\]\n\n- Proceeding with some calculations we get\n \\[\n \\mathbf{r}_\\phi \\times \\mathbf{r}_\\theta = \\begin{vmatrix}\n \\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\\n a \\cos(\\phi) \\cos(\\theta) & +a \\cos(\\phi) \\sin(\\theta) & -a \\sin(\\phi) \\\\\n -a \\sin(\\phi) \\sin(\\theta) & a \\sin(\\phi) \\cos(\\theta) & 0\n \\end{vmatrix}\n = a^2 \\sin^2(\\phi) \\cos(\\theta) \\mathbf{i} + a^2 \\sin^2(\\phi) \\sin(\\theta) \\mathbf{j} + a^2 \\sin(\\phi) \\cos(\\phi) \\mathbf{k}\n \\]\n and also\n \\[\n |\\mathbf{r}_\\phi \\times \\mathbf{r}_\\theta| = \\sqrt{a^4 \\sin^4(\\phi) \\cos^2(\\theta) + a^4 \\sin^4(\\phi) \\sin^2(\\theta) + a^4 \\sin^2(\\phi) \\cos^2(\\phi)}\n = a^2 \\sin(\\phi)\n \\]\n• Proceeding to the transformation, we may write that\n\n\\[\n\\int \\int_S z^2 \\, dS = \\int \\int_D (a \\cos(\\phi))^2 |r_\\phi \\times r_\\theta| \\, dA = \\int_0^{2\\pi} \\int_0^\\pi (a^2 \\cos^2(\\phi))(a^2 \\sin(\\phi)) \\, d\\phi d\\theta\n\\]\n\n• Now is just to solve the double integral.\n\n• At the end you should get: \\( \\frac{2a^4\\pi}{3} \\)", "id": "./materials/462.pdf" }, { "contents": "Evaluate \\( \\iint_S z \\, dS \\), where \\( S \\) is the portion of the plane \\( x + y + z = 4 \\) that lies above the square \\( 0 \\leq x \\leq 1, \\ 0 \\leq y \\leq 1 \\), in the xy-plane.\n\n- Let the parametrization be\n \\[\n \\mathbf{r}(x, y) = x \\mathbf{i} + y \\mathbf{j} + (4 - x - y) \\mathbf{k}\n \\]\n which means that we can define\n \\[\n \\mathbf{r}_x = \\mathbf{i} - \\mathbf{k} \\\\\n \\mathbf{r}_y = \\mathbf{j} - \\mathbf{k}\n \\]\n \\[\n \\Rightarrow \\mathbf{r}_x \\times \\mathbf{r}_y = \\begin{vmatrix}\n \\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\\n 1 & 0 & -1 \\\\\n 0 & 1 & -1\n \\end{vmatrix} = \\mathbf{i} + \\mathbf{j} + \\mathbf{k}\n \\]\n \\[\n \\Rightarrow |\\mathbf{r}_x \\times \\mathbf{r}_y| = \\sqrt{3}\n \\]\n\n- Proceeding to the transformation, we get that\n \\[\n \\iint_S z \\, dS = \\iint_D z \\cdot |\\mathbf{r}_x \\times \\mathbf{r}_y| \\, dA\n \\]\n \\[\n = \\int_0^1 \\int_0^1 (4 - x - y) \\sqrt{3} \\, dy \\, dx\n \\]\n \\[\n = 3 \\sqrt{3}\n \\]\n\n- Now is just to solve the double integral.\n- At the end you should get: \\( 3 \\sqrt{3} \\)", "id": "./materials/463.pdf" }, { "contents": "Evaluate the surface integral \\( \\iint_S (z - x) \\, dS \\), where \\( S \\) defined by\n\\[\nz = \\sqrt{x^2 + y^2}, \\text{ where } 0 \\leq z \\leq 1.\n\\]\n\n- Let the parametrization be\n\\[\nr(r, \\theta) = r \\cos(\\theta)i + r \\sin(\\theta)j + rk\n\\]\nwhere\n\\[\n0 \\leq r \\leq 1, \\quad 0 \\leq \\theta \\leq 2\\pi\n\\]\n- Evaluating \\( r_r \\) and \\( r_\\theta \\)\n\\[\nr_r = \\cos(\\theta)i + \\sin(\\theta)j + k\n\\]\n\\[\nr_\\theta = -r \\sin(\\theta)i + r \\cos(\\theta)j\n\\]\n- Proceeding with some calculations we get\n\\[\nr_r \\times r_\\theta = \\begin{vmatrix} i & j & k \\\\ \\cos(\\theta) & \\sin(\\theta) & 1 \\\\ -r \\sin(\\theta) & r \\cos(\\theta) & 0 \\end{vmatrix}\n= -r \\cos(\\theta)i - r \\sin(\\theta)j + rk\n\\]\nand also\n\\[\n|r_r \\times r_\\theta| = \\sqrt{(-r \\cos(\\theta))^2 + (-r \\sin(\\theta))^2 + r^2}\n= r \\sqrt{2}\n\\]\n• This means that we can proceed to the transformation\n\n\\[\n\\iint_S (z - x) \\, dS = \\iint_D (z - x) \\cdot |r_r \\times r_\\theta| \\, dA = \\int_0^{2\\pi} \\int_0^1 (r - r \\cos(\\theta))(r\\sqrt{2}) \\, dr \\, d\\theta\n\\]\n\n• Now is just to solve the double integral.\n\n• At the end you should get: \\( \\frac{2\\pi \\sqrt{2}}{3} \\)", "id": "./materials/464.pdf" }, { "contents": "Evaluate the surface integral \\( \\iint_S (x^2 \\sqrt{5 - 4z}) \\, dS \\), where \\( S \\) defined by \n\\[ z = 1 - x^2 - y^2, \\text{ where } z \\geq 0. \\]\n\n- Let the parametrization be \n\\[ \\mathbf{r}(r, \\theta) = r \\cos(\\theta) \\mathbf{i} + r \\sin(\\theta) \\mathbf{j} + (1 - r^2) \\mathbf{k} \\]\nwhere \n\\[ 0 \\leq r \\leq 1, \\quad 0 \\leq \\theta \\leq 2\\pi \\]\n\n- Evaluating \\( \\mathbf{r}_r \\) and \\( \\mathbf{r}_\\theta \\)\n\\[ \\mathbf{r}_r = \\cos(\\theta) \\mathbf{i} + \\sin(\\theta) \\mathbf{j} - 2r \\mathbf{k} \\]\n\\[ \\mathbf{r}_\\theta = -r \\sin(\\theta) \\mathbf{i} + r \\cos(\\theta) \\mathbf{j} \\]\n\n- Proceeding with some calculations we get\n\\[ \\mathbf{r}_r \\times \\mathbf{r}_\\theta = \\begin{vmatrix} \\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\ \\cos(\\theta) & \\sin(\\theta) & -2r \\\\ -r \\sin(\\theta) & r \\cos(\\theta) & 0 \\end{vmatrix} = 2r^2 \\cos(\\theta) \\mathbf{i} + 2r^2 \\sin(\\theta) \\mathbf{j} + r \\mathbf{k} \\]\nand also\n\\[ |\\mathbf{r}_r \\times \\mathbf{r}_\\theta| = \\sqrt{(2r^2 \\cos(\\theta))^2 + (2r^2 \\sin(\\theta))^2 + r^2} = r \\sqrt{1 + 4r^2} \\]\n• This means that we can proceed to the transformation\n\n\\[\n\\iint_S (x^2 \\sqrt{5 - 4z}) \\, dS\n\\]\n\n\\[\n= \\iint_D (x^2 \\sqrt{5 - 4z}) \\cdot |r_r \\times r_\\theta| \\, dA\n\\]\n\n\\[\n= \\int_0^{2\\pi} \\int_0^1 (r^2 \\cos^2(\\theta)) \\left( \\sqrt{1 + 4r^2} \\right) \\left( r \\sqrt{1 + 4r^2} \\right) \\, dr \\, d\\theta\n\\]\n\n• Now is just to solve the double integral.\n\n• At the end you should get: \\( \\frac{11\\pi}{12} \\)", "id": "./materials/465.pdf" }, { "contents": "Evaluate \\( \\int \\int_S yz \\, dS \\), where \\( S \\) is the part of the sphere \\( x^2 + y^2 + z^2 = 4 \\), that lies above the cone \\( z = \\sqrt{x^2 + y^2} \\).\n\n- Using parametric representation, we can define:\n \\[\n x = 2 \\sin(\\phi) \\cos(\\theta), \\quad y = 2 \\sin(\\phi) \\sin(\\theta), \\quad z = 2 \\cos(\\phi)\n \\]\n\n- Finding the intersection between the two surfaces, we get that\n \\[\n z = \\sqrt{x^2 + y^2} \\Rightarrow z^2 + z^2 = 4 \\Rightarrow z = \\sqrt{2} \\Rightarrow 2 \\cos(\\phi) = \\sqrt{2} \\Rightarrow \\phi = \\frac{\\pi}{4}\n \\]\n meaning that we can define the intervals\n \\[\n 0 \\leq \\phi \\leq \\frac{\\pi}{4}, \\quad 0 \\leq \\theta \\leq 2\\pi\n \\]\n\n- This means that we can also define\n \\[\n r(\\phi, \\theta) = 2 \\sin(\\phi) \\cos(\\theta) \\mathbf{i} + 2 \\sin(\\phi) \\sin(\\theta) \\mathbf{j} + 2 \\cos(\\phi) \\mathbf{k}\n \\]\n\n- Evaluating \\( r_\\phi \\) and \\( r_\\theta \\)\n \\[\n r_\\phi = 2 \\cos(\\phi) \\cos(\\theta) \\mathbf{i} + 2 \\cos(\\phi) \\sin(\\theta) \\mathbf{j} - 2 \\sin(\\phi) \\mathbf{k}\n \\]\n \\[\n r_\\theta = -2 \\sin(\\phi) \\sin(\\theta) \\mathbf{i} + 2 \\sin(\\phi) \\cos(\\theta) \\mathbf{j}\n \\]\n\n- Proceeding with some calculations we get\n \\[\n r_\\phi \\times r_\\theta = \\begin{vmatrix}\n \\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\\n 2 \\cos(\\phi) \\cos(\\theta) & +2 \\cos(\\phi) \\sin(\\theta) & -2 \\sin(\\phi) \\\\\n -2 \\sin(\\phi) \\sin(\\theta) & 2 \\sin(\\phi) \\cos(\\theta) & 0\n \\end{vmatrix}\n = 4 \\sin^2(\\phi) \\cos(\\theta) \\mathbf{i} + 4 \\sin^2(\\phi) \\sin(\\theta) \\mathbf{j} + 4 \\sin(\\phi) \\cos(\\phi) \\mathbf{k}\n \\]\n and also\n \\[\n |r_\\phi \\times r_\\theta| = \\sqrt{16 \\sin^4(\\phi) \\cos^2(\\theta) + 16 \\sin^4(\\phi) \\sin^2(\\theta) + 16 \\sin^2(\\phi) \\cos^2(\\phi)}\n = 4 \\sin(\\phi)\n \\]\n• Proceeding to the transformation, we may write that\n\n\\[\n\\int \\int_S yz \\, dS = \\int \\int_D (yz) \\cdot |r_\\phi \\times r_\\theta| \\, dA = \\int_0^{2\\pi} \\int_0^\\pi (4 \\cos(\\phi) \\sin(\\phi))(4 \\sin(\\phi)) \\, d\\phi d\\theta\n\\]\n\n• Now is just to solve the double integral.\n\n• At the end you should get: 0", "id": "./materials/466.pdf" }, { "contents": "Find the flux of the vector field \\( \\mathbf{F}(x, y, z) = z^2 \\mathbf{i} + x \\mathbf{j} - 3z \\mathbf{k} \\) across the surface cut from the parabolic cylinder \\( z = 4 - y^2 \\) by the planes \\( x = 0, x = 1, z = 0 \\), with outward orientation.\n\n- Let the parametrization be\n \\[\n \\mathbf{r}(x, y) = x \\mathbf{i} + y \\mathbf{j} + (4 - y^2) \\mathbf{k}\n \\]\n\n- We can also define the interval we are working with\n \\[\n 0 \\leq x \\leq 1 \\\\\n z = 0 \\Rightarrow 0 = 4 - y^2 \\Rightarrow y = \\pm 2\n \\]\n\n- Evaluating \\( \\mathbf{r}_x \\times \\mathbf{r}_y \\):\n \\[\n \\mathbf{r}_x = \\mathbf{i} \\\\\n \\mathbf{r}_y = \\mathbf{j} - 2y \\mathbf{k}\n \\]\n \\[\n \\Rightarrow \\mathbf{r}_x \\times \\mathbf{r}_y = 2y \\mathbf{j} + \\mathbf{k}\n \\]\n\n- Then\n \\[\n \\mathbf{F}(\\mathbf{r}(x, y)) = (4 - y^2)^2 \\mathbf{i} + x \\mathbf{j} - 3(4 - y^2) \\mathbf{k}\n \\]\n\n- Therefore\n \\[\n \\mathbf{F}(\\mathbf{r}(x, y)) \\cdot (\\mathbf{r}_x \\times \\mathbf{r}_y) = 2xy - 3(4 - y^2)\n \\]\n\n- Knowing that a flux is defined as\n \\[\n \\iint_S \\mathbf{F} \\cdot d\\mathbf{S} = \\iint_D \\mathbf{F} \\cdot (\\mathbf{r}_x \\times \\mathbf{r}_y) \\, dA\n \\]\nwe can proceed to the substitution\n\n\\[\n\\int \\int_S \\mathbf{F} \\cdot d\\mathbf{S} = \\int_0^1 \\int_{-2}^2 (2xy - 3(4 - y^2)) \\, dy \\, dx\n\\]\n\n• Now is just to solve the double integral.\n\n• At the end you should get: \\(-32\\)", "id": "./materials/467.pdf" }, { "contents": "Find the flux of the vector field \\( \\mathbf{F}(x, y, z) = x^2 \\mathbf{j} - xz \\mathbf{k} \\) across the surface cut from the parabolic cylinder \\( y = x^2, -1 \\leq x \\leq 1 \\), by the planes \\( z = 2 \\) and \\( z = 0 \\), with outward orientation.\n\n- Let the parametrization be\n \\[\n \\mathbf{r}(x, z) = xi + x^2j + zk\n \\]\n\n- We can also define the interval we are working with\n \\[\n -1 \\leq x \\leq 1 \\\\\n 0 \\leq z \\leq 2\n \\]\n\n- Evaluating \\( \\mathbf{r}_x \\times \\mathbf{r}_z \\):\n \\[\n \\mathbf{r}_x = i + 2xj \\\\\n \\mathbf{r}_z = k\n \\]\n \\[\n \\Rightarrow \\mathbf{r}_x \\times \\mathbf{r}_z = 2xi - j\n \\]\n\n- Then\n \\[\n \\mathbf{F}(\\mathbf{r}(x, z)) = x^2 \\mathbf{j} - xz \\mathbf{k}\n \\]\n\n- Therefore\n \\[\n \\mathbf{F}(\\mathbf{r}(x, z)) \\cdot (\\mathbf{r}_x \\times \\mathbf{r}_z) = -x^2\n \\]\n\n- Knowing that a flux is defined as\n \\[\n \\iint_S \\mathbf{F} \\cdot d\\mathbf{S} = \\iint_D \\mathbf{F} \\cdot (\\mathbf{r}_x \\times \\mathbf{r}_z) \\, dA\n \\]\nwe can proceed to the substitution\n\n\\[\n\\iint_S \\mathbf{F} \\cdot d\\mathbf{S} = \\int_{-1}^{1} \\int_{0}^{2} (-x^2) \\, dz \\, dx\n\\]\n\n• Now is just to solve the double integral.\n\n• At the end you should get: \\(-\\frac{4}{3}\\)", "id": "./materials/468.pdf" }, { "contents": "Find the flux of the vector field \\( \\mathbf{F}(x, y, z) = x \\mathbf{i} + y \\mathbf{j} + z \\mathbf{k} \\) across the sphere \\( x^2 + y^2 + z^2 = a^2 \\) in the direction away from the origin.\n\n- Using parametric representation, we can define:\n \\[\n x = a \\sin(\\phi) \\cos(\\theta), \\quad y = a \\sin(\\phi) \\sin(\\theta), \\quad z = a \\cos(\\phi), \\quad 0 \\leq \\phi \\leq \\pi, \\quad 0 \\leq \\theta \\leq 2\\pi\n \\]\n\n- This means that we can also define\n \\[\n \\mathbf{r}(\\phi, \\theta) = a \\sin(\\phi) \\cos(\\theta) \\mathbf{i} + a \\sin(\\phi) \\sin(\\theta) \\mathbf{j} + a \\cos(\\phi) \\mathbf{k}\n \\]\n\n- Evaluating \\( \\mathbf{r}_\\phi \\) and \\( \\mathbf{r}_\\theta \\)\n \\[\n \\mathbf{r}_\\phi = a \\cos(\\phi) \\cos(\\theta) \\mathbf{i} + a \\cos(\\phi) \\sin(\\theta) \\mathbf{j} - a \\sin(\\phi) \\mathbf{k}\n \\]\n \\[\n \\mathbf{r}_\\theta = -a \\sin(\\phi) \\sin(\\theta) \\mathbf{i} + a \\sin(\\phi) \\cos(\\theta) \\mathbf{j}\n \\]\n\n- Proceeding with some calculations we get\n \\[\n \\mathbf{r}_\\phi \\times \\mathbf{r}_\\theta = \\begin{vmatrix}\n \\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\\n a \\cos(\\phi) \\cos(\\theta) & +a \\cos(\\phi) \\sin(\\theta) & -a \\sin(\\phi) \\\\\n -a \\sin(\\phi) \\sin(\\theta) & a \\sin(\\phi) \\cos(\\theta) & 0\n \\end{vmatrix}\n = a^2 \\sin^2(\\phi) \\cos(\\theta) \\mathbf{i} + a^2 \\sin^2(\\phi) \\sin(\\theta) \\mathbf{j} + a^2 \\sin(\\phi) \\cos(\\phi) \\mathbf{k}\n \\]\n\n- Then\n \\[\n \\mathbf{F}(\\mathbf{r}(\\phi, \\theta)) = (a \\sin(\\phi) \\cos(\\theta)) \\mathbf{i} + (a \\sin(\\phi) \\sin(\\theta)) \\mathbf{j} + (a \\cos(\\phi)) \\mathbf{k}\n \\]\n\n- Therefore\n \\[\n \\mathbf{F}(\\mathbf{r}(\\phi, \\theta)) \\cdot (\\mathbf{r}_\\phi \\times \\mathbf{r}_\\theta) = a^3 \\cos^2(\\theta) \\sin^3(\\phi) + a^3 \\sin^2(\\theta) \\sin^3(\\phi) + a^3 \\sin(\\phi) \\cos^2(\\phi)\n = a^3 \\sin(\\phi)\n \\]\n• Proceeding to the transformation, we may write that\n\n\\[\n\\iint_S \\mathbf{F} \\cdot d\\mathbf{S} = \\iint_D \\mathbf{F} \\cdot (\\mathbf{r}_\\phi \\times \\mathbf{r}_\\theta) \\, dA\n\\]\n\n\\[\n= \\iint_D a^3 \\sin(\\phi) \\, d\\theta d\\phi\n\\]\n\n\\[\n= \\int_0^{2\\pi} \\int_0^\\pi a^3 \\sin(\\phi) \\, d\\phi d\\theta\n\\]\n\n• Now is just to solve the double integral.\n\n• At the end you should get: \\(4\\pi a^3\\)", "id": "./materials/469.pdf" }, { "contents": "Manipulation of Algebraic expressions\n\nSolving Logarithmic Equations\nRemember!\n\n• When solving log equations, always check that each term has the same base. If this is not the case, the change of base rule must first be used to change to a common base.\n\n• If no base is given, the equation holds true for all bases.\n\n• If $\\log_a b = \\log_a c$, then $b = c$.\n\n• If $\\log_a b = k$, then $b = a^k$\n\n• Check all solutions to make sure they do not produce logs of negative numbers as these are not defined!\nThe Laws of Logarithms\n\n1. \\( \\log_a xy = \\log_a x + \\log_a y \\)\n\n2. \\( \\log_a \\left( \\frac{x}{y} \\right) = \\log_a x - \\log_a y \\)\n\n3. \\( \\log_a x^n = n \\log_a x \\)\n\n4. \\( \\log_a a = 1 \\)\n\n5. \\( \\log_a 1 = 0 \\)\n\n6. \\( \\log_a x = \\frac{\\log_b x}{\\log_b a} \\)\nWorked Example 1\nSolving Logarithmic Equations\n\n\\[ \\log_5 x - \\log_5 (10 - x) = 1 \\]\n\n- \\[ \\log_5 x - \\log_5 (10 - x) = 1 \\]\n- \\[ \\log_5 \\frac{x}{10-x} = 1 \\]\n- \\[ \\frac{x}{10-x} = 5^1 \\]\n- \\[ \\frac{x}{10-x} = 5 \\]\n\nThis law was applied here:\n\\[ \\log_a \\left( \\frac{x}{y} \\right) = \\log_a x - \\log_a y \\]\n\nCovert to indices\n\\[ \\log_a b = k, \\text{ then } b = a^k \\]\nSolving Logarithmic Equations\n\n➢ $5(10 - x) = x$\n➢ $50 - 5x = x$\n➢ $-x - 5x + 50 = 0$\n➢ $-6x = -50$\n➢ $6x = 50$\n➢ $x = \\frac{50}{6}$\n➢ $x = 8.33$\nWorked Example 2\nSolving Logarithmic Equations\n\nExample 2\n\n\\[ \\log_3 x + 3 \\log_x 3 = 4 \\]\n\n\\[ \\log_3 x = \\left( \\frac{\\log x}{\\log 3} \\right) = \\left( \\frac{1}{\\log 3} \\right) \\]\n\n\\[ \\left( \\frac{1}{\\log 3} \\right) + 3 \\log_x 3 = 4 \\]\n\nUsing the substitution \\( \\log_x 3 = y \\)\n\n\\[ \\frac{1}{y} + 3y = 4 \\]\n\nNote both terms have different bases! We need to change base 3 to base \\( x \\) or visa versa.\n\n\\( \\log_x x = 1 \\)\nSolving Logarithmic Equations\n\nExample 2 cont’d\n\n1 + 3y^2 = 4y\n3y^2 - 4y + 1 = 0\n(3y - 1)(y - 1) = 0\nTherefore,\ny = 1 or y = \\frac{1}{3}\nRemember we substituted \\log_x 3 = y\nSolving Logarithmic Equations\n\nExample 2: cont’d\n\n\\[ \\log_x 3 = y, \\text{ where } y \\text{ is equal to } 1 \\text{ and } \\frac{1}{3} \\]\n\n\\[ \\log_x 3 = 1 \\]\n\\[ x^1 = 3 \\]\n\\[ x = 3 \\]\n\n\\[ \\log_x 3 = \\frac{1}{3} \\]\n\\[ \\frac{1}{x^3} = 3 \\]\n\\[ x = 27 \\]\n\nBoth solutions give positive logs and thus are acceptable.\n\nNote this example could also be repeated using the base 3 instead of the base \\( x \\). The same results are achieved.", "id": "./materials/47.pdf" }, { "contents": "Find the flux of the vector field \\( \\mathbf{F}(x, y, z) = x \\mathbf{i} + y \\mathbf{j} + z \\mathbf{k} \\) across the sphere \\( x^2 + y^2 + z^2 = a^2 \\) in the direction away from the origin.\n\n- Using parametric representation, we can define:\n \\[\n x = a \\sin(\\phi) \\cos(\\theta), \\quad y = a \\sin(\\phi) \\sin(\\theta), \\quad z = a \\cos(\\phi), \\quad 0 \\leq \\phi \\leq \\pi, \\quad 0 \\leq \\theta \\leq 2\\pi\n \\]\n\n- This means that we can also define\n \\[\n \\mathbf{r}(\\phi, \\theta) = a \\sin(\\phi) \\cos(\\theta) \\mathbf{i} + a \\sin(\\phi) \\sin(\\theta) \\mathbf{j} + a \\cos(\\phi) \\mathbf{k}\n \\]\n\n- Evaluating \\( \\mathbf{r}_\\phi \\) and \\( \\mathbf{r}_\\theta \\)\n \\[\n \\mathbf{r}_\\phi = a \\cos(\\phi) \\cos(\\theta) \\mathbf{i} + a \\cos(\\phi) \\sin(\\theta) \\mathbf{j} - a \\sin(\\phi) \\mathbf{k}\n \\]\n \\[\n \\mathbf{r}_\\theta = -a \\sin(\\phi) \\sin(\\theta) \\mathbf{i} + a \\sin(\\phi) \\cos(\\theta) \\mathbf{j}\n \\]\n\n- Proceeding with some calculations we get\n \\[\n \\mathbf{r}_\\phi \\times \\mathbf{r}_\\theta = \\begin{vmatrix}\n \\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\\n a \\cos(\\phi) \\cos(\\theta) & +a \\cos(\\phi) \\sin(\\theta) & -a \\sin(\\phi) \\\\\n -a \\sin(\\phi) \\sin(\\theta) & a \\sin(\\phi) \\cos(\\theta) & 0\n \\end{vmatrix}\n = a^2 \\sin^2(\\phi) \\cos(\\theta) \\mathbf{i} + a^2 \\sin^2(\\phi) \\sin(\\theta) \\mathbf{j} + a^2 \\sin(\\phi) \\cos(\\phi) \\mathbf{k}\n \\]\n\n- Then\n \\[\n \\mathbf{F}(\\mathbf{r}(\\phi, \\theta)) = (a \\sin(\\phi) \\cos(\\theta)) \\mathbf{i} + (a \\sin(\\phi) \\sin(\\theta)) \\mathbf{j} + (a \\cos(\\phi)) \\mathbf{k}\n \\]\n\n- Therefore\n \\[\n \\mathbf{F}(\\mathbf{r}(\\phi, \\theta)) \\cdot (\\mathbf{r}_\\phi \\times \\mathbf{r}_\\theta) = a^3 \\cos^2(\\theta) \\sin^3(\\phi) + a^3 \\sin^2(\\theta) \\sin^3(\\phi) + a^3 \\sin(\\phi) \\cos^2(\\phi)\n = a^3 \\sin(\\phi)\n \\]\n• Proceeding to the transformation, we may write that\n\n\\[\n\\iint_S \\mathbf{F} \\cdot d\\mathbf{S} = \\iint_D \\mathbf{F} \\cdot (\\mathbf{r}_\\phi \\times \\mathbf{r}_\\theta) \\, dA = \\iint_D a^3 \\sin(\\phi) \\, d\\theta d\\phi = \\int_0^{2\\pi} \\int_0^\\pi a^3 \\sin(\\phi) \\, d\\phi d\\theta\n\\]\n\n• Now is just to solve the double integral.\n\n• At the end you should get: \\(4\\pi a^3\\)", "id": "./materials/470.pdf" }, { "contents": "Find the flux of the vector field \\( \\mathbf{F}(x, y, z) = 2xy \\mathbf{i} + 2yz \\mathbf{j} + 2xz \\mathbf{k} \\) across the plane \\( x + y + z = 2a \\) that lies above the square \\( 0 \\leq x \\leq a, 0 \\leq y \\leq 2a \\), in the xy-plane with upward orientation.\n\n- Let the parametrization be\n \\[\n \\mathbf{r}(x, y) = x \\mathbf{i} + y \\mathbf{j} + (2a - x - y) \\mathbf{k}\n \\]\n where\n \\[\n 0 \\leq x \\leq a, \\quad 0 \\leq y \\leq a\n \\]\n- Evaluating \\( \\mathbf{r}_x \\) and \\( \\mathbf{r}_y \\)\n \\[\n \\mathbf{r}_x = \\mathbf{i} - \\mathbf{k}\n \\]\n \\[\n \\mathbf{r}_y = \\mathbf{j} - \\mathbf{k}\n \\]\n \\[\n \\Rightarrow \\mathbf{r}_x \\times \\mathbf{r}_y = \\begin{vmatrix}\n \\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\\n 1 & 0 & -1 \\\\\n 0 & 1 & -1\n \\end{vmatrix} = \\mathbf{i} + \\mathbf{j} + \\mathbf{k}\n \\]\n- Then\n \\[\n \\mathbf{F}(\\mathbf{r}(x, y)) = (2xy) \\mathbf{i} + (2y(2a - x - y)) \\mathbf{j} + (2x(2a - x - y)) \\mathbf{k}\n \\]\n- Therefore\n \\[\n \\mathbf{F}(\\mathbf{r}(x, y)) \\cdot (\\mathbf{r}_x \\times \\mathbf{r}_y) = (2xy) + (2y(2a - x - y)) + (2x(2a - x - y))\n \\]\n• Proceeding to the transformation, we may write that\n\n\\[\n\\iint_S \\mathbf{F} \\cdot d\\mathbf{S} = \\iint_D \\mathbf{F} \\cdot (\\mathbf{r}_x \\times \\mathbf{r}_y) \\, dA\n\\]\n\n\\[\n= \\iint_D ((2xy) + (2y(2a - x - y)) + (2x(2a - x - y))) \\, dy \\, dx\n\\]\n\n\\[\n= \\int_0^a \\int_0^a ((2xy) + (2y(2a - x - y)) + (2x(2a - x - y))) \\, dy \\, dx\n\\]\n\n• Now is just to solve the double integral.\n\n• At the end you should get: \\( \\frac{13a^4}{6} \\)", "id": "./materials/471.pdf" }, { "contents": "Find the flux of the vector field \\( \\mathbf{F}(x, y, z) = x \\mathbf{i} + y \\mathbf{j} + z \\mathbf{k} \\) across the portion of the cylinder \\( x^2 + y^2 = 1 \\) cut by the planes \\( z = 0 \\) and \\( z = a \\), with outward orientation.\n\n- Let the parametrization be\n \\[\n \\mathbf{r}(\\theta, z) = \\cos(\\theta) \\mathbf{i} + \\sin(\\theta) \\mathbf{j} + z \\mathbf{k}\n \\]\n\n- Evaluating \\( \\mathbf{r}_\\theta \\) and \\( \\mathbf{r}_z \\)\n \\[\n \\mathbf{r}_\\theta = -\\sin(\\theta) \\mathbf{i} + \\cos(\\theta) \\mathbf{j}\n \\]\n \\[\n \\mathbf{r}_z = \\mathbf{k}\n \\]\n\n- Proceeding with some calculations we get\n \\[\n \\mathbf{r}_\\theta \\times \\mathbf{r}_z = \\begin{vmatrix}\n \\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\\n -\\sin(\\theta) & \\cos(\\theta) & 0 \\\\\n 0 & 0 & 1\n \\end{vmatrix}\n = \\cos(\\theta) \\mathbf{i} + \\sin(\\theta) \\mathbf{j}\n \\]\n\n- Then\n \\[\n \\mathbf{F}(\\mathbf{r}(\\theta, z)) = (\\cos(\\theta)) \\mathbf{i} + (\\sin(\\theta)) \\mathbf{j} + z \\mathbf{k}\n \\]\n\n- Therefore\n \\[\n \\mathbf{F}(\\mathbf{r}(\\theta, z)) \\cdot (\\mathbf{r}_\\theta \\times \\mathbf{r}_z) = \\cos^2(\\theta) + \\sin^2(\\theta) = 1\n \\]\n• Proceeding to the transformation, we may write that\n\n\\[\n\\iint_S \\mathbf{F} \\cdot d\\mathbf{S} = \\iint_D \\mathbf{F} \\cdot (\\mathbf{r}_\\theta \\times \\mathbf{r}_z) \\, dA = \\iint_D 1 \\, dA = \\int_0^{2\\pi} \\int_0^a 1 \\, dA\n\\]\n\n• Now is just to solve the double integral.\n\n• At the end you should get: \\(2\\pi a\\)", "id": "./materials/472.pdf" }, { "contents": "Find the flux of the vector field \\( \\mathbf{F}(x, y, z) = xy \\mathbf{i} - z \\mathbf{k} \\) across the surface of the cone \\( z = \\sqrt{x^2 + y^2}, 0 \\leq z \\leq 1 \\), with outward orientation.\n\n- Let the parametrization be\n \\[\n \\mathbf{r}(r, \\theta) = r \\cos(\\theta) \\mathbf{i} + r \\sin(\\theta) \\mathbf{j} + r \\mathbf{k}\n \\]\n\n- Evaluating \\( \\mathbf{r}_\\theta \\) and \\( \\mathbf{r}_r \\)\n \\[\n \\mathbf{r}_\\theta = -r \\sin(\\theta) \\mathbf{i} + r \\cos(\\theta) \\mathbf{j}\n \\]\n \\[\n \\mathbf{r}_r = \\cos(\\theta) \\mathbf{i} + \\sin(\\theta) \\mathbf{j} + \\mathbf{k}\n \\]\n\n- Proceeding with some calculations we get\n \\[\n \\mathbf{r}_\\theta \\times \\mathbf{r}_r = \\begin{vmatrix}\n \\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\\n -r \\sin(\\theta) & r \\cos(\\theta) & 0 \\\\\n \\cos(\\theta) & \\sin(\\theta) & 1\n \\end{vmatrix}\n = r \\cos(\\theta) \\mathbf{i} + r \\sin(\\theta) \\mathbf{j} - r \\mathbf{k}\n \\]\n\n- Then\n \\[\n \\mathbf{F}(\\mathbf{r}(r, \\theta)) = (r^2 \\sin(\\theta) \\cos(\\theta)) \\mathbf{i} - r \\mathbf{k}\n \\]\n\n- Therefore\n \\[\n \\mathbf{F}(\\mathbf{r}(r, \\theta)) \\cdot (\\mathbf{r}_\\theta \\times \\mathbf{r}_r) = r^3 \\sin(\\theta) \\cos^2(\\theta) + r^2\n \\]\n\n- Proceeding to the transformation, we may write that\n \\[\n \\iint_S \\mathbf{F} \\cdot d\\mathbf{S} = \\iint_D \\mathbf{F} \\cdot (\\mathbf{r}_\\theta \\times \\mathbf{r}_r) \\, dA\n \\]\n \\[\n = \\int_0^{2\\pi} \\int_0^1 (r^3 \\sin(\\theta) \\cos^2(\\theta) + r^2) \\, dr \\, d\\theta\n \\]\n• Now is just to solve the double integral.\n\n• At the end you should get: \\( \\frac{2\\pi}{3} \\)", "id": "./materials/473.pdf" }, { "contents": "Find the flux of the vector field \\( \\mathbf{F}(x, y, z) = z\\mathbf{i} + y\\mathbf{j} + x\\mathbf{k} \\) over the unit sphere \\( x^2 + y^2 + z^2 = 1 \\).\n\n- Firstly, we can sketch the sphere we are going to work with.\n\n![Figure 1: 3D region of the unit sphere.](image)\n\n- We can resort to the divergence theorem to solve this exercise, presented as follows:\n\n\\[\n\\iint_S \\mathbf{F} \\cdot d\\mathbf{S} = \\iiint_B \\text{div}\\mathbf{F} \\, dV\n\\]\n\nwhere \\( S \\) is the boundary of the solid sphere \\( B \\).\n• For that, we need to evaluate the divergence of $\\mathbf{F}$.\n\n$$\\text{div}\\mathbf{F} = \\frac{\\partial}{\\partial x}(z) + \\frac{\\partial}{\\partial y}(y) + \\frac{\\partial}{\\partial z}(x) = 1$$\n\n• This means that we can substitute some values in the formula presented above, where:\n\n$$\\int \\int \\mathbf{F} \\cdot d\\mathbf{S} = \\int \\int \\int \\text{div}\\mathbf{F} \\, dV$$\n\n$$= \\int \\int \\int 1 \\, dV$$\n\n• Since $B$ is the solid sphere, this means that the last integral will give us the volume of $B$. By default, we know that the volume of $B$ will be given by:\n\n$$\\int \\int \\int 1 \\, dV = \\frac{4\\pi \\times 1^3}{3} = \\frac{4\\pi}{3}$$", "id": "./materials/474.pdf" }, { "contents": "Find the flux of the vector field \\( \\mathbf{F}(x, y, z) = xy \\mathbf{i} + (y^2 + e^{xz^2}) \\mathbf{j} + \\sin(xy) \\mathbf{k} \\), where \\( S \\) is the boundary of the solid region \\( E \\) bounded by \\( z = 1 - x^2 \\), \\( y = 0 \\), \\( z = 0 \\) and \\( z + y = 2 \\).\n\n- Firstly, we can sketch the region we are going to work with.\n\n![3D sketch of E.](image)\n\n- We can resort to the divergence theorem to solve this exercise, presented as follows:\n\n\\[\n\\iint_S \\mathbf{F} \\cdot d\\mathbf{S} = \\iiint_E \\text{div} \\mathbf{F} \\, dV\n\\]\n\nwhere \\( S \\) is the boundary of the solid region \\( E \\).\n• For that, we need to evaluate the divergence of $\\mathbf{F}$.\n\n$$\n\\text{div}\\mathbf{F} = \\frac{\\partial}{\\partial x}(xy) + \\frac{\\partial}{\\partial y}(y^2 + e^{xz^2}) + \\frac{\\partial}{\\partial z}(\\sin(xy))\n$$\n\n$$\n= y + 2y + 0\n$$\n\n$$\n= 3y\n$$\n\n• This means that we can substitute some values in the formula presented above, where:\n\n$$\n\\iiint_{E} \\text{div}\\mathbf{F} \\, dV = \\iiint_{E} 3y \\, dV\n$$\n\n• Now is just to define and solve the triple integral.\n\n• For that, we must define $E$, where\n\n$$\nE = \\{(x, y, z) \\mid -1 \\leq x \\leq 1, 0 \\leq z \\leq 1 - x^2, 0 \\leq y \\leq 2 - z\\}\n$$\n\n• Thus we can define the triple integral as\n\n$$\n\\iiint_{E} 3y \\, dV = \\int_{-1}^{1} \\int_{0}^{1-x^2} \\int_{0}^{2-z} 3y \\, dy \\, dz \\, dx\n$$\n\n• Now is just to solve the triple integral.\n\n• At the end you should get: $\\frac{184}{35}$", "id": "./materials/475.pdf" }, { "contents": "Using the Divergence Theorem, find the flux of the vector field\n\\[ \\mathbf{F}(x, y, z) = xy e^z \\mathbf{i} + xy^2 z^3 \\mathbf{j} - ye^z \\mathbf{k}, \\]\nwhere \\( S \\) is the surface of the box bounded by the coordinate planes and the planes \\( x = 3, y = 2 \\) and \\( z = 1 \\).\n\n- Firstly, we can sketch the region we are going to work with.\n\n![Figure 1: 3D sketch of the box E.](image)\n\n- We can resort to the divergence theorem to solve this exercise, presented as follows:\n\n\\[\n\\iint_S \\mathbf{F} \\cdot d\\mathbf{S} = \\iiint_E \\text{div} \\mathbf{F} \\, dV\n\\]\n\nwhere \\( S \\) is the boundary of the solid region \\( E \\).\n\n- For that, we need to evaluate the divergence of \\( \\mathbf{F} \\).\n\n\\[\n\\text{div} \\mathbf{F} = \\frac{\\partial}{\\partial x}(xy e^z) + \\frac{\\partial}{\\partial y}(xy^2 z^3) + \\frac{\\partial}{\\partial z}(-ye^z)\n\\]\n\\[\n= ye^z + 2xyz^3 - ye^z\n\\]\n\\[\n= 2xyz^3\n\\]\n• This means that we can substitute some values in the formula presented above, where:\n\n\\[\n\\int \\int \\int_{S} \\mathbf{F} \\cdot d\\mathbf{S} = \\int \\int \\int_{E} \\text{div}\\mathbf{F} \\, dV\n\\]\n\n\\[\n= \\int \\int \\int_{E} 2xyz^3 \\, dV\n\\]\n\n• Now is just to define and solve the triple integral.\n\n• For that, we must define E, where\n\n\\[\nE = \\{(x, y, z) | 0 \\leq x \\leq 3, 0 \\leq y \\leq 2, 0 \\leq z \\leq 1\\}\n\\]\n\n• Thus we can define the triple integral as\n\n\\[\n\\int \\int \\int_{E} 2xyz^3 \\, dV = \\int_{0}^{3} \\int_{0}^{2} \\int_{0}^{1} 2xyz^3 \\, dz \\, dy \\, dx\n\\]\n\n• Now is just to solve the triple integral.\n\n• At the end you should get: \\( \\frac{9}{2} \\)", "id": "./materials/476.pdf" }, { "contents": "Using the Divergence Theorem, find the flux of the vector field\n\\[ \\mathbf{F}(x, y, z) = 3xy^2 \\mathbf{i} + xe^z \\mathbf{j} + z^3 \\mathbf{k}, \\]\nwhere \\( S \\) is the surface of the solid bounded by \\( y^2 + z^2 = 1, x = -1 \\) and \\( x = 2 \\).\n\n- Firstly, we can sketch the region we are going to work with.\n\n![Figure 1: 3D sketch of E.](image)\n\n- We can resort to the divergence theorem to solve this exercise,\npresented as follows:\n\n\\[ \\iiint_S \\mathbf{F} \\cdot d\\mathbf{S} = \\iiint_E \\text{div} \\mathbf{F} \\, dV \\]\n\nwhere \\( S \\) is the boundary of the solid region \\( E \\).\n\n- For that, we need to evaluate the divergence of \\( \\mathbf{F} \\).\n\n\\[\n\\text{div} \\mathbf{F} = \\frac{\\partial}{\\partial x}(3xy^2) + \\frac{\\partial}{\\partial y}(xe^z) + \\frac{\\partial}{\\partial z}(z^3)\n\\]\n\n\\[\n= 3y^2 + 0 + 3z^2\n\\]\n\n\\[\n= 3(y^2 + z^2)\n\\]\n\n- This means that we can substitute some values in the formula presented above, where:\n\n\\[\n\\iiint_S \\mathbf{F} \\cdot d\\mathbf{S} = \\iiint_E \\text{div} \\mathbf{F} \\, dV\n\\]\n\n\\[\n= 3 \\iiint_E y^2 + z^2 \\, dV\n\\]\n\n- Now is just to define and solve the triple integral.\n\n- Using cylindrical coordinates, we can define that\n\n\\[\ny = r \\cos(\\theta) \\quad , \\quad z = r \\sin(\\theta) \\quad , \\quad x = x\n\\]\n\n- Thus we can define the triple integral as\n\n\\[\n3 \\iiint_E y^2 + z^2 \\, dV = 3 \\int_0^{2\\pi} \\int_0^1 \\int_{-1}^1 (r^2 \\cos(\\theta) + r^2 \\sin(\\theta))r \\, dr \\, d\\theta\n\\]\n\n- Now is just to solve the triple integral.\n\n- At the end you should get: \\( \\frac{9\\pi}{2} \\)", "id": "./materials/477.pdf" }, { "contents": "Using the Divergence Theorem, find the flux of the vector field\n\\[ \\mathbf{F}(x, y, z) = (x^3 + y^3)\\mathbf{i} + (y^3 + z^3)\\mathbf{j} + (z^3 + x^3)\\mathbf{k}, \\]\nwhere \\( S \\) is the sphere with center the origin and radius 2.\n\n- Firstly, we can sketch the region we are going to work with.\n\n![3D sketch of E.](image)\n\nFigure 1: 3D sketch of E.\n\n- We can resort to the divergence theorem to solve this exercise,\npresented as follows:\n\n\\[ \\iiint_S \\mathbf{F} \\cdot d\\mathbf{S} = \\iiint_E \\text{div} \\mathbf{F} \\, dV \\]\n\nwhere \\( S \\) is the boundary of the solid region \\( E \\).\n\n- For that, we need to evaluate the divergence of \\( \\mathbf{F} \\).\n\n\\[\n\\text{div} \\mathbf{F} = \\frac{\\partial}{\\partial x} (x^3 + y^3) + \\frac{\\partial}{\\partial y} (y^3 + z^3) + \\frac{\\partial}{\\partial z} (z^3 + x^3) \\\\\n= 3x^2 + 3y^2 + 3z^2\n\\]\n\n- This means that we can substitute some values in the formula presented above, where:\n\n\\[\n\\iiint_S \\mathbf{F} \\cdot d\\mathbf{S} = \\iiint_E \\text{div} \\mathbf{F} \\, dV \\\\\n= 3 \\iiint_E x^2 + y^2 + z^2 \\, dV\n\\]\n\n- Now is just to define and solve the triple integral.\n\n- Using spherical coordinates, we can define the triple integral as\n\n\\[\n3 \\iiint_E x^2 + y^2 + z^2 \\, dV = 3 \\int_0^\\pi \\int_0^{2\\pi} \\int_0^2 (\\rho^2) \\rho^2 \\sin (\\theta) \\, d\\rho d\\theta d\\phi\n\\]\n\n- Now is just to solve the triple integral.\n\n- At the end you should get: \\( \\frac{384\\pi}{5} \\)", "id": "./materials/478.pdf" }, { "contents": "Using the Divergence Theorem, find the flux of the vector field\n\\( \\mathbf{F}(x, y, z) = x^2 \\sin(y) \\mathbf{i} + x \\cos(y) \\mathbf{j} - xz \\sin(y) \\mathbf{k} \\), where \\( S \\) is the \"fat sphere\" defined by \\( x^8 + y^8 + z^8 = 8 \\).\n\n- We can resort to the divergence theorem to solve this exercise, presented as follows:\n\n\\[\n\\iint_S \\mathbf{F} \\cdot d\\mathbf{S} = \\iiint_E \\text{div} \\mathbf{F} \\, dV\n\\]\n\nwhere \\( S \\) is the boundary of the solid region \\( E \\).\n\n- For that, we need to evaluate the divergence of \\( \\mathbf{F} \\).\n\n\\[\n\\text{div} \\mathbf{F} = \\frac{\\partial}{\\partial x} (x^2 \\sin(y)) + \\frac{\\partial}{\\partial y} (x \\cos(y)) + \\frac{\\partial}{\\partial z} (-xz \\sin(y))\n\\]\n\n\\[\n= 2x \\sin(y) - x \\sin(y) - x \\sin(y)\n\\]\n\n\\[\n= 0\n\\]\n\n- This means that we can substitute some values in the formula presented above, where:\n\n\\[\n\\iint_S \\mathbf{F} \\cdot d\\mathbf{S} = \\iiint_E \\text{div} \\mathbf{F} \\, dV\n\\]\n\n\\[\n= \\iiint_E 0 \\, dV\n\\]\n\n\\[\n= 0\n\\]", "id": "./materials/479.pdf" }, { "contents": "Systems of linear equations\n\nExercise. Let\n\n\\[ A = \\begin{bmatrix} 1 & -2 & 0 \\\\ 1 & 0 & 1 \\\\ -5 & 2 & 3 \\end{bmatrix} \\quad \\text{and} \\quad b = \\begin{bmatrix} -1 \\\\ 1 \\\\ 0 \\end{bmatrix} \\]\n\nSolve the system of linear equations \\( AX = b \\) in three ways: (1) by Cramer rule, (2) via Gauss-Jordan elimination and (3) using the inverse matrix \\( A^{-1} \\).\n\nSolution. We observe, to begin with, that \\( \\det(A) = 14 \\), therefore \\( A \\) is invertible (hence \\( \\varrho(A) = 3 \\) and the system has a unique solution).\n\nCramer: the solution \\( X = \\begin{bmatrix} x \\\\ y \\\\ z \\end{bmatrix} \\) is given as\n\n\\[\n\\begin{align*}\nx &= \\frac{\\det \\begin{bmatrix} -1 & -2 & 0 \\\\ 1 & 0 & 1 \\\\ 0 & 2 & 3 \\end{bmatrix}}{14} = \\frac{4}{7}, \\\\\ny &= \\frac{\\det \\begin{bmatrix} 1 & -1 & 0 \\\\ 1 & 1 & 1 \\\\ -5 & 0 & 3 \\end{bmatrix}}{14} = \\frac{11}{14}, \\\\\nz &= \\frac{\\det \\begin{bmatrix} 1 & -2 & -1 \\\\ 1 & 0 & 1 \\\\ -5 & 2 & 0 \\end{bmatrix}}{14} = \\frac{3}{7}.\n\\end{align*}\n\\]\n\nGauss-Jordan elimination: by performing row operations we get\n\n\\[\n\\begin{align*}\n&\\begin{bmatrix} 1 & -2 & 0 & -1 \\\\ 1 & 0 & 1 & 1 \\\\ -5 & 2 & 3 & 0 \\end{bmatrix} \\xrightarrow{E_{21}(-1)} \\begin{bmatrix} 1 & -2 & 0 & -1 \\\\ 0 & 2 & 1 & 2 \\\\ -5 & 2 & 3 & 0 \\end{bmatrix} \\xrightarrow{E_{31}(5)} \\begin{bmatrix} 1 & -2 & 0 & -1 \\\\ 0 & 2 & 1 & 2 \\\\ 0 & -8 & 3 & -5 \\end{bmatrix}, \\\\\n&\\begin{bmatrix} 1 & -2 & 0 & -1 \\\\ 0 & 2 & 1 & 2 \\\\ 0 & 0 & 7 & 3 \\end{bmatrix} \\xrightarrow{E_{32}(4)} \\begin{bmatrix} 1 & -2 & 0 & -1 \\\\ 0 & 2 & 1 & 2 \\\\ 0 & 0 & 7 & 3 \\end{bmatrix} \\xrightarrow{E_{32}(1/2)} \\begin{bmatrix} 1 & -2 & 0 & -1 \\\\ 0 & 1 & 1/2 & 1 \\\\ 0 & 0 & 7 & 3 \\end{bmatrix} \\xrightarrow{E_{31}(1/7)} \\begin{bmatrix} 1 & -2 & 0 & -1 \\\\ 0 & 1 & 1/2 & 1 \\\\ 0 & 0 & 1 & 3/7 \\end{bmatrix}, \\\\\n&\\begin{bmatrix} 1 & 0 & 1 & 1 \\\\ 0 & 1 & 1/2 & 1 \\\\ 0 & 0 & 1 & 3/7 \\end{bmatrix} \\xrightarrow{E_{31}(-1)} \\begin{bmatrix} 1 & 0 & 0 & 4/7 \\\\ 0 & 1 & 1/2 & 1 \\\\ 0 & 0 & 1 & 3/7 \\end{bmatrix} \\xrightarrow{E_{32}(-1/2)} \\begin{bmatrix} 1 & 0 & 0 & 4/7 \\\\ 0 & 1 & 1/2 & 1 \\\\ 0 & 0 & 1 & 3/7 \\end{bmatrix} \\xrightarrow{E_{31}(1)} \\begin{bmatrix} 1 & 0 & 0 & 4/7 \\\\ 0 & 1 & 1/2 & 1 \\\\ 0 & 0 & 1 & 3/7 \\end{bmatrix}.\n\\end{align*}\n\\]\n\nThe linear system associated to the last matrix gives the same solution as before.\n\nUsing the inverse: the inverse matrix of \\( A \\) can be obtained via the (classical) adjoint matrix, as well as, again, via Gauss-Jordan reduction. Using the adjoint matrix:\n\n\\[\nA^* = \\begin{bmatrix} -2 & -8 & 2 \\\\ 6 & 3 & 8 \\\\ -2 & -1 & 2 \\end{bmatrix},\n\\]\n\nwe get\n\n\\[\nA^{-1} = \\frac{\\text{adj}(A)}{\\det(A)} = \\begin{bmatrix} -1/7 & 3/7 & -1/7 \\\\ -4/7 & 3/14 & -1/14 \\\\ 1/7 & 4/7 & 1/7 \\end{bmatrix}.\n\\]\n\nThus\n\n\\[\nX = A^{-1}b = \\begin{bmatrix} -1/7 & 3/7 & -1/7 \\\\ -4/7 & 3/14 & -1/14 \\\\ 1/7 & 4/7 & 1/7 \\end{bmatrix} \\cdot \\begin{bmatrix} -1 \\\\ 1 \\\\ 0 \\end{bmatrix} = \\begin{bmatrix} 4/7 \\\\ 11/14 \\\\ 3/7 \\end{bmatrix}.\n\\]", "id": "./materials/48.pdf" }, { "contents": "Using the Divergence Theorem, find the flux of the vector field\n\\[ \\mathbf{F}(x, y, z) = (\\cos(z) + xy^2)\\mathbf{i} + xe^z\\mathbf{j} + (\\sin(y) + x^2z)\\mathbf{k}, \\]\nwhere \\( S \\) is the surface of the solid bounded by \\( z = x^2 + y^2 \\) and \\( z = 4 \\).\n\n- Firstly, we can sketch the region we are going to work with.\n\n![Figure 1: 3D sketch of E.](image)\n\n- We can resort to the divergence theorem to solve this exercise,\npresented as follows:\n\n\\[ \\iiint_S \\mathbf{F} \\cdot d\\mathbf{S} = \\iiint_E \\text{div} \\mathbf{F} \\, dV \\]\n\nwhere \\( S \\) is the boundary of the solid region \\( E \\).\n\n- For that, we need to evaluate the divergence of \\( \\mathbf{F} \\).\n\n\\[\n\\text{div} \\mathbf{F} = \\frac{\\partial}{\\partial x} (\\cos(z) + xy^2) + \\frac{\\partial}{\\partial y} (xe^{-z}) + \\frac{\\partial}{\\partial z} (\\sin(y) + x^2z)\n\\]\n\n\\[\n= y^2 + 0 + x^2\n\\]\n\n\\[\n= y^2 + x^2\n\\]\n\n- This means that we can substitute some values in the formula presented above, where:\n\n\\[\n\\iiint_S \\mathbf{F} \\cdot d\\mathbf{S} = \\iiint_E \\text{div} \\mathbf{F} \\, dV\n\\]\n\n\\[\n= \\iiint_E x^2 + y^2 \\, dV\n\\]\n\n- Using cylindrical coordinates we can define the triple integral as\n\n\\[\n\\iiint_E x^2 + y^2 \\, dV = \\int_0^{2\\pi} \\int_0^2 \\int_0^4 r^2 \\cdot r \\, dz \\, dr \\, d\\theta\n\\]\n\n- Now is just to solve the triple integral.\n\n- At the end you should get: \\( \\frac{32\\pi}{3} \\)", "id": "./materials/480.pdf" }, { "contents": "Using the Divergence Theorem, find the flux of the vector field\n\\( \\mathbf{F}(x, y, z) = x^4 \\mathbf{i} - x^3 z^2 \\mathbf{j} + 4xy^2 z \\mathbf{k} \\), where \\( S \\) is the surface of the solid bounded by \\( x^2 + y^2 = 1 \\), \\( z = x + 2 \\) and \\( z = 0 \\).\n\n- We can resort to the divergence theorem to solve this exercise, presented as follows:\n\n\\[\n\\iint_S \\mathbf{F} \\cdot d\\mathbf{S} = \\iiint_E \\text{div} \\mathbf{F} \\, dV\n\\]\n\nwhere \\( S \\) is the boundary of the solid region \\( E \\).\n\n- For that, we need to evaluate the divergence of \\( \\mathbf{F} \\).\n\n\\[\n\\text{div} \\mathbf{F} = \\frac{\\partial}{\\partial x} (x^4) + \\frac{\\partial}{\\partial y} (-x^3 z^2) + \\frac{\\partial}{\\partial z} (4xy^2 z)\n\\]\n\n\\[\n= 4x^3 + 0 + 4xy^2\n\\]\n\n\\[\n= 4x^3 + 4xy^2\n\\]\n\n- This means that we can substitute some values in the formula presented above, where:\n\n\\[\n\\iint_S \\mathbf{F} \\cdot d\\mathbf{S} = \\iiint_E \\text{div} \\mathbf{F} \\, dV\n\\]\n\n\\[\n= \\iiint_E 4x^3 + 4xy^2 \\, dV\n\\]\n\n- Using cylindrical coordinates we can define the triple integral as\n\n\\[\n\\iiint_E x^2 + y^2 \\, dV = \\int_0^{2\\pi} \\int_0^1 \\int_0^{r \\cos(\\theta) + 2} (4r^3 \\cos(\\theta)) r \\, dz \\, dr \\, d\\theta\n\\]\n• Now is just to solve the triple integral.\n\n• At the end you should get: \\( \\frac{2\\pi}{3} \\)", "id": "./materials/481.pdf" }, { "contents": "Using the Divergence Theorem, find the flux of the vector field\n\\( \\mathbf{F}(x, y, z) = e^x \\sin(y) \\mathbf{i} + e^x \\cos(y) \\mathbf{j} + yz^2 \\mathbf{k} \\), where \\( S \\) is the surface of the box where \\( 0 \\leq x \\leq 1, 0 \\leq y \\leq 1, 0 \\leq z \\leq 2 \\).\n\n- We can resort to the divergence theorem to solve this exercise, presented as follows:\n\n\\[\n\\iint_S \\mathbf{F} \\cdot d\\mathbf{S} = \\iiint_E \\text{div} \\mathbf{F} \\, dV\n\\]\n\nwhere \\( S \\) is the boundary of the solid region \\( E \\).\n\n- For that, we need to evaluate the divergence of \\( \\mathbf{F} \\).\n\n\\[\n\\text{div} \\mathbf{F} = \\frac{\\partial}{\\partial x}(e^x \\sin(y)) + \\frac{\\partial}{\\partial y}(e^x \\cos(y)) + \\frac{\\partial}{\\partial z}(yz^2)\n\\]\n\n\\[\n= e^x \\sin(y) - e^x \\sin(y) + 2zy\n\\]\n\n\\[\n= 2zy\n\\]\n\n- This means that we can substitute some values in the formula presented above, where:\n\n\\[\n\\iint_S \\mathbf{F} \\cdot d\\mathbf{S} = \\iiint_E \\text{div} \\mathbf{F} \\, dV\n\\]\n\n\\[\n= \\iiint_E 2zy \\, dV\n\\]\n\n- Thus, we can define the triple integral as\n\n\\[\n\\iiint_E 2zy \\, dV = \\int_0^1 \\int_0^1 \\int_0^2 2zy \\, dz \\, dy \\, dx\n\\]\n\n- Now is just to solve the triple integral.\n\n- At the end you should get: 2", "id": "./materials/482.pdf" }, { "contents": "Teaching material. Systems of linear equations\n\nExercise. Perform Gauss-Jordan elimination on the following matrices:\n\n\\[\n\\begin{bmatrix}\n0 & 1 & 0 & 1 \\\\\n0 & 0 & 1 & 2 \\\\\n0 & 1 & 2 & 0\n\\end{bmatrix},\n\\begin{bmatrix}\n1 & 1 & 1 & 1 \\\\\n1 & 1 & 0 & 2 \\\\\n1 & 0 & 0 & 3\n\\end{bmatrix},\n\\begin{bmatrix}\n1 & 1 & -1 & 1 \\\\\n2 & 2 & 1 & 0 \\\\\n1 & 1 & 2 & -1\n\\end{bmatrix}.\n\\]\n\nThen, for each matrix, solve the system of linear equation whose augmented matrix is the given matrix.\n\nSolution.\n\n(1) By row operations (i.e. left multiplication by elementary matrices) we reduce the matrix in row echelon form:\n\n\\[\n\\begin{bmatrix}\n0 & 1 & 0 & 1 \\\\\n0 & 0 & 1 & 2 \\\\\n0 & 1 & 2 & 0\n\\end{bmatrix}\n\\]\n\nTo achieve its reduced row echelon form, we divide each non zero row by its pivot and we perform row operations to set to zero the entries above each pivot:\n\n\\[\n\\begin{bmatrix}\n0 & 1 & 0 & 1 \\\\\n0 & 0 & 1 & 2 \\\\\n0 & 0 & 0 & -1\n\\end{bmatrix}\n\\]\n\nThe corresponding system is therefore equivalent to\n\n\\[\n\\begin{cases}\ny = 0 \\\\\nz = 0 \\\\\n0 = 1\n\\end{cases}\n\\]\n\ninconsistent: solution set \\(\\emptyset\\).\n\n(2) As before:\n\n\\[\n\\begin{bmatrix}\n1 & 1 & 1 & 1 \\\\\n1 & 1 & 0 & 2 \\\\\n1 & 0 & 0 & 3\n\\end{bmatrix}\n\\]\n\nWhich is already in reduced row echelon form. The corresponding system is\n\n\\[\n\\begin{cases}\nx = 3 \\\\\ny = -1 \\\\\nz = -1\n\\end{cases}\n\\]\n\nsolution set: \\(\\{(3; -1; -1)\\}\\).\n\n(3) Row echelon form:\n\n\\[\n\\begin{bmatrix}\n1 & 1 & -1 & 1 \\\\\n2 & 2 & 1 & 0 \\\\\n1 & 1 & 2 & -1\n\\end{bmatrix}\n\\]\n\nReduced row echelon form:\n\n\\[\n\\begin{bmatrix}\n1 & 1 & -1 & 1 \\\\\n0 & 0 & 3 & -2 \\\\\n0 & 0 & 3 & -2\n\\end{bmatrix}\n\\]\n\nThe corresponding system is\n\n\\[\n\\begin{cases}\nx + y = \\frac{1}{3} \\\\\nz = -\\frac{2}{3}\n\\end{cases}\n\\]\n\n(choosing \\(y\\) as free variable)\n\n\\[\n\\begin{cases}\nx = \\frac{1}{3} - y \\\\\ny = y \\\\\nz = -\\frac{2}{3}\n\\end{cases}\n\\]\n\nThus, its solution set is \\(\\{(\\frac{1}{3} - y; y; -\\frac{2}{3}) \\mid y \\in \\mathbb{R}\\}\\).", "id": "./materials/49.pdf" }, { "contents": "Systems of linear equations\n\nExercise. Let\n\n\\[ A_\\alpha = \\begin{pmatrix} 1 & 1 & -1 & \\alpha \\\\ 2 & 3 & -2 & \\alpha \\\\ \\alpha & \\alpha + 1 & 0 & 0 \\end{pmatrix} \\in M_{34}(\\mathbb{R}) \\]\n\nwhere \\( \\alpha \\in \\mathbb{R} \\).\n\n1. Find the rank \\( \\varrho(A_\\alpha) \\) for each \\( \\alpha \\in \\mathbb{R} \\).\n\n2. Let \\( \\alpha = 0 \\): decide whether the system of linear equations\n\n\\[ A_0 \\cdot \\begin{pmatrix} x \\\\ y \\\\ z \\\\ t \\end{pmatrix} = \\begin{pmatrix} -1 \\\\ 1 \\\\ 2 \\end{pmatrix} \\]\n\nis consistent or not and, in positive case, how many solutions it has.\n\n3. Let \\( \\alpha = 1 \\): decide whether the system of linear equations\n\n\\[ A_1 \\cdot \\begin{pmatrix} x \\\\ y \\\\ z \\\\ t \\end{pmatrix} = \\begin{pmatrix} 1 \\\\ 1 \\\\ 1 \\end{pmatrix} \\]\n\nis consistent or not and, in positive case, how many solutions it has.\n\nSolution.\n\n1. We reduce \\( A_\\alpha \\) in row echelon form:\n\n\\[ A_\\alpha = \\begin{pmatrix} 1 & 1 & -1 & \\alpha \\\\ 2 & 3 & -2 & \\alpha \\\\ \\alpha & \\alpha + 1 & 0 & 0 \\end{pmatrix} \\rightarrow \\begin{pmatrix} 1 & 1 & -1 & \\alpha \\\\ 0 & 1 & 0 & -\\alpha \\\\ 0 & 1 & \\alpha & -\\alpha^2 \\end{pmatrix} \\rightarrow \\begin{pmatrix} 1 & 1 & -1 & \\alpha \\\\ 0 & 1 & 0 & -\\alpha \\\\ 0 & 0 & \\alpha & \\alpha(1-\\alpha) \\end{pmatrix} \\]\n\nIf \\( \\alpha = 0 \\), the matrix becomes\n\n\\[ A_0 = \\begin{pmatrix} 1 & 1 & -1 & 0 \\\\ 0 & 1 & 0 & 0 \\\\ 0 & 0 & 0 & 0 \\end{pmatrix} \\]\n\nhence \\( \\varrho(A_0) = 2 \\). If, otherwise, \\( \\alpha \\neq 0 \\), then the \\( \\alpha \\) in the last row is a pivot. Therefore:\n\n\\[ \\begin{pmatrix} 1 & 1 & -1 & \\alpha \\\\ 0 & 1 & 0 & -\\alpha \\\\ 0 & 0 & \\alpha & \\alpha(1-\\alpha) \\end{pmatrix} \\]\n\nhence \\( \\varrho(A_\\alpha) = 3 \\) for each \\( \\alpha \\neq 0 \\).\n\n2. We found above \\( \\varrho(A_0) = 2 \\). For the augmented matrix we get instead:\n\n\\[ \\begin{pmatrix} 1 & 1 & -1 & 0 & -1 \\\\ 2 & 3 & -2 & 0 & 1 \\\\ 0 & 1 & 0 & 0 & 2 \\end{pmatrix} \\rightarrow \\begin{pmatrix} 1 & 1 & -1 & 0 & -1 \\\\ 0 & 1 & 0 & 0 & 3 \\\\ 0 & 1 & 0 & 0 & 2 \\end{pmatrix} \\rightarrow \\begin{pmatrix} 1 & 1 & -1 & 0 & -1 \\\\ 0 & 1 & 0 & 0 & 3 \\\\ 0 & 0 & 0 & 0 & -1 \\end{pmatrix} \\]\n\nhence the corresponding linear system is not consistent (as the last row translates into the impossible equation \\( 0 = -1 \\)).\n3. By Gauss-Jordan elimination we find the reduced row echelon form:\n\n\\[\n\\begin{pmatrix}\n1 & 1 & -1 & 1 & | & 1 \\\\\n2 & 3 & -2 & 1 & | & 1 \\\\\n1 & 2 & 0 & 0 & | & 1\n\\end{pmatrix}\n\\rightarrow\n\\begin{pmatrix}\n1 & 1 & -1 & 1 & | & 1 \\\\\n0 & 1 & 0 & -1 & | & -1 \\\\\n0 & 1 & 1 & -1 & | & 0\n\\end{pmatrix}\n\\rightarrow\n\\begin{pmatrix}\n1 & 1 & -1 & 1 & | & 1 \\\\\n0 & 1 & 0 & -1 & | & -1 \\\\\n0 & 0 & 1 & 0 & | & 1\n\\end{pmatrix}\n\\rightarrow\n\\begin{pmatrix}\n1 & 1 & 0 & 1 & | & 2 \\\\\n0 & 1 & 0 & -1 & | & -1 \\\\\n0 & 0 & 1 & 0 & | & 1\n\\end{pmatrix}\n\\rightarrow\n\\begin{pmatrix}\n1 & 0 & 0 & 2 & | & 3 \\\\\n0 & 1 & 0 & -1 & | & -1 \\\\\n0 & 0 & 1 & 0 & | & 1\n\\end{pmatrix}\n\\]\n\nwhose associated linear system is\n\n\\[\n\\begin{align*}\nx + 2t &= 3 \\\\\ny - t &= -1 \\\\\nz &= 1\n\\end{align*}\n\\Rightarrow\n\\begin{align*}\nx &= 3 - 2t \\\\\ny &= -1 + t \\\\\nz &= 1\n\\end{align*}\n\\]\n\nHence, its solution set is\n\n\\[\n\\{(3 - 2t; t - 1; 1; t) \\mid t \\in \\mathbb{R}\\}.\n\\]", "id": "./materials/50.pdf" }, { "contents": "Systems of linear equations\n\nExample. Consider the system\n\\[\n\\begin{align*}\n x + y + z &= a \\\\\n ax + y + 2z &= 2 \\\\\n x + ay + z &= 4\n\\end{align*}\n\\]\nDecide whether the system is consistent and find the number of solutions in dependence on parameter \\(a \\in \\mathbb{R}\\). Find its solution set for each \\(a \\in \\mathbb{R}\\).\n\nSolution\n\nThe system has augmented matrix \\(\\overline{A}\\), where \\(A\\) is the coefficient matrix, and \\(b\\) is the constant column term:\n\\[\n\\overline{A} := (A|b) = \\begin{pmatrix}\n 1 & 1 & 1 & a \\\\\n 0 & 1 & 2 & 2 \\\\\n 1 & a & 1 & 4\n\\end{pmatrix}\n\\]\nperforming row operations \\((R_2 \\rightarrow R_2 - aR_1, R_3 \\rightarrow R_3 - R_1, R'_3 \\rightarrow R'_3 + R'_2)\\) we get a row echelon form\n\\[\n\\overline{A} \\rightarrow \\begin{pmatrix}\n 1 & 1 & 1 & a \\\\\n 0 & 1 - a & 2 - a & 2 - a^2 \\\\\n 0 & a - 1 & 0 & 4 - a\n\\end{pmatrix} \\rightarrow \\begin{pmatrix}\n 1 & 1 & 1 & a \\\\\n 0 & 1 - a & 2 - a & 2 - a^2 \\\\\n 0 & 0 & 2 - a & 6 - a - a^2\n\\end{pmatrix}\n\\]\nthat is (changing sign in the last two rows):\n\\[\n\\overline{A} \\rightarrow \\begin{pmatrix}\n 1 & 1 & 1 & a \\\\\n 0 & a - 1 & a - 2 & a^2 - 2 \\\\\n 0 & 0 & a - 2 & (a + 3)(a - 2)\n\\end{pmatrix}\n\\]\nNow, if \\(a \\neq 1 \\land a \\neq 2\\), then \\(A\\) and \\(\\overline{A}\\) have equal rank \\(\\varrho(A) = \\varrho(\\overline{A}) = 3\\), hence the system is consistent and it has exactly one solution, which can be found by Gauss-Jordan elimination:\n\\[\n\\overline{A} \\rightarrow \\begin{pmatrix}\n 1 & 1 & 1 & a \\\\\n 0 & a - 1 & a - 2 & a^2 - 2 \\\\\n 0 & 0 & a - 2 & (a + 3)(a - 2)\n\\end{pmatrix} \\rightarrow \\begin{pmatrix}\n 1 & 1 & 1 & a \\\\\n 0 & 1 & 1 & a - 2 \\\\\n 0 & 0 & 1 & a + 3\n\\end{pmatrix} \\rightarrow \\begin{pmatrix}\n 1 & 1 & 1 & a \\\\\n 0 & 1 & 1 & a - 2 \\\\\n 0 & 0 & 1 & a + 3\n\\end{pmatrix}\n\\]\nHence the solution set of the associated system is \\(\\{ \\left( \\frac{2a + 1}{a - 1}, \\frac{a - 4}{a - 1}, a + 3 \\right) \\}\\).\n\nThe exceptional cases \\(a = 1\\) and \\(a = 2\\) must be studied directly.\n\nIf \\(a = 1\\), \\(A\\) becomes \\((R_3 \\rightarrow R_3 - R_2)\\):\n\\[\n\\overline{A} \\rightarrow \\begin{pmatrix}\n 1 & 1 & 1 & 1 \\\\\n 0 & 0 & -1 & -1 \\\\\n 0 & 0 & -1 & 4\n\\end{pmatrix} \\rightarrow \\begin{pmatrix}\n 1 & 1 & 1 & 1 \\\\\n 0 & 0 & -1 & -1 \\\\\n 0 & 0 & 0 & 5\n\\end{pmatrix},\n\\]\nhence the system is not consistent, as \\(\\varrho(A) = 2\\) while \\(\\varrho(\\overline{A}) = 3\\) (the last equation, \\(0 = 5\\), has no solution).\nIf $a = 2$, $\\overline{A}$ becomes (in reduced row echelon form)\n\n$$\n\\overline{A} \\rightarrow \\begin{pmatrix} 1 & 1 & 1 & 2 \\\\ 0 & 1 & 0 & 2 \\\\ 0 & 0 & 0 & 0 \\end{pmatrix} \\rightarrow \\begin{pmatrix} 1 & 0 & 1 & 0 \\\\ 0 & 1 & 0 & 2 \\\\ 0 & 0 & 0 & 0 \\end{pmatrix} \\Rightarrow \\begin{cases} x + z = 0 \\\\ y = 2 \\end{cases}\n$$\n\ntherefore $A$ and $\\overline{A}$ have equal rank $\\varrho(A) = \\varrho(\\overline{A}) = 2$, hence the system is consistent and it has $\\infty^{3-2} = \\infty^1$ solutions:\n\n$$\\{(-z; 2; z) \\mid z \\in \\mathbb{R}\\}.$$\n\nObserve that, for $a = 2$, we have $\\left(\\frac{2a+1}{1-a}, \\frac{a-4}{1-a}, a+3\\right) = (-5; 2; 5)$ and, on the other hand\n\n$$\\begin{cases} -\\frac{2a+1}{1-a} = a + 3 \\\\ \\frac{a-4}{1-a} = 2 \\end{cases} \\iff a = 2.$$\n\nHence, the generic solution we found for $a \\neq 1 \\land a \\neq 2$ is coherent with that found for $a = 2$. \n\n2", "id": "./materials/51.pdf" }, { "contents": "GLOSSARY: A DICTIONARY FOR LINEAR ALGEBRA\n\nAdjacency matrix of a graph. Square matrix with \\( a_{ij} = 1 \\) when there is an edge from node \\( i \\) to node \\( j \\); otherwise \\( a_{ij} = 0 \\). \\( A = A^T \\) for an undirected graph.\n\nAffine transformation \\( T(v) = Av + v_0 \\) = linear transformation plus shift.\n\nAssociative Law \\((AB)C = A(BC)\\). Parentheses can be removed to leave \\( ABC \\).\n\nAugmented matrix \\([ A \\ b ]\\). \\( Ax = b \\) is solvable when \\( b \\) is in the column space of \\( A \\); then \\([ A \\ b ]\\) has the same rank as \\( A \\). Elimination on \\([ A \\ b ]\\) keeps equations correct.\n\nBack substitution. Upper triangular systems are solved in reverse order \\( x_n \\) to \\( x_1 \\).\n\nBasis for \\( V \\). Independent vectors \\( v_1, \\ldots, v_d \\) whose linear combinations give every \\( v \\) in \\( V \\). A vector space has many bases!\n\nBig formula for \\( n \\) by \\( n \\) determinants. \\( \\det(A) \\) is a sum of \\( n! \\) terms, one term for each permutation \\( P \\) of the columns. That term is the product \\( a_{1\\alpha} \\cdots a_{n\\omega} \\) down the diagonal of the reordered matrix, times \\( \\det(P) = \\pm 1 \\).\n\nBlock matrix. A matrix can be partitioned into matrix blocks, by cuts between rows and/or between columns. Block multiplication of \\( AB \\) is allowed if the block shapes permit (the columns of \\( A \\) and rows of \\( B \\) must be in matching blocks).\n\nCayley-Hamilton Theorem. \\( p(\\lambda) = \\det(A - \\lambda I) \\) has \\( p(A) = \\text{zero matrix} \\).\n\nChange of basis matrix \\( M \\). The old basis vectors \\( v_j \\) are combinations \\( \\sum m_{ij} w_i \\) of the new basis vectors. The coordinates of \\( c_1 v_1 + \\cdots + c_n v_n = d_1 w_1 + \\cdots + d_n w_n \\) are related by \\( d = M c \\). (For \\( n = 2 \\) set \\( v_1 = m_{11} w_1 + m_{21} w_2, \\ v_2 = m_{12} w_1 + m_{22} w_2 \\).)\n\nCharacteristic equation \\( \\det(A - \\lambda I) = 0 \\). The \\( n \\) roots are the eigenvalues of \\( A \\).\n\nCholesky factorization \\( A = CC^T = (L\\sqrt{D})(L\\sqrt{D})^T \\) for positive definite \\( A \\).\n\nCirculant matrix \\( C \\). Constant diagonals wrap around as in cyclic shift \\( S \\). Every circulant is \\( c_0 I + c_1 S + \\cdots + c_{n-1} S^{n-1} \\). \\( Cx = \\text{convolution} \\ c \\ast x \\). Eigenvectors in \\( F \\).\n\nCofactor \\( C_{ij} \\). Remove row \\( i \\) and column \\( j \\); multiply the determinant by \\((-1)^{i+j}\\).\n\nColumn picture of \\( Ax = b \\). The vector \\( b \\) becomes a combination of the columns of \\( A \\). The system is solvable only when \\( b \\) is in the column space \\( C(A) \\).\n\nColumn space \\( C(A) = \\text{space of all combinations of the columns of } A \\).\n\nCommuting matrices \\( AB = BA \\). If diagonalizable, they share \\( n \\) eigenvectors.\n\nCompanion matrix. Put \\( c_1, \\ldots, c_n \\) in row \\( n \\) and put \\( n - 1 \\) 1’s along diagonal 1. Then \\( \\det(A - \\lambda I) = \\pm (c_1 + c_2 \\lambda + c_3 \\lambda^2 + \\cdots) \\).\n\nComplete solution \\( x = x_p + x_n \\) to \\( Ax = b \\). (Particular \\( x_p \\) ) + (\\( x_n \\) in nullspace).\n\nComplex conjugate \\( \\bar{z} = a - ib \\) for any complex number \\( z = a + ib \\). Then \\( z\\bar{z} = |z|^2 \\).\nCondition number \\( \\text{cond}(A) = \\kappa(A) = \\|A\\|\\|A^{-1}\\| = \\sigma_{\\text{max}}/\\sigma_{\\text{min}} \\). In \\( Ax = b \\), the relative change \\( \\|\\delta x\\|/\\|x\\| \\) is less than \\( \\text{cond}(A) \\) times the relative change \\( \\|\\delta b\\|/\\|b\\| \\). Condition numbers measure the sensitivity of the output to change in the input.\n\nConjugate Gradient Method. A sequence of steps (end of Chapter 9) to solve positive definite \\( Ax = b \\) by minimizing \\( \\frac{1}{2}x^TAx - x^Tb \\) over growing Krylov subspaces.\n\nCovariance matrix \\( \\Sigma \\). When random variables \\( x_i \\) have mean = average value = 0, their covariances \\( \\Sigma_{ij} \\) are the averages of \\( x_i x_j \\). With means \\( \\bar{x}_i \\), the matrix \\( \\Sigma = \\text{mean of } (x - \\bar{x})(x - \\bar{x})^T \\) is positive (semi)definite; it is diagonal if the \\( x_i \\) are independent.\n\nCramer’s Rule for \\( Ax = b \\). \\( B_j \\) has \\( b \\) replacing column \\( j \\) of \\( A \\), and \\( x_j = |B_j|/|A| \\).\n\nCross product \\( u \\times v \\) in \\( \\mathbb{R}^3 \\). Vector perpendicular to \\( u \\) and \\( v \\), length \\( \\|u\\|\\|v\\|\\sin \\theta \\) = parallelogram area, computed as the “determinant” of \\([i \\ j \\ k; u_1 \\ u_2 \\ u_3; v_1 \\ v_2 \\ v_3]\\).\n\nCyclic shift \\( S \\). Permutation with \\( s_{21} = 1, s_{32} = 1, \\ldots \\), finally \\( s_{1n} = 1 \\). Its eigenvalues are \\( n \\)th roots \\( e^{2\\pi ik/n} \\) of 1; eigenvectors are columns of the Fourier matrix \\( F \\).\n\nDeterminant \\( |A| = \\det(A) \\). Defined by det \\( I = 1 \\), sign reversal for row exchange, and linearity in each row. Then \\( |A| = 0 \\) when \\( A \\) is singular. Also \\( |AB| = |A||B| \\) and \\( |A^{-1}| = 1/|A| \\) and \\( |A^T| = |A| \\). The big formula for \\( \\det(A) \\) has a sum of \\( n! \\) terms, the cofactor formula uses determinants of size \\( n - 1 \\), volume of box = \\( |\\det(A)| \\).\n\nDiagonal matrix \\( D \\). \\( d_{ij} = 0 \\) if \\( i \\neq j \\). Block-diagonal: zero outside square blocks \\( D_{ii} \\).\n\nDiagonalizable matrix \\( A \\). Must have \\( n \\) independent eigenvectors (in the columns of \\( S \\); automatic with \\( n \\) different eigenvalues). Then \\( S^{-1}AS = \\Lambda = \\text{eigenvalue matrix} \\).\n\nDiagonalization \\( \\Lambda = S^{-1}AS \\). \\( \\Lambda = \\text{eigenvalue matrix} \\) and \\( S = \\text{eigenvector matrix} \\). \\( A \\) must have \\( n \\) independent eigenvectors to make \\( S \\) invertible. All \\( A^k = SA^kS^{-1} \\).\n\nDimension of vector space \\( \\dim(V) = \\text{number of vectors in any basis for } V \\).\n\nDistributive Law \\( A(B + C) = AB + AC \\). Add then multiply, or multiply then add.\n\nDot product \\( x^Ty = x_1y_1 + \\cdots + x_ny_n \\). Complex dot product is \\( \\bar{x}^Ty \\). Perpendicular vectors have zero dot product. \\( (AB)_{ij} = (\\text{row } i \\text{ of } A) \\cdot (\\text{column } j \\text{ of } B) \\).\n\nEchelon matrix \\( U \\). The first nonzero entry (the pivot) in each row comes after the pivot in the previous row. All zero rows come last.\n\nEigenvalue \\( \\lambda \\) and eigenvector \\( x \\). \\( Ax = \\lambda x \\) with \\( x \\neq 0 \\) so \\( \\det(A - \\lambda I) = 0 \\).\n\nEigshow. Graphical 2 by 2 eigenvalues and singular values (MATLAB or Java).\n\nElimination. A sequence of row operations that reduces \\( A \\) to an upper triangular \\( U \\) or to the reduced form \\( R = \\text{rref}(A) \\). Then \\( A = LU \\) with multipliers \\( \\ell_{ij} \\) in \\( L \\), or \\( PA = LU \\) with row exchanges in \\( P \\), or \\( EA = R \\) with an invertible \\( E \\).\n\nElimination matrix = Elementary matrix \\( E_{ij} \\). The identity matrix with an extra \\(-\\ell_{ij}\\) in the \\( i, j \\) entry \\( (i \\neq j) \\). Then \\( E_{ij}A \\) subtracts \\( \\ell_{ij} \\) times row \\( j \\) of \\( A \\) from row \\( i \\).\n\nEllipse (or ellipsoid) \\( x^TAx = 1 \\). \\( A \\) must be positive definite; the axes of the ellipse are eigenvectors of \\( A \\), with lengths \\( 1/\\sqrt{\\lambda} \\). (For \\( \\|x\\| = 1 \\) the vectors \\( y = Ax \\) lie on the ellipse \\( \\|A^{-1}y\\|^2 = y^T(AA^T)^{-1}y = 1 \\) displayed by eigshow; axis lengths \\( \\sigma_i \\).)\n\nExponential \\( e^{At} = I + At + (At)^2/2! + \\cdots \\) has derivative \\( Ae^{At} \\); \\( e^{At}u(0) \\) solves \\( u' = Au \\).\nFactorization $A = LU$. If elimination takes $A$ to $U$ without row exchanges, then the lower triangular $L$ with multipliers $\\ell_{ij}$ (and $\\ell_{ii} = 1$) brings $U$ back to $A$.\n\nFast Fourier Transform (FFT). A factorization of the Fourier matrix $F_n$ into $\\ell = \\log_2 n$ matrices $S_i$ times a permutation. Each $S_i$ needs only $n/2$ multiplications, so $F_n x$ and $F_n^{-1} c$ can be computed with $n\\ell/2$ multiplications. Revolutionary.\n\nFibonacci numbers $0, 1, 1, 2, 3, 5, \\ldots$ satisfy $F_n = F_{n-1} + F_{n-2} = (\\lambda_1^n - \\lambda_2^n)/(\\lambda_1 - \\lambda_2)$. Growth rate $\\lambda_1 = (1 + \\sqrt{5})/2$ is the largest eigenvalue of the Fibonacci matrix $\\begin{bmatrix} 1 & 1 \\\\ 1 & 0 \\end{bmatrix}$.\n\nFour fundamental subspaces of $A = C(A), N(A), C(A^T), N(A^T)$.\n\nFourier matrix $F$. Entries $F_{jk} = e^{2\\pi ijk/n}$ give orthogonal columns $F^T F = nI$. Then $y = F c$ is the (inverse) Discrete Fourier Transform $y_j = \\sum c_k e^{2\\pi ijk/n}$.\n\nFree columns of $A$. Columns without pivots; combinations of earlier columns.\n\nFree variable $x_i$. Column $i$ has no pivot in elimination. We can give the $n - r$ free variables any values, then $A x = b$ determines the $r$ pivot variables (if solvable!).\n\nFull column rank $r = n$. Independent columns, $N(A) = \\{0\\}$, no free variables.\n\nFull row rank $r = m$. Independent rows, at least one solution to $A x = b$, column space is all of $\\mathbb{R}^m$. Full rank means full column rank or full row rank.\n\nFundamental Theorem. The nullspace $N(A)$ and row space $C(A^T)$ are orthogonal complements (perpendicular subspaces of $\\mathbb{R}^n$ with dimensions $r$ and $n - r$) from $A x = 0$. Applied to $A^T$, the column space $C(A)$ is the orthogonal complement of $N(A^T)$.\n\nGauss-Jordan method. Invert $A$ by row operations on $[A \\ I]$ to reach $[I \\ A^{-1}]$.\n\nGram-Schmidt orthogonalization $A = QR$. Independent columns in $A$, orthonormal columns in $Q$. Each column $q_j$ of $Q$ is a combination of the first $j$ columns of $A$ (and conversely, so $R$ is upper triangular). Convention: $\\text{diag}(R) > 0$.\n\nGraph $G$. Set of $n$ nodes connected pairwise by $m$ edges. A complete graph has all $n(n-1)/2$ edges between nodes. A tree has only $n-1$ edges and no closed loops. A directed graph has a direction arrow specified on each edge.\n\nHankel matrix $H$. Constant along each antidiagonal; $h_{ij}$ depends on $i + j$.\n\nHermitian matrix $A^H = A^T = A$. Complex analog of a symmetric matrix: $\\overline{a_{ij}} = a_{ij}$.\n\nHessenberg matrix $H$. Triangular matrix with one extra nonzero adjacent diagonal.\n\nHilbert matrix $\\text{hilb}(n)$. Entries $H_{ij} = 1/(i+j-1) = \\int_0^1 x^{i-1} x^{j-1} dx$. Positive definite but extremely small $\\lambda_{\\text{min}}$ and large condition number.\n\nHypercube matrix $P_L^2$. Row $n+1$ counts corners, edges, faces, . . . of a cube in $\\mathbb{R}^n$.\n\nIdentity matrix $I$ (or $I_n$). Diagonal entries $= 1$, off-diagonal entries $= 0$.\n\nIncidence matrix of a directed graph. The $m$ by $n$ edge-node incidence matrix has a row for each edge (node $i$ to node $j$), with entries $-1$ and $1$ in columns $i$ and $j$.\n\nIndefinite matrix. A symmetric matrix with eigenvalues of both signs (+ and −).\n\nIndependent vectors $v_1, \\ldots, v_k$. No combination $c_1 v_1 + \\cdots + c_k v_k = 0$ unless all $c_i = 0$. If the $v$’s are the columns of $A$, the only solution to $A x = 0$ is $x = 0$. \n\nGlossary\nInverse matrix $A^{-1}$. Square matrix with $A^{-1}A = I$ and $AA^{-1} = I$. No inverse if $\\det A = 0$ and $\\text{rank}(A) < n$ and $Ax = 0$ for a nonzero vector $x$. The inverses of $AB$ and $A^T$ are $B^{-1}A^{-1}$ and $(A^{-1})^T$. Cofactor formula $(A^{-1})_{ij} = C_{ji}/\\det A$.\n\nIterative method. A sequence of steps intended to approach the desired solution.\n\nJordan form $J = M^{-1}AM$. If $A$ has $s$ independent eigenvectors, its “generalized” eigenvector matrix $M$ gives $J = \\text{diag}(J_1, \\ldots, J_s)$. The block $J_k$ is $\\lambda_k I_k + N_k$ where $N_k$ has 1’s on diagonal 1. Each block has one eigenvalue $\\lambda_k$ and one eigenvector $(1, 0, \\ldots, 0)$.\n\nKirchhoff’s Laws. Current law: net current (in minus out) is zero at each node. Voltage law: Potential differences (voltage drops) add to zero around any closed loop.\n\nKronecker product (tensor product) $A \\otimes B$. Blocks $a_{ij}B$, eigenvalues $\\lambda_p(A)\\lambda_q(B)$.\n\nKrylov subspace $K_j(A, b)$. The subspace spanned by $b, Ab, \\ldots, A^{j-1}b$. Numerical methods approximate $A^{-1}b$ by $x_j$ with residual $b - Ax_j$ in this subspace. A good basis for $K_j$ requires only multiplication by $A$ at each step.\n\nLeast squares solution $\\hat{x}$. The vector $\\hat{x}$ that minimizes the error $\\|e\\|^2$ solves $A^T A \\hat{x} = A^T b$. Then $e = b - A \\hat{x}$ is orthogonal to all columns of $A$.\n\nLeft inverse $A^+$. If $A$ has full column rank $n$, then $A^+ = (A^T A)^{-1} A^T$ has $A^+ A = I_n$.\n\nLeft nullspace $\\mathcal{N}(A^T)$. Nullspace of $A^T$ is “left nullspace” of $A$ because $y^T A = 0^T$.\n\nLength $\\|x\\|$. Square root of $x^T x$ (Pythagoras in $n$ dimensions).\n\nLinear combination $cv + dw$ or $\\sum c_j v_j$. Vector addition and scalar multiplication.\n\nLinear transformation $T$. Each vector $v$ in the input space transforms to $T(v)$ in the output space, and linearity requires $T(cv + dw) = cT(v) + dT(w)$. Examples: Matrix multiplication $Av$, differentiation in function space.\n\nLinearly dependent $v_1, \\ldots, v_n$. A combination other than all $c_i = 0$ gives $\\sum c_i v_i = 0$.\n\nLucas numbers $L_n = 2, 1, 3, 4, \\ldots$ satisfy $L_n = L_{n-1} + L_{n-2} = \\lambda_1^n + \\lambda_2^n$, with eigenvalues $\\lambda_1, \\lambda_2 = (1 \\pm \\sqrt{5})/2$ of the Fibonacci matrix $\\begin{bmatrix} 1 & 1 \\\\ 1 & 0 \\end{bmatrix}$. Compare $L_0 = 2$ with Fibonacci.\n\nMarkov matrix $M$. All $m_{ij} \\geq 0$ and each column sum is 1. Largest eigenvalue $\\lambda = 1$. If $m_{ij} > 0$, the columns of $M^k$ approach the steady state eigenvector $Ms = s > 0$.\n\nMatrix multiplication $AB$. The $i, j$ entry of $AB$ is $(\\text{row } i \\text{ of } A) \\cdot (\\text{column } j \\text{ of } B) = \\sum a_{ik} b_{kj}$.\n\nBy columns: Column $j$ of $AB = A$ times column $j$ of $B$. By rows: row $i$ of $A$ multiplies $B$. Columns times rows: $AB = \\text{sum of } (\\text{column } k)(\\text{row } k)$. All these equivalent definitions come from the rule that $AB$ times $x$ equals $A$ times $Bx$.\n\nMinimal polynomial of $A$. The lowest degree polynomial with $m(A) = \\text{zero matrix}$. The roots of $m$ are eigenvalues, and $m(\\lambda)$ divides $\\det(A - \\lambda I)$.\n\nMultiplication $Ax = x_1(\\text{column } 1) + \\cdots + x_n(\\text{column } n) = \\text{combination of columns}$.\n\nMultiplicities $AM$ and $GM$. The algebraic multiplicity $AM$ of an eigenvalue $\\lambda$ is the number of times $\\lambda$ appears as a root of $\\det(A - \\lambda I) = 0$. The geometric multiplicity $GM$ is the number of independent eigenvectors (= dimension of the eigenspace for $\\lambda$).\nMultiplier $\\ell_{ij}$. The pivot row $j$ is multiplied by $\\ell_{ij}$ and subtracted from row $i$ to eliminate the $i, j$ entry: $\\ell_{ij} = (\\text{entry to eliminate})/(j\\text{th pivot})$.\n\nNetwork. A directed graph that has constants $c_1, \\ldots, c_m$ associated with the edges.\n\nNilpotent matrix $N$. Some power of $N$ is the zero matrix, $N^k = 0$. The only eigenvalue is $\\lambda = 0$ (repeated $n$ times). Examples: triangular matrices with zero diagonal.\n\nNorm $\\|A\\|$ of a matrix. The “$\\ell^2$ norm” is the maximum ratio $\\|Ax\\|/\\|x\\| = \\sigma_{\\text{max}}$. Then $\\|Ax\\| \\leq \\|A\\|\\|x\\|$ and $\\|AB\\| \\leq \\|A\\|\\|B\\|$ and $\\|A + B\\| \\leq \\|A\\| + \\|B\\|$. Frobenius norm $\\|A\\|_F = \\sum a_{ij}^2$; $\\ell^1$ and $\\ell^\\infty$ norms are largest column and row sums of $|a_{ij}|$.\n\nNormal equation $A^TA\\hat{x} = A^Tb$. Gives the least squares solution to $Ax = b$ if $A$ has full rank $n$. The equation says that (columns of $A$)$\\cdot(b - A\\hat{x}) = 0$.\n\nNormal matrix $N$. $NN^T = N^TN$, leads to orthonormal (complex) eigenvectors.\n\nNullspace $N(A) = \\text{Solutions to } Ax = 0$. Dimension $n - r = (# \\text{ columns}) - \\text{rank}$.\n\nNullspace matrix $N$. The columns of $N$ are the $n - r$ special solutions to $As = 0$. Orthogonal matrix $Q$. Square matrix with orthonormal columns, so $Q^TQ = I$ implies $Q^T = Q^{-1}$. Preserves length and angles, $\\|Qx\\| = \\|x\\|$ and $(Qx)^T(Qy) = x^Ty$. All $|\\lambda| = 1$, with orthogonal eigenvectors. Examples: Rotation, reflection, permutation.\n\nOrthogonal subspaces. Every $v$ in $V$ is orthogonal to every $w$ in $W$.\n\nOrthonormal vectors $q_1, \\ldots, q_n$. Dot products are $q_i^Tq_j = 0$ if $i \\neq j$ and $q_i^Tq_i = 1$. The matrix $Q$ with these orthonormal columns has $Q^TQ = I$. If $m = n$ then $Q^T = Q^{-1}$ and $q_1, \\ldots, q_n$ is an orthonormal basis for $\\mathbb{R}^n$: every $v = \\sum(v^Tq_j)q_j$.\n\nOuter product $uv^T = \\text{column times row} = \\text{rank one matrix}$.\n\nPartial pivoting. In elimination, the $j$th pivot is chosen as the largest available entry (in absolute value) in column $j$. Then all multipliers have $|\\ell_{ij}| \\leq 1$. Roundoff error is controlled (depending on the condition number of $A$).\n\nParticular solution $x_p$. Any solution to $Ax = b$; often $x_p$ has free variables = 0.\n\nPascal matrix $P_S = \\text{pascal}(n)$. The symmetric matrix with binomial entries $\\binom{i+j-2}{i-1}$. $P_S = P_iP_j$ all contain Pascal’s triangle with det = 1 (see index for more properties).\n\nPermutation matrix $P$. There are $n!$ orders of $1, \\ldots, n$; the $n!$ $P$’s have the rows of $I$ in those orders. $PA$ puts the rows of $A$ in the same order. $P$ is a product of row exchanges $P_{ij}$; $P$ is even or odd (det $P = 1$ or $-1$) based on the number of exchanges.\n\nPivot columns of $A$. Columns that contain pivots after row reduction; not combinations of earlier columns. The pivot columns are a basis for the column space.\n\nPivot $d$. The diagonal entry (first nonzero) when a row is used in elimination.\n\nPlane (or hyperplane) in $\\mathbb{R}^n$. Solutions to $a^T x = 0$ give the plane (dimension $n - 1$) perpendicular to $a \\neq 0$.\n\nPolar decomposition $A = QH$. Orthogonal $Q$, positive (semi)definite $H$.\n\nPositive definite matrix $A$. Symmetric matrix with positive eigenvalues and positive pivots. Definition: $x^TAx > 0$ unless $x = 0$. \nProjection $p = a(a^T b / a^T a)$ onto the line through $a$. $P = a a^T / a^T a$ has rank 1.\n\nProjection matrix $P$ onto subspace $S$. Projection $p = Pb$ is the closest point to $b$ in $S$, error $e = b - Pb$ is perpendicular to $S$. $P^2 = P = P^T$, eigenvalues are 1 or 0, eigenvectors are in $S$ or $S^\\perp$. If columns of $A$ is basis for $S$ then $P = A(A^T A)^{-1} A^T$.\n\nPseudoinverse $A^+$ (Moore-Penrose inverse). The $n$ by $m$ matrix that “inverts” $A$ from column space back to row space, with $N(A^+) = N(A^T)$. $A^+ A$ and $A A^+$ are the projection matrices onto the row space and column space. $\\text{Rank}(A^+) = \\text{rank}(A)$.\n\nRandom matrix $\\text{rand}(n)$ or $\\text{randn}(n)$. MATLAB creates a matrix with random entries, uniformly distributed on $[0 \\ 1]$ for $\\text{rand}$ and standard normal distribution for $\\text{randn}$.\n\nRank one matrix $A = u v^T \\neq 0$. Column and row spaces = lines $c u$ and $c v$.\n\nRank $r(A) =$ number of pivots = dimension of column space = dimension of row space.\n\nRayleigh quotient $q(x) = x^T A x / x^T x$ for symmetric $A$: $\\lambda_{\\min} \\leq q(x) \\leq \\lambda_{\\max}$. Those extremes are reached at the eigenvectors $x$ for $\\lambda_{\\min}(A)$ and $\\lambda_{\\max}(A)$.\n\nReduced row echelon form $R = \\text{rref}(A)$. Pivots = 1; zeros above and below pivots; $r$ nonzero rows of $R$ give a basis for the row space of $A$.\n\nReflection matrix $Q = I - 2 u u^T$. The unit vector $u$ is reflected to $Q u = -u$. All vectors $x$ in the plane mirror $u^T x = 0$ are unchanged because $Q x = x$. The “Householder matrix” has $Q^T = Q^{-1} = Q$.\n\nRight inverse $A^+$. If $A$ has full row rank $m$, then $A^+ = A^T (A A^T)^{-1}$ has $A A^+ = I_m$.\n\nRotation matrix $R = \\begin{bmatrix} \\cos \\theta & -\\sin \\theta \\\\ \\sin \\theta & \\cos \\theta \\end{bmatrix}$ rotates the plane by $\\theta$ and $R^{-1} = R^T$ rotates back by $-\\theta$. Orthogonal matrix, eigenvalues $e^{i\\theta}$ and $e^{-i\\theta}$, eigenvectors $(1, \\pm i)$.\n\nRow picture of $A x = b$. Each equation gives a plane in $\\mathbb{R}^n$; planes intersect at $x$.\n\nRow space $C(A^T) =$ all combinations of rows of $A$. Column vectors by convention.\n\nSaddle point of $f(x_1, \\ldots, x_n)$. A point where the first derivatives of $f$ are zero and the second derivative matrix $(\\partial^2 f / \\partial x_i \\partial x_j) =$ Hessian matrix is indefinite.\n\nSchur complement $S = D - C A^{-1} B$. Appears in block elimination on $[\\begin{bmatrix} A & B \\\\ C & D \\end{bmatrix}]$.\n\nSchwarz inequality $|v \\cdot w| \\leq \\|v\\| \\|w\\|$. Then $|v^T A w|^2 \\leq (v^T A v)(w^T A w)$ if $A = C^T C$.\n\nSemidefinite matrix $A$. (Positive) semidefinite means symmetric with $x^T A x \\geq 0$ for all vectors $x$. Then all eigenvalues $\\lambda \\geq 0$; no negative pivots.\n\nSimilar matrices $A$ and $B$. Every $B = M^{-1} A M$ has the same eigenvalues as $A$.\n\nSimplex method for linear programming. The minimum cost vector $x^*$ is found by moving from corner to lower cost corner along the edges of the feasible set (where the constraints $A x = b$ and $x \\geq 0$ are satisfied). Minimum cost at a corner!\n\nSingular matrix $A$. A square matrix that has no inverse: $\\det(A) = 0$.\n\nSingular Value Decomposition (SVD) $A = U \\Sigma V^T =$ (orthogonal $U$) times (diagonal $\\Sigma$) times (orthogonal $V^T$). First $r$ columns of $U$ and $V$ are orthonormal bases of $C(A)$ and $C(A^T)$ with $A v_i = \\sigma_i u_i$ and singular value $\\sigma_i > 0$. Last columns of $U$ and $V$ are orthonormal bases of the nullspaces of $A^T$ and $A$. \n\n\nSkew-symmetric matrix $K$. The transpose is $-K$, since $K_{ij} = -K_{ji}$. Eigenvalues are pure imaginary, eigenvectors are orthogonal, $e^{Kt}$ is an orthogonal matrix.\n\nSolvable system $Ax = b$. The right side $b$ is in the column space of $A$.\n\nSpanning set $v_1, \\ldots, v_m$ for $V$. Every vector in $V$ is a combination of $v_1, \\ldots, v_m$.\n\nSpecial solutions to $AS = 0$. One free variable is $s_i = 1$, other free variables $= 0$.\n\nSpectral theorem $A = QΛQ^T$. Real symmetric $A$ has real $λ_i$ and orthonormal $q_i$ with $Aq_i = λ_iq_i$. In mechanics the $q_i$ give the principal axes.\n\nSpectrum of $A$ = the set of eigenvalues $\\{λ_1, \\ldots, λ_n\\}$. Spectral radius $= |λ_{\\text{max}}|$.\n\nStandard basis for $R^n$. Columns of $n$ by $n$ identity matrix (written $i, j, k$ in $R^3$).\n\nStiffness matrix $K$. If $x$ gives the movements of the nodes in a discrete structure, $Kx$ gives the internal forces. Often $K = A^TCA$ where $C$ contains spring constants from Hooke’s Law and $Ax = \\text{stretching (strains)}$ from the movements $x$.\n\nSubspace $S$ of $V$. Any vector space inside $V$, including $V$ and $Z = \\{\\text{zero vector}\\}$.\n\nSum $V + W$ of subspaces. Space of all $(v \\text{ in } V) + (w \\text{ in } W)$. Direct sum: $\\dim(V + W) = \\dim V + \\dim W$ when $V$ and $W$ share only the zero vector.\n\nSymmetric factorizations $A = LDL^T$ and $A = QΛQ^T$. The number of positive pivots in $D$ and positive eigenvalues in $Λ$ is the same.\n\nSymmetric matrix $A$. The transpose is $A^T = A$, and $a_{ij} = a_{ji}$. $A^{-1}$ is also symmetric. All matrices of the form $R^T R$ and $LDL^T$ and $QΛQ^T$ are symmetric. Symmetric matrices have real eigenvalues in $Λ$ and orthonormal eigenvectors in $Q$.\n\nToeplitz matrix $T$. Constant-diagonal matrix, so $t_{ij}$ depends only on $j - i$. Toeplitz matrices represent linear time-invariant filters in signal processing.\n\nTrace of $A$ = sum of diagonal entries = sum of eigenvalues of $A$. $\\text{Tr} AB = \\text{Tr} BA$.\n\nTranspose matrix $A^T$. Entries $A^T_{ij} = A_{ji}$. $A^T$ is $n$ by $m$, $A^T A$ is square, symmetric, positive semidefinite. The transposes of $AB$ and $A^{-1}$ are $B^T A^T$ and $(A^T)^{-1}$.\n\nTriangle inequality $\\|u + v\\| \\leq \\|u\\| + \\|v\\|$. For matrix norms $\\|A + B\\| \\leq \\|A\\| + \\|B\\|$.\n\nTridiagonal matrix $T$: $t_{ij} = 0$ if $|i - j| > 1$. $T^{-1}$ has rank 1 above and below diagonal.\n\nUnitary matrix $U^H = U^T = U^{-1}$. Orthonormal columns (complex analog of $Q$).\n\nVandermonde matrix $V$. $Vc = b$ gives the polynomial $p(x) = c_0 + \\cdots + c_{n-1}x^{n-1}$ with $p(x_i) = b_i$ at $n$ points. $V_{ij} = (x_i)^{j-1}$ and $\\det V = \\text{product of } (x_k - x_i)$ for $k > i$.\n\nVector $v$ in $R^n$. Sequence of $n$ real numbers $v = (v_1, \\ldots, v_n) = \\text{point in } R^n$.\n\nVector addition. $v + w = (v_1 + w_1, \\ldots, v_n + w_n) = \\text{diagonal of parallelogram}$.\n\nVector space $V$. Set of vectors such that all combinations $cv + dw$ remain in $V$. Eight required rules are given in Section 3.1 for $cv + dw$.\n\nVolume of box. The rows (or columns) of $A$ generate a box with volume $|\\det(A)|$.\n\nWavelets $w_{jk}(t)$ or vectors $w_{jk}$. Stretch and shift the time axis to create $w_{jk}(t) = w_{00}(2^j t - k)$. Vectors from $w_{00} = (1, 1, -1, -1)$ would be $(1, -1, 0, 0)$ and $(0, 0, 1, -1)$.", "id": "./materials/52.pdf" }, { "contents": "Chapter 9\nMatrices and Determinants\n\n9.1 Introduction:\nIn many economic analysis, variables are assumed to be related by sets of linear equations. Matrix algebra provides a clear and concise notation for the formulation and solution of such problems, many of which would be complicated in conventional algebraic notation. The concept of determinant and is based on that of matrix. Hence we shall first explain a matrix.\n\n9.2 Matrix:\nA set of mn numbers (real or complex), arranged in a rectangular formation (array or table) having m rows and n columns and enclosed by a square bracket [ ] is called m×n matrix (read “m by n matrix”).\n\nAn m×n matrix is expressed as\n\n\\[\nA = \\begin{bmatrix}\n a_{11} & a_{12} & \\cdots & a_{1n} \\\\\n a_{21} & a_{22} & \\cdots & a_{2n} \\\\\n \\vdots & \\vdots & \\ddots & \\vdots \\\\\n a_{m1} & a_{m2} & \\cdots & a_{mn}\n\\end{bmatrix}\n\\]\n\nThe letters \\(a_{ij}\\) stand for real numbers. Note that \\(a_{ij}\\) is the element in the \\(i\\)th row and \\(j\\)th column of the matrix. Thus the matrix \\(A\\) is sometimes denoted by simplified form as \\((a_{ij})\\) or by \\(\\{a_{ij}\\}\\) i.e., \\(A = (a_{ij})\\).\n\nMatrices are usually denoted by capital letters \\(A, B, C\\) etc and its elements by small letters \\(a, b, c\\) etc.\n\nOrder of a Matrix:\nThe order or dimension of a matrix is the ordered pair having as first component the number of rows and as second component the number of columns in the matrix. If there are 3 rows and 2 columns in a matrix, then its order is written as \\((3, 2)\\) or \\((3 \\times 2)\\) read as three by two. In general if \\(m\\) are rows and \\(n\\) are columns of a matrix, then its order is \\((m \\times n)\\).\n\nExamples:\n9.3 Some types of matrices:\n\n1. Row Matrix and Column Matrix:\n A matrix consisting of a single row is called a row matrix or a row vector, whereas a matrix having single column is called a column matrix or a column vector.\n\n2. Null or Zero Matrix:\n A matrix in which each element is ‘0’ is called a Null or Zero matrix. Zero matrices are generally denoted by the symbol O. This distinguishes zero matrix from the real number 0.\n\n For example, \\( O = \\begin{bmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\end{bmatrix} \\) is a zero matrix of order 2 x 4.\n\n The matrix \\( O_{mxn} \\) has the property that for every matrix \\( A_{mxn} \\),\n \\[ A + O = O + A = A \\]\n\n3. Square matrix:\n A matrix \\( A \\) having same numbers of rows and columns is called a square matrix. A matrix \\( A \\) of order \\( m \\times n \\) can be written as \\( A_{mxn} \\). If \\( m = n \\), then the matrix is said to be a square matrix. A square matrix of order \\( n \\times n \\), is simply written as \\( A_n \\).\n\n Thus \\( \\begin{bmatrix} 2 & 5 \\\\ 1 & 3 \\end{bmatrix} \\) and \\( \\begin{bmatrix} a & b & c \\\\ d & e & f \\\\ g & h & i \\end{bmatrix} \\) are square matrix of order 2 and 3\n\nMain or Principal (leading)Diagonal:\n The principal diagonal of a square matrix is the ordered set of elements \\( a_{ij} \\), where \\( i = j \\), extending from the upper left-hand corner to the lower right-hand corner of the matrix. Thus, the principal diagonal contains elements \\( a_{11}, a_{22}, a_{33} \\) etc.\n\n For example, the principal diagonal of\nconsists of elements 1, 2 and 0, in that order.\n\n**Particular cases of a square matrix:**\n\n(a) **Diagonal matrix:**\n\nA square matrix in which all elements are zero except those in the main or principal diagonal is called a diagonal matrix. Some elements of the principal diagonal may be zero but not all.\n\nFor example\n\n\\[\n\\begin{bmatrix}\n1 & 0 & 0 \\\\\n0 & 1 & 0 \\\\\n0 & 0 & 0\n\\end{bmatrix}\n\\]\n\nare diagonal matrices.\n\nIn general\n\n\\[\nA = \\begin{bmatrix}\na_{11} & a_{12} & \\cdots & a_{1n} \\\\\na_{21} & a_{22} & \\cdots & a_{2n} \\\\\n\\vdots & \\vdots & \\ddots & \\vdots \\\\\na_{n1} & a_{n2} & \\cdots & a_{nn}\n\\end{bmatrix} = (a_{ij})_{n \\times n}\n\\]\n\nis a diagonal matrix if and only if\n\n\\[\na_{ij} = 0 \\quad \\text{for } i \\neq j\n\\]\n\n\\[\na_{ij} \\neq 0 \\quad \\text{for at least one } i = j\n\\]\n\n(b) **Scalar Matrix:**\n\nA diagonal matrix in which all the diagonal elements are same, is called a scalar matrix i.e.\n\nThus\n\n\\[\n\\begin{bmatrix}\n3 & 0 \\\\\n0 & 3\n\\end{bmatrix}\n\\quad \\text{and} \\quad\n\\begin{bmatrix}\nk & 0 & 0 \\\\\n0 & k & 0 \\\\\n0 & 0 & k\n\\end{bmatrix}\n\\]\n\nare scalar matrices\n\n(c) **Identity Matrix or Unit matrix:**\n\nA scalar matrix in which each diagonal element is 1(unity) is called a unit matrix. An identity matrix of order n is denoted by \\(I_n\\).\nThus \\( I_2 = \\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix} \\) and \\( I_3 = \\begin{bmatrix} 1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 1 \\end{bmatrix} \\)\n\nare the identity matrices of order 2 and 3.\n\nIn general, \\( A = \\begin{bmatrix} a_{11} & a_{12} & \\cdots & a_{1n} \\\\ a_{21} & a_{22} & \\cdots & a_{2n} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ a_{m1} & a_{m2} & \\cdots & a_{mn} \\end{bmatrix} = [a_{ij}]_{mxn} \\)\n\nis an identity matrix if and only if\n\n\\( a_{ij} = 0 \\) for \\( i \\neq j \\) and \\( a_{ij} = 1 \\) for \\( i = j \\)\n\nNote: If a matrix \\( A \\) and identity matrix \\( I \\) are comformable for multiplication, then \\( I \\) has the property that\n\n\\( AI = IA = A \\) i.e., \\( I \\) is the identity matrix for multiplication.\n\n4. **Equal Matrices:**\n\nTwo matrices \\( A \\) and \\( B \\) are said to be equal if and only if they have the same order and each element of matrix \\( A \\) is equal to the corresponding element of matrix \\( B \\) i.e for each \\( i, j \\), \\( a_{ij} = b_{ij} \\)\n\nThus \\( A = \\begin{bmatrix} 2 \\\\ 3 \\\\ 0 \\end{bmatrix} \\) and \\( B = \\begin{bmatrix} \\frac{4}{2} \\\\ \\frac{2}{\\sqrt{9}} \\\\ 2 - 1 \\\\ 0 \\end{bmatrix} \\)\n\nthen \\( A = B \\) because the order of matrices \\( A \\) and \\( B \\) is same and \\( a_{ij} = b_{ij} \\) for every \\( i, j \\).\n\n**Example 1:** Find the values of \\( x, y, z \\) and \\( a \\) which satisfy the matrix equation\n\n\\[\n\\begin{bmatrix} x + 3 & 2y + x \\\\ z - 1 & 4a - 6 \\end{bmatrix} = \\begin{bmatrix} 0 & -7 \\\\ 3 & 2a \\end{bmatrix}\n\\]\n\n**Solution:** By the definition of equality of matrices, we have\n\n\\[\n\\begin{align*}\nx + 3 &= 0 \\quad \\text{..........................(1)} \\\\\n2y + x &= -7 \\quad \\text{..........................(2)} \\\\\nz - 1 &= 3 \\quad \\text{..........................(3)} \\\\\n4a - 6 &= 2a \\quad \\text{..........................(4)}\n\\end{align*}\n\\]\n\nFrom (1) \\( x = -3 \\)\nPut the value of \\( x \\) in (2), we get \\( y = -2 \\)\nFrom (3) \\( z = 4 \\)\nFrom (4) \\( a = 3 \\)\n\n5. **The Negative of a Matrix:**\n\nThe negative of the matrix \\( A_{mxn} \\), denoted by \\(-A_{mxn}\\), is the matrix formed by replacing each element in the matrix \\( A_{mxn} \\) with its additive inverse. For example,\n\n\\[\nA_{3x2} = \\begin{bmatrix}\n3 & -1 \\\\\n2 & -2 \\\\\n-4 & 5\n\\end{bmatrix}\n\\]\n\nThen \\(-A_{3x2} = \\begin{bmatrix}\n-3 & 1 \\\\\n-2 & 2 \\\\\n4 & -5\n\\end{bmatrix}\\)\n\nfor every matrix \\( A_{mxn} \\), the matrix \\(-A_{mxn}\\) has the property that\n\\( A + (-A) = (-A) + A = 0 \\)\ni.e., \\((-A)\\) is the additive inverse of \\( A \\).\n\nThe sum \\( B_{m-n} + (-A_{mxn}) \\) is called the difference of \\( B_{mxn} \\) and \\( A_{mxn} \\) and is denoted by \\( B_{mxn} - A_{mxn} \\).\n\n9.4 **Operations on matrices:**\n\n(a) **Multiplication of a Matrix by a Scalar:**\n\nIf \\( A \\) is a matrix and \\( k \\) is a scalar (constant), then \\( kA \\) is a matrix whose elements are the elements of \\( A \\), each multiplied by \\( k \\)\n\nFor example, if \\( A = \\begin{bmatrix}\n4 & -3 \\\\\n8 & -2 \\\\\n-1 & 0\n\\end{bmatrix}\\) then for a scalar \\( k \\),\n\n\\[\nkA = \\begin{bmatrix}\n4k & -3k \\\\\n8k & -2k \\\\\n-k & 0\n\\end{bmatrix}\n\\]\nAlso, \\[\n\\begin{bmatrix}\n3 & 0 & 3 & -5 \\\\\n3 & -1 & 4 & 0\n\\end{bmatrix}\n= \\begin{bmatrix}\n15 & -24 & 12 \\\\\n9 & -3 & 12\n\\end{bmatrix}\n\\]\n\n(b) **Addition and subtraction of Matrices:**\n\nIf A and B are two matrices of same order \\(m \\times n\\) then their sum \\(A + B\\) is defined as \\(C\\), \\(m \\times n\\) matrix such that each element of \\(C\\) is the sum of the corresponding elements of \\(A\\) and \\(B\\).\n\nFor example\n\nIf \\(A = \\begin{bmatrix} 3 & 1 & 2 \\\\ 2 & 1 & 4 \\end{bmatrix}\\) and \\(B = \\begin{bmatrix} 1 & 0 & 2 \\\\ -1 & 3 & 0 \\end{bmatrix}\\)\n\nThen \\(C = A + B = \\begin{bmatrix} 3+1 & 1+0 & 2+2 \\\\ 2-1 & 1+3 & 4+0 \\end{bmatrix} = \\begin{bmatrix} 4 & 1 & 4 \\\\ 1 & 4 & 4 \\end{bmatrix}\\)\n\nSimilarly, the difference \\(A - B\\) of the two matrices \\(A\\) and \\(B\\) is a matrix each element of which is obtained by subtracting the elements of \\(B\\) from the corresponding elements of \\(A\\).\n\nThus if \\(A = \\begin{bmatrix} 6 & 2 \\\\ 7 & -5 \\end{bmatrix}\\), \\(B = \\begin{bmatrix} 8 & 1 \\\\ 3 & 4 \\end{bmatrix}\\)\n\nthen \\(A - B = \\begin{bmatrix} 6 & 2 \\\\ 7 & -5 \\end{bmatrix} - \\begin{bmatrix} 8 & 1 \\\\ 3 & 4 \\end{bmatrix} = \\begin{bmatrix} 6-8 & 2-1 \\\\ 7-3 & -5-4 \\end{bmatrix} = \\begin{bmatrix} -2 & 1 \\\\ 4 & -9 \\end{bmatrix}\\)\n\nIf \\(A\\), \\(B\\) and \\(C\\) are the matrices of the same order \\(m \\times n\\)\n\nthen \\(A + B = B + A\\)\n\nand \\((A + B) + C = A + (B + C)\\) i.e., the addition of matrices is commutative and associative respectively.\n\n**Note:** The sum or difference of two matrices of different order is not defined.\n\n(c) **Product of Matrices:**\n\nTwo matrices \\(A\\) and \\(B\\) are said to be conformable for the product \\(AB\\) if the number of columns of \\(A\\) is equal to the number of rows of \\(B\\).\n\nThen the product matrix \\(AB\\) has the same number of rows as \\(A\\) and the same number of columns as \\(B\\).\n\nThus the product of the matrices \\(A_{mxp}\\) and \\(B_{pxn}\\) is the matrix \\((AB)_{mxn}\\). The elements of \\(AB\\) are determined as follows:\nThe element $C_{ij}$ in the $i$th row and $j$th column of $(AB)_{mxn}$ is found by\n\n$$c_{ij} = a_{i1}b_{1j} + a_{i2}b_{2j} + a_{i3}b_{3j} + \\ldots + a_{in}b_{nj}$$\n\nfor example, consider the matrices\n\n$$A_{2x2} = \\begin{bmatrix} a_{11} & a_{12} \\\\ a_{21} & a_{22} \\end{bmatrix} \\quad \\text{and} \\quad B_{2x2} = \\begin{bmatrix} b_{11} & b_{12} \\\\ b_{21} & b_{22} \\end{bmatrix}$$\n\nSince the number of columns of $A$ is equal to the number of rows of $B$, the product $AB$ is defined and is given as\n\n$$AB = \\begin{bmatrix} a_{11}b_{11} + a_{12}b_{21} & a_{11}b_{12} + a_{12}b_{22} \\\\ a_{21}b_{11} + a_{22}b_{21} & a_{21}b_{12} + a_{22}b_{22} \\end{bmatrix}$$\n\nThus $c_{11}$ is obtained by multiplying the elements of the first row of $A$ i.e., $a_{11}, a_{12}$ by the corresponding elements of the first column of $B$ i.e., $b_{11}, b_{21}$ and adding the product.\n\nSimilarly, $c_{12}$ is obtained by multiplying the elements of the first row of $A$ i.e., $a_{11}, a_{12}$ by the corresponding elements of the second column of $B$ i.e., $b_{12}, b_{22}$ and adding the product. Similarly for $c_{21}, c_{22}$.\n\n**Note:**\n\n1. Multiplication of matrices is not commutative i.e., $AB \\neq BA$ in general.\n2. For matrices $A$ and $B$ if $AB = BA$ then $A$ and $B$ commute to each other.\n3. A matrix $A$ can be multiplied by itself if and only if it is a square matrix. The product $A.A$ in such cases is written as $A^2$.\n Similarly we may define higher powers of a square matrix i.e., $A^2 = A^3$, $A^2 = A^4$.\n4. In the product $AB$, $A$ is said to be pre multiple of $B$ and $B$ is said to be post multiple of $A$.\n\n**Example 1:** If $A = \\begin{bmatrix} 1 & 2 \\\\ -1 & 3 \\end{bmatrix}$ and $B = \\begin{bmatrix} 2 & 1 \\\\ 1 & 1 \\end{bmatrix}$ Find $AB$ and $BA$.\n\n**Solution:**\n\n$$AB = \\begin{bmatrix} 1 & 2 \\\\ -1 & 3 \\end{bmatrix} \\begin{bmatrix} 2 & 1 \\\\ 1 & 1 \\end{bmatrix} = \\begin{bmatrix} 2+2 & 1+2 \\\\ -2+3 & -1+3 \\end{bmatrix} = \\begin{bmatrix} 4 & 3 \\\\ 1 & 2 \\end{bmatrix}$$\nThis example shows very clearly that multiplication of matrices in general, is not commutative i.e., \\( AB \\neq BA \\).\n\nExample 2: If\n\n\\[\nA = \\begin{bmatrix}\n3 & 1 & 2 \\\\\n1 & 0 & 1\n\\end{bmatrix}\n\\quad \\text{and} \\quad\nB = \\begin{bmatrix}\n1 & -1 \\\\\n2 & 1 \\\\\n3 & 1\n\\end{bmatrix},\n\\]\n\nfind \\( AB \\).\n\nSolution:\n\nSince \\( A \\) is a \\((2 \\times 3)\\) matrix and \\( B \\) is a \\((3 \\times 2)\\) matrix, they are conformable for multiplication. We have\n\n\\[\nAB = \\begin{bmatrix}\n3 & 1 & 2 \\\\\n1 & 0 & 1\n\\end{bmatrix}\n\\begin{bmatrix}\n1 & -1 \\\\\n2 & 1 \\\\\n3 & 1\n\\end{bmatrix}\n= \\begin{bmatrix}\n3+2+6 & -3+1+2 \\\\\n1+0+3 & -1+0+1\n\\end{bmatrix}\n= \\begin{bmatrix}\n11 & 0 \\\\\n4 & 0\n\\end{bmatrix}\n\\]\n\nRemark:\n\nIf \\( A, B \\) and \\( C \\) are the matrices of order \\((m \\times p), (p \\times q)\\) and \\((q \\times n)\\) respectively, then\n\ni. \\((AB)C = A(BC)\\) i.e., Associative law holds.\n\n\\[\nC(A+B) = CA + CB\n\\]\n\nand \\((A + B)C = AC + BC\\) i.e distributive laws holds.\n\nNote: that if a matrix \\( A \\) and identity matrix \\( I \\) are conformable for multiplication, then \\( I \\) has the property that\n\n\\( AI = IA = A \\) i.e, \\( I \\) is the identity matrix for multiplication.\n\nExercise 9.1\n\nQ.No. 1 Write the following matrices in tabular form:\n\ni. \\( A = [a_{ij}] \\), where \\( i = 1, 2, 3 \\) and \\( j = 1, 2, 3, 4 \\)\n\nii. \\( B = [b_{ij}] \\), where \\( i = 1 \\) and \\( j = 1, 2, 3, 4 \\)\n\niii. \\( C = [c_{jk}] \\), where \\( j = 1, 2, 3 \\) and \\( k = 1 \\)\nQ.No.2 Write each sum as a single matrix:\n\ni. \\[\n\\begin{bmatrix}\n2 & 1 & 4 \\\\\n3 & -1 & 0\n\\end{bmatrix}\n+ \n\\begin{bmatrix}\n6 & 3 & 0 \\\\\n-2 & 1 & 0\n\\end{bmatrix}\n\\]\n\nii. \\[\n\\begin{bmatrix}\n1 & 3 & 5 & 6 \\\\\n4 & 6 & 2 & 1\n\\end{bmatrix}\n+ \n\\begin{bmatrix}\n0 & -2 & 1 & 3 \\\\\n0 & 0 & 0 & 0\n\\end{bmatrix}\n\\]\n\niii. \\[\n\\begin{bmatrix}\n3 \\\\\n-1\n\\end{bmatrix}\n+ \n\\begin{bmatrix}\n0 \\\\\n-2\n\\end{bmatrix}\n\\]\n\niv. \\[\n\\begin{bmatrix}\n2 & 3 & 4 \\\\\n-1 & 6 & 2\n\\end{bmatrix}\n+ \n\\begin{bmatrix}\n0 & 0 & 0 \\\\\n1 & 0 & 3\n\\end{bmatrix}\n\\]\n\nv. \\[\n\\begin{bmatrix}\n6 & 1 \\\\\n-1 & 2\n\\end{bmatrix}\n+ \n\\begin{bmatrix}\n4 & 2 \\\\\n-5 & -1\n\\end{bmatrix}\n\\]\n\nQ.3 Show that \\[\n\\begin{bmatrix}\nb_{11} - a_{11} & b_{12} - a_{12} \\\\\nb_{21} - a_{21} & b_{22} - a_{22}\n\\end{bmatrix}\n\\]\nis a solution of the matrix equation \\(X + A = B\\), where \\(A = \\begin{bmatrix} a_{11} & a_{12} \\\\ a_{21} & a_{22} \\end{bmatrix}\\) and \\(B = \\begin{bmatrix} b_{11} & b_{12} \\\\ b_{21} & b_{22} \\end{bmatrix}\\).\n\nQ.4 Solve each of the following matrix equations:\n\ni. \\[\nX + \\begin{bmatrix} 3 & -1 \\\\ 2 & 2 \\end{bmatrix} = \\begin{bmatrix} 5 & 1 \\\\ -3 & 1 \\end{bmatrix}\n\\]\n\nii. \\[\nX + \\begin{bmatrix} -1 & 0 \\\\ 0 & 2 \\end{bmatrix} = \\begin{bmatrix} 2 & 6 \\\\ 1 & 5 \\end{bmatrix} + \\begin{bmatrix} -4 & -8 \\\\ -2 & 0 \\end{bmatrix}\n\\]\n\niii. \\[\n3X + \\begin{bmatrix} 2 & 1 & 3 \\\\ 4 & -1 & 5 \\end{bmatrix} = \\begin{bmatrix} -2 & 3 & 1 \\\\ 0 & 1 & 5 \\end{bmatrix}\n\\]\n\niv. \\[\nX + 2I = \\begin{bmatrix} 3 & -1 \\\\ 1 & 2 \\end{bmatrix}\n\\]\nQ.5 Write each product as a single matrix:\n\ni. \\[\n\\begin{bmatrix}\n3 & 1 & -1 \\\\\n0 & -1 & 2 \\\\\n1 & 0 & 0\n\\end{bmatrix}\n\\begin{bmatrix}\n1 & -1 \\\\\n0 & 2 \\\\\n1 & 0\n\\end{bmatrix}\n\\]\n\nii. \\[\n\\begin{bmatrix}\n3 & -2 & 2 \\\\\n2 & -2 & -2\n\\end{bmatrix}\n\\begin{bmatrix}\n1 \\\\\n-2\n\\end{bmatrix}\n\\]\n\niii. \\[\n\\begin{bmatrix}\n1 & 1 & -2 \\\\\n1 & 0 & -1 \\\\\n-1 & -2 & 5\n\\end{bmatrix}\n\\begin{bmatrix}\n-1 & -2 & 5 \\\\\n2 & -2 & -1\n\\end{bmatrix}\n\\]\n\niv. \\[\n\\begin{bmatrix}\n-1 & -1 & 3 \\\\\n-1 & -2 & 4 \\\\\n1 & 0 & -1\n\\end{bmatrix}\n\\begin{bmatrix}\n1 & 1 & -2 \\\\\n1 & 1 & -2 \\\\\n1 & 0 & -1\n\\end{bmatrix}\n\\]\n\nQ.6 If \\( A = \\begin{bmatrix} 1 & 4 \\\\ 2 & 1 \\end{bmatrix} \\), \\( B = \\begin{bmatrix} -3 & 2 \\\\ 4 & 0 \\end{bmatrix} \\), \\( C = \\begin{bmatrix} 1 & 0 \\\\ 0 & 2 \\end{bmatrix} \\), find \\( A^2 + BC \\).\n\nQ.7 Show that if \\( A = \\begin{bmatrix} -1 & 2 \\\\ 0 & 1 \\end{bmatrix} \\) and \\( B = \\begin{bmatrix} 1 & 0 \\\\ -1 & 2 \\end{bmatrix} \\), then\n\n(a) \\( (A + B)(A + B) \\neq A^2 + 2AB + B^2 \\)\n\n(b) \\( (A + B)(A - B) \\neq A^2 - B^2 \\)\n\nQ.8 Show that:\n\n(i) \\[\n\\begin{bmatrix}\n-1 & 2 & 3 \\\\\n2 & 1 & 0 \\\\\n3 & 5 & -1\n\\end{bmatrix}\n\\begin{bmatrix}\na \\\\\nb \\\\\nc\n\\end{bmatrix}\n= \\begin{bmatrix}\n-a + 2b + 3c \\\\\n2a + b \\\\\n3a + 5b - c\n\\end{bmatrix}\n\\]\n\n(ii) \\[\n\\begin{bmatrix}\n\\cos \\theta & 0 & -\\sin \\theta \\\\\n0 & 1 & 0 \\\\\n+\\sin \\theta & 0 & \\cos \\theta\n\\end{bmatrix}\n\\begin{bmatrix}\n\\cos \\theta & 0 & +\\sin \\theta \\\\\n0 & 1 & 0 \\\\\n-\\sin \\theta & 0 & \\cos \\theta\n\\end{bmatrix}\n= \\begin{bmatrix} 1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 1 \\end{bmatrix}\n\\]\n\nQ.9 If \\( A = \\begin{bmatrix} 2 & -2\\sqrt{2} \\\\ \\sqrt{2} & 2 \\end{bmatrix} \\) and \\( B = \\begin{bmatrix} 2 & 2\\sqrt{2} \\\\ -\\sqrt{2} & 2 \\end{bmatrix} \\), show that \\( A \\) and \\( B \\) commute.\nAnswers 9.1\n\nQ.1 (i) \\[\n\\begin{bmatrix}\na_{11} & a_{12} & a_{13} & a_{14} \\\\\na_{21} & a_{22} & a_{23} & a_{24} \\\\\na_{31} & a_{32} & a_{33} & a_{34}\n\\end{bmatrix}\n\\]\n\n(ii) \\[\n\\begin{bmatrix}\nb_{11} & b_{12} & b_{13} & b_{14}\n\\end{bmatrix}\n\\]\n\n(iii) \\[\n\\begin{bmatrix}\nc_{11} \\\\\nc_{21} \\\\\nc_{31}\n\\end{bmatrix}\n\\]\n\nQ.2 (i) \\[\n\\begin{bmatrix}\n8 & 4 & 4 \\\\\n1 & 0 & 0\n\\end{bmatrix}\n\\]\n\n(ii) \\[\n\\begin{bmatrix}\n1 & 1 & 6 & 9\n\\end{bmatrix}\n\\]\n\n(iii) \\[\n\\begin{bmatrix}\n10 \\\\\n3 \\\\\n-3\n\\end{bmatrix}\n\\]\n\n(iv) \\[\n\\begin{bmatrix}\n2 & 3 & 4 \\\\\n-1 & 6 & 2 \\\\\n1 & 0 & 3\n\\end{bmatrix}\n\\]\n\n(v) \\[\n\\begin{bmatrix}\n0 & 4 & 0 \\\\\n-9 & 13 & 7\n\\end{bmatrix}\n\\]\n\nQ.4 (i) \\[\n\\begin{bmatrix}\n2 & 2 \\\\\n-5 & -1\n\\end{bmatrix}\n\\]\n\n(ii) \\[\n\\begin{bmatrix}\n-1 & -2 \\\\\n-1 & 3\n\\end{bmatrix}\n\\]\n\n(iii) \\[\n\\begin{bmatrix}\n-1 & 1 & -\\frac{1}{3} \\\\\n-1 & -1 & -1 \\\\\n\\frac{4}{3} & \\frac{2}{3} & 0\n\\end{bmatrix}\n\\]\n\n(iv) \\[\n\\begin{bmatrix}\n1 & -1 \\\\\n1 & 0\n\\end{bmatrix}\n\\]\n\nQ.5 (i) \\[\n\\begin{bmatrix}\n2 & -1 \\\\\n2 & -2\n\\end{bmatrix}\n\\]\n\n(ii) \\[\n\\begin{bmatrix}\n-1\n\\end{bmatrix}\n\\]\n\n(iii) \\[\n\\begin{bmatrix}\n1 & 0 & 0 \\\\\n0 & 1 & 0 \\\\\n0 & 0 & 1\n\\end{bmatrix}\n\\]\n\n(iv) \\[\n\\begin{bmatrix}\n1 & 0 & 0 \\\\\n0 & 1 & 0 \\\\\n0 & 0 & 1\n\\end{bmatrix}\n\\]\n9.5 Determinants:\n\nThe Determinant of a Matrix:\n\nThe determinant of a matrix is a scalar (number), obtained from the elements of a matrix by specified operations, which is characteristic of the matrix. The determinants are defined only for square matrices. It is denoted by det A or |A| for a square matrix A.\n\nThe determinant of the (2 x 2) matrix\n\n\\[\nA = \\begin{bmatrix}\na_{11} & a_{12} \\\\\na_{21} & a_{22}\n\\end{bmatrix}\n\\]\n\nis given by\n\n\\[\n\\text{det } A = |A| = \\begin{vmatrix}\na_{11} & a_{12} \\\\\na_{21} & a_{22}\n\\end{vmatrix} = a_{11}a_{22} - a_{12}a_{21}\n\\]\n\nExample 3: If \\( A = \\begin{bmatrix} 3 & 1 \\\\ -2 & 3 \\end{bmatrix} \\) find \\( |A| \\)\n\nSolution:\n\n\\[\n|A| = \\begin{vmatrix} 3 & 1 \\\\ -2 & 3 \\end{vmatrix} = 9 - (-2) = 9 + 2 = 11\n\\]\n\nThe determinant of the (3 x 3) matrix\n\n\\[\nA = \\begin{bmatrix}\na_{11} & a_{12} & a_{13} \\\\\na_{21} & a_{22} & a_{23} \\\\\na_{31} & a_{32} & a_{33}\n\\end{bmatrix}, \\text{ denoted by } |A| = \\begin{vmatrix}\na_{11} & a_{12} & a_{13} \\\\\na_{21} & a_{22} & a_{23} \\\\\na_{31} & a_{32} & a_{33}\n\\end{vmatrix}\n\\]\n\nis given as,\n\n\\[\n\\text{det } A = |A| = a_{11} \\begin{vmatrix} a_{22} & a_{23} \\\\ a_{32} & a_{33} \\end{vmatrix} - a_{12} \\begin{vmatrix} a_{21} & a_{23} \\\\ a_{31} & a_{33} \\end{vmatrix} + a_{13} \\begin{vmatrix} a_{21} & a_{22} \\\\ a_{31} & a_{32} \\end{vmatrix}\n\\]\n\n\\[\n= a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31})\n\\]\n\nNote: Each determinant in the sum (In the R.H.S) is the determinant of a submatrix of A obtained by deleting a particular row and column of A.\nThese determinants are called minors. We take the sign + or −, according to \\((-1)^{i+j} a_{ij}\\)\nWhere i and j represent row and column.\n\n9.6 Minor and Cofactor of Element:\nThe minor \\(M_{ij}\\) of the element \\(a_{ij}\\) in a given determinant is the determinant of order \\((n - 1 \\times n - 1)\\) obtained by deleting the ith row and jth column of \\(A_{nxn}\\).\n\nFor example in the determinant\n\n\\[\n\\begin{vmatrix}\n11 & 12 & 13 \\\\\n21 & 22 & 23 \\\\\n31 & 32 & 33\n\\end{vmatrix}\n\\]\n\nThe minor of the element \\(a_{11}\\) is \\(M_{11} = \\begin{vmatrix} 22 & 23 \\\\ 32 & 33 \\end{vmatrix}\\)\n\nThe minor of the element \\(a_{12}\\) is \\(M_{12} = \\begin{vmatrix} 21 & 23 \\\\ 31 & 33 \\end{vmatrix}\\)\n\nThe minor of the element \\(a_{13}\\) is \\(M_{13} = \\begin{vmatrix} 21 & 22 \\\\ 31 & 32 \\end{vmatrix}\\) and so on.\n\nThe scalars \\(C_{ij} = (-1)^{i+j} M_{ij}\\) are called the cofactor of the element \\(a_{ij}\\) of the matrix \\(A\\).\n\nNote: The value of the determinant in equation (1) can also be found by its minor elements or cofactors, as\n\n\\[a_{11}M_{11} - a_{12}M_{12} + a_{13}M_{13}\\] \nOr \n\\[a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13}\\]\n\nHence the \\(\\det A\\) is the sum of the elements of any row or column multiplied by their corresponding cofactors.\n\nThe value of the determinant can be found by expanding it from any row or column.\n\nExample 4: If \\(A = \\begin{bmatrix} 3 & 2 & 1 \\\\ 0 & 1 & -2 \\\\ 1 & 3 & 4 \\end{bmatrix}\\)\n\nfind \\(\\det A\\) by expansion about (a) the first row (b) the first column.\n\nSolution (a)\n\n\\[\n\\begin{vmatrix}\n3 & 2 & 1 \\\\\n0 & 1 & -2 \\\\\n1 & 3 & 4\n\\end{vmatrix}\n\\]\n\\[\n\\begin{vmatrix}\n1 & -2 \\\\\n3 & 4\n\\end{vmatrix}\n- \\begin{vmatrix}\n0 & -2 \\\\\n1 & 4\n\\end{vmatrix}\n+ \\begin{vmatrix}\n0 & 1 \\\\\n1 & 3\n\\end{vmatrix}\n= 3(4 + 6) - 2(0 + 2) + 1(0 - 1)\n= 30 - 4 - 1\n= 25\n\\]\n\n(b) \\[\n\\begin{vmatrix}\n1 & -2 \\\\\n3 & 4\n\\end{vmatrix}\n- \\begin{vmatrix}\n2 & 1 \\\\\n3 & 4\n\\end{vmatrix}\n+ \\begin{vmatrix}\n0 & 1 \\\\\n1 & 3\n\\end{vmatrix}\n= 3(4 + 6) + 1(-4 - 1)\n= 30 - 5\n= 25\n\\]\n\n9.7 Properties of the Determinant:\nThe following properties of determinants are frequently useful in their evaluation:\n\n1. Interchanging the corresponding rows and columns of a determinant does not change its value (i.e., \\(|A| = |A'|\\)). For example, consider a determinant\n\n\\[\n\\begin{vmatrix}\na & b & c \\\\\na & b & c \\\\\na & b & c\n\\end{vmatrix}\n= a_1(b_2c_3 - b_3c_2) - b_1(a_2c_3 - a_3c_2) + c_1(a_2b_3 - a_3b_2) \\quad (2)\n\\]\n\nNow again consider\n\n\\[\n\\begin{vmatrix}\na & a & a \\\\\nb & b & b \\\\\nc & c & c\n\\end{vmatrix}\n= a_1(b_2c_3 - b_3c_2) - b_1(a_2c_3 - a_3c_2) + c_1(a_2b_3 - a_3b_2)\n\\]\n\nwhich is same as equation (2)\n\nso \\[\n\\begin{vmatrix}\na & b & c \\\\\na & b & c \\\\\na & b & c\n\\end{vmatrix}\n= \\begin{vmatrix}\na & b & c \\\\\na & b & c \\\\\na & b & c\n\\end{vmatrix}\n\\]\n\nor \\[\n\\begin{vmatrix}\na & b & c \\\\\na & b & c \\\\\na & b & c\n\\end{vmatrix}\n= |A|\n\\]\n\n2. If two rows or two columns of a determinant are interchanged, the sign of the determinant is changed but its absolute value is unchanged.\n\nFor example if\nConsider the determinant,\n\n\\[ |B| = \\begin{vmatrix} a & b & c \\\\ a & b & c \\\\ a & b & c \\end{vmatrix} \\]\n\nexpand by second row,\n\n\\[ |B| = -a_1(b_2c_3 - b_3c_2) + b_1(a_2c_3 - a_3c_2) - c_1(a_2b_3 - a_3b_2) \\]\n\nThe term in the bracket is same as the equation (2)\n\nSo \\[ |B| = -|A| \\]\n\n3. If every element of a row or column of a determinant is zero, the value of the determinant is zero. For example\n\n\\[ |A| = \\begin{vmatrix} 0 & 0 & 0 \\\\ a_2 & b_2 & c_2 \\\\ a_3 & b_3 & c_3 \\end{vmatrix} = 0(b_2c_3 - b_3c_2) - 0(a_2c_3 - a_3c_2) + 0(a_2b_3 - a_3b_3) \\]\n\n\\[ |A| = 0 \\]\n\n4. If two rows or columns of a determinant are identical, the value of the determinant is zero. For example, if\n\n\\[ |A| = \\begin{vmatrix} a_1 & b_1 & c_1 \\\\ a_1 & b_1 & c_1 \\\\ a_3 & b_3 & c_3 \\end{vmatrix} = a_1(b_1c_3 - b_3c_1) - b_1(a_1c_3 - a_3c_1) + c_1(a_1b_3 - a_3b_1) \\]\n\n\\[ |A| = 0 \\]\n\n5. If every element of a row or column of a determinant is multiplied by the same constant K, the value of the determinant is multiplied by that constant. For example if,\nConsider a determinant, \\( |B| = \\begin{vmatrix} a_1 & b_1 & c_1 \\\\ a_2 & b_2 & c_2 \\\\ a_3 & b_3 & c_3 \\end{vmatrix} \\)\n\n\\[ |B| = k(a_1(b_2c_3 - b_3c_2) - b_1(a_2c_3 - a_3c_2) + c_1(a_2b_3 - a_3b_2)) \\]\n\nSo \\( |B| = k \\begin{vmatrix} a_1 & b_1 & c_1 \\\\ a_2 & b_2 & c_2 \\\\ a_3 & b_3 & c_3 \\end{vmatrix} \\)\n\nOr \\( |B| = K|A| \\)\n\n6. The value of a determinant is not changed if each element of any row or of any column is added to (or subtracted from) a constant multiple of the corresponding element of another row or column. For example, if\n\n\\[ |A| = \\begin{vmatrix} a_1 & b_1 & c_1 \\\\ a_2 & b_2 & c_2 \\\\ a_3 & b_3 & c_3 \\end{vmatrix} \\]\n\nConsider a matrix,\n\n\\[ |B| = \\begin{vmatrix} a_1 + ka_2 & b_1 + kb_2 & c_1 + kc_2 \\\\ a_2 & b_2 & c_2 \\\\ a_3 & b_3 & c_3 \\end{vmatrix} \\]\n\n\\[ = (a_1+ka_2)(b_2c_3-b_3c_2)-(b_1+kb_2)(a_2c_3-a_3c_2)+(c_1+kc_2)(a_2b_3-a_3b_2) \\]\n\n\\[ = [a_1(b_2c_3-b_3c_2)-b_1(a_2c_3-a_3c_2)+c_1(a_2b_3-a_3b_2)] \\]\n\n\\[ = [ka_2(b_2c_3-b_3c_2)-kb_2(a_2c_3-a_3c_2)+kc_2(a_2b_3-a_3b_2)] \\]\n\n\\[ = a_2 \\begin{vmatrix} b_2 & c_2 \\\\ b_3 & c_3 \\end{vmatrix} + k \\begin{vmatrix} a_2 & b_2 & c_2 \\\\ a_3 & b_3 & c_3 \\end{vmatrix} \\]\nThe determinant of a diagonal matrix is equal to the product of its diagonal elements. For example, if\n\n\\[ |A| = \\begin{vmatrix} 2 & 0 & 0 \\\\ 0 & -5 & 0 \\\\ 0 & 0 & 3 \\end{vmatrix} = 2(-15 - 0) - (0 - 0) + 0(0 - 0) = 30, \\text{ which is the product of diagonal elements.} \\]\n\ni.e., \\(2(-5)3 = -30\\)\n\n8. The determinant of the product of two matrices is equal to the product of the determinants of the two matrices, that is \\(|AB| = |A||B|\\). For example, if\n\n\\[ A = \\begin{bmatrix} a_{11} & a_{12} \\\\ a_{21} & a_{22} \\end{bmatrix}, \\quad B = \\begin{bmatrix} b_{11} & b_{12} \\\\ b_{21} & b_{22} \\end{bmatrix} \\]\n\nThen \\(AB = \\begin{vmatrix} a_{11}b_{11} + a_{12}b_{21} & a_{11}b_{12} + a_{12}b_{22} \\\\ a_{21}b_{11} + a_{22}b_{21} & a_{21}b_{12} + a_{22}b_{22} \\end{vmatrix}\\)\n\n\\[ |AB| = (a_{11}b_{11} + a_{12}b_{21})(a_{21}b_{12} + a_{22}b_{22}) - (a_{11}b_{12} + a_{12}b_{22} - a_{11}b_{22})(a_{21}b_{11} + a_{22}b_{21}) \\]\n\n\\[ = a_{11}b_{11}a_{21}b_{12} + a_{11}b_{11}a_{22}b_{22} + a_{12}b_{21}a_{21}b_{12} + a_{12}b_{21}a_{22}b_{22} - a_{11}b_{12}a_{21}b_{11} - a_{11}b_{12}a_{22}b_{21} - a_{12}b_{22}a_{21}b_{11} - a_{12}b_{22}a_{22}b_{21} \\]\n\n\\[ |AB| = a_{11}b_{11}a_{22}b_{22} + a_{12}b_{21}a_{21}b_{12} - a_{11}b_{12}a_{22}b_{21} - a_{12}b_{22}a_{21}b_{11} \\]\n\n\\[ \\text{and} \\quad |A| = a_{11}a_{22} - a_{12}a_{21} \\]\n\n\\[ |B| = b_{11}b_{22} - b_{12}b_{21} \\]\n\n\\[ |A||B| = a_{11}b_{11}a_{22}b_{22} + a_{12}b_{21}a_{21}b_{12} - a_{11}b_{12}a_{22}b_{21} - a_{12}b_{22}a_{21}b_{11} \\]\n\nR.H.S of equations (A) and (B) are equal, so\n\n\\[ |AB| = |A||B| \\]\n\n9. The determinant in which each element in any row, or column, consists of two terms, then the determinant can be expressed as the sum of two other determinants\nExpand by first column.\n\nProof:\nL.H.S = \\((a_1 + \\alpha_1)(b_2c_3 - b_3c_2) - (a_2 + \\alpha_2)(b_1c_3 - b_3c_1) + (a_3 + \\alpha_3)(b_1c_2 - b_2c_1)\\)\n\n\\[= [(a_1(b_2c_3 - b_3c_2) - a_2(b_1c_3 - b_3c_1) + a_3(b_1c_2 - b_2c_1)]\n+ [(\\alpha_1(b_2c_3 - b_3c_2) - \\alpha_2(b_1c_2 - b_3c_1) + \\alpha_3(b_1c_2 - b_2c_1)]\\]\n\n\\[= \\begin{vmatrix} a_1 & b_1 & c_1 \\\\ a_2 & b_2 & c_2 \\\\ a_3 & b_3 & c_3 \\end{vmatrix} + \\begin{vmatrix} \\alpha_1 & b_1 & c_1 \\\\ \\alpha_2 & b_2 & c_2 \\\\ \\alpha_3 & b_3 & c_3 \\end{vmatrix}\\]\n\n= R.H.S\n\nSimilarly\n\\[\\begin{vmatrix} \\alpha_1 + a_1 & b_1 + \\beta_1 & c_1 \\\\ \\alpha_2 + a_2 & b_2 + \\beta_2 & c_2 \\\\ \\alpha_3 + a_3 & b_3 + \\beta_3 & c_3 \\end{vmatrix} = \\begin{vmatrix} a_1 & b_1 & c_1 \\\\ a_2 & b_2 & c_2 \\\\ a_3 & b_3 & c_3 \\end{vmatrix} + \\begin{vmatrix} \\alpha_1 & b_1 & c_1 \\\\ \\alpha_2 & b_2 & c_2 \\\\ \\alpha_3 & b_3 & c_3 \\end{vmatrix}\\]\n\nAnd,\n\\[\\begin{vmatrix} a_1 & b_1 & c_1 \\\\ a_2 & b_2 & c_2 \\\\ a_3 & b_3 & c_3 \\end{vmatrix} + \\text{sum of six determinant} + \\begin{vmatrix} \\alpha_1 & \\beta_1 & \\gamma_1 \\\\ \\alpha_2 & \\beta_2 & \\gamma_2 \\\\ \\alpha_3 & \\beta_3 & \\gamma_3 \\end{vmatrix}\\]\n\nAlso\n\\[\\begin{vmatrix} a_1 + \\alpha_1 & b_1 + \\beta_1 & c_1 + \\gamma_1 \\\\ a_2 & b_2 & c_2 \\\\ a_3 & b_3 & c_3 \\end{vmatrix} = \\begin{vmatrix} a_1 & b_1 & c_1 \\\\ a_2 & b_2 & c_2 \\\\ a_3 & b_3 & c_3 \\end{vmatrix} + \\begin{vmatrix} \\alpha_1 & \\beta_1 & \\gamma_1 \\\\ \\alpha_2 & \\beta_2 & \\gamma_2 \\\\ \\alpha_3 & \\beta_3 & \\gamma_3 \\end{vmatrix}\\]\nExample 5: Verify that\n\\[\n\\begin{vmatrix}\n1 & a & bc \\\\\n1 & b & ca \\\\\n1 & c & ab \\\\\n\\end{vmatrix}\n= \\begin{vmatrix}\n1 & a & a^2 \\\\\n1 & b & b^2 \\\\\n1 & c & c^2 \\\\\n\\end{vmatrix}\n\\]\n\nSolution:\nMultiply row first, second and third by a, b and c respectively, in the L.H.S., then\n\\[\n\\text{L.H.S} = \\frac{1}{abc} \\begin{vmatrix}\na & a^2 & abc \\\\\nb & b^2 & abc \\\\\nc & c^2 & abc \\\\\n\\end{vmatrix}\n\\]\nTake abc common from 3rd column\n\\[\n= \\frac{abc}{abc} \\begin{vmatrix}\na & a^2 & 1 \\\\\nb & b^2 & 1 \\\\\nc & c^2 & 1 \\\\\n\\end{vmatrix}\n\\]\nInterchange column first and third\n\\[\n= - \\begin{vmatrix}\n1 & a^2 & a \\\\\n1 & b^2 & b \\\\\n1 & c^2 & c \\\\\n\\end{vmatrix}\n\\]\nAgain interchange column second and third\n\\[\n= \\begin{vmatrix}\n1 & a & a^2 \\\\\n1 & b & b^2 \\\\\n1 & c & c^2 \\\\\n\\end{vmatrix}\n= \\text{R.H.S}\n\\]\n\nExample 6: Show that\n\\[\n\\begin{vmatrix}\n1 & a & a^2 \\\\\n1 & b & b^2 \\\\\n1 & c & c^2 \\\\\n\\end{vmatrix}\n= (b - c) (c - a) (a - b)\n\\]\nSolution:\n\n\\[\n\\begin{vmatrix}\n1 & a & a^2 \\\\\n1 & b & b^2 \\\\\n1 & c & c^2 \\\\\n\\end{vmatrix}\n\\]\n\nsubtracting row first from second and third row\n\n\\[\n\\begin{vmatrix}\n1 & a & a^2 \\\\\n0 & b-a & b^2-a^2 \\\\\n0 & c-a & c^2-a^2 \\\\\n\\end{vmatrix}\n\\]\n\nfrom row second and third taking \\((b-a)\\) and \\((c-a)\\) common.\n\n\\[\n= (b-a)(c-a)\n\\]\n\nexpand from first column\n\n\\[\n= (b-a)(c-a)(c+a-b-a)\n\\]\n\nOr L.H.S\n\n\\[\n= (b-c)(c-a)(a-b)(-1)(-1)\n\\]\n\n\\[\n= (b-c)(c-a)(a-b) = R.H.S\n\\]\n\nExample 7: Without expansion, show that\n\n\\[\n\\begin{vmatrix}\n6 & 1 & 3 & 2 \\\\\n-2 & 0 & 1 & 4 \\\\\n3 & 6 & 1 & 2 \\\\\n-4 & 0 & 2 & 8 \\\\\n\\end{vmatrix} = 0\n\\]\n\nSolution:\n\nIn the L.H.S Taking 2 common from fourth row, so\n\n\\[\nL.H.S = 2\n\\]\n\nSince rows 2nd and 3rd are identical, so\n\n\\[\n= 2(0) = 0\n\\]\n\nL.H.S = R.H.S\n9.8 Solution of Linear Equations by Determinants: (Cramer’s Rule)\n\nConsider a system of linear equations in two variables \\( x \\) and \\( y \\),\n\n\\[\n\\begin{align*}\n a_1x + b_1y &= c_1 \\\\\n a_2x + b_2y &= c_2\n\\end{align*}\n\\]\n\nMultiply equation (1) by \\( b_2 \\) and equation (2) by \\( b_1 \\) and subtracting, we get\n\n\\[\nx(a_1b_2 - a_2b_1) = b_2c_1 - b_1c_2\n\\]\n\n\\[\nx = \\frac{b_2c_1 - b_1c_2}{a_1b_2 - a_2b_1} \\quad (3)\n\\]\n\nAgain multiply eq. (1) by \\( a_2 \\) and eq. (2) by \\( a_1 \\) and subtracting, we get\n\n\\[\ny(a_2b_1 - a_1b_2) = a_2c_1 - a_1c_2\n\\]\n\n\\[\ny = \\frac{a_2c_1 - a_1c_2}{a_2b_1 - a_1b_2} \\quad (4)\n\\]\n\nNote that \\( x \\) and \\( y \\) from equations (3) and (4) has the same denominator \\( a_1b_2 - a_2b_1 \\). So the system of equations (1) and (2) has solution only when \\( a_1b_2 - a_2b_1 \\neq 0 \\).\n\nThe solutions for \\( x \\) and \\( y \\) of the system of equations (1) and (2) can be written directly in terms of determinants without any algebraic operations, as\n\n\\[\nx = \\frac{\\begin{vmatrix} c_1 & b_1 \\\\ c_2 & b_2 \\end{vmatrix}}{\\begin{vmatrix} a_1 & c_1 \\\\ a_2 & c_2 \\end{vmatrix}} \\quad \\text{and} \\quad y = \\frac{\\begin{vmatrix} a_1 & c_1 \\\\ a_2 & c_2 \\end{vmatrix}}{\\begin{vmatrix} a_1 & b_1 \\\\ a_2 & b_2 \\end{vmatrix}}\n\\]\n\nThis result is called Cramer’s Rule.\n\nHere \\( \\begin{vmatrix} a_1 & b_1 \\\\ a_2 & b_2 \\end{vmatrix} = |A| \\) is the determinant of the coefficient of \\( x \\) and \\( y \\) in equations (1) and (2).\n\nIf \\( \\begin{vmatrix} c_1 & b_1 \\\\ c_2 & b_2 \\end{vmatrix} = |A| \\)\nand \\[ \\begin{vmatrix} a_1 & c_1 \\\\ a_2 & c_2 \\end{vmatrix} = |A| \\]\n\nThen \\( x = \\frac{|A_x|}{|A|} \\) and \\( y = \\frac{|A_y|}{|A|} \\)\n\n**Solution for a system of Linear Equations in Three Variables:**\n\nConsider the linear equations:\n\\[\n\\begin{align*}\n a_1x + b_1y + c_1z &= d_1 \\\\\n a_2x + b_2y + c_2z &= d_2 \\\\\n a_3x + b_3y + c_3z &= d_3\n\\end{align*}\n\\]\n\nHence the determinant of coefficients is\n\\[ |A| = \\begin{vmatrix} a_1 & b_1 & c_1 \\\\ a_2 & b_2 & c_2 \\\\ a_3 & b_3 & c_3 \\end{vmatrix}, \\text{ if } |A| \\neq 0 \\]\n\nThen by Cramer’s Rule the value of variables is:\n\\[\n\\begin{align*}\n x &= \\frac{|d_1 b_1 c_1|}{|A|} = \\frac{|A_x|}{|A|} \\\\\n y &= \\frac{|a_1 d_1 c_1|}{|A|} = \\frac{|A_y|}{|A|} \\\\\n z &= \\frac{|a_1 b_1 d_1|}{|A|} = \\frac{|A_z|}{|A|}\n\\end{align*}\n\\]\n\n**Example 8:** Use Cramer’s rule to solve the system\n\\[\n\\begin{align*}\n -4x + 2y - 9z &= 2 \\\\\n 3x + 4y + z &= 5 \\\\\n x - 3y + 2z &= 8\n\\end{align*}\n\\]\n\n**Solution:**\nHere the determinant of the coefficients is:\n\n\\[\n|A| = \\begin{vmatrix}\n-4 & 2 & -9 \\\\\n3 & 4 & 1 \\\\\n1 & -3 & 2 \\\\\n\\end{vmatrix}\n\\]\n\n\\[\n= -4(8 + 3) - 2(6 - 1) - 9(-9 - 4)\n\\]\n\n\\[\n= -44 - 10 + 117\n\\]\n\n\\[\n|A| = 63\n\\]\n\nfor \\(|A_x|\\), replacing the first column of \\(|A|\\) with the corresponding constants 2, 5 and 8, we have\n\n\\[\n|A_x| = \\begin{vmatrix}\n2 & 2 & -9 \\\\\n5 & 4 & 1 \\\\\n8 & -3 & 2 \\\\\n\\end{vmatrix}\n\\]\n\n\\[\n= 2(11) - 2(2) - 9(-47) = 22 - 4 + 423\n\\]\n\n\\[\n|A_x| = 441\n\\]\n\nSimilarly,\n\n\\[\n|A_y| = \\begin{vmatrix}\n-4 & 2 & -9 \\\\\n3 & 5 & 1 \\\\\n1 & 8 & 2 \\\\\n\\end{vmatrix}\n\\]\n\n\\[\n= -4(2) - 2(5) - 9(19)\n\\]\n\n\\[\n= -8 - 10 - 171\n\\]\n\n\\[\n|A_y| = -189\n\\]\n\nand\n\n\\[\n|A_z| = \\begin{vmatrix}\n-4 & 2 & 2 \\\\\n3 & 4 & 5 \\\\\n1 & -3 & 8 \\\\\n\\end{vmatrix}\n\\]\n\n\\[\n= -4(47) - 2(19) + 2(-13)\n\\]\n\n\\[\n= -188 - 38 - 26\n\\]\n\n\\[\n|A_z| = -252\n\\]\nHence \\( x = \\frac{|A_x|}{|A|} = \\frac{441}{63} = 7 \\)\n\n\\( y = \\frac{|A_y|}{|A|} = \\frac{-189}{63} = -3 \\)\n\n\\( z = \\frac{|A_z|}{|A|} = \\frac{-252}{63} = -4 \\)\n\nSo the solution set of the system is \\{(7, -3, -4)\\}\n\n**Exercise 9.2**\n\nQ.1 Expand the determinants\n\n(i) \\[\n\\begin{vmatrix}\n1 & 2 & 0 \\\\\n3 & -1 & 4 \\\\\n-2 & 1 & 3\n\\end{vmatrix}\n\\]\n\n(ii) \\[\n\\begin{vmatrix}\na & b & 1 \\\\\na & b & 1 \\\\\n1 & 1 & 1\n\\end{vmatrix}\n\\]\n\n(iii) \\[\n\\begin{vmatrix}\nx & 0 & 0 \\\\\n0 & x & 0 \\\\\n0 & 0 & x\n\\end{vmatrix}\n\\]\n\nQ.2 Without expansion, verify that\n\n(i) \\[\n\\begin{vmatrix}\n-2 & 1 & 0 \\\\\n3 & 4 & 1 \\\\\n-4 & 2 & 0\n\\end{vmatrix} = 0\n\\]\n\n(ii) \\[\n\\begin{vmatrix}\n1 & 2 & 1 \\\\\n0 & 2 & 3 \\\\\n2 & -1 & 2\n\\end{vmatrix} = 0\n\\]\n\n(iii) \\[\n\\begin{vmatrix}\na-b & b-c & c-a \\\\\nb-c & c-a & a-b \\\\\nc-a & a-b & b-c\n\\end{vmatrix} = 0\n\\]\n\n(iv) \\[\n\\begin{vmatrix}\nbc & ca & ab \\\\\na^3 & b^3 & c^3 \\\\\n\\frac{1}{a} & \\frac{1}{b} & \\frac{1}{c}\n\\end{vmatrix} = 0\n\\]\n\n(v) \\[\n\\begin{vmatrix}\nx + 1 & x + 2 & x + 3 \\\\\nx + 4 & x + 5 & x + 6 \\\\\nx + 7 & x + 8 & x + 9\n\\end{vmatrix} = 0\n\\]\nChapter 9 Matrices and Determinants\n\n(vi) \\[\n\\begin{vmatrix}\n a & b & c \\\\\n d & e & f \\\\\n g & h & k \\\\\n\\end{vmatrix} = \\begin{vmatrix}\n e & b & h \\\\\n d & a & g \\\\\n f & c & k \\\\\n\\end{vmatrix}\n\\]\n\nQ.3 Show that\n\n\\[\n\\begin{vmatrix}\n a_1 & a_2 & a_3 \\\\\n b_1 & b_2 & b_3 \\\\\n c_1x+d_1 & c_2x+d_2 & c_3x+d_3 \\\\\n\\end{vmatrix} = x \\begin{vmatrix}\n a_1 & a_2 & a_3 \\\\\n b_1 & b_2 & b_3 \\\\\n c_1 & c_2 & c_3 \\\\\n\\end{vmatrix} + \\begin{vmatrix}\n b_1 & b_2 & b_3 \\\\\n c_1 & c_2 & c_3 \\\\\n d_1 & d_2 & d_3 \\\\\n\\end{vmatrix}\n\\]\n\nQ.4 Show that\n\n(i) \\[\n\\begin{vmatrix}\n 0 & a & b \\\\\n -a & 0 & c \\\\\n -b & -c & 0 \\\\\n\\end{vmatrix} = 0\n\\]\n\n(ii) \\[\n\\begin{vmatrix}\n a & a+b & a+b+c \\\\\n a & 2a+b & 3a+2b+c \\\\\n\\end{vmatrix} = a^3\n\\]\n\n(iii) \\[\n\\begin{vmatrix}\n a-b-c & 2a & 2a \\\\\n 2b & b-c-a & 2b \\\\\n 2c & 2c & c-a-b \\\\\n\\end{vmatrix} = (a + b + c)^3\n\\]\n\n(iv) \\[\n\\begin{vmatrix}\n bc & ca & ab \\\\\n b+c & c+a & a+b \\\\\n\\end{vmatrix} = (b - c)(c - a)(a - b)\n\\]\n\nQ.5 Show that:\n\n(i) \\[\n\\begin{vmatrix}\n \\ell & a & a \\\\\n a & \\ell & a \\\\\n a & a & \\ell \\\\\n\\end{vmatrix} = (2a + \\ell)(\\ell - a)^2\n\\]\n\n(ii) \\[\n\\begin{vmatrix}\n a & a + \\ell & a \\\\\n a & a & a + \\ell \\\\\n\\end{vmatrix} = \\ell^2(3a + \\ell)\n\\]\n\nQ.6 prove that:\n\n(i) \\[\n\\begin{vmatrix}\n a & b+c & a+b \\\\\n b & c+a & b+c \\\\\n c & a+b & c+a \\\\\n\\end{vmatrix} = a^3 + b^3 + c^3 - 3 abc\n\\]\n(ii) \\[\n\\begin{vmatrix}\n a + \\lambda & b & c \\\\\n a & b + \\lambda & c \\\\\n a & b & c + \\lambda \\\\\n\\end{vmatrix} = \\lambda^2 (a + b + c + \\lambda)\n\\]\n\n(iii) \\[\n\\begin{vmatrix}\n \\sin \\alpha & \\cos \\alpha & 0 \\\\\n -\\sin \\beta & \\cos \\beta & \\sin \\gamma \\\\\n \\cos \\beta & \\sin \\beta & \\cos \\gamma \\\\\n\\end{vmatrix} = \\sin (\\alpha + \\beta + \\gamma)\n\\]\n\nQ.7 Find values of \\( x \\) if\n\n(i) \\[\n\\begin{vmatrix}\n 3 & 1 & x \\\\\n -1 & 3 & 4 \\\\\n x & 1 & 0 \\\\\n\\end{vmatrix} = -30\n\\]\n\n(ii) \\[\n\\begin{vmatrix}\n 1 & 2 & 1 \\\\\n 2 & x & 2 \\\\\n 3 & 6 & x \\\\\n\\end{vmatrix} = 0\n\\]\n\nQ.8 Use Cramer’s rule to solve the following system of equations.\n\n(i) \\[\n\\begin{align*}\nx - y &= 2 \\\\\nx + 4y &= 5\n\\end{align*}\n\\]\n\n(ii) \\[\n\\begin{align*}\n3x - 4y &= -2 \\\\\nx + y &= 6\n\\end{align*}\n\\]\n\n(iii) \\[\n\\begin{align*}\nx - 2y + z &= -1 \\\\\n3x + y - 2z &= 4 \\\\\ny - z &= 1\n\\end{align*}\n\\]\n\n(iv) \\[\n\\begin{align*}\n2x + 2y + z &= 1 \\\\\nx - y + 6z &= 21 \\\\\n3x + 2y - z &= -4\n\\end{align*}\n\\]\n\n(v) \\[\n\\begin{align*}\nx + y + z &= 0 \\\\\n2x - y - 4z &= 15 \\\\\nx - 2y - z &= 7\n\\end{align*}\n\\]\n\n(vi) \\[\n\\begin{align*}\nx - 2y - 2z &= 3 \\\\\n2x - 4y + 4z &= 1 \\\\\n3x - 3y - 3z &= 4\n\\end{align*}\n\\]\n\nAnswers 9.2\n\nQ.1 (i) \\(-41\\) (ii) \\(0\\) (iii) \\(x^3\\)\n\nQ.7 (i) \\(x = -2, 3\\) (ii) \\(x = 3, 4\\)\n\nQ.8 (i) \\(\\left\\{ \\left( \\frac{13}{5}, \\frac{3}{5} \\right) \\right\\}\\) (ii) \\(\\left\\{ \\left( \\frac{22}{7}, \\frac{20}{7} \\right) \\right\\}\\)\n\n(iii) \\(\\{(1, 1, 0)\\}\\) (iv) \\(\\{(1, -2, 3)\\}\\)\n\n(v) \\(\\{(3, -1, -2)\\}\\) (vi) \\(\\left\\{ \\left( -\\frac{1}{3}, -\\frac{25}{24}, -\\frac{5}{8} \\right) \\right\\}\\)\n9.9 Special Matrices:\n\n1. Transpose of a Matrix\n If \\( A = [a_{ij}] \\) is \\( m \\times n \\) matrix, then the matrix of order \\( n \\times m \\) obtained by interchanging the rows and columns of \\( A \\) is called the transpose of \\( A \\). It is denoted \\( A^t \\) or \\( A' \\).\n\n Example if \\( A = \\begin{bmatrix} 1 & 2 & 3 \\\\ 4 & 5 & 6 \\\\ 7 & 8 & 9 \\end{bmatrix} \\), then \\( A^t = \\begin{bmatrix} 1 & 4 & 7 \\\\ 2 & 5 & 8 \\\\ 3 & 6 & 9 \\end{bmatrix} \\)\n\n2. Symmetric Matrix:\n A square matrix \\( A \\) is called symmetric if \\( A = A^t \\) for example if\n \\[\n A = \\begin{bmatrix} a & b & c \\\\ b & d & e \\\\ c & e & f \\end{bmatrix}, \\quad \\text{then} \\quad A^t = \\begin{bmatrix} a & b & c \\\\ b & d & e \\\\ c & e & f \\end{bmatrix} = A\n \\]\n Thus \\( A \\) is symmetric\n\n3. Skew Symmetric:\n A square matrix \\( A \\) is called skew symmetric if \\( A = -A^t \\) for example if\n \\[\n B = \\begin{bmatrix} 0 & -4 & 1 \\\\ 4 & 0 & -3 \\\\ -1 & 3 & 0 \\end{bmatrix}, \\text{then} \\quad B^t = \\begin{bmatrix} 0 & 4 & -1 \\\\ -4 & 0 & 3 \\\\ 1 & -3 & 0 \\end{bmatrix} = (-1) \\begin{bmatrix} 0 & -4 & 1 \\\\ 4 & 0 & -3 \\\\ -1 & 3 & 0 \\end{bmatrix}\n \\]\n Thus matrix \\( B \\) is skew symmetric.\n\n4. Singular and Non-singular Matrices:\n A square matrix \\( A \\) is called singular if \\( |A| = 0 \\) and is non-singular if \\( |A| \\neq 0 \\), for example if\n \\[\n A = \\begin{bmatrix} 3 & 2 \\\\ 9 & 6 \\end{bmatrix}, \\text{then} \\quad |A| = 0, \\text{Hence} \\ A \\text{ is singular}\n \\]\nand if \\( A = \\begin{bmatrix} 3 & 1 & 6 \\\\ -1 & 3 & 2 \\\\ 1 & 0 & 0 \\end{bmatrix} \\), then \\( |A| \\neq 0 \\),\n\nHence \\( A \\) is non-singular.\n\n**Example:** Find \\( k \\) if \\( A = \\begin{bmatrix} k-2 & 1 \\\\ 5 & k+2 \\end{bmatrix} \\) is singular\n\n**Solution:** Since \\( A \\) is singular so \\( \\begin{vmatrix} k-2 & 1 \\\\ 5 & k+2 \\end{vmatrix} = 0 \\)\n\n\\[\n(k - 2)(k + 2) - 5 = 0 \\\\\nk^2 - 4 - 5 = 0 \\\\\nk^2 - 9 = 0 \\Rightarrow K = \\pm 3\n\\]\n\n5. **Adjoint of a Matrix:**\n\nLet \\( A = (a_{ij}) \\) be a square matrix of order \\( n \\times n \\) and \\( (c_{ij}) \\) is a matrix obtained by replacing each element \\( a_{ij} \\) by its corresponding cofactor \\( c_{ij} \\) then \\( (c_{ij})^t \\) is called the adjoint of \\( A \\). It is written as \\( \\text{adj. } A \\).\n\nFor example, if\n\n\\[\nA = \\begin{bmatrix} 1 & 0 & -1 \\\\ 1 & 3 & 1 \\\\ 0 & 1 & 2 \\end{bmatrix}\n\\]\n\nCofactor of \\( A \\) are:\n\n\\[\nA_{11} = 5, \\quad A_{12} = -2, \\quad A_{13} = +1 \\\\\nA_{21} = -1, \\quad A_{22} = 2, \\quad A_{23} = -1 \\\\\nA_{31} = 3, \\quad A_{32} = -2, \\quad A_{33} = 3\n\\]\n\nMatrix of cofactors is\n\n\\[\nC = \\begin{bmatrix} 5 & -2 & +1 \\\\ -1 & 2 & -1 \\\\ 3 & -2 & 3 \\end{bmatrix}\n\\]\n\n\\[\nC^t = \\begin{bmatrix} 5 & -1 & 3 \\\\ -2 & 2 & -2 \\\\ +1 & -1 & 3 \\end{bmatrix}\n\\]\nHence \\( \\text{adj } A = C^t = \\begin{bmatrix} 5 & -1 & 3 \\\\ -2 & 2 & -2 \\\\ +1 & -1 & 3 \\end{bmatrix} \\)\n\n**Note:** Adjoint of a 2x2 Matrix:\n\nThe adjoint of matrix \\( A = \\begin{bmatrix} a & b \\\\ c & d \\end{bmatrix} \\) is denoted by \\( \\text{adj } A \\) is defined as\n\n\\[\n\\text{adj } A = \\begin{bmatrix} d & -b \\\\ -c & a \\end{bmatrix}\n\\]\n\n6. **Inverse of a Matrix:**\n\nIf \\( A \\) is a non-singular square matrix, then \\( A^{-1} = \\frac{\\text{adj } A}{|A|} \\)\n\nFor example, if matrix \\( A = \\begin{bmatrix} 3 & 4 \\\\ 1 & 2 \\end{bmatrix} \\)\n\nThen \\( \\text{adj } A = \\begin{bmatrix} 2 & -4 \\\\ -1 & 3 \\end{bmatrix} \\)\n\n\\[\n|A| = \\begin{bmatrix} 3 & 4 \\\\ 1 & 2 \\end{bmatrix} = 6 - 4 = 2\n\\]\n\nHence \\( A^{-1} = \\frac{\\text{adj } A}{|A|} = \\frac{1}{2} \\begin{bmatrix} 2 & -4 \\\\ -1 & 3 \\end{bmatrix} \\)\n\n**Alternately:**\n\nFor a non-singular matrix \\( A \\) of order \\((n \\times n)\\) if there exist another matrix \\( B \\) of order \\((n \\times n)\\) such that their product is the identity matrix \\( I \\) of order \\((n \\times n)\\) i.e., \\( AB = BA = I \\)\n\nThen \\( B \\) is said to be the inverse (or reciprocal) of \\( A \\) and is written as \\( B = A^{-1} \\)\n\n**Example 9:** If \\( A = \\begin{bmatrix} 1 & -3 \\\\ -2 & 7 \\end{bmatrix} \\) and \\( B = \\begin{bmatrix} 7 & 3 \\\\ 2 & 1 \\end{bmatrix} \\) then show that \\( AB = BA = I \\) and therefore, \\( B = A^{-1} \\)\n\n**Solution:**\n\n\\[\nAB = \\begin{bmatrix} 1 & -3 \\\\ -2 & 7 \\end{bmatrix} \\begin{bmatrix} 7 & 3 \\\\ 2 & 1 \\end{bmatrix} = \\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix}\n\\]\nand \\[ BA = \\begin{bmatrix} 7 & 3 \\\\ 2 & 1 \\end{bmatrix} \\begin{bmatrix} 1 & -3 \\\\ -2 & 7 \\end{bmatrix} = \\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix} \\]\n\nHence \\( AB = BA = I \\)\n\nand therefore \\( B = A^{-1} = \\begin{bmatrix} 7 & 3 \\\\ 2 & 1 \\end{bmatrix} \\)\n\n**Example 10:** Find the inverse, if it exists, of the matrix.\n\n\\[ A = \\begin{bmatrix} 0 & -2 & -3 \\\\ 1 & 3 & 3 \\\\ -1 & -2 & -2 \\end{bmatrix} \\]\n\n**Solution:**\n\n\\[ |A| = 0 + 2(-2 + 3) - 3(-2 + 3) = 2 - 3 \\]\n\n\\[ |A| = -1, \\text{ Hence solution exists.} \\]\n\nCofactor of \\( A \\) are:\n\n\\[ A_{11} = 0, \\quad A_{12} = 1, \\quad A_{13} = 1 \\]\n\\[ A_{21} = 2, \\quad A_{22} = -3, \\quad A_{23} = 2 \\]\n\\[ A_{31} = 3, \\quad A_{32} = -3, \\quad A_{33} = 2 \\]\n\nMatrix of transpose of the cofactors is\n\n\\[ \\text{adj } A = C' = \\begin{bmatrix} 0 & 2 & 3 \\\\ -1 & -3 & -3 \\\\ 1 & 2 & 2 \\end{bmatrix} \\]\n\nSo\n\n\\[ A^{-1} = \\frac{1}{|A|} \\text{adj } A = \\frac{1}{-1} \\begin{bmatrix} 0 & 2 & 3 \\\\ -1 & -3 & -3 \\\\ 1 & 2 & 2 \\end{bmatrix} \\]\n\n\\[ A^{-1} = \\begin{bmatrix} 0 & -2 & -3 \\\\ 1 & 3 & 3 \\\\ -1 & -2 & -2 \\end{bmatrix} \\]\n9.11 Solution of Linear Equations by Matrices:\n\nConsider the linear system:\n\n\\[\n\\begin{align*}\n a_{11}x_1 + a_{12}x_2 + \\cdots + a_{1n}x_n &= b_1 \\\\\n a_{21}x_1 + a_{22}x_2 + \\cdots + a_{2n}x_n &= b_2 \\\\\n \\vdots & \\quad \\vdots \\\\\n a_{n1}x_1 + a_{n2}x_2 + \\cdots + a_{nn}x_n &= b_n\n\\end{align*}\n\\]\n\n\\[\\text{............... (1)}\\]\n\nIt can be written as the matrix equation\n\n\\[\n\\begin{bmatrix}\n a_{11} & a_{12} & \\cdots & a_{1n} \\\\\n a_{21} & a_{22} & \\cdots & a_{2n} \\\\\n \\vdots & \\vdots & \\ddots & \\vdots \\\\\n a_{n1} & a_{n2} & \\cdots & a_{nn}\n\\end{bmatrix}\n\\begin{bmatrix}\n x_1 \\\\\n x_2 \\\\\n \\vdots \\\\\n x_n\n\\end{bmatrix}\n= \n\\begin{bmatrix}\n b_1 \\\\\n b_2 \\\\\n \\vdots \\\\\n b_n\n\\end{bmatrix}\n\\]\n\nLet \\( A = \\begin{bmatrix}\n a_{11} & a_{12} & \\cdots & a_{1n} \\\\\n a_{21} & a_{22} & \\cdots & a_{2n} \\\\\n \\vdots & \\vdots & \\ddots & \\vdots \\\\\n a_{n1} & a_{n2} & \\cdots & a_{nn}\n\\end{bmatrix} \\), \\( X = \\begin{bmatrix}\n x_1 \\\\\n x_2 \\\\\n \\vdots \\\\\n x_n\n\\end{bmatrix} \\)\n\n\\[\nB = \\begin{bmatrix}\n b_1 \\\\\n b_2 \\\\\n \\vdots \\\\\n b_n\n\\end{bmatrix}\n\\]\n\nThen latter equation can be written as,\n\n\\[AX = B\\]\n\nIf \\( B \\neq 0 \\), then (1) is called non-homogenous system of linear equations and if \\( B = 0 \\), it is called a system of homogenous linear equations.\n\nIf now \\( B \\neq 0 \\) and \\( A \\) is non-singular then \\( A^{-1} \\) exists.\n\nMultiply both sides of \\( AX = B \\) on the left by \\( A^{-1} \\), we get\n\n\\[\nA^{-1}(AX) = A^{-1}B\n\\]\n\n\\[\n(A^{-1}A)X = A^{-1}B\n\\]\n\n\\[\n1X = A^{-1}B\n\\]\n\nOr \\( X = A^{-1}B \\)\n\nWhere \\( A^{-1}B \\) is an \\( n \\times 1 \\) column matrix. Since \\( X \\) and \\( A^{-1}B \\) are equal, each element in \\( X \\) is equal to the corresponding element in \\( A^{-1}B \\). These elements of \\( X \\) constitute the solution of the given linear equations.\nIf $A$ is a singular matrix, then of course it has no inverse, and either the system has no solution or the solution is not unique.\n\n**Example 11:** Use matrices to find the solution set of\n\n\\[\n\\begin{align*}\n x + y - 2z &= 3 \\\\\n 3x - y + z &= 5 \\\\\n 3x + 3y - 6z &= 9\n\\end{align*}\n\\]\n\n**Solution:**\n\nLet\n\n\\[\nA = \\begin{bmatrix}\n 1 & 1 & -2 \\\\\n 3 & -1 & 1 \\\\\n 3 & 3 & -6\n\\end{bmatrix}\n\\]\n\nSince $|A| = 3 + 21 - 24 = 0$\n\nHence the solution of the given linear equations does not exists.\n\n**Example 12:** Use matrices to find the solution set of\n\n\\[\n\\begin{align*}\n 4x + 8y + z &= -6 \\\\\n 2x - 3y + 2z &= 0 \\\\\n x + 7y - 3z &= -8\n\\end{align*}\n\\]\n\n**Solution:**\n\nLet\n\n\\[\nA = \\begin{bmatrix}\n 4 & 8 & 1 \\\\\n 2 & -3 & 2 \\\\\n 1 & 7 & -3\n\\end{bmatrix}\n\\]\n\nSince $|A| = -32 + 48 + 17 = 61$\n\nSo $A^{-1}$ exists.\n\n\\[\nA^{-1} = \\frac{1}{|A|} \\text{adj } A\n\\]\n\n\\[\n= \\frac{1}{61} \\begin{bmatrix}\n -5 & 31 & 19 \\\\\n 8 & -13 & -16 \\\\\n 17 & -20 & -28\n\\end{bmatrix}\n\\]\n\nNow since,\n\n\\[\nX = A^{-1} B, \\text{ we have}\n\\]\n\n\\[\n\\begin{bmatrix}\n x \\\\\n y \\\\\n z\n\\end{bmatrix} = \\frac{1}{61} \\begin{bmatrix}\n -5 & 31 & 19 \\\\\n 8 & -13 & -16 \\\\\n 17 & -20 & -28\n\\end{bmatrix} \\begin{bmatrix}\n -6 \\\\\n 0 \\\\\n -8\n\\end{bmatrix}\n\\]\n\\[\n\\begin{bmatrix}\n30 + 152 \\\\\n-48 + 48 \\\\\n-102 + 224\n\\end{bmatrix}\n= \\frac{1}{61}\n\\begin{bmatrix}\n-2 \\\\\n0 \\\\\n2\n\\end{bmatrix}\n\\]\n\nHence Solution set: \\{(x, y, z)\\} = \\{(-2, 0, 2)\\}\n\n**Exercise 9.3**\n\nQ.1 Which of the following matrices are singular or non-singular.\n\n(i) \\[\n\\begin{bmatrix}\n1 & 2 & 1 \\\\\n3 & 1 & -2 \\\\\n0 & 1 & -1\n\\end{bmatrix}\n\\]\n\n(ii) \\[\n\\begin{bmatrix}\n1 & 2 & -1 \\\\\n-3 & 4 & 5 \\\\\n-4 & 2 & 6\n\\end{bmatrix}\n\\]\n\n(iii) \\[\n\\begin{bmatrix}\n1 & 1 & -2 \\\\\n3 & -1 & 1 \\\\\n3 & 3 & -6\n\\end{bmatrix}\n\\]\n\nQ.2 Which of the following matrices are symmetric and skew-symmetric\n\n(i) \\[\n\\begin{bmatrix}\n2 & 6 & 7 \\\\\n6 & -2 & 3 \\\\\n7 & 3 & 0\n\\end{bmatrix}\n\\]\n\n(ii) \\[\n\\begin{bmatrix}\n0 & 3 & -5 \\\\\n-3 & 0 & 6 \\\\\n5 & -6 & 0\n\\end{bmatrix}\n\\]\n\n(iii) \\[\n\\begin{bmatrix}\na & b & c \\\\\nb & d & e \\\\\nc & e & f\n\\end{bmatrix}\n\\]\n\nQ.3 Find K such that the following matrices are singular\n\n(i) \\[\n\\begin{bmatrix}\nK & 6 \\\\\n4 & 3\n\\end{bmatrix}\n\\]\n\n(ii) \\[\n\\begin{bmatrix}\n1 & 2 & -1 \\\\\n-3 & 4 & K \\\\\n-4 & 2 & 6\n\\end{bmatrix}\n\\]\n\n(iii) \\[\n\\begin{bmatrix}\n1 & 1 & -2 \\\\\n3 & -1 & 1 \\\\\nk & 3 & -6\n\\end{bmatrix}\n\\]\n\nQ.4 Find the inverse if it exists, of the following matrices\n\n(i) \\[\n\\begin{bmatrix}\n1 & 3 \\\\\n2 & -1\n\\end{bmatrix}\n\\]\n\n(ii) \\[\n\\begin{bmatrix}\n0 & -2 & -3 \\\\\n1 & 3 & 3 \\\\\n-1 & -2 & -2\n\\end{bmatrix}\n\\]\n\n(iii) \\[\n\\begin{bmatrix}\n1 & 2 & 3 \\\\\n-1 & 0 & 4 \\\\\n0 & 2 & 2\n\\end{bmatrix}\n\\]\n\n(iv) \\[\n\\begin{bmatrix}\n1 & 2 & -1 \\\\\n-3 & 4 & 5 \\\\\n-4 & 2 & 6\n\\end{bmatrix}\n\\]\nQ.5 Find the solution set of the following system by means of matrices:\n\n(i) \\[2x - 3y = -1\\] \\[x + 4y = 5\\] \n(ii) \\[x + y = 2\\] \n(iii) \\[x - 2y + z = -1\\] \\[2x - z = 1\\] \\[3x + y - 2z = 4\\] \\[2y - 3z = -1\\] \\[y - z = 1\\] \n(iv) \\[-4x + 2y - 9z = 2\\] \\[3x + 4y + z = 5\\] \\[x - 3y + 2z = 8\\] \n(v) \\[x + y - 2z = 3\\] \\[3x - y + z = 0\\] \\[3x + 3y - 6z = 8\\] \n\n**Answers 9.3**\n\nQ.1 (i) Non-singular (ii) Singular (iii) Singular\n\nQ.2 (i) Symmetric (ii) Skew-symmetric (iii) Symmetric\n\nQ.3 (i) 8 (ii) 5 (iii) 3\n\nQ.4 (i) \\[\n\\begin{bmatrix}\n1 & 3 \\\\\n7 & 7 \\\\\n2 & 1 \\\\\n7 & -7\n\\end{bmatrix}\n\\] (ii) \\[\n\\begin{bmatrix}\n0 & 2 & 3 \\\\\n-1 & -3 & -3 \\\\\n1 & 2 & 2\n\\end{bmatrix}\n\\] (iii) \\[\n\\begin{bmatrix}\n4 & -1 & -4 \\\\\n5 & 5 & 5 \\\\\n1 & 1 & 7 \\\\\n5 & 5 & 10 \\\\\n1 & 1 & 1 \\\\\n5 & 5 & 5\n\\end{bmatrix}\n\\]\n\n(iv) \\[A^{-1} \\text{ does not exist.}\\]\n\nQ.5 (i) \\{(1, 1)\\} (ii) \\{(1, 1)\\} (iii) \\{(1, 1, 0)\\} (iv) \\{(7, -3, -4)\\} (v) no solution\nSummary\n\n1. If $A = [a_{ij}]$, $A = [b_{ij}]$ of order $m \\times n$. Then $A + B = [a_{ij} + b_{ij}]$ is also $m \\times n$ order.\n\n2. The product $AB$ of two matrices $A$ and $B$ is conformable for multiplication if No of columns in $A$ = No. of rows in $B$.\n\n3. If $A = [a_{ij}]$ is $m \\times n$ matrix, then the $n \\times m$ matrix obtained by interchanging the rows and columns of $A$ is called the transpose of $A$. It is denoted by $A^t$.\n\n4. Symmetric Matrix:\n A square matrix $A$ is symmetric if $A^t = A$.\n\n5. If $A = \\begin{bmatrix} a_{11} & a_{12} & a_{13} \\\\ a_{21} & a_{22} & a_{23} \\\\ a_{31} & a_{32} & a_{33} \\end{bmatrix}$ Then,\n \\[\n \\text{adj } A = \\begin{bmatrix} c_{11} & c_{21} & c_{31} \\\\ c_{12} & c_{22} & c_{32} \\\\ c_{13} & c_{23} & c_{33} \\end{bmatrix}, \\text{ } a_{ij} \\text{ are the co-factor elements.}\n \\]\n And inverse of $A$ is:\n \\[\n A^{-1} = \\frac{\\text{adj } A}{|A|}\n \\]\n\n6. A square matrix $A$ is singular if $|A| = 0$. \nShort Questions\n\nWrite the short answers of the following:\n\nQ.1: Define row and column vectors.\nQ.2: Define identity matrix.\nQ.3: Define symmetric matrix.\nQ.4: Define diagonal matrix.\nQ.5: Define scalar matrix.\nQ.6: Define rectangular matrix.\n\nQ.7: Show that \\( A = \\begin{vmatrix} 2 & 3 & -1 \\\\ 1 & 1 & 0 \\\\ 2 & -3 & 5 \\end{vmatrix} \\) is singular matrix\n\nQ.8: Show that \\( A = \\begin{vmatrix} 2 & 4 & -3 \\\\ 3 & -3 & 6 \\end{vmatrix} \\) is symmetric\n\nQ.9: Show that \\( \\begin{vmatrix} b & -1 & a \\\\ a & b & 0 \\\\ 1 & a & b \\end{vmatrix} = b^3 + a^3 \\)\n\nQ.10: Evaluate \\( \\begin{vmatrix} 1 & 2 & -2 \\\\ -1 & 1 & -3 \\\\ 2 & 4 & -1 \\end{vmatrix} \\)\n\nQ.11: Without expansion show that \\( \\begin{vmatrix} 1 & 2 & 3 \\\\ 4 & 5 & 6 \\\\ 7 & 8 & 9 \\end{vmatrix} = 0 \\)\n\nQ.12: Find \\( x \\) and \\( y \\) if \\( \\begin{bmatrix} 2 \\\\ -3 \\\\ 1 \\end{bmatrix} = \\begin{bmatrix} x + 3 \\\\ -3 \\\\ 3y - 4 \\end{bmatrix} \\)\n\nQ.13: Find \\( x \\) and \\( y \\) if \\( \\begin{bmatrix} x + 3 \\\\ -3 \\\\ 3y - 4 \\end{bmatrix} = \\begin{bmatrix} y \\\\ -3 \\\\ 2x \\end{bmatrix} \\)\nQ.14: If \\( A = \\begin{bmatrix} 1 & -1 & 2 \\\\ 3 & 2 & 5 \\\\ -1 & 0 & 4 \\end{bmatrix} \\) and \\( B = \\begin{bmatrix} 2 & 1 & -1 \\\\ 1 & 3 & 4 \\\\ -1 & 2 & 1 \\end{bmatrix} \\), find \\( A - B \\).\n\nQ.15: Find inverse of \\( \\begin{bmatrix} 2 \\\\ 6 \\\\ 1 \\\\ 3 \\end{bmatrix} \\).\n\nQ.16: If \\( A \\) is non-singular, then show that \\( (A^{-1})^{-1} = A \\).\n\nQ.17: If \\( A \\) is any square matrix then show that \\( AA^t \\) is symmetric.\n\nQ.18: Find \\( K \\) if \\( A = \\begin{bmatrix} 4 & k & 3 \\\\ 7 & 3 & 6 \\\\ 2 & 3 & 1 \\end{bmatrix} \\) is singular matrix.\n\nQ.19: Define the minor of an element of a matrix.\n\nQ.20: Define a co-factor of an element of a matrix.\n\nQ.21: Without expansion verify that\n\\[\n\\begin{vmatrix}\n\\alpha & \\beta + \\gamma & 1 \\\\\n\\beta & \\gamma + \\alpha & 1 \\\\\n\\gamma & \\alpha + \\beta & 1\n\\end{vmatrix} = 0\n\\]\n\nQ.22: What are the minor and cofactor of 3 in matrix.\n\\[\n\\begin{bmatrix}\n3 & 1 & -4 \\\\\n2 & 5 & 6 \\\\\n1 & 4 & 8\n\\end{bmatrix}\n\\]\n\nQ.23: What are the minor and cofactor of 4 in matrix.\n\\[\n\\begin{bmatrix}\n3 & 1 & -4 \\\\\n2 & 5 & 6 \\\\\n1 & 4 & 8\n\\end{bmatrix}\n\\]\n\nQ.24: If \\( \\begin{vmatrix} k-2 & 1 \\\\ 5 & k+2 \\end{vmatrix} = 0 \\), then find \\( k \\).\n\nQ.25: If \\( A = \\begin{bmatrix} 1 & 2 \\\\ 3 & 4 \\end{bmatrix} \\), \\( B = \\begin{bmatrix} 2 & 3 \\\\ 4 & 5 \\end{bmatrix} \\), then find \\( A + B \\).\nQ.26: If \\( A = \\begin{bmatrix} 1 & 2 \\\\ 3 & 4 \\end{bmatrix} \\), \\( B = \\begin{bmatrix} 2 & 3 \\\\ 4 & 5 \\end{bmatrix} \\), Then find \\( A - B \\)\n\nQ.27: If \\( A = \\begin{bmatrix} 1 & 2 \\\\ 3 & 4 \\end{bmatrix} \\), \\( B = \\begin{bmatrix} 2 & 3 \\\\ 4 & 5 \\end{bmatrix} \\), Then find \\( AB \\)\n\nQ.28: If \\( \\begin{vmatrix} 2 & 3 \\\\ 4 & k \\end{vmatrix} \\) is singular, Then find \\( k \\).\n\nQ.29: Find \\( A^{-1} \\) if \\( A = \\begin{vmatrix} 5 & 3 \\\\ 1 & 1 \\end{vmatrix} \\)\n\n**Answers**\n\nQ10. 9 Q12. \\( x = -1, y = 2 \\) Q13. \\( x = -5, y = -2 \\)\n\nQ14. \\( \\begin{vmatrix} -1 & -2 & 3 \\\\ 2 & -1 & 1 \\\\ 0 & -2 & 3 \\end{vmatrix} \\) Q15. \\( D^{-1} \\) does not exist Q18. \\( k = 3 \\)\n\nQ22. \\( M_{11} = 16, C_{11} = 16 \\) Q23. \\( M_{32} = 26, C_{32} = -26 \\)\n\nQ24. \\( K = \\pm 3 \\) Q25. \\( \\begin{bmatrix} 3 & 5 \\\\ 7 & 9 \\end{bmatrix} \\) Q26. \\( \\begin{bmatrix} -1 & -1 \\\\ -1 & -1 \\end{bmatrix} \\)\n\nQ27. \\( \\begin{bmatrix} 10 & 13 \\\\ 22 & 29 \\end{bmatrix} \\) Q28. \\( k = 6 \\) Q29. \\( \\frac{1}{2} \\begin{bmatrix} 1 & -3 \\\\ -1 & 5 \\end{bmatrix} \\)\nObjective type Exercise\n\nQ.1 Each question has four possible answers. Choose the correct answer and encircle it.\n\n___1. The order of the matrix \\[\n\\begin{bmatrix}\n2 \\\\\n3 \\\\\n4\n\\end{bmatrix}\n\\] is\n(a) 2 x 1 (b) 2 x 2 (c) 3 x 1 (d) 1 x 3\n\n___2. The order of the matrix \\[\n\\begin{bmatrix}\n1 & 2 & 3\n\\end{bmatrix}\n\\] is\n(a) 1 x 3 (b) 3 x 1 (c) 3 x 3 (d) 2 x 3\n\n___3. The matrix \\[\n\\begin{bmatrix}\n0 & 0 \\\\\n0 & 0\n\\end{bmatrix}\n\\] is called\n(a) Identity (b) scalar (c) diagonal (d) Null\n\n___4. Two matrices A and B are conformable for multiplication if\n(a) No of columns in A = No of rows in B\n(b) No of columns in A = No of columns in B\n(c) No of rows in A = No of rows in B\n(d) None of these\n\n___5. If the order of the matrix A is pxq and order of B is qxr, then order of AB will be:\n(a) pxq (b) qxp (c) pxr (d) rxp\n\n___6. In an identity matrix all the diagonal elements are:\n(a) zero (b) 2 (c) 1 (d) none of these\n\n___7. The value of determinant \\[\n\\begin{bmatrix}\n2 & 0 \\\\\n1 & 3\n\\end{bmatrix}\n\\] is:\n(a) 6 (b) –6 (c) 1 (d) 0\n\n___8. If two rows of a determinant are identical then its value is\n(a) 1 (b) zero (c) –1 (d) None of these\n\n___9. If \\[\nA = \\begin{bmatrix}\n2 & 3 & 4 \\\\\n0 & 1 & -1 \\\\\n2 & 0 & 1\n\\end{bmatrix}\n\\] is a matrix, then Cofactor of 4 is\n(a) –2 (b) 2 (c) 3 (d) 4\n\n___10. If all the elements of a row or a column are zero, then value of the determinant is:\n(a) 1 (b) 2 (c) zero (d) None of these\n11. Value of m for which matrix \\[\n\\begin{bmatrix}\n2 & 3 \\\\\n6 & m\n\\end{bmatrix}\n\\] is singular.\n(a) 6 (b) 3 (c) 8 (d) 9\n\n12. If \\([a_{ij}]\\) and \\([b_{ij}]\\) are of the same order and \\(a_{ij} = b_{ij}\\) then the matrix will be\n(a) Singular (b) Null (c) unequal (d) equal\n\n13. Matrix \\([a_{ij}]_{mxn}\\) is a row matrix if:\n(a) \\(i = 1\\) (b) \\(j = 1\\) (c) \\(m = 1\\) (d) \\(n = 1\\)\n\n14. Matrix \\([c_{ij}]_{mxn}\\) is a rectangular if:\n(a) \\(i \\neq j\\) (b) \\(i = j\\) (c) \\(m = n\\) (d) \\(m - n \\neq 0\\)\n\n15. If \\(A = [a_{ij}]_{mxn}\\) is a scalar matrix if:\n(a) \\(a_{ij} = 0\\) \\(\\forall i \\neq j\\) (b) \\(a_{ij} = k\\) \\(\\forall i = j\\)\n(c) \\(a_{ij} = k\\) \\(\\forall i \\neq j\\) (d) (a) and (b)\n\n16. Matrix \\(A = [a_{ij}]_{mxn}\\) is an identity matrix if:\n(a) \\(\\forall i = j, a_{ij} = 0\\) (b) \\(\\forall i = j, a_{ij} = 1\\)\n(c) \\(\\forall i \\neq j, a_{ij} = 0\\) (d) both (b) and (c)\n\n17. Which matrix can be a rectangular matrix?\n(a) Diagonal (b) Identity (c) Scalar (d) None\n\n18. If \\(A = [a_{ij}]_{mxn}\\) then order \\(kA\\) is:\n(a) \\(m \\times n\\) (b) \\(km \\times kn\\) (c) \\(km \\times n\\) (d) \\(m \\times kn\\)\n\n19. \\((A - B)^2 = A^2 - 2AB + B^2\\), if and only if:\n(a) \\(A + B = 0\\) (b) \\(AB - BA = 0\\) (c) \\(A^2 + B^2 = 0\\) (d) (a) and (c)\n\n20. If \\(A\\) and \\(B\\) ARE symmetric, then \\(AB =\\)\n(a) \\(BA\\) (b) \\(A^tB^t\\) (c) \\(B^tA^t\\) (d) (a) and (c)\n\n**Answers**\n\nQ.1 (1) c (2) a (3) d (4) a (5) c (6) c (7) a (8) b (9) a (10) c (11) d (12) d (13) c (14) d (15) d (16) d (17) d (18) a (19) b (20) d", "id": "./materials/54.pdf" }, { "contents": "2.4 Matrix Inverses\n\nThree basic operations on matrices, addition, multiplication, and subtraction, are analogs for matrices of the same operations for numbers. In this section we introduce the matrix analog of numerical division.\n\nTo begin, consider how a numerical equation $ax = b$ is solved when $a$ and $b$ are known numbers. If $a = 0$, there is no solution (unless $b = 0$). But if $a \\neq 0$, we can multiply both sides by the inverse $a^{-1} = \\frac{1}{a}$ to obtain the solution $x = a^{-1}b$. Of course multiplying by $a^{-1}$ is just dividing by $a$, and the property of $a^{-1}$ that makes this work is that $a^{-1}a = 1$. Moreover, we saw in Section 2.2 that the role that 1 plays in arithmetic is played in matrix algebra by the identity matrix $I$. This suggests the following definition.\n\n**Definition 2.11 Matrix Inverses**\n\nIf $A$ is a square matrix, a matrix $B$ is called an inverse of $A$ if and only if\n\n$$AB = I \\quad \\text{and} \\quad BA = I$$\n\nA matrix $A$ that has an inverse is called an invertible matrix.\\(^8\\)\n\n---\n\n**Example 2.4.1**\n\nShow that $B = \\begin{bmatrix} -1 & 1 \\\\ 1 & 0 \\end{bmatrix}$ is an inverse of $A = \\begin{bmatrix} 0 & 1 \\\\ 1 & 1 \\end{bmatrix}$.\n\n**Solution.** Compute $AB$ and $BA$.\n\n$$AB = \\begin{bmatrix} 0 & 1 \\\\ 1 & 1 \\end{bmatrix} \\begin{bmatrix} -1 & 1 \\\\ 1 & 0 \\end{bmatrix} = \\begin{bmatrix} 1 & 0 \\\\ 1 & 0 \\end{bmatrix} \\quad BA = \\begin{bmatrix} -1 & 1 \\\\ 1 & 0 \\end{bmatrix} \\begin{bmatrix} 0 & 1 \\\\ 1 & 1 \\end{bmatrix} = \\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix}$$\n\nHence $AB = I = BA$, so $B$ is indeed an inverse of $A$.\n\n---\n\n**Example 2.4.2**\n\nShow that $A = \\begin{bmatrix} 0 & 0 \\\\ 1 & 3 \\end{bmatrix}$ has no inverse.\n\n**Solution.** Let $B = \\begin{bmatrix} a & b \\\\ c & d \\end{bmatrix}$ denote an arbitrary $2 \\times 2$ matrix. Then\n\n$$AB = \\begin{bmatrix} 0 & 0 \\\\ 1 & 3 \\end{bmatrix} \\begin{bmatrix} a & b \\\\ c & d \\end{bmatrix} = \\begin{bmatrix} 0 & 0 \\\\ a+3c & b+3d \\end{bmatrix}$$\n\nso $AB$ has a row of zeros. Hence $AB$ cannot equal $I$ for any $B$.\n\n---\n\n\\(^8\\)Only square matrices have inverses. Even though it is plausible that nonsquare matrices $A$ and $B$ could exist such that $AB = I_m$ and $BA = I_n$, where $A$ is $m \\times n$ and $B$ is $n \\times m$, we claim that this forces $n = m$. Indeed, if $m < n$ there exists a nonzero column $x$ such that $Ax = 0$ (by Theorem 1.3.1), so $x = I_n x = (BA)x = B(Ax) = B(0) = 0$, a contradiction. Hence $m \\geq n$. Similarly, the condition $AB = I_m$ implies that $n \\geq m$. Hence $m = n$ so $A$ is square.\nThe argument in Example 2.4.2 shows that no zero matrix has an inverse. But Example 2.4.2 also shows that, unlike arithmetic, it is possible for a nonzero matrix to have no inverse. However, if a matrix does have an inverse, it has only one.\n\n**Theorem 2.4.1**\n\nIf $B$ and $C$ are both inverses of $A$, then $B = C$.\n\n**Proof.** Since $B$ and $C$ are both inverses of $A$, we have $CA = I = AB$. Hence\n\n$$B = IB = (CA)B = C(AB) = CI = C$$\n\nIf $A$ is an invertible matrix, the (unique) inverse of $A$ is denoted $A^{-1}$. Hence $A^{-1}$ (when it exists) is a square matrix of the same size as $A$ with the property that\n\n$$AA^{-1} = I \\quad \\text{and} \\quad A^{-1}A = I$$\n\nThese equations characterize $A^{-1}$ in the following sense:\n\n**Inverse Criterion:** If somehow a matrix $B$ can be found such that $AB = I$ and $BA = I$, then $A$ is invertible and $B$ is the inverse of $A$; in symbols, $B = A^{-1}$.\n\nThis is a way to verify that the inverse of a matrix exists. Example 2.4.3 and Example 2.4.4 offer illustrations.\n\n**Example 2.4.3**\n\nIf $A = \\begin{bmatrix} 0 & -1 \\\\ 1 & -1 \\end{bmatrix}$, show that $A^3 = I$ and so find $A^{-1}$.\n\n**Solution.** We have $A^2 = \\begin{bmatrix} 0 & -1 \\\\ 1 & -1 \\end{bmatrix} \\begin{bmatrix} 0 & -1 \\\\ 1 & -1 \\end{bmatrix} = \\begin{bmatrix} -1 & 1 \\\\ -1 & 0 \\end{bmatrix}$, and so\n\n$$A^3 = A^2A = \\begin{bmatrix} -1 & 1 \\\\ -1 & 0 \\end{bmatrix} \\begin{bmatrix} 0 & -1 \\\\ 1 & -1 \\end{bmatrix} = \\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix} = I$$\n\nHence $A^3 = I$, as asserted. This can be written as $A^2A = I = AA^2$, so it shows that $A^2$ is the inverse of $A$. That is, $A^{-1} = A^2 = \\begin{bmatrix} -1 & 1 \\\\ -1 & 0 \\end{bmatrix}$.\n\nThe next example presents a useful formula for the inverse of a $2 \\times 2$ matrix $A = \\begin{bmatrix} a & b \\\\ c & d \\end{bmatrix}$ when it exists. To state it, we define the **determinant** $\\det A$ and the **adjugate** $\\text{adj} A$ of the matrix $A$ as follows:\n\n$$\\det \\begin{bmatrix} a & b \\\\ c & d \\end{bmatrix} = ad - bc, \\quad \\text{and} \\quad \\text{adj} \\begin{bmatrix} a & b \\\\ c & d \\end{bmatrix} = \\begin{bmatrix} d & -b \\\\ -c & a \\end{bmatrix}$$\nExample 2.4.4\n\nIf \\( A = \\begin{bmatrix} a & b \\\\ c & d \\end{bmatrix} \\), show that \\( A \\) has an inverse if and only if \\( \\det A \\neq 0 \\), and in this case\n\n\\[\nA^{-1} = \\frac{1}{\\det A} \\text{adj} A\n\\]\n\n**Solution.** For convenience, write \\( e = \\det A = ad - bc \\) and \\( B = \\text{adj} A = \\begin{bmatrix} d & -b \\\\ -c & a \\end{bmatrix} \\). Then\n\n\\[\nAB = eI = BA\n\\]\n\nas the reader can verify. So if \\( e \\neq 0 \\), scalar multiplication by \\( \\frac{1}{e} \\) gives\n\n\\[\nA\\left(\\frac{1}{e}B\\right) = I = \\left(\\frac{1}{e}B\\right)A\n\\]\n\nHence \\( A \\) is invertible and \\( A^{-1} = \\frac{1}{e}B \\). Thus it remains only to show that if \\( A^{-1} \\) exists, then \\( e \\neq 0 \\). We prove this by showing that assuming \\( e = 0 \\) leads to a contradiction. In fact, if \\( e = 0 \\), then\n\n\\[\nAB = eI = 0,\n\\]\n\nso left multiplication by \\( A^{-1} \\) gives \\( A^{-1}AB = A^{-1}0 \\); that is, \\( IB = 0 \\), so \\( B = 0 \\). But this implies that \\( a, b, c, \\) and \\( d \\) are all zero, so \\( A = 0 \\), contrary to the assumption that \\( A^{-1} \\) exists.\n\nAs an illustration, if \\( A = \\begin{bmatrix} 2 & 4 \\\\ -3 & 8 \\end{bmatrix} \\) then \\( \\det A = 2 \\cdot 8 - 4 \\cdot (-3) = 28 \\neq 0 \\). Hence \\( A \\) is invertible and\n\n\\[\nA^{-1} = \\frac{1}{\\det A} \\text{adj} A = \\frac{1}{28} \\begin{bmatrix} 8 & -4 \\\\ 3 & 2 \\end{bmatrix},\n\\]\n\nas the reader is invited to verify.\n\nThe determinant and adjugate will be defined in Chapter 3 for any square matrix, and the conclusions in Example 2.4.4 will be proved in full generality.\n\n**Inverses and Linear Systems**\n\nMatrix inverses can be used to solve certain systems of linear equations. Recall that a *system* of linear equations can be written as a *single* matrix equation\n\n\\[\nAx = b\n\\]\n\nwhere \\( A \\) and \\( b \\) are known and \\( x \\) is to be determined. If \\( A \\) is invertible, we multiply each side of the equation on the left by \\( A^{-1} \\) to get\n\n\\[\nA^{-1}Ax = A^{-1}b\n\\]\n\n\\[\nIx = A^{-1}b\n\\]\n\n\\[\nx = A^{-1}b\n\\]\n\nThis gives the solution to the system of equations (the reader should verify that \\( x = A^{-1}b \\) really does satisfy \\( Ax = b \\)). Furthermore, the argument shows that if \\( x \\) is any solution, then necessarily \\( x = A^{-1}b \\), so the solution is unique. Of course the technique works only when the coefficient matrix \\( A \\) has an inverse. This proves Theorem 2.4.2.\nTheorem 2.4.2\n\nSuppose a system of $n$ equations in $n$ variables is written in matrix form as\n\n$$Ax = b$$\n\nIf the $n \\times n$ coefficient matrix $A$ is invertible, the system has the unique solution\n\n$$x = A^{-1}b$$\n\nExample 2.4.5\n\nUse Example 2.4.4 to solve the system\n\n$$\\begin{align*}\n5x_1 - 3x_2 &= -4 \\\\\n7x_1 + 4x_2 &= 8\n\\end{align*}$$\n\n**Solution.** In matrix form this is $Ax = b$ where $A = \\begin{bmatrix} 5 & -3 \\\\ 7 & 4 \\end{bmatrix}$, $x = \\begin{bmatrix} x_1 \\\\ x_2 \\end{bmatrix}$, and $b = \\begin{bmatrix} -4 \\\\ 8 \\end{bmatrix}$. Then $\\det A = 5 \\cdot 4 - (-3) \\cdot 7 = 41$, so $A$ is invertible and $A^{-1} = \\frac{1}{41} \\begin{bmatrix} 4 & 3 \\\\ -7 & 5 \\end{bmatrix}$ by Example 2.4.4. Thus Theorem 2.4.2 gives\n\n$$x = A^{-1}b = \\frac{1}{41} \\begin{bmatrix} 4 & 3 \\\\ -7 & 5 \\end{bmatrix} \\begin{bmatrix} -4 \\\\ 8 \\end{bmatrix} = \\frac{1}{41} \\begin{bmatrix} 8 \\\\ 68 \\end{bmatrix}$$\n\nso the solution is $x_1 = \\frac{8}{41}$ and $x_2 = \\frac{68}{41}$.\n\nAn Inversion Method\n\nIf a matrix $A$ is $n \\times n$ and invertible, it is desirable to have an efficient technique for finding the inverse. The following procedure will be justified in Section 2.5.\n\nMatrix Inversion Algorithm\n\nIf $A$ is an invertible (square) matrix, there exists a sequence of elementary row operations that carry $A$ to the identity matrix $I$ of the same size, written $A \\rightarrow I$. This same series of row operations carries $I$ to $A^{-1}$; that is, $I \\rightarrow A^{-1}$. The algorithm can be summarized as follows:\n\n$$\\begin{bmatrix} A & I \\end{bmatrix} \\rightarrow \\begin{bmatrix} I & A^{-1} \\end{bmatrix}$$\n\nwhere the row operations on $A$ and $I$ are carried out simultaneously.\nExample 2.4.6\n\nUse the inversion algorithm to find the inverse of the matrix\n\n\\[\nA = \\begin{bmatrix}\n2 & 7 & 1 \\\\\n1 & 4 & -1 \\\\\n1 & 3 & 0\n\\end{bmatrix}\n\\]\n\n**Solution.** Apply elementary row operations to the double matrix\n\n\\[\n\\begin{bmatrix}\nA & I\n\\end{bmatrix} = \\begin{bmatrix}\n2 & 7 & 1 & 1 & 0 & 0 \\\\\n1 & 4 & -1 & 0 & 1 & 0 \\\\\n1 & 3 & 0 & 0 & 0 & 1\n\\end{bmatrix}\n\\]\n\nso as to carry \\( A \\) to \\( I \\). First interchange rows 1 and 2.\n\n\\[\n\\begin{bmatrix}\n1 & 4 & -1 & 0 & 1 & 0 \\\\\n2 & 7 & 1 & 1 & 0 & 0 \\\\\n1 & 3 & 0 & 0 & 0 & 1\n\\end{bmatrix}\n\\]\n\nNext subtract 2 times row 1 from row 2, and subtract row 1 from row 3.\n\n\\[\n\\begin{bmatrix}\n1 & 4 & -1 & 0 & 1 & 0 \\\\\n0 & -1 & 3 & 1 & -2 & 0 \\\\\n0 & -1 & 1 & 0 & -1 & 1\n\\end{bmatrix}\n\\]\n\nContinue to reduced row-echelon form.\n\n\\[\n\\begin{bmatrix}\n1 & 0 & 11 & 4 & -7 & 0 \\\\\n0 & 1 & -3 & -1 & 2 & 0 \\\\\n0 & 0 & -2 & -1 & 1 & 1\n\\end{bmatrix}\n\\]\n\n\\[\n\\begin{bmatrix}\n1 & 0 & 0 & \\frac{-3}{2} & \\frac{-3}{2} & \\frac{11}{2} \\\\\n0 & 1 & 0 & \\frac{1}{2} & \\frac{1}{2} & \\frac{-3}{2} \\\\\n0 & 0 & 1 & \\frac{1}{2} & \\frac{-1}{2} & \\frac{-1}{2}\n\\end{bmatrix}\n\\]\n\nHence \\( A^{-1} = \\frac{1}{2} \\begin{bmatrix}\n-3 & -3 & 11 \\\\\n1 & 1 & -3 \\\\\n1 & -1 & -1\n\\end{bmatrix} \\), as is readily verified.\n\nGiven any \\( n \\times n \\) matrix \\( A \\), Theorem 1.2.1 shows that \\( A \\) can be carried by elementary row operations to a matrix \\( R \\) in reduced row-echelon form. If \\( R = I \\), the matrix \\( A \\) is invertible (this will be proved in the next section), so the algorithm produces \\( A^{-1} \\). If \\( R \\neq I \\), then \\( R \\) has a row of zeros (it is square), so no system of linear equations \\( Ax = b \\) can have a unique solution. But then \\( A \\) is not invertible by Theorem 2.4.2. Hence, the algorithm is effective in the sense conveyed in Theorem 2.4.3.\nTheorem 2.4.3\n\nIf $A$ is an $n \\times n$ matrix, either $A$ can be reduced to $I$ by elementary row operations or it cannot. In the first case, the algorithm produces $A^{-1}$; in the second case, $A^{-1}$ does not exist.\n\nProperties of Inverses\n\nThe following properties of an invertible matrix are used everywhere.\n\nExample 2.4.7: Cancellation Laws\n\nLet $A$ be an invertible matrix. Show that:\n\n1. If $AB = AC$, then $B = C$.\n2. If $BA = CA$, then $B = C$.\n\nSolution. Given the equation $AB = AC$, left multiply both sides by $A^{-1}$ to obtain $A^{-1}AB = A^{-1}AC$. Thus $IB = IC$, that is $B = C$. This proves (1) and the proof of (2) is left to the reader.\n\nProperties (1) and (2) in Example 2.4.7 are described by saying that an invertible matrix can be “left cancelled” and “right cancelled”, respectively. Note however that “mixed” cancellation does not hold in general: If $A$ is invertible and $AB = CA$, then $B$ and $C$ may not be equal, even if both are $2 \\times 2$. Here is a specific example:\n\n$$A = \\begin{bmatrix} 1 & 1 \\\\ 0 & 1 \\end{bmatrix}, \\quad B = \\begin{bmatrix} 0 & 0 \\\\ 1 & 2 \\end{bmatrix}, \\quad C = \\begin{bmatrix} 1 & 1 \\\\ 1 & 1 \\end{bmatrix}$$\n\nSometimes the inverse of a matrix is given by a formula. Example 2.4.4 is one illustration; Example 2.4.8 and Example 2.4.9 provide two more. The idea is the Inverse Criterion: If a matrix $B$ can be found such that $AB = I = BA$, then $A$ is invertible and $A^{-1} = B$.\n\nExample 2.4.8\n\nIf $A$ is an invertible matrix, show that the transpose $A^T$ is also invertible. Show further that the inverse of $A^T$ is just the transpose of $A^{-1}$; in symbols, $(A^T)^{-1} = (A^{-1})^T$.\n\nSolution. $A^{-1}$ exists (by assumption). Its transpose $(A^{-1})^T$ is the candidate proposed for the inverse of $A^T$. Using the inverse criterion, we test it as follows:\n\n$$A^T(A^{-1})^T = (A^{-1}A)^T = I^T = I$$\n\n$$(A^{-1})^TA^T = (AA^{-1})^T = I^T = I$$\n\nHence $(A^{-1})^T$ is indeed the inverse of $A^T$; that is, $(A^T)^{-1} = (A^{-1})^T$. \nExample 2.4.9\n\nIf $A$ and $B$ are invertible $n \\times n$ matrices, show that their product $AB$ is also invertible and $(AB)^{-1} = B^{-1}A^{-1}$.\n\nSolution. We are given a candidate for the inverse of $AB$, namely $B^{-1}A^{-1}$. We test it as follows:\n\n$$(B^{-1}A^{-1})(AB) = B^{-1}(A^{-1}A)B = B^{-1}IB = B^{-1}B = I$$\n\n$$(AB)(B^{-1}A^{-1}) = A(AB^{-1})A^{-1} = AIA^{-1} = AA^{-1} = I$$\n\nHence $B^{-1}A^{-1}$ is the inverse of $AB$; in symbols, $(AB)^{-1} = B^{-1}A^{-1}$.\n\nWe now collect several basic properties of matrix inverses for reference.\n\nTheorem 2.4.4\n\nAll the following matrices are square matrices of the same size.\n\n1. $I$ is invertible and $I^{-1} = I$.\n2. If $A$ is invertible, so is $A^{-1}$, and $(A^{-1})^{-1} = A$.\n3. If $A$ and $B$ are invertible, so is $AB$, and $(AB)^{-1} = B^{-1}A^{-1}$.\n4. If $A_1, A_2, \\ldots, A_k$ are all invertible, so is their product $A_1A_2\\cdots A_k$, and\n\n$$(A_1A_2\\cdots A_k)^{-1} = A_k^{-1}\\cdots A_2^{-1}A_1^{-1}.$$ \n\n5. If $A$ is invertible, so is $A^k$ for any $k \\geq 1$, and $(A^k)^{-1} = (A^{-1})^k$.\n6. If $A$ is invertible and $a \\neq 0$ is a number, then $aA$ is invertible and $(aA)^{-1} = \\frac{1}{a}A^{-1}$.\n7. If $A$ is invertible, so is its transpose $A^T$, and $(A^T)^{-1} = (A^{-1})^T$.\n\nProof.\n\n1. This is an immediate consequence of the fact that $I^2 = I$.\n2. The equations $AA^{-1} = I = A^{-1}A$ show that $A$ is the inverse of $A^{-1}$; in symbols, $(A^{-1})^{-1} = A$.\n3. This is Example 2.4.9.\n4. Use induction on $k$. If $k = 1$, there is nothing to prove, and if $k = 2$, the result is property 3. If $k > 2$, assume inductively that $(A_1A_2\\cdots A_{k-1})^{-1} = A_{k-1}^{-1}\\cdots A_2^{-1}A_1^{-1}$. We apply this fact together with property 3 as follows:\n\n$$[A_1A_2\\cdots A_{k-1}A_k]^{-1} = [(A_1A_2\\cdots A_{k-1})A_k]^{-1}$$\n\n$$= A_k^{-1}(A_1A_2\\cdots A_{k-1})^{-1}$$\n\n$$= A_k^{-1}(A_{k-1}^{-1}\\cdots A_2^{-1}A_1^{-1})$$\nSo the proof by induction is complete.\n\n5. This is property 4 with \\( A_1 = A_2 = \\cdots = A_k = A \\).\n\n6. This is left as Exercise 2.4.29.\n\n7. This is Example 2.4.8.\n\nThe reversal of the order of the inverses in properties 3 and 4 of Theorem 2.4.4 is a consequence of the fact that matrix multiplication is not commutative. Another manifestation of this comes when matrix equations are dealt with. If a matrix equation \\( B = C \\) is given, it can be left-multiplied by a matrix \\( A \\) to yield \\( AB = AC \\). Similarly, right-multiplication gives \\( BA = CA \\). However, we cannot mix the two: If \\( B = C \\), it need not be the case that \\( AB = CA \\) even if \\( A \\) is invertible, for example, \\( A = \\begin{bmatrix} 1 & 1 \\\\ 0 & 1 \\end{bmatrix}, B = \\begin{bmatrix} 0 & 0 \\\\ 1 & 0 \\end{bmatrix} = C \\).\n\nPart 7 of Theorem 2.4.4 together with the fact that \\((A^T)^T = A\\) gives\n\n**Corollary 2.4.1**\n\nA square matrix \\( A \\) is invertible if and only if \\( A^T \\) is invertible.\n\n**Example 2.4.10**\n\nFind \\( A \\) if \\((A^T - 2I)^{-1} = \\begin{bmatrix} 2 & 1 \\\\ -1 & 0 \\end{bmatrix}\\).\n\n**Solution.** By Theorem 2.4.4(2) and Example 2.4.4, we have\n\n\\[\n(A^T - 2I) = \\left( (A^T - 2I)^{-1} \\right)^{-1} = \\begin{bmatrix} 2 & 1 \\\\ -1 & 0 \\end{bmatrix}^{-1} = \\begin{bmatrix} 0 & -1 \\\\ 1 & 2 \\end{bmatrix}\n\\]\n\nHence \\( A^T = 2I + \\begin{bmatrix} 0 & -1 \\\\ 1 & 2 \\end{bmatrix} = \\begin{bmatrix} 2 & -1 \\\\ 1 & 4 \\end{bmatrix} \\), so \\( A = \\begin{bmatrix} 2 & 1 \\\\ -1 & 4 \\end{bmatrix} \\) by Theorem 2.4.4(7).\n\nThe following important theorem collects a number of conditions all equivalent\\(^9\\) to invertibility. It will be referred to frequently below.\n\n**Theorem 2.4.5: Inverse Theorem**\n\nThe following conditions are equivalent for an \\( n \\times n \\) matrix \\( A \\):\n\n1. \\( A \\) is invertible.\n\n2. The homogeneous system \\( Ax = 0 \\) has only the trivial solution \\( x = 0 \\).\n\n3. \\( A \\) can be carried to the identity matrix \\( I_n \\) by elementary row operations.\n\n\\(^9\\)If \\( p \\) and \\( q \\) are statements, we say that \\( p \\) implies \\( q \\) (written \\( p \\Rightarrow q \\)) if \\( q \\) is true whenever \\( p \\) is true. The statements are called equivalent if both \\( p \\Rightarrow q \\) and \\( q \\Rightarrow p \\) (written \\( p \\Leftrightarrow q \\), spoken “\\( p \\) if and only if \\( q \\)”). See Appendix B.\n4. The system $Ax = b$ has at least one solution $x$ for every choice of column $b$.\n\n5. There exists an $n \\times n$ matrix $C$ such that $AC = I_n$.\n\n**Proof.** We show that each of these conditions implies the next, and that (5) implies (1).\n\n(1) $\\Rightarrow$ (2). If $A^{-1}$ exists, then $Ax = 0$ gives $x = I_n x = A^{-1} Ax = A^{-1} 0 = 0$.\n\n(2) $\\Rightarrow$ (3). Assume that (2) is true. Certainly $A \\rightarrow R$ by row operations where $R$ is a reduced, row-echelon matrix. It suffices to show that $R = I_n$. Suppose that this is not the case. Then $R$ has a row of zeros (being square). Now consider the augmented matrix $[ A \\mid 0 ]$ of the system $Ax = 0$. Then $[ A \\mid 0 ] \\rightarrow [ R \\mid 0 ]$ is the reduced form, and $[ R \\mid 0 ]$ also has a row of zeros. Since $R$ is square there must be at least one nonleading variable, and hence at least one parameter. Hence the system $Ax = 0$ has infinitely many solutions, contrary to (2). So $R = I_n$ after all.\n\n(3) $\\Rightarrow$ (4). Consider the augmented matrix $[ A \\mid b ]$ of the system $Ax = b$. Using (3), let $A \\rightarrow I_n$ by a sequence of row operations. Then these same operations carry $[ A \\mid b ] \\rightarrow [ I_n \\mid c ]$ for some column $c$. Hence the system $Ax = b$ has a solution (in fact unique) by gaussian elimination. This proves (4).\n\n(4) $\\Rightarrow$ (5). Write $I_n = [ e_1 \\ e_2 \\ \\cdots \\ e_n ]$ where $e_1$, $e_2$, $\\ldots$, $e_n$ are the columns of $I_n$. For each $j = 1$, $2$, $\\ldots$, $n$, the system $Ax = e_j$ has a solution $c_j$ by (4), so $Ac_j = e_j$. Now let $C = [ c_1 \\ c_2 \\ \\cdots \\ c_n ]$ be the $n \\times n$ matrix with these matrices $c_j$ as its columns. Then Definition 2.9 gives (5):\n\n$$AC = A [ c_1 \\ c_2 \\ \\cdots \\ c_n ] = [ Ac_1 \\ Ac_2 \\ \\cdots \\ Ac_n ] = [ e_1 \\ e_2 \\ \\cdots \\ e_n ] = I_n$$\n\n(5) $\\Rightarrow$ (1). Assume that (5) is true so that $AC = I_n$ for some matrix $C$. Then $Cx = 0$ implies $x = 0$ (because $x = I_n x = ACx = A0 = 0$). Thus condition (2) holds for the matrix $C$ rather than $A$. Hence the argument above that (2) $\\Rightarrow$ (3) $\\Rightarrow$ (4) $\\Rightarrow$ (5) (with $A$ replaced by $C$) shows that a matrix $C'$ exists such that $CC' = I_n$.\n\nBut then\n\n$$A = AI_n = A(CC') = (AC)C' = I_n C' = C'$$\n\nThus $CA = CC' = I_n$ which, together with $AC = I_n$, shows that $C$ is the inverse of $A$. This proves (1). \\qed\n\nThe proof of (5) $\\Rightarrow$ (1) in Theorem 2.4.5 shows that if $AC = I$ for square matrices, then necessarily $CA = I$, and hence that $C$ and $A$ are inverses of each other. We record this important fact for reference.\n\n**Corollary 2.4.1**\n\nIf $A$ and $C$ are square matrices such that $AC = I$, then also $CA = I$. In particular, both $A$ and $C$ are invertible, $C = A^{-1}$, and $A = C^{-1}$.\n\nHere is a quick way to remember Corollary 2.4.1. If $A$ is a square matrix, then\n\n1. If $AC = I$ then $C = A^{-1}$.\n2. If $CA = I$ then $C = A^{-1}$.\n\nObserve that Corollary 2.4.1 is false if $A$ and $C$ are not square matrices. For example, we have\n\n$$\\begin{bmatrix} 1 & 2 & 1 \\\\ 1 & 1 & 1 \\end{bmatrix} \\begin{bmatrix} -1 & 1 \\\\ 1 & -1 \\\\ 0 & 1 \\end{bmatrix} = I_2 \\quad \\text{but} \\quad \\begin{bmatrix} -1 & 1 \\\\ 1 & -1 \\\\ 0 & 1 \\end{bmatrix} \\begin{bmatrix} 1 & 2 & 1 \\\\ 1 & 1 & 1 \\end{bmatrix} \\neq I_3$$\nIn fact, it is verified in the footnote on page 80 that if $AB = I_m$ and $BA = I_n$, where $A$ is $m \\times n$ and $B$ is $n \\times m$, then $m = n$ and $A$ and $B$ are (square) inverses of each other.\n\nAn $n \\times n$ matrix $A$ has rank $n$ if and only if (3) of Theorem 2.4.5 holds. Hence\n\n**Corollary 2.4.2**\n\nAn $n \\times n$ matrix $A$ is invertible if and only if rank $A = n$.\n\nHere is a useful fact about inverses of block matrices.\n\n**Example 2.4.11**\n\nLet $P = \\begin{bmatrix} A & X \\\\ 0 & B \\end{bmatrix}$ and $Q = \\begin{bmatrix} A & 0 \\\\ Y & B \\end{bmatrix}$ be block matrices where $A$ is $m \\times m$ and $B$ is $n \\times n$ (possibly $m \\neq n$).\n\na. Show that $P$ is invertible if and only if $A$ and $B$ are both invertible. In this case, show that\n\n$$P^{-1} = \\begin{bmatrix} A^{-1} & -A^{-1}XB^{-1} \\\\ 0 & B^{-1} \\end{bmatrix}$$\n\nb. Show that $Q$ is invertible if and only if $A$ and $B$ are both invertible. In this case, show that\n\n$$Q^{-1} = \\begin{bmatrix} A^{-1} & 0 \\\\ -B^{-1}YA^{-1} & B^{-1} \\end{bmatrix}$$\n\n**Solution.** We do (a.) and leave (b.) for the reader.\n\na. If $A^{-1}$ and $B^{-1}$ both exist, write $R = \\begin{bmatrix} A^{-1} & -A^{-1}XB^{-1} \\\\ 0 & B^{-1} \\end{bmatrix}$. Using block multiplication, one verifies that $PR = I_{m+n} = RP$, so $P$ is invertible, and $P^{-1} = R$. Conversely, suppose that $P$ is invertible, and write $P^{-1} = \\begin{bmatrix} C & V \\\\ W & D \\end{bmatrix}$ in block form, where $C$ is $m \\times m$ and $D$ is $n \\times n$.\n\nThen the equation $PP^{-1} = I_{n+m}$ becomes\n\n$$\\begin{bmatrix} A & X \\\\ 0 & B \\end{bmatrix} \\begin{bmatrix} C & V \\\\ W & D \\end{bmatrix} = \\begin{bmatrix} AC+XW & AV+XD \\\\ BW & BD \\end{bmatrix} = I_{m+n} = \\begin{bmatrix} I_m & 0 \\\\ 0 & I_n \\end{bmatrix}$$\n\nusing block notation. Equating corresponding blocks, we find\n\n$$AC+XW = I_m, \\quad BW = 0, \\quad \\text{and} \\quad BD = I_n$$\n\nHence $B$ is invertible because $BD = I_n$ (by Corollary 2.4.1), then $W = 0$ because $BW = 0$, and finally, $AC = I_m$ (so $A$ is invertible, again by Corollary 2.4.1).\nInverses of Matrix Transformations\n\nLet \\( T = T_A : \\mathbb{R}^n \\to \\mathbb{R}^n \\) denote the matrix transformation induced by the \\( n \\times n \\) matrix \\( A \\). Since \\( A \\) is square, it may very well be invertible, and this leads to the question:\n\nWhat does it mean geometrically for \\( T \\) that \\( A \\) is invertible?\n\nTo answer this, let \\( T' = T_{A^{-1}} : \\mathbb{R}^n \\to \\mathbb{R}^n \\) denote the transformation induced by \\( A^{-1} \\). Then\n\n\\[\nT'[T(x)] = A^{-1}[Ax] = Ix = x\n\\]\n\nfor all \\( x \\) in \\( \\mathbb{R}^n \\) \\hspace{1cm} (2.8)\n\n\\[\nT[T'(x)] = A[A^{-1}x] = Ix = x\n\\]\n\nThe first of these equations asserts that, if \\( T \\) carries \\( x \\) to a vector \\( T(x) \\), then \\( T' \\) carries \\( T(x) \\) right back to \\( x \\); that is \\( T' \\) “reverses” the action of \\( T \\). Similarly \\( T \\) “reverses” the action of \\( T' \\). Conditions (2.8) can be stated compactly in terms of composition:\n\n\\[\nT' \\circ T = 1_{\\mathbb{R}^n} \\quad \\text{and} \\quad T \\circ T' = 1_{\\mathbb{R}^n}\n\\]\n\n(2.9)\n\nWhen these conditions hold, we say that the matrix transformation \\( T' \\) is an inverse of \\( T \\), and we have shown that if the matrix \\( A \\) of \\( T \\) is invertible, then \\( T \\) has an inverse (induced by \\( A^{-1} \\)).\n\nThe converse is also true: If \\( T \\) has an inverse, then its matrix \\( A \\) must be invertible. Indeed, suppose \\( S : \\mathbb{R}^n \\to \\mathbb{R}^n \\) is any inverse of \\( T \\), so that \\( S \\circ T = 1_{\\mathbb{R}^n} \\) and \\( T \\circ S = 1_{\\mathbb{R}^n} \\). It can be shown that \\( S \\) is also a matrix transformation. If \\( B \\) is the matrix of \\( S \\), we have\n\n\\[\nBAx = S[T(x)] = (S \\circ T)(x) = 1_{\\mathbb{R}^n}(x) = x = I_n x \\quad \\text{for all} \\ x \\ \\text{in} \\ \\mathbb{R}^n\n\\]\n\nIt follows by Theorem 2.2.6 that \\( BA = I_n \\), and a similar argument shows that \\( AB = I_n \\). Hence \\( A \\) is invertible with \\( A^{-1} = B \\). Furthermore, the inverse transformation \\( S \\) has matrix \\( A^{-1} \\), so \\( S = T' \\) using the earlier notation. This proves the following important theorem.\n\n**Theorem 2.4.6**\n\nLet \\( T : \\mathbb{R}^n \\to \\mathbb{R}^n \\) denote the matrix transformation induced by an \\( n \\times n \\) matrix \\( A \\). Then\n\n\\[\nA \\text{ is invertible if and only if } T \\text{ has an inverse.}\n\\]\n\nIn this case, \\( T \\) has exactly one inverse (which we denote as \\( T^{-1} \\)), and \\( T^{-1} : \\mathbb{R}^n \\to \\mathbb{R}^n \\) is the transformation induced by the matrix \\( A^{-1} \\). In other words\n\n\\[\n(T_A)^{-1} = T_{A^{-1}}\n\\]\n\nThe geometrical relationship between \\( T \\) and \\( T^{-1} \\) is embodied in equations (2.8) above:\n\n\\[\nT^{-1}[T(x)] = x \\quad \\text{and} \\quad T[T^{-1}(x)] = x \\quad \\text{for all} \\ x \\ \\text{in} \\ \\mathbb{R}^n\n\\]\n\nThese equations are called the fundamental identities relating \\( T \\) and \\( T^{-1} \\). Loosely speaking, they assert that each of \\( T \\) and \\( T^{-1} \\) “reverses” or “undoes” the action of the other.\n\nThis geometric view of the inverse of a linear transformation provides a new way to find the inverse of a matrix \\( A \\). More precisely, if \\( A \\) is an invertible matrix, we proceed as follows:\n1. Let $T$ be the linear transformation induced by $A$.\n\n2. Obtain the linear transformation $T^{-1}$ which “reverses” the action of $T$.\n\n3. Then $A^{-1}$ is the matrix of $T^{-1}$.\n\nHere is an example.\n\n**Example 2.4.12**\n\nFind the inverse of $A = \\begin{bmatrix} 0 & 1 \\\\ 1 & 0 \\end{bmatrix}$ by viewing it as a linear transformation $\\mathbb{R}^2 \\to \\mathbb{R}^2$.\n\n**Solution.** If $x = \\begin{bmatrix} x \\\\ y \\end{bmatrix}$ the vector $Ax = \\begin{bmatrix} 0 & 1 \\\\ 1 & 0 \\end{bmatrix} \\begin{bmatrix} x \\\\ y \\end{bmatrix} = \\begin{bmatrix} y \\\\ x \\end{bmatrix}$ is the result of reflecting $x$ in the line $y = x$ (see the diagram). Hence, if $Q_1 : \\mathbb{R}^2 \\to \\mathbb{R}^2$ denotes reflection in the line $y = x$, then $A$ is the matrix of $Q_1$. Now observe that $Q_1$ reverses itself because reflecting a vector $x$ twice results in $x$. Consequently $Q_1^{-1} = Q_1$.\n\nSince $A^{-1}$ is the matrix of $Q_1^{-1}$ and $A$ is the matrix of $Q$, it follows that $A^{-1} = A$. Of course this conclusion is clear by simply observing directly that $A^2 = I$, but the geometric method can often work where these other methods may be less straightforward.\n\n**Exercises for 2.4**\n\n**Exercise 2.4.1** In each case, show that the matrices are inverses of each other.\n\na. $\\begin{bmatrix} 1 & -1 \\\\ -1 & 3 \\end{bmatrix}$\n\nb. $\\begin{bmatrix} 4 & 1 \\\\ 3 & 2 \\end{bmatrix}$\n\nc. $\\begin{bmatrix} 1 & 0 & -1 \\\\ 3 & 2 & 0 \\\\ -1 & -1 & 0 \\end{bmatrix}$\n\nd. $\\begin{bmatrix} 1 & -1 & 2 \\\\ -5 & 7 & -11 \\\\ -2 & 3 & -5 \\end{bmatrix}$\n\ne. $\\begin{bmatrix} 3 & 5 & 0 \\\\ 3 & 7 & 1 \\\\ 1 & 2 & 1 \\end{bmatrix}$\n\nf. $\\begin{bmatrix} 3 & 1 & -1 \\\\ 2 & 1 & 0 \\\\ 1 & 5 & -1 \\end{bmatrix}$\n\ng. $\\begin{bmatrix} 2 & 4 & 1 \\\\ 3 & 3 & 2 \\\\ 4 & 1 & 4 \\end{bmatrix}$\n\nh. $\\begin{bmatrix} 3 & 1 & -1 \\\\ 5 & 2 & 0 \\\\ 1 & 1 & -1 \\end{bmatrix}$\n\ni. $\\begin{bmatrix} 3 & 1 & 2 \\\\ 1 & -1 & 3 \\\\ 1 & 2 & 4 \\end{bmatrix}$\n\nj. $\\begin{bmatrix} -1 & 4 & 5 & 2 \\\\ 0 & 0 & 0 & -1 \\\\ 1 & -2 & -2 & 0 \\\\ 0 & -1 & -1 & 0 \\end{bmatrix}$\n\n**Exercise 2.4.2** Find the inverse of each of the following matrices.\nExercise 2.4.3 In each case, solve the systems of equations by finding the inverse of the coefficient matrix.\n\na. \\[3x - y = 5\\]\n\\[2x + 2y = 1\\]\nb. \\[2x - 3y = 0\\]\n\\[x - 4y = 1\\]\nc. \\[x + y + 2z = 5\\]\n\\[x + y + z = 0\\]\n\\[x + 2y + 4z = -2\\]\nd. \\[x + 4y + 2z = 1\\]\n\\[2x + 3y + 3z = -1\\]\n\\[4x + y + 4z = 0\\]\n\nExercise 2.4.4 Given \\(A^{-1} = \\begin{bmatrix} 1 & -1 & 3 \\\\ 2 & 0 & 5 \\\\ -1 & 1 & 0 \\end{bmatrix}\\):\n\na. Solve the system of equations \\(Ax = \\begin{bmatrix} 1 \\\\ -1 \\\\ 3 \\end{bmatrix}\\).\n\nb. Find a matrix \\(B\\) such that \\(AB = \\begin{bmatrix} 1 & -1 & 2 \\\\ 0 & 1 & 1 \\\\ 1 & 0 & 0 \\end{bmatrix}\\).\n\nc. Find a matrix \\(C\\) such that \\(CA = \\begin{bmatrix} 1 & 2 & -1 \\\\ 3 & 1 & 1 \\end{bmatrix}\\).\n\nExercise 2.4.5 Find \\(A\\) when\n\na. \\((3A)^{-1} = \\begin{bmatrix} 1 & -1 \\\\ 0 & 1 \\end{bmatrix}\\)\nb. \\((2A)^T = \\begin{bmatrix} 1 & -1 \\\\ 2 & 3 \\end{bmatrix}\\)^{-1}\n\nc. \\((I + 3A)^{-1} = \\begin{bmatrix} 2 & 0 \\\\ 1 & -1 \\end{bmatrix}\\)\n\nd. \\((I - 2A)^{-1} = \\begin{bmatrix} 2 & 1 \\\\ 1 & 1 \\end{bmatrix}\\)\n\ne. \\(A \\begin{bmatrix} 1 & -1 \\\\ 0 & 1 \\end{bmatrix}^{-1} = \\begin{bmatrix} 2 & 3 \\\\ 1 & 1 \\end{bmatrix}\\)\n\nf. \\(\\begin{bmatrix} 1 & 0 \\\\ 2 & 1 \\end{bmatrix}A^{-1} = \\begin{bmatrix} 1 & 0 \\\\ 2 & 2 \\end{bmatrix}\\)\n\ng. \\((A^T - 2I)^{-1} = 2 \\begin{bmatrix} 1 & 1 \\\\ 2 & 3 \\end{bmatrix}\\)\n\nh. \\((A^{-1} - 2I)^T = -2 \\begin{bmatrix} 1 & 1 \\\\ 1 & 0 \\end{bmatrix}\\)\n\nExercise 2.4.6 Find \\(A\\) when:\n\na. \\(A^{-1} = \\begin{bmatrix} 1 & -1 & 3 \\\\ 2 & 1 & 1 \\\\ 0 & 2 & -2 \\end{bmatrix}\\)\nb. \\(A^{-1} = \\begin{bmatrix} 0 & 1 & -1 \\\\ 1 & 2 & 1 \\\\ 0 & 1 & 0 \\end{bmatrix}\\)\n\nExercise 2.4.7 Given \\(\\begin{bmatrix} x_1 \\\\ x_2 \\\\ x_3 \\end{bmatrix} = \\begin{bmatrix} 3 & -1 & 2 \\\\ 1 & 0 & 4 \\\\ 2 & 1 & 0 \\end{bmatrix} \\begin{bmatrix} y_1 \\\\ y_2 \\\\ y_3 \\end{bmatrix}\\), express the variables \\(x_1, x_2,\\) and \\(x_3\\) in terms of \\(z_1, z_2,\\) and \\(z_3\\).\n\nExercise 2.4.8\n\na. In the system \\(3x + 4y = 7\\)\n\\(4x + 5y = 1\\), substitute the new variables \\(x' = -5x' + 4y'\\)\n\\(y' = 4x' - 3y'\\). Then find \\(x\\) and \\(y\\).\n\nb. Explain part (a) by writing the equations as \\(A \\begin{bmatrix} x \\\\ y \\end{bmatrix} = \\begin{bmatrix} 7 \\\\ 1 \\end{bmatrix}\\) and \\(\\begin{bmatrix} x \\\\ y \\end{bmatrix} = B \\begin{bmatrix} x' \\\\ y' \\end{bmatrix}\\). What is the relationship between \\(A\\) and \\(B\\)?\n\nExercise 2.4.9 In each case either prove the assertion or give an example showing that it is false.\n\na. If \\(A \\neq 0\\) is a square matrix, then \\(A\\) is invertible.\n\nb. If \\(A\\) and \\(B\\) are both invertible, then \\(A + B\\) is invertible.\n\nc. If \\(A\\) and \\(B\\) are both invertible, then \\((A^{-1}B)^T\\) is invertible.\n\nd. If \\(A^4 = 3I\\), then \\(A\\) is invertible.\n\ne. If \\(A^2 = A\\) and \\(A \\neq 0\\), then \\(A\\) is invertible.\n\nf. If \\(AB = B\\) for some \\(B \\neq 0\\), then \\(A\\) is invertible.\n\ng. If \\(A\\) is invertible and skew symmetric \\((A^T = -A)\\), the same is true of \\(A^{-1}\\).\n\nh. If \\(A^2\\) is invertible, then \\(A\\) is invertible.\n\ni. If \\(AB = I\\), then \\(A\\) and \\(B\\) commute.\nExercise 2.4.10\n\na. If $A$, $B$, and $C$ are square matrices and $AB = I$, $I = CA$, show that $A$ is invertible and $B = C = A^{-1}$.\n\nb. If $C^{-1} = A$, find the inverse of $C^T$ in terms of $A$.\n\nExercise 2.4.11 Suppose $CA = I_m$, where $C$ is $m \\times n$ and $A$ is $n \\times m$. Consider the system $Ax = b$ of $n$ equations in $m$ variables.\n\na. Show that this system has a unique solution $CB$ if it is consistent.\n\nb. If $C = \\begin{bmatrix} 0 & -5 & 1 \\\\ 3 & 0 & -1 \\end{bmatrix}$ and $A = \\begin{bmatrix} 2 & -3 \\\\ 1 & -2 \\\\ 6 & -10 \\end{bmatrix}$, find $x$ (if it exists) when\n\n(i) $b = \\begin{bmatrix} 1 \\\\ 0 \\end{bmatrix}$; and (ii) $b = \\begin{bmatrix} 7 \\\\ 4 \\\\ 22 \\end{bmatrix}$.\n\nExercise 2.4.12 Verify that $A = \\begin{bmatrix} 1 & -1 \\\\ 0 & 2 \\end{bmatrix}$ satisfies $A^2 - 3A + 2I = 0$, and use this fact to show that $A^{-1} = \\frac{1}{2}(3I - A)$.\n\nExercise 2.4.13 Let $Q = \\begin{bmatrix} a & -b & -c & -d \\\\ b & a & -d & c \\\\ c & d & a & -b \\\\ d & -c & b & a \\end{bmatrix}$. Compute $QQ^T$ and so find $Q^{-1}$ if $Q \\neq 0$.\n\nExercise 2.4.14 Let $U = \\begin{bmatrix} 0 & 1 \\\\ 1 & 0 \\end{bmatrix}$. Show that each of $U$, $-U$, and $-I_2$ is its own inverse and that the product of any two of these is the third.\n\nExercise 2.4.15 Consider $A = \\begin{bmatrix} 1 & 1 \\\\ -1 & 0 \\end{bmatrix}$, $B = \\begin{bmatrix} 0 & -1 \\\\ 1 & 0 \\end{bmatrix}$, $C = \\begin{bmatrix} 0 & 1 & 0 \\\\ 0 & 0 & 1 \\\\ 5 & 0 & 0 \\end{bmatrix}$. Find the inverses by computing (a) $A^6$; (b) $B^4$; and (c) $C^3$.\n\nExercise 2.4.16 Find the inverse of $\\begin{bmatrix} 1 & 0 & 1 \\\\ c & 1 & c \\\\ 3 & c & 2 \\end{bmatrix}$ in terms of $c$.\n\nExercise 2.4.17 If $c \\neq 0$, find the inverse of $\\begin{bmatrix} 1 & -1 & 1 \\\\ 2 & -1 & 2 \\\\ 0 & 2 & c \\end{bmatrix}$ in terms of $c$.\n\nExercise 2.4.18 Show that $A$ has no inverse when:\n\na. $A$ has a row of zeros.\n\nb. $A$ has a column of zeros.\n\nc. each row of $A$ sums to 0.\n\n[Hint: Theorem 2.4.5(2).]\n\nd. each column of $A$ sums to 0.\n\n[Hint: Corollary 2.4.1, Theorem 2.4.4.]\n\nExercise 2.4.19 Let $A$ denote a square matrix.\n\na. Let $YA = 0$ for some matrix $Y \\neq 0$. Show that $A$ has no inverse. [Hint: Corollary 2.4.1, Theorem 2.4.4.]\n\nb. Use part (a) to show that (i) $\\begin{bmatrix} 1 & -1 & 1 \\\\ 0 & 1 & 1 \\\\ 1 & 0 & 2 \\end{bmatrix}$; and\n\n(ii) $\\begin{bmatrix} 2 & 1 & -1 \\\\ 1 & 1 & 0 \\\\ 1 & 0 & -1 \\end{bmatrix}$ have no inverse.\n\n[Hint: For part (ii) compare row 3 with the difference between row 1 and row 2.]\n\nExercise 2.4.20 If $A$ is invertible, show that\n\na. $A^2 \\neq 0$.\n\nb. $A^k \\neq 0$ for all $k = 1, 2, \\ldots$.\n\nExercise 2.4.21 Suppose $AB = 0$, where $A$ and $B$ are square matrices. Show that:\n\na. If one of $A$ and $B$ has an inverse, the other is zero.\n\nb. It is impossible for both $A$ and $B$ to have inverses.\n\nc. $(BA)^2 = 0$.\n\nExercise 2.4.22 Find the inverse of the $x$-expansion in Example 2.2.16 and describe it geometrically.\n\nExercise 2.4.23 Find the inverse of the shear transformation in Example 2.2.17 and describe it geometrically.\nExercise 2.4.24 In each case assume that $A$ is a square matrix that satisfies the given condition. Show that $A$ is invertible and find a formula for $A^{-1}$ in terms of $A$.\n\na. $A^3 - 3A + 2I = 0$.\n\nb. $A^4 + 2A^3 - A - 4I = 0$.\n\nExercise 2.4.25 Let $A$ and $B$ denote $n \\times n$ matrices.\n\na. If $A$ and $AB$ are invertible, show that $B$ is invertible using only (2) and (3) of Theorem 2.4.4.\n\nb. If $AB$ is invertible, show that both $A$ and $B$ are invertible using Theorem 2.4.5.\n\nExercise 2.4.26 In each case find the inverse of the matrix $A$ using Example 2.4.11.\n\na. $A = \\begin{bmatrix} -1 & 1 & 2 \\\\ 0 & 2 & -1 \\\\ 0 & 1 & -1 \\end{bmatrix}$\n\nb. $A = \\begin{bmatrix} 3 & 1 & 0 \\\\ 5 & 2 & 0 \\\\ 1 & 3 & -1 \\end{bmatrix}$\n\nc. $A = \\begin{bmatrix} 3 & 4 & 0 & 0 \\\\ 2 & 3 & 0 & 0 \\\\ 1 & -1 & 1 & 3 \\\\ 3 & 1 & 1 & 4 \\end{bmatrix}$\n\nd. $A = \\begin{bmatrix} 2 & 1 & 5 & 2 \\\\ 1 & 1 & -1 & 0 \\\\ 0 & 0 & 1 & -1 \\\\ 0 & 0 & 1 & -2 \\end{bmatrix}$\n\nExercise 2.4.27 If $A$ and $B$ are invertible symmetric matrices such that $AB = BA$, show that $A^{-1}$, $AB^{-1}$, and $A^{-1}B^{-1}$ are also invertible and symmetric.\n\nExercise 2.4.28 Let $A$ be an $n \\times n$ matrix and let $I$ be the $n \\times n$ identity matrix.\n\na. If $A^2 = 0$, verify that $(I - A)^{-1} = I + A$.\n\nb. If $A^3 = 0$, verify that $(I - A)^{-1} = I + A + A^2$.\n\nc. Find the inverse of $\\begin{bmatrix} 1 & 2 & -1 \\\\ 0 & 1 & 3 \\\\ 0 & 0 & 1 \\end{bmatrix}$.\n\nd. If $A^n = 0$, find the formula for $(I - A)^{-1}$.\n\nExercise 2.4.29 Prove property 6 of Theorem 2.4.4: If $A$ is invertible and $a \\neq 0$, then $aA$ is invertible and $(aA)^{-1} = \\frac{1}{a}A^{-1}$.\n\nExercise 2.4.30 Let $A$, $B$, and $C$ denote $n \\times n$ matrices. Using only Theorem 2.4.4, show that:\n\na. If $A$, $C$, and $ABC$ are all invertible, $B$ is invertible.\n\nb. If $AB$ and $BA$ are both invertible, $A$ and $B$ are both invertible.\n\nExercise 2.4.31 Let $A$ and $B$ denote invertible $n \\times n$ matrices.\n\na. If $A^{-1} = B^{-1}$, does it mean that $A = B$? Explain.\n\nb. Show that $A = B$ if and only if $A^{-1}B = I$.\n\nExercise 2.4.32 Let $A$, $B$, and $C$ be $n \\times n$ matrices, with $A$ and $B$ invertible. Show that:\n\na. If $A$ commutes with $C$, then $A^{-1}$ commutes with $C$.\n\nb. If $A$ commutes with $B$, then $A^{-1}$ commutes with $B^{-1}$.\n\nExercise 2.4.33 Let $A$ and $B$ be square matrices of the same size.\n\na. Show that $(AB)^2 = A^2B^2$ if $AB = BA$.\n\nb. If $A$ and $B$ are invertible and $(AB)^2 = A^2B^2$, show that $AB = BA$.\n\nc. If $A = \\begin{bmatrix} 1 & 0 \\\\ 0 & 0 \\end{bmatrix}$ and $B = \\begin{bmatrix} 1 & 1 \\\\ 0 & 0 \\end{bmatrix}$, show that $(AB)^2 = A^2B^2$ but $AB \\neq BA$.\n\nExercise 2.4.34 Let $A$ and $B$ be $n \\times n$ matrices for which $AB$ is invertible. Show that $A$ and $B$ are both invertible.\n\nExercise 2.4.35 Consider $A = \\begin{bmatrix} 1 & 3 & -1 \\\\ 2 & 1 & 5 \\\\ 1 & -7 & 13 \\end{bmatrix}$, $B = \\begin{bmatrix} 1 & 1 & 2 \\\\ 3 & 0 & -3 \\\\ -2 & 5 & 17 \\end{bmatrix}$.\n\na. Show that $A$ is not invertible by finding a nonzero $1 \\times 3$ matrix $Y$ such that $YA = 0$.\n\n[Hint: Row 3 of $A$ equals $2$(row 2) $- 3$(row 1).]\nb. Show that $B$ is not invertible.\n\n[Hint: Column 3 = 3(column 2) − column 1.]\n\nExercise 2.4.36 Show that a square matrix $A$ is invertible if and only if it can be left-cancelled: $AB = AC$ implies $B = C$.\n\nExercise 2.4.37 If $U^2 = I$, show that $I + U$ is not invertible unless $U = I$.\n\nExercise 2.4.38\n\na. If $J$ is the $4 \\times 4$ matrix with every entry 1, show that $I - \\frac{1}{2}J$ is self-inverse and symmetric.\n\nb. If $X$ is $n \\times m$ and satisfies $X^TX = I_m$, show that $I_n - 2XX^T$ is self-inverse and symmetric.\n\nExercise 2.4.39 An $n \\times n$ matrix $P$ is called an idempotent if $P^2 = P$. Show that:\n\na. $I$ is the only invertible idempotent.\n\nb. $P$ is an idempotent if and only if $I - 2P$ is self-inverse.\n\nc. $U$ is self-inverse if and only if $U = I - 2P$ for some idempotent $P$.\n\nd. $I - aP$ is invertible for any $a \\neq 1$, and that $(I - aP)^{-1} = I + \\left(\\frac{a}{1-a}\\right)P$.\n\nExercise 2.4.40 If $A^2 = kA$, where $k \\neq 0$, show that $A$ is invertible if and only if $A = kI$.\n\nExercise 2.4.41 Let $A$ and $B$ denote $n \\times n$ invertible matrices.\n\na. Show that $A^{-1} + B^{-1} = A^{-1}(A + B)B^{-1}$.\n\nb. If $A + B$ is also invertible, show that $A^{-1} + B^{-1}$ is invertible and find a formula for $(A^{-1} + B^{-1})^{-1}$.\n\nExercise 2.4.42 Let $A$ and $B$ be $n \\times n$ matrices, and let $I$ be the $n \\times n$ identity matrix.\n\na. Verify that $A(I + BA) = (I + AB)A$ and that $(I + BA)B = B(I + AB)$.\n\nb. If $I + AB$ is invertible, verify that $I + BA$ is also invertible and that $(I + BA)^{-1} = I - B(I + AB)^{-1}A$. \n\\[ \\begin{bmatrix} 0 & 0 & 1 & -2 \\\\ -1 & -2 & -1 & -3 \\\\ 1 & 2 & 1 & 2 \\\\ 0 & -1 & 0 & 0 \\end{bmatrix} \\]\n\n\\[ \\begin{bmatrix} 1 & -2 & 6 & -30 & 210 \\\\ 0 & 1 & -3 & 15 & -105 \\\\ 0 & 0 & 1 & -5 & 35 \\\\ 0 & 0 & 0 & 1 & -7 \\\\ 0 & 0 & 0 & 0 & 1 \\end{bmatrix} \\]\n\n2.4.3 \n\\[ \\begin{bmatrix} x \\\\ y \\end{bmatrix} = \\frac{1}{3} \\begin{bmatrix} 4 & -3 \\\\ 1 & -2 \\end{bmatrix} \\begin{bmatrix} 0 \\\\ 1 \\end{bmatrix} = \\frac{1}{3} \\begin{bmatrix} -3 \\\\ -2 \\end{bmatrix} \\]\n\n\\[ \\begin{bmatrix} x \\\\ y \\\\ z \\end{bmatrix} = \\frac{1}{3} \\begin{bmatrix} 9 & -14 & 6 \\\\ 4 & -4 & 1 \\\\ -10 & 15 & -5 \\end{bmatrix} \\begin{bmatrix} 1 \\\\ -1 \\\\ 0 \\end{bmatrix} = \\frac{1}{3} \\begin{bmatrix} 23 \\\\ 8 \\\\ -25 \\end{bmatrix} \\]\n\n2.4.4 \n\\[ B = A^{-1}AB = \\begin{bmatrix} 4 & -2 & 1 \\\\ 7 & -2 & 4 \\\\ -1 & 2 & -1 \\end{bmatrix} \\]\n\n2.4.5 \n\\[ \\begin{bmatrix} 3 & -2 \\\\ 1 & 1 \\end{bmatrix} \\]\n\n\\[ \\begin{bmatrix} 0 & 1 \\\\ 1 & -1 \\end{bmatrix} \\]\n\n\\[ \\begin{bmatrix} 2 & 0 \\\\ -6 & 1 \\end{bmatrix} \\]\n\n\\[ \\begin{bmatrix} 1 & 1 \\\\ 1 & 0 \\end{bmatrix} \\]\n\n2.4.6 \n\\[ A = \\frac{1}{2} \\begin{bmatrix} 2 & -1 & 3 \\\\ 0 & 1 & -1 \\\\ -2 & 1 & -1 \\end{bmatrix} \\]\n\n2.4.8 \n\\[ A \\text{ and } B \\text{ are inverses.} \\]\n\n2.4.9 \n\\[ \\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix} + \\begin{bmatrix} 1 & 0 \\\\ 0 & -1 \\end{bmatrix} \\]\n\n\\[ \\text{d. True. } A^{-1} = \\frac{1}{3}A^3 \\]\n\n\\[ \\text{f. False. } A = B = \\begin{bmatrix} 1 & 0 \\\\ 0 & 0 \\end{bmatrix} \\]\n\n\\[ \\text{h. True. If } (A^2)B = I, \\text{ then } A(AB) = I; \\text{ use Theorem 2.4.5.} \\]\n\n2.4.10 \n\\[ (C^T)^{-1} = (C^{-1})^T = A^T \\text{ because } C^{-1} = (A^{-1})^{-1} = A. \\]\n2.4.11 b. (i) Inconsistent.\n (ii) \\[\n \\begin{bmatrix}\n x_1 \\\\\n x_2\n \\end{bmatrix} = \\begin{bmatrix}\n 2 \\\\\n -1\n \\end{bmatrix}\n \\]\n\n2.4.15 b. \\(B^4 = I\\), so \\(B^{-1} = B^3 = \\begin{bmatrix}\n 0 & 1 \\\\\n -1 & 0\n \\end{bmatrix}\\)\n\n2.4.16 \\[\n\\begin{bmatrix}\n c^2 - 2 & -c & 1 \\\\\n -c & 1 & 0 \\\\\n 3 - c^2 & c & -1\n\\end{bmatrix}\n\\]\n\n2.4.18 b. If column \\(j\\) of \\(A\\) is zero, \\(Ay = 0\\) where \\(y\\) is column \\(j\\) of the identity matrix. Use Theorem 2.4.5.\n d. If each column of \\(A\\) sums to 0, \\(XA = 0\\) where \\(X\\) is the row of 1s. Hence \\(A^TX^T = 0\\) so \\(A\\) has no inverse by Theorem 2.4.5 (\\(X^T \\neq 0\\)).\n\n2.4.19 b. (ii) \\((-1, 1, 1)A = 0\\)\n\n2.4.20 b. Each power \\(A^k\\) is invertible by Theorem 2.4.4 (because \\(A\\) is invertible). Hence \\(A^k\\) cannot be 0.\n\n2.4.21 b. By (a), if one has an inverse the other is zero and so has no inverse.\n\n2.4.22 If \\(A = \\begin{bmatrix}\n a & 0 \\\\\n 0 & 1\n\\end{bmatrix}\\), \\(a > 1\\), then \\(A^{-1} = \\begin{bmatrix}\n \\frac{1}{a} & 0 \\\\\n 0 & 1\n\\end{bmatrix}\\) is an x-compression because \\(\\frac{1}{a} < 1\\).\n\n2.4.24 b. \\(A^{-1} = \\frac{1}{2}(A^3 + 2A^2 - 1)\\)\n\n2.4.25 b. If \\(Bx = 0\\), then \\((AB)x = (A)Bx = 0\\), so \\(x = 0\\) because \\(AB\\) is invertible. Hence \\(B\\) is invertible by Theorem 2.4.5. But then \\(A = (AB)B^{-1}\\) is invertible by Theorem 2.4.4.\n\n2.4.26 b. \\[\n\\begin{bmatrix}\n 2 & -1 & 0 \\\\\n -5 & 3 & 0 \\\\\n -13 & 8 & -1\n\\end{bmatrix}\n\\]\n d. If \\(A^n = 0\\), \\((I - A)^{-1} = I + A + \\cdots + A^{n-1}\\).\n\n2.4.28 b. \\(A[B(AB)^{-1}] = I = [(BA)^{-1}B]A\\), so \\(A\\) is invertible by Exercise 2.4.10.\n\n2.4.30 a. Have \\(AC = CA\\). Left-multiply by \\(A^{-1}\\) to get \\(C = A^{-1}CA\\). Then right-multiply by \\(A^{-1}\\) to get \\(CA^{-1} = A^{-1}C\\).", "id": "./materials/55.pdf" }, { "contents": "3.4 Properties of the Determinant\n\nAS YOU READ . . .\n\n1. Having the choice to compute the determinant of a matrix using cofactor expansion along any row or column is most useful when there are lots of what in a row or column?\n\n2. Which elementary row operation does not change the determinant of a matrix?\n\n3. Why do mathematicians rarely smile?\n\n4. T/F: When computers are used to compute the determinant of a matrix, cofactor expansion is rarely used.\n\nIn the previous section we learned how to compute the determinant. In this section we learn some of the properties of the determinant, and this will allow us to compute determinants more easily. In the next section we will see one application of determinants.\n\nWe start with a theorem that gives us more freedom when computing determinants.\n\nTheorem 14\n\nCofactor Expansion Along Any Row or Column\n\nLet $A$ be an $n \\times n$ matrix. The determinant of $A$ can be computed using cofactor expansion along any row or column of $A$. \n3.4 Properties of the Determinant\n\nWe alluded to this fact way back after Example 70. We had just learned what cofactor expansion was and we practiced along the second row and down the third column. Later, we found the determinant of this matrix by computing the cofactor expansion along the first row. In all three cases, we got the number 0. This wasn’t a coincidence. The above theorem states that all three expansions were actually computing the determinant.\n\nHow does this help us? By giving us freedom to choose any row or column to use for the expansion, we can choose a row or column that looks “most appealing.” This usually means “It has lots of zeros.” We demonstrate this principle below.\n\nExample 74 Find the determinant of\n\n\\[\nA = \\begin{bmatrix}\n1 & 2 & 0 & 9 \\\\\n2 & -3 & 0 & 5 \\\\\n7 & 2 & 3 & 8 \\\\\n-4 & 1 & 0 & 2\n\\end{bmatrix}.\n\\]\n\nSolution Our first reaction may well be “Oh no! Not another 4 × 4 determinant!” However, we can use cofactor expansion along any row or column that we choose. The third column looks great; it has lots of zeros in it. The cofactor expansion along this column is\n\n\\[\n\\det(A) = a_{1,3}C_{1,3} + a_{2,3}C_{2,3} + a_{3,3}C_{3,3} + a_{4,3}C_{4,3}\n\\]\n\n\\[\n= 0 \\cdot C_{1,3} + 0 \\cdot C_{2,3} + 3 \\cdot C_{3,3} + 0 \\cdot C_{4,3}\n\\]\n\nThe wonderful thing here is that three of our cofactors are multiplied by 0. We won’t bother computing them since they will not contribute to the determinant. Thus\n\n\\[\n\\det(A) = 3 \\cdot C_{3,3}\n\\]\n\n\\[\n= 3 \\cdot (-1)^{3+3} \\cdot \\begin{vmatrix}\n1 & 2 & 9 \\\\\n2 & -3 & 5 \\\\\n-4 & 1 & 2\n\\end{vmatrix}\n\\]\n\n\\[\n= 3 \\cdot (-147) \\quad \\text{(we computed the determinant of the 3 × 3 matrix without showing our work; it is −147)}\n\\]\n\n\\[\n= -447\n\\]\n\nWow. That was a lot simpler than computing all that we did in Example 73. Of course, in that example, we didn’t really have any shortcuts that we could have employed.\n\nExample 75 Find the determinant of\n\n\\[\nA = \\begin{bmatrix}\n1 & 2 & 3 & 4 & 5 \\\\\n0 & 6 & 7 & 8 & 9 \\\\\n0 & 0 & 10 & 11 & 12 \\\\\n0 & 0 & 0 & 13 & 14 \\\\\n0 & 0 & 0 & 0 & 15\n\\end{bmatrix}.\n\\]\nChapter 3 Operations on Matrices\n\n**Solution** At first glance, we think “I don’t want to find the determinant of a $5 \\times 5$ matrix!” However, using our newfound knowledge, we see that things are not that bad. In fact, this problem is very easy.\n\nWhat row or column should we choose to find the determinant along? There are two obvious choices: the first column or the last row. Both have 4 zeros in them. We choose the first column.\\(^{19}\\) We omit most of the cofactor expansion, since most of it is just 0:\n\n$$\\det (A) = 1 \\cdot (-1)^{1+1} \\cdot \\begin{vmatrix} 6 & 7 & 8 & 9 \\\\ 0 & 10 & 11 & 12 \\\\ 0 & 0 & 13 & 14 \\\\ 0 & 0 & 0 & 15 \\end{vmatrix}.$$ \n\nSimilarly, this determinant is not bad to compute; we again choose to use cofactor expansion along the first column. Note: technically, this cofactor expansion is $6 \\cdot (-1)^{1+1} A_{11}$; we are going to drop the $(-1)^{1+1}$ terms from here on out in this example (it will show up a lot...).\n\n$$\\det (A) = 1 \\cdot 6 \\cdot \\begin{vmatrix} 10 & 11 & 12 \\\\ 0 & 13 & 14 \\\\ 0 & 0 & 15 \\end{vmatrix}.$$ \n\nYou can probably can see a trend. We’ll finish out the steps without explaining each one.\n\n$$\\det (A) = 1 \\cdot 6 \\cdot 10 \\cdot \\begin{vmatrix} 13 & 14 \\\\ 0 & 15 \\end{vmatrix} = 1 \\cdot 6 \\cdot 10 \\cdot 13 \\cdot 15 = 11700$$\n\nWe see that the final determinant is the product of the diagonal entries. This works for any triangular matrix (and since diagonal matrices are triangular, it works for diagonal matrices as well). This is an important enough idea that we’ll put it into a box.\n\n**Key Idea 12**\n\n**The Determinant of Triangular Matrices**\n\nThe determinant of a triangular matrix is the product of its diagonal elements.\n\n---\n\n\\(^{19}\\)We do not choose this because it is the better choice; both options are good. We simply had to make a choice.\n3.4 Properties of the Determinant\n\nknow?” So now we ask, “If we change a matrix in some way, how is it’s determinant changed?”\n\nThe standard way that we change matrices is through elementary row operations. If we perform an elementary row operation on a matrix, how will the determinant of the new matrix compare to the determinant of the original matrix?\n\nLet’s experiment first and then we’ll officially state what happens.\n\nExample 76 Let\n\n\\[\nA = \\begin{bmatrix} 1 & 2 \\\\ 3 & 4 \\end{bmatrix}.\n\\]\n\nLet \\( B \\) be formed from \\( A \\) by doing one of the following elementary row operations:\n\n1. \\( 2R_1 + R_2 \\rightarrow R_2 \\)\n2. \\( 5R_1 \\rightarrow R_1 \\)\n3. \\( R_1 \\leftrightarrow R_2 \\)\n\nFind \\( \\det(A) \\) as well as \\( \\det(B) \\) for each of the row operations above.\n\n**Solution** It is straightforward to compute \\( \\det(A) = -2 \\).\n\nLet \\( B \\) be formed by performing the row operation in 1) on \\( A \\); thus\n\n\\[\nB = \\begin{bmatrix} 1 & 2 \\\\ 5 & 8 \\end{bmatrix}.\n\\]\n\nIt is clear that \\( \\det(B) = -2 \\), the same as \\( \\det(A) \\).\n\nNow let \\( B \\) be formed by performing the elementary row operation in 2) on \\( A \\); that is,\n\n\\[\nB = \\begin{bmatrix} 5 & 10 \\\\ 3 & 4 \\end{bmatrix}.\n\\]\n\nWe can see that \\( \\det(B) = -10 \\), which is \\( 5 \\cdot \\det(A) \\).\n\nFinally, let \\( B \\) be formed by the third row operation given; swap the two rows of \\( A \\). We see that\n\n\\[\nB = \\begin{bmatrix} 3 & 4 \\\\ 1 & 2 \\end{bmatrix}\n\\]\n\nand that \\( \\det(B) = 2 \\), which is \\( (-1) \\cdot \\det(A) \\).\n\nWe’ve seen in the above example that there seems to be a relationship between the determinants of matrices “before and after” being changed by elementary row operations. Certainly, one example isn’t enough to base a theory on, and we have not proved anything yet. Regardless, the following theorem is true.\nChapter 3 Operations on Matrices\n\nTheorem 15 The Determinant and Elementary Row Operations\n\nLet $A$ be an $n \\times n$ matrix and let $B$ be formed by performing one elementary row operation on $A$.\n\n1. If $B$ is formed from $A$ by adding a scalar multiple of one row to another, then $\\det(B) = \\det(A)$.\n\n2. If $B$ is formed from $A$ by multiplying one row of $A$ by a scalar $k$, then $\\det(B) = k \\cdot \\det(A)$.\n\n3. If $B$ is formed from $A$ by interchanging two rows of $A$, then $\\det(B) = -\\det(A)$.\n\nLet’s put this theorem to use in an example.\n\nExample 77 Let\n\n$$A = \\begin{bmatrix} 1 & 2 & 1 \\\\ 0 & 1 & 1 \\\\ 1 & 1 & 1 \\end{bmatrix}.$$ \n\nCompute $\\det(A)$, then find the determinants of the following matrices by inspection using Theorem 15.\n\n$$B = \\begin{bmatrix} 1 & 1 & 1 \\\\ 1 & 2 & 1 \\\\ 0 & 1 & 1 \\end{bmatrix} \\quad C = \\begin{bmatrix} 1 & 2 & 1 \\\\ 0 & 1 & 1 \\\\ 7 & 7 & 7 \\end{bmatrix} \\quad D = \\begin{bmatrix} 1 & -1 & -2 \\\\ 0 & 1 & 1 \\\\ 1 & 1 & 1 \\end{bmatrix}$$\n\n**SOLUTION** Computing $\\det(A)$ by cofactor expansion down the first column or along the second row seems like the best choice, utilizing the one zero in the matrix. We can quickly confirm that $\\det(A) = 1$.\n\nTo compute $\\det(B)$, notice that the rows of $A$ were rearranged to form $B$. There are different ways to describe what happened; saying $R_1 \\leftrightarrow R_2$ was followed by $R_1 \\leftrightarrow R_3$ produces $B$ from $A$. Since there were two row swaps, $\\det(B) = (-1)(-1)\\det(A) = \\det(A) = 1$.\n\nNotice that $C$ is formed from $A$ by multiplying the third row by 7. Thus $\\det(C) = 7 \\cdot \\det(A) = 7$.\n\nIt takes a little thought, but we can form $D$ from $A$ by the operation $-3R_2 + R_1 \\rightarrow R_1$. This type of elementary row operation does not change determinants, so $\\det(D) = \\det(A)$. \n\n150\nLet’s continue to think like mathematicians; mathematicians tend to remember “problems” they’ve encountered in the past, and when they learn something new, in the backs of their minds they try to apply their new knowledge to solve their old problem.\n\nWhat “problem” did we recently uncover? We stated in the last chapter that even computers could not compute the determinant of large matrices with cofactor expansion. How then can we compute the determinant of large matrices?\n\nWe just learned two interesting and useful facts about matrix determinants. First, the determinant of a triangular matrix is easy to compute: just multiply the diagonal elements. Secondly, we know how elementary row operations affect the determinant. Put these two ideas together: given any square matrix, we can use elementary row operations to put the matrix in triangular form, find the determinant of the new matrix (which is easy), and then adjust that number by recalling what elementary operations we performed. Let’s practice this.\n\n**Example 78** Find the determinant of $A$ by first putting $A$ into a triangular form, where\n\n$$A = \\begin{bmatrix} 2 & 4 & -2 \\\\ -1 & -2 & 5 \\\\ 3 & 2 & 1 \\end{bmatrix}.$$ \n\n**Solution** In putting $A$ into a triangular form, we need not worry about getting leading 1s, but it does tend to make our life easier as we work out a problem by hand. So let’s scale the first row by $1/2$:\n\n$$\\frac{1}{2}R_1 \\rightarrow R_1 \\quad \\begin{bmatrix} 1 & 2 & -1 \\\\ -1 & -2 & 5 \\\\ 3 & 2 & 1 \\end{bmatrix}.$$ \n\nNow let’s get 0s below this leading 1:\n\n$$R_1 + R_2 \\rightarrow R_2 \\quad \\begin{bmatrix} 1 & 2 & -1 \\\\ 0 & 0 & 4 \\\\ 3 & 2 & 1 \\end{bmatrix},$$\n\n$$-3R_1 + R_3 \\rightarrow R_3 \\quad \\begin{bmatrix} 1 & 2 & -1 \\\\ 0 & 0 & 4 \\\\ 0 & -4 & 4 \\end{bmatrix}.$$ \n\nWe can finish in one step; by interchanging rows 2 and 3 we’ll have our matrix in triangular form.\n\n$$R_2 \\leftrightarrow R_3 \\quad \\begin{bmatrix} 1 & 2 & -1 \\\\ 0 & -4 & 4 \\\\ 0 & 0 & 4 \\end{bmatrix}.$$ \n\nLet’s name this last matrix $B$. The determinant of $B$ is easy to compute as it is triangular; $\\det(B) = -16$. We can use this to find $\\det(A)$.\n\nRecall the steps we used to transform $A$ into $B$. They are:\n\n---\n\n20 which is why mathematicians rarely smile: they are remembering their problems\n\n21 or echelon form\nChapter 3 Operations on Matrices\n\n\\[\n\\begin{align*}\n\\frac{1}{2}R_1 & \\rightarrow R_1 \\\\\nR_1 + R_2 & \\rightarrow R_2 \\\\\n-3R_1 + R_3 & \\rightarrow R_3 \\\\\nR_2 & \\leftrightarrow R_3\n\\end{align*}\n\\]\n\nThe first operation multiplied a row of \\( A \\) by \\( \\frac{1}{2} \\). This means that the resulting matrix had a determinant that was \\( \\frac{1}{2} \\) the determinant of \\( A \\).\n\nThe next two operations did not affect the determinant at all. The last operation, the row swap, changed the sign. Combining these effects, we know that\n\n\\[-16 = \\det(B) = (-1) \\frac{1}{2} \\det(A).\\]\n\nSolving for \\( \\det(A) \\) we have that \\( \\det(A) = 32 \\).\n\nIn practice, we don’t need to keep track of operations where we add multiples of one row to another; they simply do not affect the determinant. Also, in practice, these steps are carried out by a computer, and computers don’t care about leading 1s. Therefore, row scaling operations are rarely used. The only things to keep track of are row swaps, and even then all we care about are the number of row swaps. An odd number of row swaps means that the original determinant has the opposite sign of the triangular form matrix; an even number of row swaps means they have the same determinant.\n\nLet’s practice this again.\n\n**Example 79** The matrix \\( B \\) was formed from \\( A \\) using the following elementary row operations, though not necessarily in this order. Find \\( \\det(A) \\).\n\n\\[\nB = \\begin{bmatrix}\n1 & 2 & 3 \\\\\n0 & 4 & 5 \\\\\n0 & 0 & 6\n\\end{bmatrix}\n\\]\n\n\\[\n\\begin{align*}\n2R_1 & \\rightarrow R_1 \\\\\n\\frac{1}{3}R_3 & \\rightarrow R_3 \\\\\nR_1 & \\leftrightarrow R_2 \\\\\n6R_1 + R_2 & \\rightarrow R_2\n\\end{align*}\n\\]\n\n**Solution** It is easy to compute \\( \\det(B) = 24 \\). In looking at our list of elementary row operations, we see that only the first three have an effect on the determinant. Therefore\n\n\\[\n24 = \\det(B) = 2 \\cdot \\frac{1}{3} \\cdot (-1) \\cdot \\det(A)\n\\]\n\nand hence\n\n\\[\n\\det(A) = -36.\n\\]\n\nIn the previous example, we may have been tempted to “rebuild” \\( A \\) using the elementary row operations and then computing the determinant. This can be done, but in general it is a bad idea; it takes too much work and it is too easy to make a mistake.\n\nLet’s think some more like a mathematician. How does the determinant work with other matrix operations that we know? Specifically, how does the determinant interact with matrix addition, scalar multiplication, matrix multiplication, the transpose\nand the trace? We’ll again do an example to get an idea of what is going on, then give a theorem to state what is true.\n\n**Example 80** \nLet\n\n\\[\nA = \\begin{bmatrix} 1 & 2 \\\\ 3 & 4 \\end{bmatrix} \\quad \\text{and} \\quad B = \\begin{bmatrix} 2 & 1 \\\\ 3 & 5 \\end{bmatrix}.\n\\]\n\nFind the determinants of the matrices \\(A, B, A + B, 3A, AB, A^T, A^{-1}\\), and compare the determinant of these matrices to their trace.\n\n**Solution** \nWe can quickly compute that \\(\\det(A) = -2\\) and that \\(\\det(B) = 7\\).\n\n\\[\n\\det(A - B) = \\det \\left( \\begin{bmatrix} 1 & 2 \\\\ 3 & 4 \\end{bmatrix} - \\begin{bmatrix} 2 & 1 \\\\ 3 & 5 \\end{bmatrix} \\right) = \\begin{vmatrix} -1 & 1 \\\\ 0 & -1 \\end{vmatrix} = 1\n\\]\n\nIt’s tough to find a connection between \\(\\det(A - B)\\), \\(\\det(A)\\) and \\(\\det(B)\\).\n\n\\[\n\\det(3A) = \\begin{vmatrix} 3 & 6 \\\\ 9 & 12 \\end{vmatrix} = -18\n\\]\n\nWe can figure this one out; multiplying one row of \\(A\\) by 3 increases the determinant by a factor of 3; doing it again (and hence multiplying both rows by 3) increases the determinant again by a factor of 3. Therefore \\(\\det(3A) = 3 \\cdot 3 \\cdot \\det(A)\\), or \\(3^2 \\cdot A\\).\n\n\\[\n\\det(AB) = \\det \\left( \\begin{bmatrix} 1 & 2 \\\\ 3 & 4 \\end{bmatrix} \\begin{bmatrix} 2 & 1 \\\\ 3 & 5 \\end{bmatrix} \\right) = \\begin{vmatrix} 8 & 11 \\\\ 18 & 23 \\end{vmatrix} = -14\n\\]\n\nThis one seems clear; \\(\\det(AB) = \\det(A) \\det(B)\\).\n\n\\[\n\\det(A^T) = \\begin{vmatrix} 1 & 3 \\\\ 2 & 4 \\end{vmatrix} = -2\n\\]\nChapter 3 Operations on Matrices\n\nObviously \\( \\det (A^T) = \\det (A) \\); is this always going to be the case? If we think about it, we can see that the cofactor expansion along the first row of \\( A \\) will give us the same result as the cofactor expansion along the first column of \\( A^T \\).\\(^{22}\\)\n\n\\[\n\\det (A^{-1}) = \\begin{vmatrix} -2 & 1 \\\\ 3/2 & -1/2 \\end{vmatrix} = 1 - 3/2 = -1/2\n\\]\n\nIt seems as though\n\n\\[\n\\det (A^{-1}) = \\frac{1}{\\det (A)}.\n\\]\n\nWe end by remarking that there seems to be no connection whatsoever between the trace of a matrix and its determinant. We leave it to the reader to compute the trace for some of the above matrices and confirm this statement.\n\nWe now state a theorem which will confirm our conjectures from the previous example.\n\n**Theorem 16**\n\n**Determinant Properties**\n\nLet \\( A \\) and \\( B \\) be \\( n \\times n \\) matrices and let \\( k \\) be a scalar. The following are true:\n\n1. \\( \\det (kA) = k^n \\cdot \\det (A) \\)\n2. \\( \\det (A^T) = \\det (A) \\)\n3. \\( \\det (AB) = \\det (A) \\det (B) \\)\n4. If \\( A \\) is invertible, then\n \\[\n \\det (A^{-1}) = \\frac{1}{\\det (A)}.\n \\]\n5. A matrix \\( A \\) is invertible if and only if \\( \\det (A) \\neq 0 \\).\n\nThis last statement of the above theorem is significant: what happens if \\( \\det (A) = 0 \\)? It seems that \\( \\det (A^{-1}) = \\frac{1}{0} \\), which is undefined. There actually isn’t a problem here; it turns out that if \\( \\det (A) = 0 \\), then \\( A \\) is not invertible (hence part 5 of Theorem 16). This allows us to add on to our Invertible Matrix Theorem.\n\n\\(^{22}\\)This can be a bit tricky to think out in your head. Try it with a \\( 3 \\times 3 \\) matrix \\( A \\) and see how it works. All the \\( 2 \\times 2 \\) submatrices that are created in \\( A^T \\) are the transpose of those found in \\( A \\); this doesn’t matter since it is easy to see that the determinant isn’t affected by the transpose in a \\( 2 \\times 2 \\) matrix.\n3.4 Properties of the Determinant\n\nTheorem 17\n\n**Invertible Matrix Theorem**\n\nLet $A$ be an $n \\times n$ matrix. The following statements are equivalent.\n\n(a) $A$ is invertible.\n\n(g) $\\det(A) \\neq 0$.\n\nThis new addition to the Invertible Matrix Theorem is very useful; we’ll refer back to it in Chapter 4 when we discuss eigenvalues.\n\nWe end this section with a shortcut for computing the determinants of $3 \\times 3$ matrices. Consider the matrix $A$:\n\n$$\n\\begin{bmatrix}\n1 & 2 & 3 \\\\\n4 & 5 & 6 \\\\\n7 & 8 & 9\n\\end{bmatrix}.\n$$\n\nWe can compute its determinant using cofactor expansion as we did in Example 71. Once one becomes proficient at this method, computing the determinant of a $3 \\times 3$ isn’t all that hard. A method many find easier, though, starts with rewriting the matrix without the brackets, and repeating the first and second columns at the end as shown below.\n\n$$\n\\begin{array}{cccc}\n1 & 2 & 3 & 1 & 2 \\\\\n4 & 5 & 6 & 4 & 5 \\\\\n7 & 8 & 9 & 7 & 8\n\\end{array}\n$$\n\nIn this $3 \\times 5$ array of numbers, there are 3 full “upper left to lower right” diagonals, and 3 full “upper right to lower left” diagonals, as shown below with the arrows.\n\n$$\n\\begin{array}{cccc}\n1 & 2 & 3 & 1 & 2 \\\\\n4 & 5 & 6 & 4 & 5 \\\\\n7 & 8 & 9 & 7 & 8\n\\end{array}\n$$\n\nThe numbers that appear at the ends of each of the arrows are computed by multiplying the numbers found along the arrows. For instance, the 105 comes from multiplying $3 \\cdot 5 \\cdot 7 = 105$. The determinant is found by adding the numbers on the right, and subtracting the sum of the numbers on the left. That is,\n\n$$\n\\det(A) = (45 + 84 + 96) - (105 + 48 + 72) = 0.\n$$\n\nTo help remind ourselves of this shortcut, we’ll make it into a Key Idea.\nChapter 3 Operations on Matrices\n\nKey Idea 13\n\n3 × 3 Determinant Shortcut\n\nLet A be a 3 × 3 matrix. Create a 3 × 5 array by repeating the first 2 columns and consider the products of the 3 “right hand” diagonals and 3 “left hand” diagonals as shown previously. Then\n\n\\[\n\\det(A) = \\text{“(the sum of the right hand numbers) - (the sum of the left hand numbers)”}\n\\]\n\nWe’ll practice once more in the context of an example.\n\nExample 81 Find the determinant of A using the previously described shortcut, where\n\n\\[\nA = \\begin{bmatrix}\n1 & 3 & 9 \\\\\n-2 & 3 & 4 \\\\\n-5 & 7 & 2\n\\end{bmatrix}\n\\]\n\nSolution Rewriting the first 2 columns, drawing the proper diagonals, and multiplying, we get:\n\n\\[\n\\begin{array}{ccc}\n1 & 3 & 9 \\\\\n-2 & 3 & 4 \\\\\n-5 & 7 & 2\n\\end{array}\n\\]\n\nSumming the numbers on the right and subtracting the sum of the numbers on the left, we get\n\n\\[\n\\det(A) = (6 - 60 - 126) - (-135 + 28 - 12) = -61.\n\\]\n\nIn the next section we’ll see how the determinant can be used to solve systems of linear equations.\n\nExercises 3.4\n\nIn Exercises 1 – 14, find the determinant of the given matrix using cofactor expansion along any row or column you choose.\n\n1. \\[\n\\begin{bmatrix}\n1 & 2 & 3 \\\\\n-5 & 0 & 3 \\\\\n4 & 0 & 6\n\\end{bmatrix}\n\\]\n\n2. \\[\n\\begin{bmatrix}\n-4 & 4 & -4 \\\\\n0 & 0 & -3 \\\\\n-2 & -2 & -1\n\\end{bmatrix}\n\\]\n\n3. \\[\n\\begin{bmatrix}\n-4 & 1 & 1 \\\\\n0 & 0 & 0 \\\\\n-1 & -2 & -5\n\\end{bmatrix}\n\\]\n\n156\n3.4 Properties of the Determinant\n\n4. \\[\n\\begin{bmatrix}\n0 & -3 & 1 \\\\\n0 & 0 & 5 \\\\\n-4 & 1 & 0\n\\end{bmatrix}\n\\]\n\nperforming operations on \\( M \\). Determine the determinants of \\( A, B \\) and \\( C \\) using Theorems 15 and 16, and indicate the operations used to form \\( A, B \\) and \\( C \\).\n\n15. \\( M = \\begin{bmatrix} 0 & 3 & 5 \\\\ 3 & 1 & 0 \\\\ -2 & -4 & -1 \\end{bmatrix} \\)\n\\[ \\det(M) = -41. \\]\n\n(a) \\( A = \\begin{bmatrix} 0 & 3 & 5 \\\\ -2 & -4 & -1 \\\\ 3 & 1 & 0 \\end{bmatrix} \\)\n\n(b) \\( B = \\begin{bmatrix} 0 & 3 & 5 \\\\ 3 & 1 & 0 \\\\ 8 & 16 & 4 \\end{bmatrix} \\)\n\n(c) \\( C = \\begin{bmatrix} 3 & 4 & 5 \\\\ 3 & 1 & 0 \\\\ -2 & -4 & -1 \\end{bmatrix} \\)\n\n16. \\( M = \\begin{bmatrix} 9 & 7 & 8 \\\\ 1 & 3 & 7 \\\\ 6 & 3 & 3 \\end{bmatrix} \\)\n\\[ \\det(M) = 45. \\]\n\n10. \\( A = \\begin{bmatrix} 18 & 14 & 16 \\\\ 1 & 3 & 7 \\\\ 6 & 3 & 3 \\end{bmatrix} \\)\n\n11. \\( B = \\begin{bmatrix} 9 & 7 & 8 \\\\ 1 & 3 & 7 \\\\ 96 & 73 & 83 \\end{bmatrix} \\)\n\n12. \\( C = \\begin{bmatrix} 9 & 1 & 6 \\\\ 7 & 3 & 3 \\\\ 8 & 7 & 3 \\end{bmatrix} \\)\n\n17. \\( M = \\begin{bmatrix} 5 & 1 & 5 \\\\ 4 & 0 & 2 \\\\ 0 & 0 & 4 \\end{bmatrix} \\)\n\\[ \\det(M) = -16. \\]\n\n(a) \\( A = \\begin{bmatrix} 0 & 0 & 4 \\\\ 5 & 1 & 5 \\\\ 4 & 0 & 2 \\end{bmatrix} \\)\n\n(b) \\( B = \\begin{bmatrix} -5 & -1 & -5 \\\\ -4 & 0 & -2 \\\\ 0 & 0 & 4 \\end{bmatrix} \\)\n\n(c) \\( C = \\begin{bmatrix} 15 & 3 & 15 \\\\ 12 & 0 & 6 \\\\ 0 & 0 & 12 \\end{bmatrix} \\)\n\nIn Exercises 15 – 18, a matrix \\( M \\) and \\( \\det(M) \\) are given. Matrices \\( A, B \\) and \\( C \\) are formed by\nChapter 3 Operations on Matrices\n\n18. \\( M = \\begin{bmatrix} 5 & 4 & 0 \\\\ 7 & 9 & 3 \\\\ 1 & 3 & 9 \\end{bmatrix} \\), \\( \\det(M) = 120. \\)\n\n\\( B = \\begin{bmatrix} 0 & 0 \\\\ 4 & -4 \\end{bmatrix} \\)\n\nIn Exercises 23 – 30, find the determinant of the given matrix using Key Idea 13.\n\n23. \\( \\begin{bmatrix} 3 & 2 & 3 \\\\ -6 & 1 & -10 \\\\ -8 & -9 & -9 \\end{bmatrix} \\)\n\n24. \\( \\begin{bmatrix} 8 & -9 & -2 \\\\ -9 & 9 & -7 \\\\ 5 & -1 & 9 \\end{bmatrix} \\)\n\n25. \\( \\begin{bmatrix} -4 & 3 & -4 \\\\ -4 & -5 & 3 \\\\ 3 & -4 & 5 \\end{bmatrix} \\)\n\n26. \\( \\begin{bmatrix} 1 & -2 & 1 \\\\ 5 & 5 & 4 \\\\ 4 & 0 & 0 \\end{bmatrix} \\)\n\n27. \\( \\begin{bmatrix} 1 & -4 & 1 \\\\ 0 & 3 & 0 \\\\ 1 & 2 & 2 \\end{bmatrix} \\)\n\n28. \\( \\begin{bmatrix} 3 & -1 & 0 \\\\ -3 & 0 & -4 \\\\ 0 & -1 & -4 \\end{bmatrix} \\)\n\n29. \\( \\begin{bmatrix} -5 & 0 & -4 \\\\ 2 & 4 & -1 \\\\ -5 & 0 & -4 \\end{bmatrix} \\)\n\n30. \\( \\begin{bmatrix} 1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ -1 & 1 & 1 \\end{bmatrix} \\)\n\nIn Exercises 19 – 22, matrices \\( A \\) and \\( B \\) are given. Verify part 3 of Theorem 16 by computing \\( \\det(A) \\), \\( \\det(B) \\) and \\( \\det(AB) \\).\n\n19. \\( A = \\begin{bmatrix} 2 & 0 \\\\ 1 & 2 \\end{bmatrix} \\)\n\n\\( B = \\begin{bmatrix} 0 & -4 \\\\ 1 & 3 \\end{bmatrix} \\)\n\n20. \\( A = \\begin{bmatrix} 3 & -1 \\\\ 4 & 1 \\end{bmatrix} \\)\n\n\\( B = \\begin{bmatrix} -4 & -1 \\\\ -5 & 3 \\end{bmatrix} \\)\n\n21. \\( A = \\begin{bmatrix} -4 & 4 \\\\ 5 & -2 \\end{bmatrix} \\)\n\n\\( B = \\begin{bmatrix} -3 & -4 \\\\ 5 & -3 \\end{bmatrix} \\)\n\n22. \\( A = \\begin{bmatrix} -3 & -1 \\\\ 2 & -3 \\end{bmatrix} \\)", "id": "./materials/56.pdf" }, { "contents": "Angle between two lines\n\n**Definition:** The angle between two lines is defined as the smallest angle between their directions.\n\nIn the figure to the side we can see that:\n\n- The angle of the straight lines \\( s \\) and \\( t \\) belonging to the \\( ABC \\) plane measures \\( 30^\\circ \\).\n- The angle of the reverse lines \\( r \\) and \\( s \\) is of \\( 90^\\circ \\) (equal to the angle between lines \\( BC \\) and \\( s \\) in the same plane).\n\n**Definition:** The angle between two reverse lines (which do not intersect and are not parallel to each other) is the acute angle that one forms with a line parallel to the other.\n\n**Example:** Let us consider the lines\n\n\\[\n\\begin{align*}\n r : (x, y, z) &= (1, 2, 0) + k(2, 1, 3), \\quad k \\in \\mathbb{R} \\\\\n s : (x, y, z) &= (0, -1, -1) + t(3, 2, 1), \\quad t \\in \\mathbb{R}\n\\end{align*}\n\\]\n\nof \\( \\mathbb{R}^3 \\), whose directions are those of the non-collinear vectors \\( u = (2, 1, 3) \\) and \\( v = (3, 2, 1) \\), respectively. We can see that \\( r \\) and \\( s \\) do not intersect. In fact,\n\n\\[\n(1, 2, 0) + k(2, 1, 3) = (0, -1, -1) + t(3, 2, 1) \\iff \\begin{cases} \n 2k - 3t = -1 \\\\\n k - 2t = -3 \\\\\n 3k - t = -1\n\\end{cases} \\iff \\begin{cases} \n k = \\frac{-2}{7} \\\\\n k = \\frac{1}{5} \\\\\n t = 3k + 1\n\\end{cases}\n\\]\n\nSo \\( r \\) and \\( s \\) are reverse lines.\n\nBesides that, \\( \\cos(\\hat{r}s) = \\frac{|\\cos(\\hat{u}v)|}{|u||v|} = \\frac{6 + 2 + 3}{\\sqrt{14}\\sqrt{14}} = \\frac{11}{14} \\), that is, \\( \\hat{r}s = 23, 6^\\circ \\).", "id": "./materials/233.pdf" }, { "contents": "Distance from a point to a plane, from a line to a parallel plane or between two parallel planes\n\nDistance from a point to a plane:\n\nWe can determine the distance from point $P$ to plane $p$, by performing:\n\n- Calculate the line $r$ that contains the point $P$ and is normal to the plane $p$;\n- Calculate $P' = r \\cap p$;\n- Determine the distance from $P$ to $P'$.\n\n**Example:** To calculate the distance from $P = (1, 2, -1)$ to the plane $p : x - y + z = 0$, we can take $n = (2, -2, 1) \\perp p$ and the line $r$ that contains $P$ and is normal to the plane $p$ is $r : \\frac{x - 1}{2} = \\frac{y - 2}{-2} = z + 1$.\n\n$$P' = r \\cap p = \\begin{cases} x - 1 = -y + 2 \\\\ -y + 2 = z + 1 \\\\ x - y + z = 1 \\end{cases} \\iff \\begin{cases} x = 2 \\\\ y = 0 \\\\ z = 1 \\end{cases}$$\n\nThen, $d(P, p) = d(P, P') = \\sqrt{(1 - 2)^2 + (2 - 0)^2 + (-1 - 1)^2} = 3$\n\nDistance from a straight line to a parallel plane:\n\nGiven a line $r$ parallel to a plane $p$, the distance $d$ from the line $r$ is the distance from any point $p$ on the line to the plane, that is,\n\n$$d(r, p) = d(P, p)$$\n\n**Example:** To calculate the distance from $r : \\frac{x - 1}{2} = -y = \\frac{z + 1}{-3}$ to the plane $p : x - y + z = 0$ is the distance from $P = (1, 0, -1) \\in r$ to the plane $p$. \nDistance between two parallel planes:\n\nTo calculate the distance between two planes $\\alpha$ and $\\beta$ parallel to each other, we can perform:\n\n- Calculate the line $r$ that is normal to the planes $\\alpha$ and $\\beta$;\n- Calculate $A = r \\cap \\alpha$;\n- Calculate $B = r \\cap \\beta$;\n- Determine the distance from $A$ to $B$.\n\n**Example:** To calculate the distance from $p_1 : x - 2y + z + 6 = 0$ to the plane $p_2 : 2x - 4y + 2z + 6 = 0$, we consider:\n\n- $r : x = -\\frac{y}{2} = z$ that is normal $p_1$ and $p_2$;\n- $A = r \\cap p_1 : \\begin{cases} x = -\\frac{y}{2} \\\\ x = z \\end{cases} \\iff \\begin{cases} y = -2x \\\\ z = x \\end{cases} \\iff \\begin{cases} y = \\frac{1}{2} \\\\ z = -1 \\end{cases}$;\n- $B = r \\cap p_2 : \\begin{cases} x = -\\frac{y}{2} \\\\ x = z \\end{cases} \\iff \\begin{cases} y = -2x \\\\ z = x \\end{cases} \\iff \\begin{cases} y = \\frac{1}{2} \\\\ z = -\\frac{1}{2} \\end{cases}$.\n\nThen, $A = (-1, \\frac{1}{2}, -1), B = (-\\frac{1}{2}, 1, -\\frac{1}{2})$ and\n\n$$d(p_1, p_2) = AB = \\sqrt{(-\\frac{1}{2} + 1)^2 + (1 - \\frac{1}{2})^2 + (-\\frac{1}{2} + 1)^2} = \\frac{\\sqrt{3}}{2}.$$", "id": "./materials/234.pdf" }, { "contents": "Definitions and basic properties\n\nA graph $G$ consists of two finite sets $V(G)$ and $E(G)$, where $V(G)$ is nonempty and each element of $E(G)$ is an unordered pair of distinct elements of $V(G)$. The elements of $V(G)$ are called vertices and the elements of $E$ are called edges. Thus, if $e$ is an edge, then $e$ is a set of the form $\\{v, w\\}$, where $v$ and $w$ are different elements of $V$ called the end vertices or endpoints of edge $e$. We often omit the braces and refer to the edge $vw$, and this is of course the same as the edge $wv$.\n\nExample 1. Consider the sets\n\n$$V_1 = \\{1, 2, 3, 4, 5, 6\\} \\text{ and } E_1 = \\{\\{2, 3\\}, \\{3, 5\\}, \\{3, 6\\}, \\{2, 6\\}, \\{4, 6\\}, \\{2, 5\\}\\}.$$ \n\n$G_1 = (V, E)$ is a (simple) graph with vertices $V$ and edges $E$ and could be pictured as shown below.\n\n![Graph G1](image)\n\nWe can also write the vertex set $E_1 = \\{23, 35, 36, 26, 46, 25\\}$.\n\nA directed graph, or digraph, consists of two finite sets: a nonempty set $V(G)$ of vertices and a set $D(G)$ of directed edges, where each is associated with an ordered pair of vertices, its endpoints. If edge $e$ is associated with the pair $(v, w)$ of vertices, then $e$ is said to be the (directed) edge or arc from $v$ to $w$.\n\nExample 2. Consider the set of the vertices $V_2 = \\{u, v, w\\}$ and the set of the edges $D_2 = \\{(u, v); (v, w); (u, w)\\}$. $G_2 = (V_2, D_2)$ is a directed graph.\n\n![Graph G2](image)\nA graph that have two or more edges connecting the same pair of vertices, multiple edges or parallel edges, is called multigraph. An edge with just one endpoint is called a loop.\n\n**Example 3.** The following graphs are not simple, because three is two different edges connecting the same pair of vertices. They are multigraph.\n\n![Graphs G3 and G4](image)\n\nA simple graph is a graph that does not have any loops or multiple (parallel) edges. The order of a graph is the number of vertices in the graph and the size of a graph is the number of edges in the graph.\n\nAn edge $vw$ is said to be incident on each of its endpoints, $v$ and $w$, and two edges incident on the same endpoint are called adjacent.\n\nTwo vertices are called adjacent if they are the endpoints of same edge. A vertex on which no edges are incident is called isolated vertex.\n\n**Example 4.** In the graph $G_1$ in the Example ?? the edge 23 is adjacent to the edge 35 and the vertex 4 is adjacent to vertex 6.\n\nThe degree of $v$, denoted $\\deg(v)$, equals the number of edges that are incident on $v$, with an edge that is a loop counted twice. The total degree of $G$ is the sum of the degrees of all the vertices of $G$. A vertex with degree zero is an isolated vertex.\n\n**Example 5.** In the graph $G_1$ in the Example ?? $\\deg(6) = 3$ and $\\deg(1) = 0$. The vertex 1 is an isolated vertex.\n\nIn a digraph, each vertex has two kinds of degree, the indegree of vertex $v$ is the number of edges which are coming into the vertex $v$ and outdegree of vertex $v$ is the number of edges which are going out from the vertex $v$, with notation $\\deg^-(v)$ and $\\deg^+(v)$, respectively.\n\n**Example 6.** In the graph $G_2$ in the Example ?? $\\deg^-(u) = 0$ and $\\deg^+(u) = 2$ and the $\\deg^-(w) = \\deg^+(w) = 1$.\n\n**Proposition 1.** Let $G = (V, E)$ be a undirected graph. Then the sum of the vertex degrees of the graph is equal to twice of the number of edges, i.e,\n\n$$\\sum_{v \\in V} \\deg(v) = 2|E|.$$\nProof. Adding the degrees of all the vertices involves counting one for each edge incident with each vertex. If it is not a loop, it is incident with two different vertices and so gets counted twice, once at each vertex. On the other hand, a loop at a vertex is also counted twice, by convention, in the degree of that vertex.\n\nCorollary 1. The number of vertices with odd degree is always even.\n\nProof. The total degree of a graph $G = (V, E)$ is equal two times the number of edges, that is,\n\n$$\\sum_{v \\in V} \\deg(v) = 2|E| \\Rightarrow \\sum_{v \\in V} \\deg(v) \\text{ is even} \\Rightarrow \\text{the number of vertices with odd degree is always even.}$$\n\nExample 7. A graph has six vertices each of degree 3. Since $\\sum_{v \\in V} \\deg(v) = 6 \\times 3 = 18 = 2 \\times 9$, the graph must have 9 edges.\n\nA degree sequence of the vertex degrees of a graph $G$ is a nonincreasing sequence of degrees of its graph vertices.\n\nExample 8. $(3, 2, 2, 1)$ is the degree sequence of the graph $G_3$ in the Example ??\n\nExample 9. A graph with degree sequence $(3, 2, 2, 1)$ has 4 edges because $\\sum_{v \\in V} \\deg(v) = 3 + 2 \\times 2 + 1 = 8$. (You can check in the graph!)\n\nReferences\n\n[1] Edgar Goodair and Michael Parmenter. *Discrete Mathematics with Graph Theory*. (3rd Ed.) Pearson, 2006.\n\n[2] Susanna Epp. *Discrete Mathematics and Applications*. (4th Ed.) Brooks/Cole CENGAGE Learning, 2011.\n\nExercises in MathE platform", "id": "./materials/235.pdf" }, { "contents": "Complementary graphs\n\nThe complementary graph $\\bar{G}$ of a simple graph $G$ has the same vertices as $G$ and two vertices are adjacent in $\\bar{G}$ if and only if they are not adjacent in $G$. Describe each of these graphs.\n\nExample 1. The following graphs are complementary.\n\nIsomorphic graphs\n\nThere is a distinction between a graph and its picture and it is importance to know when two graphs are the same or different, meaning that they differ only in the way they are labeled or drawn.\n\nLet $G$ and $G'$ be graphs with vertex sets $V(G)$ and $V(G')$ and edge sets $E(G)$ and $E(G')$, respectively.\n\n$G$ is isomorphic to $G'$ if, and only if, there exist one-to-one correspondences $f : V(G) \\rightarrow V(G')$ and $g : E(G) \\rightarrow E(G')$ that preserve the edge endpoint functions of $G$ and $G'$ in the sense that for all $v \\in V(G)$ and $e \\in E(G)$, $v$ is an endpoint of $e \\iff f(v)$ is an endpoint of $g(e)$.\n\nTwo graphs isomorphic are equal, meaning the same graph.\n\nExample 2. The following graphs are isomorphic.\n\nExercise 1. The graphs pictured are isomorphic? Justify.\nSolution:\nTo solve this problem, you must find functions \\( f : V(G) \\to V(G') \\) and \\( g : E(G) \\to E(G') \\) such that for all \\( v \\in V(G) \\) and \\( e \\in E(G) \\), \\( v \\) is an endpoint of \\( e \\) if, and only if, \\( f(v) \\) is an endpoint of \\( g(e) \\). Setting up such functions is partly a matter of trial and error and partly a matter of deduction.\n\nThe function \\( f \\) with \\( f(1) = t \\), \\( f(2) = x \\), \\( f(3) = z \\), and \\( f(4) = y \\) is a one-to-one correspondence between \\( V \\) and \\( V' \\). To see that this correspondence preserves adjacency, note that adjacent vertices in \\( G \\) are 1 and 2, 1 and 3, 2 and 3, 2 and 4, and 3 and 4, and each of the pairs \\( f(1) = t \\) and \\( f(2) = x \\), \\( f(1) = t \\) and \\( f(3) = z \\), \\( f(2) = x \\) and \\( f(3) = z \\), \\( f(2) = x \\) and \\( f(4) = y \\) and \\( f(3) = z \\) and \\( f(4) = y \\) consists of two adjacent vertices in \\( G' \\), thus, or all \\( v \\in V(G) \\) and \\( e \\in E(G) \\), \\( v \\) is an endpoint of \\( e \\) \\( \\iff \\) \\( f(v) \\) is an endpoint of \\( g(e) \\).\n\nExample 3. The following graph are complementary and isomorphic graphs.\n\n![Graphs](image)\n\nReferences\n\n[1] Domingos Cardoso, Jerzy Szymanski, and Mohammad Rostami. *Matemática Discreta: Combinatória, Teoria dos Grafos, Algoritmos*. Escolar Editora, 2009.\n\n[2] Edgar Goodair and Michael Parmenter. *Discrete Mathematics with Graph Theory*. (3rd Ed.) Pearson, 2006.\n\nExercises in MathE platform", "id": "./materials/236.pdf" }, { "contents": "Particular Graphs\n\nA simple graph is called **regular** if every vertex of this graph has the same degree. A regular graph is called **n-regular** if every vertex in this graph has degree $n$.\n\n**Example 1.** The graph pictured is called Petersen graph and it is a 3-regular graph.\n\nA simple graph is called **complete** on $n$ vertices, denoted by $K_n$, if it contains exactly one edge between each pair of distinct vertices.\n\nA **cycle** $C_n$, $n \\geq 3$, consists of $n$ vertices $v_1, v_2, \\ldots, v_n$ and edges $\\{v_1, v_2\\}, \\{v_2, v_3\\}, \\ldots, \\{v_{n-1}, v_n\\}$ and $\\{v_n, v_1\\}$.\n\nA simple graph $G$ is called **bipartite** if its vertex set $V$ can be partitioned into two disjoint sets $V_1$ and $V_2$ such that every edge in the graph connects a vertex in $V_1$ and a vertex in $V_2$ (so that no edge in $G$ connects either two vertices in $V_1$ or two vertices in $V_2$). When this condition holds, we call the pair $(V_1, V_2)$ a bipartition of the vertex set $V$ of $G$.\n\n**Example 2.** The following graph is bipartite.\nExercise 1. The cycle $C_6$ is bipartite?\nSolution:\nYes, the cycle $C_6$ is a bipartite graph because its vertex set can be partitioned into the two sets $V_1 = \\{v_1, v_3, v_5\\}$ and $V_2 = \\{v_2, v_4, v_6\\}$, and every edge of $C_6$ connects a vertex in $V_1$ and a vertex in $V_2$.\n\nExercise 2. Show that $K_3$ is not bipartite.\nSolution:\nTo show that $K_3$ is not bipartite, note that if we divide the vertex set of $K_3$ into two disjoint sets, one of the two sets must contain two vertices. If the graph were bipartite, these two vertices could not be connected by an edge, but in $K_3$ each vertex is connected to every other vertex by an edge.\n\nA complete bipartite graph $K_{m,n}$ is a graph that has its vertex set partitioned into two subsets of $m$ and $n$ vertices, respectively with an edge between two vertices if and only if one vertex is in the first subset and the other vertex is in the second subset.\n\nExample 3. The following graph displayed is the complete bipartite graph $K_{2,3}$\n\nSubgraphs\n\nA graph $H$ is a subgraph of another graph $G$ if, and only if, every vertex in $H$ is also a vertex in $G$, every edge in $H$ is also an edge in $G$, and every edge in $H$ has the same endpoints as it has in $G$.\n\nExample 4. The graph $H$ is a subgraph of the graph $G$.\n\nLet $G = (V, E)$ be a simple graph. The subgraph induced by a subset $W$ of the vertex set $V$ is the graph $H = (W, F)$, where the edge set $F$ contains an edge in $E$ if and only if both endpoints of this edge are in $W$. \n**Example 5.** Considering the complete graph $K_5$ the subgraph induced by $W = \\{v_2, v_3, v_4, v_5\\}$ is the graph $H$ pictured\n\n![Graph K5 and H](image)\n\n**Exercise 3.** In the Example 4 the subgraph $H$ is not an induced subgraph of graph $G = (V, E)$, why?\n\n**Solution:**\n\n*Considering the subset $W = \\{1, 2, 3, 4, 6\\}$ from the vertex set $V$ is missing an edge, the edge 23.*\n\n**References**\n\n[1] Domingos Cardoso, Jerzy Szymanski, and Mohammad Rostami. *Matemática Discreta: Combinatória, Teoria dos Grafos, Algoritmos*. Escolar Editora, 2009.\n\n[2] Susanna Epp. *Discrete Mathematics and Applications*. (4th Ed.) Brooks/Cole CENGAGE Learning, 2011.\n\nExercises in MathE platform", "id": "./materials/237.pdf" }, { "contents": "Matrix representation of graphs\n\nA graph $G$ (or multigraph, directed or not) with $n$ vertices $V(G) = \\{v_1, v_2, \\ldots, v_n\\}$ can be represented by a square matrix of order $n$, $A_G = [a_{ij}]$ called adjacency matrix, such us $a_{ij}$ is equal to the number of edges between vertices $v_i$ and $v_j$.\n\nObserve that if the graph is simple then $a_{ij} = \\begin{cases} 1 & \\text{if } \\{v_i, v_j\\} \\text{ is an edge of } G \\\\ 0 & \\text{otherwise} \\end{cases}$\n\n**Example 1.** A adjacency matrix of the graph\n\n![Graph](image)\n\nis\n\n$$\n\\begin{bmatrix}\n1 & 2 & 3 \\\\\n1 & 0 & 1 \\\\\n2 & 1 & 0 \\\\\n3 & 1 & 1 & 0\n\\end{bmatrix}\n$$\n\n**Exercise 1.** Find the adjacency matrix for the following graph.\n\n![Graph](image)\n\n**Solution:**\n\n$$\n\\begin{bmatrix}\n1 & 2 & 3 & 4 \\\\\n1 & 0 & 0 & 1 & 1 \\\\\n2 & 0 & 0 & 1 & 1 \\\\\n3 & 1 & 1 & 0 & 1 \\\\\n4 & 1 & 1 & 1 & 0\n\\end{bmatrix}\n$$\n\n*Considering the same order of vertices considered in the labels*\n\n*Notice that the sum of entries in each line (or column, since the matrix is symmetric) is the degree of the vertex, that is, for example $\\text{deg}(1) = 2$ and $\\text{deg}(3) = 3$.*\n\n**Example 2.** A adjacency matrix of the direct graph\n\n![Graph](image)\n\nis\n\n$$\n\\begin{bmatrix}\n1 & 2 & 3 & 4 \\\\\n1 & 0 & 1 & 0 & 2 \\\\\n2 & 1 & 0 & 1 & 0 \\\\\n3 & 0 & 1 & 1 & 1 \\\\\n4 & 2 & 0 & 1 & 0\n\\end{bmatrix}\n$$\n\nThe incidence matrix of a undirected graph $G$ with $n$ vertices $V(G) = \\{v_1, v_2, \\ldots, v_n\\}$ and $m$ edges, $A(G) = \\{a_1, a_2, \\ldots, a_m\\}$ is a matrix of type $n \\times m$, $M_G = [m_{ij}]$, $1 \\leq i \\leq n$, $1 \\leq j \\leq m$\nsuch that\n\n\\[ a_{ij} = \\begin{cases} \n0, & \\text{se } a_j = v_pv_q, \\text{ com } i \\notin \\{p, q\\} \\\\\n1, & \\text{se } a_j = v_kv_i, \\text{ com } k \\neq i \\\\\n2, & \\text{se } a_j = v_iv_i \n\\end{cases} \\]\n\n**Example 3.** The incidence matrix of the following graph\n\n\\[\n\\begin{pmatrix}\n1 & 0 & 1 & 0 & 0 \\\\\n0 & 0 & 1 & 0 & 1 \\\\\n1 & 1 & 0 & 1 & 0 \\\\\n0 & 1 & 1 & 0 & 1 \n\\end{pmatrix}\n\\]\n\nThe incidence matrix of a directed graph is similar to the undirected graphs but considering the direction of each edge, that is, the edge \\(v_1v_2\\) is different from the edge \\(v_2v_1\\). Thus, the incidence matrix of a directed graph without loops is a matrix \\(M_G = [m_{ij}]\\) where\n\n\\[ m_{ij} = \\begin{cases} \n0, & \\text{if } a_j = v_pv_q, \\text{ with } i \\notin \\{p, q\\} \\\\\n-1, & \\text{if } a_j = v_kv_i, \\text{ for some vertex } v_k \\\\\n1, & \\text{if } a_j = v_jv_k, \\text{ for some vertex } v_k \n\\end{cases} \\]\n\n**Example 4.** A incidence matrix of the graph\n\n\\[\n\\begin{pmatrix}\n1 & 1 & 0 \\\\\n2 & -1 & 1 \\\\\n3 & 0 & -1 & -1 \n\\end{pmatrix}\n\\]\n\n**Exercise 2.** Consider the adjacency matrix \\(A_G = \\begin{pmatrix}\n0 & 1 & 1 & 0 \\\\\n1 & 0 & 0 & 1 \\\\\n1 & 0 & 0 & 1 \\\\\n0 & 1 & 1 & 0 \n\\end{pmatrix}\\).\n\na. Draw a simple graph \\(G\\) represented by \\(A_G\\);\n\nb. Draw the complementary graph \\(\\bar{G}\\);\n\nc. Draw an directed graph \\(D\\) represented by \\(A_G\\);\n\nd. Determine the incidence matrix of the directed graph \\(D\\).\n\n**Solution:**\nGraph theory 4: matrix representation of graphs\n\na. Nominating the vertices by \\( v_1, v_2, v_3 \\) and \\( v_4 \\) the graph \\( G \\) could be\n\n\\[\n\\begin{array}{c}\n\\text{\\( v_1 \\)} \\\\\n\\text{\\( v_2 \\)} \\\\\n\\text{\\( v_3 \\)} \\\\\n\\text{\\( v_4 \\)}\n\\end{array}\n\\]\n\nb. Considering the graph \\( G \\) pictured in the previous question, the complementary graph \\( \\bar{G} \\) is\n\n\\[\n\\begin{array}{c}\n\\text{\\( v_1 \\)} \\\\\n\\text{\\( v_2 \\)} \\\\\n\\text{\\( v_3 \\)} \\\\\n\\text{\\( v_4 \\)}\n\\end{array}\n\\]\n\nc. An directed graph \\( D \\) represented by \\( A_G \\) could be\n\n\\[\n\\begin{array}{c}\n\\text{\\( v_1 \\)} \\\\\n\\text{\\( v_2 \\)} \\\\\n\\text{\\( v_3 \\)} \\\\\n\\text{\\( v_4 \\)}\n\\end{array}\n\\]\n\nd. Considering the directed graph \\( D \\) pictured in the previous question, and that the edges \\( a_1 = v_1v_2, a_2 = v_2v_4, a_3 = v_4v_3, a_4 = v_3v_1 \\), then the incidence matrix is\n\n\\[\n\\begin{bmatrix}\n1 & 0 & 0 & -1 \\\\\n-1 & 1 & 0 & 0 \\\\\n0 & 0 & -1 & 1 \\\\\n0 & -1 & 1 & 0\n\\end{bmatrix}\n\\]\n\nReferences\n\n[1] Domingos Cardoso, Jerzy Szymański, and Mohammad Rostami. Matemática Discreta: Combinatória, Teoria dos Grafos, Algoritmos. Escolar Editora, 2009.\n\n[2] Edgar Goodair and Michael Parmenter. Discrete Mathematics with Graph Theory. (3rd Ed.) Pearson, 2006.\n\nExercises in MathE platform", "id": "./materials/238.pdf" }, { "contents": "Walks, trails and paths\n\nA walk in a graph $G$ consists of an alternating (non empty) sequence of vertices and edges. $P = v_0e_1v_1e_2v_2\\ldots e_kv_k$, such that $v_0, v_1, \\ldots, v_k \\in V$, $e_1, e_2, \\ldots, e_k \\in E$ and vertices $v_{i-1}$ and $v_i$ are endpoints for edge $e_i$, with $i = 1, \\ldots, k$. The vertex $v_0$ is called initial vertex, the vertex $v_k$ is called end vertex and the other $v_1, \\ldots, v_{k-1}$ are intermediate vertices of the walk $P$.\n\nThe length of the walk $P$ is the number of edges in the sequence, eventually repeated and it is denoted by $\\text{comp}(P)$.\n\nA trail is a walk in which all edges are distinct; a path is a walk in which all vertices are distinct.\n\nA closed trail is called a circuit. A circuit in which the first vertex appears exactly twice (at the beginning and the end) and in which no other vertex appears more than once is a cycle.\n\nExample 1. $v_1v_3v_4v_2v_3$ is a walk with length four in the following graph.\n\nBecause the graph is simple its enough to refer the vertices. The walk is also a trail because don’t repeat edges but is not a path because the vertex $v_3$ appear twice.\n\nExercise 1. Considering the connected Petersen graph pictured, give an example of:\n\na. a walk that is not a path starting at vertex $v_3$;\n\nb. a walk that is not a trail;\n\nc. a cycle with length five.\nSolution\n\na. To be a walk that is not a path starting at vertex 3 it must have repeated vertices, for example, \\(v_3v_8v_6v_9v_7v_2v_3v_8v_10\\).\n\nb. To be a walk but not a trail, the previous example works, because the edge \\(v_3v_8\\) is repeated.\n\nc. An example of a cycle with length five, could be, \\(v_6v_8v_10v_7v_9v_6\\) and also, the more evident, \\(v_1v_2v_3v_4v_5v_1\\).\n\nConnectivity\n\nLet \\(G\\) be a graph. Two vertices \\(v\\) and \\(w\\) of \\(G\\) are connected if, and only if, there is a walk from \\(v\\) to \\(w\\). The graph \\(G\\) is connected if, and only if, given any two vertices \\(v\\) and \\(w\\) in \\(G\\), there is a walk from \\(v\\) to \\(w\\).\n\nExample 2. The graph \\(G_1\\) is connected because there is a walk between all pairs of vertices.\n\nExample 3. The graph pictured is not connected, because there is no walk, for example, between the vertex \\(a\\) and the vertex \\(b\\).\n\nA graph \\(H\\) is a connected component of a graph \\(G\\) if, and only if,\n\n- \\(H\\) is subgraph of \\(G\\);\n- \\(H\\) is connected; and\n- no connected subgraph of \\(G\\) has \\(H\\) as a subgraph and contains vertices or edges that are not in \\(H\\).\n\nA bridge is an edge that when deleted from the graph is obtained two connected component of the original graphs.\nExample 4. The graph pictured is a connected graph with a bright, the edge $v_3v_4$.\n\nIf we delete the edge $v_3v_4$ we obtain the following subgraph with two connected components.\n\nReferences\n\n[1] Edgar Goodair and Michael Parmenter. *Discrete Mathematics with Graph Theory*. (3rd Ed.) Pearson, 2006.\n\n[2] Kenneth H. Rosen. *Discrete Mathematics and its Application*. (7th Ed.) McGraw Hill, 2012.\n\nExercises in MathE platform", "id": "./materials/239.pdf" }, { "contents": "Eulerian graphs\n\nLet $G$ be a graph. An **Euler circuit** for $G$ is a circuit that contains every vertex and every edge of $G$. That is, an Euler circuit for $G$ is a sequence of adjacent vertices and edges in $G$ that has at least one edge, starts and ends at the same vertex, uses every vertex of $G$ at least once, and uses every edge of $G$ exactly once. We can also say that $G$ is an **Eulerian graph**.\n\n**Theorem 1.** A graph (with at least two vertices) is Eulerian if and only if it is connected and every vertex is even.\n\n**Proof.** $\\Rightarrow$ In walking along an Eulerian circuit, every time we meet a vertex (other than the one where we started), either we leave on a loop and return immediately, never traversing that loop again, or we leave on an edge different from that by which we entered and traverse neither edge again. So the edges (other than loops) incident with any vertex in the middle of the circuit can be paired. So also can the edges incident with the first (and last) vertex since the edge by which we left it at the beginning can be paired with the edge by which we returned at the end. Thus, an Eulerian graph must not only be connected, but also have vertices of even degree.\n\nConversely, a connected graph all of whose vertices are even must be Eulerian.\n\n$\\Leftarrow$ For the converse, suppose that $G$ is a connected pseudograph with all vertices of even degree. We must prove that $G$ has an Eulerian circuit. Let $v$ be any vertex of $G$. If there are any loops incident with $v$, follow these first, one after the other without repetition. Then, since we are assuming that $G$ has at least two vertices and since $G$ is connected, there must be an edge $vv_1$ (with $v_1 \\neq v$) incident with $v$. If there are loops incident with $v_1$, follow these one after the other without repetition. Then, since $\\deg(v_1)$ is even and bigger than 0, there must be an edge $v_1v_2$ different from $vv_1$. Thus we have a trail from $v$ to $v_2$ which we continue if possible. Each time we arrive at a vertex not encountered before, follow all the loops without repetition. Since the degree of each vertex is even, we can leave any vertex different from $v$ on an edge not yet covered. Remembering that we are considering that graphs are always finite, we see that the process just described cannot continue indefinitely; eventually, we must return to $v$, having traced a circuit $C_1$. Notice that every vertex in $C_1$ is even since we entered and left on different edges each time it was encountered. At this point, it may happen that every edge has been covered; in other words, that $C_1$ is an Eulerian circuit, in which case we are done. If $C_1$ is not Eulerian, as in the preceding example, we delete from $G$ all the edges of $C_1$ and all the vertices of $G$ which are left isolated (that is, acquire degree 0) by this procedure. All vertices of the remaining graph $G_1$ are even (since both $g$ and $C_1$ have only even vertices) and of positive degree. Also, $G_1$ and $C_1$ have a vertex $u$ in common, because $G$ is connected. Starting at $u$, and proceeding in $G_1$ as we did in $G$, we construct a circuit $C$ in $G_1$ which returns to $u$. Now combine $C$ and $C_1$ by starting at $v$, moving along $C_1$ to $u$, then through $C$ back to $u$, and then back to $v$ on the remaining edges of $C_1$. We obtain a circuit $C_2$ in $G$ which contains more edges than $C_1$. If it contains all the edges of $G$, it is Eulerian; otherwise, we repeat the process, obtaining a sequence of larger and larger circuits. Since our graph is finite, the process must eventually stop, and it stops only\nwith a circuit through all edges and vertices, that is, with an Eulerian circuit.\n\n**Example 1.** The complete graph $K_5$ is an Eulerian graph.\n\n**Proposition 1.** A graph $G$ possesses an Eulerian trail between two (different) vertices $u$ and $v$ if and only if $G$ is connected and all vertices except $u$ and $v$ are even.\n\n**Proof.** $\\Rightarrow$ If $G$ possesses an Eulerian trail that is not a circuit, then because the star vertex and the end vertex are different, only that two vertices have odd degree. The other vertices have even degree because if they belong to the trail we use one edge to get in the vertex and another (different) edge to get out. $\\Leftarrow$ Suppose that all, but exactly two vertices in $G$ have even degree, then from the previous theorem the is no Euler circuit. Consider that $u$ and $v$ are vertices from $G$ with odd degree and consider the graph $G'$, $E(G') = E(G) \\cup \\{uv\\}$, thus all vertices in $G'$ have even degree, then $G'$ admits an Euler circuit. If we add the edge $uv$ to the circuit we obtain a trail.\n\n**Example 2.** The graph pictured is not an Eulerian graph, because there exists vertices with odd degree, but admits an eulerian trail: $v_1v_2v_3v_4v_5v_3v_1v_4v_2$.\n\n**Hamiltonian graphs**\n\nAn **Hamilton path** in a graph is a path which contains every vertex of the graph. $G$. If the path is close, that is the star vertex and the final vertex are the same, is a **Hamiltonian cycle**. A **Hamiltonian graph** is one with a Hamiltonian cycle.\n\n**Example 3.** The following graph is Hamiltonian, because it admits an Hamiltonian cycle: $abcdefa$\nProperties of cycles\nSuppose $H$ is a cycle in a graph $G$.\n\n- For each vertex $v$ of $H$, precisely two edges incident with $v$ are in $H$; hence, if $H$ is a Hamiltonian cycle of $G$ and a vertex $v$ in $G$ has degree 2, then both edges incident with $v$ must be part of $H$.\n\n- The only cycle contained in $H$ is $H$ itself. (We say that $H$ contains no proper cycles.)\n\n**Theorem 2** (Ore). If $G$ is a simple graph with $n$ vertices with $n \\geq 3$ such that $\\text{deg}(u) + \\text{deg}(v) \\geq n$ for every pair of nonadjacent vertices $u$ and $v$ in $G$, then $G$ is Hamiltonian.\n\n**Proof.** (for reduction to absurdity) Let $G$ be a graph that satisfies the hypotheses of the theorem and suppose that $G$ is not Hamiltonian. Suppose further that $G$ is such that adding an edge gives a Hamilton cycle (which contains all the vertices). Since $G$ it is not complete (otherwise it would admit a Hamilton cycle) there is a pair of non-adjacent vertices $u$ and $w$ such that adding the $uw$ edge to $G$ gives a cycle of Hamilton. Thus $G$ will contain a path between $u$ and $w$ that traverses all other vertices of $G$, that is, $u = v_1 \\ v_2 \\ldots \\ v_{n-1} \\ v_n = w$. If $u$ is adjacent to $v_1$, $w$ is not adjacent to $v_{i-1}$, otherwise there would be a cycle of Hamilton in the form $w \\ v_{i-1} \\ v_{i-2} \\ldots \\ v_1 (= u)v_i \\ v_{i+1} \\ldots \\ v_{n-1} \\ v_n (= w)$. If $v_1 = u$ is adjacent to $r$ vertices, $v_n = w$ can only be adjacent to $(n - 1) - r$ vertices and therefore, as $\\text{deg}(u) + \\text{deg}(v) = r + (n - 1) - r \\leq n$, which contradicts the theorem.\n\n**Theorem 3** (Dirac). If a graph $G$ has $n > 3$ vertices and every vertex has degree at least $n$ then $G$ is Hamiltonian.\n\n**Proof.** Among the possible paths in $G$, consider $P = v_1v_2\\ldots v_t$, the longest path, in the sense that it contains the greater number of vertices. So there will be no path in $G$ that uses more than $t$ vertices. If there was a vertex $w$ adjacent to $v_1$ that was not in $P$ then $P$ would not be the biggest path in $G$. Then all vertices adjacent to $v_1$ are in $P$. If $\\text{deg}(v_1) \\geq \\frac{n}{2}$ then $t \\geq \\frac{n}{2} + 1$, where “$+1$” is relative to the vertex $v_1$.\n\nNote that there is a pair of vertices $v_k, v_{k+1}$ in $P$ ($1 \\leq k \\leq t$) such that $v_1$ is adjacent to $v_{k+1}$ and $v_t$ is adjacent to $v_k$. If not, then each $P$ vertex adjacent to $v_1$ will determine a non-vertex adjacent to $v_t$. Like all vertices $v_2, \\ldots, v_t$ are different exist in $G$ at least $\\frac{n}{2}$ vertices that are not adjacent to $v_t$. These vertices together with the vertices adjacent to $v_t$ will be at least $n$. With the apex $v_t$, $G$ will then have more than $n$ vertices, which is false. So the observation is true. It follows that $G$ has a cycle $C = v_1v_{k+1}v_{k+2}\\ldots v_t\\ldots v_kv_{k-1}\\ldots v_1$.\n\nIt remains to be shown that $C$ contains all vertices of $G$ and is therefore a Hamiltonian cycle. We know that $C$ contains at least $\\frac{n}{2} + 1$ vertices, and therefore may exist $\\frac{n}{2}$ vertices of $G$ that are not in $C$. If so some vertex $w$ that is not in $C$ will be adjacent to a vertex $v_s$ of $C$. So $w, v_s$, and the rest vertices of the cycle will define a path greater than $P$, which contradicts the choice of $P$. \n\n\\[\\square\\]\nExample 4. The graph pictured is not Hamiltonian. $v_3v_1v_2v_3$ is a cycle and also $v_4v_6v_7v_5v_4$. The vertex $v_3$ and $v_4$ already have two edges incident in each cycle, so it is not possible to connect this two cycles in order to obtain an Hamiltonian cycle.\n\nReferences\n\n[1] Domingos Cardoso, Jerzy Szymański, and Mohammad Rostami. Matemática Discreta: Combinatória, Teoria dos Grafos, Algoritmos. Escolar Editora, 2009.\n\nExercises in MathE platform", "id": "./materials/240.pdf" }, { "contents": "Planar graphs\n\nA graph is called **plane** if it is drawn in the plane without any edges crossing, where a crossing of edges is the intersection of the lines or arcs representing them at a point other than their common endpoint. A graph is called **planar** if it can be drawn in the plane without any edges crossing. Such a drawing is called a **planar representation** of the graph.\n\n**Example 1.** The complete graph $K_4$ is a planar graph because it can be drawn without crossings.\n\n![K4](image)\n\nA planar graph divides the plane into various connected regions, one of which is called the **exterior region**. Every region, including the exterior, is bounded by edges.\n\nPlanar graphs were first studied by Euler because of their connections with polyhedra. A convex regular polyhedron is a geometric solid all of whose faces are congruent. There are in all just five of these—the cube, the tetrahedron, the octahedron, the icosahedron, and the dodecahedron—and they are popularly known as the Platonic solids because they were regarded by Plato as symbolizing earth, fire, air, water, and the universe, respectively.\n\n**Example 2.** Planar representation of tetrahedro.\n\n![Tetrahedron](image)\n\nThis representation splits the plane into four region, three in the interior and one exterior region.\n\n**Exercise 1.** Draw a planar representation of a cube.\n\n**Solution:**\n\n![Cube](image)\n\nIn 1752, Euler published the remarkable formula $V - E + F = 2$, which holds for any convex polyhedron with $V$ vertices, $E$ edges, and $F$ faces. (A polygon is convex if the line joining\nany pair of nonadjacent vertices lies entirely within the polygon.) Euler showed that all planar representations of a graph split the plane into the same number of regions. He accomplished this by finding a relationship among the number of regions, the number of vertices, and the number of edges of a planar graph.\n\n**Theorem 1** (Euler’s Formula). Let $G$ be a connected planar simple graph with $e$ edges and $v$ vertices. Let $r$ be the number of regions in a planar representation of $G$. Then $v - e + r = 2$.\n\n**Proof.** We use induction on $e$, the number of edges.\n\nIf $e = 0$, then $v = r = 1$ (because $G$ is connected) and the formula is true.\n\nNow assume the formula holds for connected plane graphs $G'$ with $e' = e - 1$ edges, where $e \\geq 1$.\n\nWe must show that $G$ is a connected plane graph with $e$ edges, $v$ vertices, and $r$ regions. We must show that $v - e + r = 2$. We have two cases:\n\n1st case: the edge connect to existent vertices, then the number $v$ remains the same but the number of the regions increase one unity; $v - e + r = v' - (e' + 1) + r' + 1 = v' - e' + r'$ and the formula is true because $G'$ is a planar graph;\n\n2nd case: the edge connect one existent vertex to another new vertex, then $e = e' + 1$ and $v = v' + 1$, but $r = r'$, $v - e + r = v' + 1 - (e' + 1) + r' = v' - e' + r' = 2$, again because $G'$ is planar.\n\n**Example 3.** The following graph has 12 edges, 8 vertices and 6 regions, thus the formula is verified: $8 - 12 + 6 = 2$.\n\n![Graph](image)\n\n**Corollary 1.** If $G$ is a connected planar simple graph with $e$ edges and $v$ vertices, where $v \\geq 3$, then $e \\leq 3v - 6$.\n\n**Proof.** A connected planar simple graph drawn in the plane divides the plane into regions, say $r$ of them. Each region has at least three edges on the boundary. (Because the graphs discussed here are simple graphs, no multiple edges that could produce regions of degree two, or loops that could produce regions of degree one, are permitted.) In particular, note that of the unbounded region is at least three because there are at least three vertices in the graph.\n\nNote that the sum $S$ of the number of boundary edges of the regions is exactly twice the number of edges in the graph, because each edge occurs on the boundary of a region exactly twice (either in two different regions, or twice in the same region). Because each region has at least three edges on the boundary, it follows that $2e = S \\geq 3r$. Hence, $\\frac{2}{3}e \\geq r$. Using $r = e - v + 2$ (Euler’s formula), we obtain $e - v + 2 \\leq \\frac{2}{3}e$. It follows that $\\frac{5}{3} \\leq v - 2$. This shows that $e \\leq 3v - 6$. \n\n2\nExercise 2. Show that $K_5$ is nonplanar using Corollary 1.\n\nSolution:\nThe graph $K_5$ has five vertices and 10 edges. However, the inequality $e \\leq 3v - 6$ is not satisfied for this graph because $e = 10$ and, because $v = 5$, $3 \\times 5 - 6 = 9$. Therefore, $K_5$ is not planar.\n\nCorollary 2. If a connected planar, bipartite and simple graph has $e$ edges and $v$ vertices with $v \\geq 3$, then $e \\leq 2v - 4$.\n\nProof. Notice that if the graph is bipartite the graph contains no cycles of odd length. Thus, each region is bounded at least for four edges. Analogous to the proof of Corollary 1 we obtain $2e \\geq 4r$ and by the Euler’s Formula $4r = 4(e - v + 2)$, then $e \\leq 2v - 4$.\n\nCorollary 3. If $G$ is a connected planar simple graph, then $G$ has a vertex of degree not exceeding five.\n\nProof. If $G$ has one or two vertices, the result is true. If $G$ has at least three vertices, by Corollary 1 we know that $e \\leq 3v - 6$, so $2e \\leq 6v - 12$. If the degree of every vertex were at least six, then because $2e = S$ ($S$ is the sum of the number of boundary edges of all regions), we would have $2e \\geq 6v$. But this contradicts the inequality $2e \\leq 6v - 12$. It follows that there must be a vertex with degree no greater than five.\n\nIt was the Polish mathematician Kazimierz Kuratowski (1896-1980) who discovered the crucial role played by $K_{3,3}$ and $K_5$ in determining whether or not a graph is planar. He defined that two graphs are homeomorphic if and only if each can be obtained from the same graph by adding vertices (necessarily of degree 2) to edges.\n\nExample 4. The following graphs are homeomorphic.\n\nTheorem 2 (Kuratowski). A graph is planar if and only if it has no subgraph homeomorphic to $K_5$ or $K_{3,3}$.\nExercise 3. Is the Petersen graph planar?\n\nSolution:\nThe subgraph $H$ of the Petersen graph is obtained deleting the vertex 2 and the three edges that have 2 as an endpoint.\n\nThis subgraph is homeomorphic to $K_{3,3}$ considering the vertex sets $\\{6, 4, 10\\}$ and $\\{5, 9, 8\\}$ because it can be obtained by a sequence of elementary subdivisions, deleting $\\{4, 8\\}$ and adding $\\{3, 8\\}$ and $\\{3, 4\\}$, deleting $\\{5, 6\\}$ and adding $\\{1, 5\\}$ and $\\{1, 6\\}$, and deleting $\\{9, 10\\}$ and adding $\\{7, 9\\}$ and $\\{7, 10\\}$.\n\nThus, the Petersen graph is not a planar graph.\n\nReferences\n\nExercises in MathE platform", "id": "./materials/241.pdf" }, { "contents": "Coloring (the vertices) of a graph\n\nA coloring of a graph is an assignment of colors to the vertices so that adjacent vertices have different colors. An $n$-coloring is a coloring with $n$ colors. The chromatic number of a graph $G$, denoted $\\chi(G)$, is the minimum value of $n$ for which an $n$-coloring of $G$ exists.\n\nTo realize a coloring of a graph, in order to determine the chromatic number, we must:\n\n1. start for the vertex with maximum degree, $v_1$, color it with a color;\n2. use the same color to coloring the vertices non adjacents to $v_1$;\n3. choose the non colored vertex with maximum degree, $v_2$, and color with a color not already used;\n4. use the same color to coloring all vertices non adjacents to $v_2$;\n5. Repeat that procedure until all vertices are colored.\n\nExample 1. The chromatic number of graph following graph $G$ is 3, that is, $\\chi(G) = 3$.\n\nExercise 1. Determine the chromatic number of the following graph.\n\nSolution:\nLet’s start coloring a vertex with maximum degree. We can choose vertex $b$, $d$, $f$ or $h$. Let’s choose the vertex $b$ and color it and the non adjacent vertices with red.\nNow, we can choose between the vertices with maximum degree, $d$, $f$ or $h$. Let’s choose the vertex $d$ and color it and the non adjacent vertices with green.\n\nContinuing with the procedure we obtain\n\nThen, the chromatic number is 4, that is, $\\chi = 4$\n\n**Theorem 1.** Let $\\Delta(G)$ be the maximum of the degrees of the vertices of a graph $G$. Then $\\chi(G) < 1 + \\Delta(G)$.\n\n*Proof.* The proof is by induction on $V$, the number of vertices of the graph. When $V = 1$, $\\Delta(G) = 0$ and $\\chi(G) = 1$, so the result clearly holds. Now let $k$ be an integer, $k > 1$, and assume that the result holds for all graphs with $|V| = k$ vertices. Suppose $G$ is a graph with $k + 1$ vertices. Let $v$ be any vertex of $G$ and let $G_0 = G \\setminus \\{v\\}$ be the subgraph with $v$ (and all edges incident with it) deleted. Note that $\\Delta(G_0) \\leq \\Delta(G)$. Now $G_0$ can be colored with $\\chi(G_0)$ colors. Since $G_0$ has $k$ vertices, we can use the induction hypothesis to conclude that $\\chi(G_0) \\leq 1 + \\Delta(G_0)$. Thus, $\\chi(G_0) \\leq 1 + \\Delta(G)$, so $G_0$ can be colored with at most $1 + \\Delta(G)$ colors. Since there are at most $\\Delta(G)$ vertices adjacent to $v$, one of the available $1 + \\Delta(G)$ colors remains for $v$. Thus, $G$ can be colored with at most $1 + \\Delta(G)$ colors. \\qed\nTheorem 2 (Four-Color Theorem). For any planar graph $G$, $\\chi(G) \\leq 4$.\n\nExample 2. The planar representation of a cube has chromatic number 2\n\nReferences\n\n[1] Edgar Goodair and Michael Parmenter. *Discrete Mathematics with Graph Theory*. (3rd Ed.) Pearson, 2006.\n\n[2] Susanna Epp. *Discrete Mathematics and Applications*. (4th Ed.) Brooks/Cole CENGAGE Learning, 2011.\n\nExercises in MathE platform", "id": "./materials/242.pdf" }, { "contents": "Tree\n\nA simple graph $G$ is called a **forest** if it contains no circuits. A **tree** is a connected graph which contains no circuits. In other words a tree is a connected component of a forest. A **trivial tree** is a graph that consists of a single vertex. The vertices with degree one are called **leafs**.\n\n**Example 1.** The graph picture is a forest with two trees.\n\n![Graph](image)\n\n**Proposition 1.** Let $G = (V, E)$ be a graph with $n$ vertices, then the following statements are equivalent:\n\n(a) $G$ is a tree;\n\n(b) $G$ is an acyclic graph and has $n - 1$ edges;\n\n(c) $G$ is connected and has $n - 1$ edges;\n\n(d) $G$ is connected and each edge is a bridge;\n\n(e) $\\forall v, w \\in V$ there is precisely one path between $v$ and $w$;\n\n(f) $G$ is an acyclic graph but adding an edge we obtain a cycle.\n\n**Proof.** Let $G = (V, E)$ be a graph with $n$ vertices.\n\n(a) $\\Rightarrow$ (b) We will prove by mathematical induction on the number of the vertices $n$. If $n = 1$, the only tree with one vertex is the trivial tree and $0 = n - 1$ edges, then the implication is true. Suppose now, that the implication is true for all trees with less than $n \\geq 2$ vertices. Since, by definition, $G$ does not contain cycles, the removal of any edge subdivides the graph into two components $G_1$ and $G_2$, each of which is a tree. Considering that $G_1$ has $n_1$ vertices nad $G_2$ has $n_2$ vertices, in witch $n = n_1 + n_2$, by induction hypothesis, $G_1$ has $n_1 - 1$ edges, $G_2$ has $n_2 - 1$ edges, thus $G$ has $n_1 - 1 + n_2 - 1 + 1 = n - 1$ edges.\n\n(b) $\\Rightarrow$ (c) Suppose that $G$ is not connected. Then, each component of $G$ is a connected graph without circuits, so, by hypothesis, the number of vertices of each component exceeds the number of edges by one unit. Hence, the total number of vertices of $G$, exceeds that in a total number of edges of $G$ by at least two units, contradicting the hypothesis that $G$ has $n - 1$ edges.\n\n(c) $\\Rightarrow$ (d) As $G$ is connected, with $n - 1$ edges, the removal of any edge produces a graph with $n$ vertices and $n - 2$ edges and, consequently, this graph is not connected, since a connected graph of order $n$ has at least $n - 1$ edges.\n\n(d) $\\Rightarrow$ (e) Given two arbitrary vertices $u$ and $v$, by definition of the connected graph, there is a\npath between \\( u \\) and \\( v \\). Since, by hypothesis, any edge of that path is a bridge, we can conclude that the path is unique.\n\n(e) \\( \\Rightarrow \\) (f) Assuming that \\( G \\) contains a cycle, then any two vertices of that cycle are connected by at least two paths and, consequently, there are vertices of \\( G \\) that are connected by more than one path. Therefore, if there is a single path between any two vertices of \\( G \\), then \\( G \\) does not contain cycles. However, adding an edge between two vertices \\( u \\) and \\( v \\), as, for hypothesis, there is already a path between \\( u \\) and \\( v \\), we create a cycle\n\n(f) \\( \\Rightarrow \\) (a) Note that it is sufficient to prove that if \\( G \\) satisfies the hypothesis then it is connected. Suppose \\( G \\) satisfies the hypothesis, but it is not connected. If we add an edge to \\( G \\), connecting two vertices belonging to different components, no cycle is created, which is a contradiction.\n\n\\[ \\square \\]\n\n**Example 2.** The following connected graph is a tree with 10 vertices and 9 edges. Adding an edge we obtain a cycle. Deleting an edge we obtain a disconnected graph\n\n![Graph](image)\n\n**Proposition 2.** Each non-trivial tree contains at least two vertices of degree one (which are called leaf).\n\n**Proof.** Let \\( G = (V, E) \\) be a non-trivial tree with \\( n \\) vertices. As the tree is connected then for all vertex \\( v \\in V \\) has degree \\( \\deg(v) \\geq 1 \\). Recall that \\( \\sum_{v \\in V} \\deg(v) = 2|E| \\) nad because \\( G \\) is a tree \\( |E| = n - 1 \\). Thus\n\n\\[\n\\sum_{v \\in V} \\deg(v) = 2n - 2.\n\\]\n\nAs a consequence, at least two vertices are grade one (otherwise, \\( \\sum_{v \\in V} \\deg(v) \\geq 2n - 2 \\)). \\[ \\square \\]\n\n**Example 3.** A non trivial tree has, at least, 2 vertices connected by an edge.\n\n![Graph](image)\nA spanning tree of a simple graph $G$ is a subgraph of $G$ that is a tree containing every vertex of $G$.\n\n**Example 4.** Considering the graph $G$ pictured\n\n![Graph G](image)\n\nthe following graphs are spanning trees of $G$\n\n![Spanning Trees ST1 and ST2](image)\n\n**References**\n\n[1] Domingos Cardoso, Jerzy Szymanski, and Mohammad Rostami. *Matemática Discreta: Combinatória, Teoria dos Grafos, Algoritmos*. Escolar Editora, 2009.\n\nExercises in MathE platform", "id": "./materials/243.pdf" }, { "contents": "Weighted graph\n\nA weighted graph $G$ is a graph that have a number assigned to each edge $e$ and that number is called the weight of the edge $e$ and noted by $w(e)$. The weight of the graph $G$, $w(G)$, is the sum of the weights of all edges.\n\nExample 1. The graph $G$ pictured is a weighted graph with $W(G) = 41$.\n\n![Graph](image)\n\nThe edge $de$ has weight $w(de) = 5$.\n\nShortest path\n\nA shortest path between two vertices in a weighted graph is a path of least weight.\n\nDijkstra’s algorithm\n\nTo find a shortest path from vertex $v_1$ to vertex $v_n$ in a weighted graph, carry out the following procedure.\n\nStep 1 Assign to $v_1$ the label $(-, 0)$.\n\nStep 2 Until $v_n$ is labeled or no further labels can be assigned, do the following:\n\n(a) For each labeled vertex $u(x, d)$ and for each unlabeled vertex $v$ adjacent to $u$, compute $d + w(e)$, where $e = uv$.\n\n(b) For each labeled vertex $u$ and adjacent unlabeled vertex $v$ giving minimum $d' = d + w(e)$, assign to $v$ the label $(u, d')$. If a vertex can be labeled $(x, d')$ for various vertices $x$, make any choice.\n\nExample 2. Applying the Dijkstra algorithm to determine the shortest path between the vertex $a$ and the vertex $f$ in the graph pictured:\n\n![Graph](image)\n\nThen, the shortest path between the vertex $a$ and the vertex $f$ is $acef$ with weight 11.\nMinimum spanning tree\n\nAs we know, a spanning tree of a connected graph $G$ is a subgraph which is a tree and which includes every vertex of $G$. A minimum spanning tree of a weighted graph is a spanning tree of least weight, that is, a spanning tree for which the sum of the weights of all its edges is least among all spanning trees.\n\nKruskal’s algorithm\n\nTo find a minimum spanning tree in a connected weighted graph with $n > 1$ vertices, carry out the following procedure.\n\n**Step 1** Find an edge of least weight and call this $e_1$. Set $k = 1$.\n\n**Step 2** While $k < n$:\n\n- if there exists an edge $e$ such that $\\{e\\} \\cup \\{e_1, e_2, \\ldots, e_k\\}$ does not contain a circuit\n- then let $e_{k+1}$ be such an edge of least weight and replace $k$ by $k + 1$;\n- else output $e_1, e_2, \\ldots, e_k$ and stop.\n\n**end while**\n\n**Example 3.** To determine the minimum spanning tree, applying the Kruskal’s algorithm, in the connected weighted graph pictured:\n\n![Graph](image)\n\nthe edge $e_1 = df$ because is the lowest weight, $w(df) = 1$. Then $e_2 = ac$ with $w(ac) = 2$; and know we can choose between $ab$ or $bc$ because they have the same weight $w(ab) = w(bc) = 3$. Let’s consider $e_3 = bc$ and the next edge is $e_4 = de$ with $w(de) = 5$. We obtained two disconnected spanning graphs, but to be a tree we need a connected graph. So, add $e_6 = ce$ which $w(ce) = 6$. Our minimum spanning tree is\n\n![Graph](image)\n\nwith weight $W(T) = 17$. \n\n2\nPrim’s algorithm\nTo find a minimum spanning tree in a connected weighted graph with \\( n > 1 \\) vertices, proceed as follows.\n\n**Step 1** Choose any vertex \\( v \\) and let \\( e_1 \\) be an edge of least weight incident with \\( v \\). Set \\( k = 1 \\).\n\n**Step 2** While \\( k < n \\):\n\n- if there exists a vertex which is not in the subgraph \\( T \\) whose edges are \\( e_1, e_2, \\ldots, e_k \\)\n- then let \\( e_{k+1} \\) be an edge of least weight among all edges of the form \\( ux \\), where \\( u \\) is a vertex of \\( T \\) and \\( x \\) is a vertex not in \\( T \\);\n- replace \\( k \\) by \\( k + 1 \\);\n- else output \\( e_1, e_2, \\ldots, e_k \\) and stop.\n\nend while\n\n**Example 4.** To determine the minimum spanning tree, applying the Prim’s algorithm, in the connected weighted graph pictured:\n\n![Graph](image)\n\nconsider for example \\( v = c \\); the edge with least weight incident with \\( v \\) is \\( ac \\) with \\( w(ac) = 2 \\), then \\( e_1 = ac \\); now we can choose \\( ab \\) or \\( bc \\) because both edges as the same weight. Let’s consider \\( e_2 = ab \\), and now because we can’t have a circuit, \\( e_3 = ce \\) because \\( w(ce) = 6 \\), \\( e_4 = ed \\) because \\( w(ed) = 4 \\) and \\( e_5 = df \\) because \\( w(df) = 1 \\). Our minimum spanning tree is\n\n![Graph](image)\n\nwith weight \\( W(T) = 17 \\).\n\n**References**\n\n[1] Edgar Goodair and Michael Parmenter. *Discrete Mathematics with Graph Theory*. (3rd Ed.) Pearson, 2006.\nExercises in MathE platform", "id": "./materials/244.pdf" }, { "contents": "Circumference and spherical surface\n\nCircumference\n\nA circumference is a two-dimensional shape made by drawing a curve that is the same distance all around from the center.\n\nThe circumference centered in \\( C = (c_1, c_2) \\) with radius \\( r \\) is the set of points \\( P = (x, y) \\) (the locus) that are distant from \\( C \\) the measure \\( r \\), that is,\n\n\\[\n||\\overrightarrow{CP}|| = r \\iff (x - c_1)^2 + (y - c_2)^2 = r^2.\n\\]\n\nThe distance between the midpoint and the circumference is called the radius.\n\nExample: Let us consider, on the Cartesian plane, the circumference that contains points \\( A = (-1, 4) \\) and \\( B(3, 1) \\) and whose diameter measures \\( AB = 5 \\). Then the midpoint of \\( [AB] \\), \\( M = (1, \\frac{5}{2}) \\), corresponds to the center of the circumference and the radius is equal to \\( \\frac{AB}{2} = \\frac{5}{2} \\). Thus, the cartesian equation for this circumference is as follows:\n\n\\[\n(x - 1)^2 + (y - \\frac{5}{2})^2 = \\frac{25}{4}.\n\\]\n\nSpherical surface\n\nA Spherical surface is a three-dimensional shape where any of its points is at the same distance from a fixed point, called the center of the spherical surface.\n\nThe Spherical surface centered in \\( C = (c_1, c_2, c_3) \\) with radius \\( r \\) is the set of points \\( P = (x, y, z) \\) (the locus) that are distant from \\( C \\) the measure \\( r \\), that is,\n\n\\[\n||\\overrightarrow{CP}|| = r \\iff (x - c_1)^2 + (y - c_2)^2 + (z - c_3)^2 = r^2.\n\\]", "id": "./materials/245.pdf" }, { "contents": "Classical Definition of Probability (*Laplace*)\n\nIf an event can have $N_A$ different outcomes, within a total of $N$ possible outcomes (mutually exclusive and equally possible), then the probability of event $A$ is:\n\n$$P(A) = \\frac{N_A}{N}$$\n\n**Example:**\n\nConsider rolling a six sided die.\n\nEvent $A$: rolling a multiple of 3. $N_A$ outcomes: $A = \\{3, 6\\}$\n\nThe set of possible outcomes is $S = \\{1, 2, 3, 4, 5, 6\\}$, thus $N = 6$.\n\nTherefore, $P(A) = \\frac{N_A}{N} = \\frac{2}{6} = \\frac{1}{3}$", "id": "./materials/246.pdf" }, { "contents": "Example 2 (Complementary events, reunion, intersection)\n\nConsider A and B, two events from a sample space S.\n\nKnowing that \\( P(A) = 0.5 \\), \\( P(\\bar{A} \\cap \\bar{B}) = 0.1 \\) and \\( P(A \\cap B) = P(B \\cap \\bar{A}) \\), calculate \\( P(B) \\).\n\n**Answer:**\n\nIf \\( P(\\bar{A} \\cap \\bar{B}) = 0.1 \\), then \\( P(A \\cup B) = 1 - P(\\bar{A} \\cap \\bar{B}) = 1 - 0.1 = 0.9 \\)\n\nFor any two events A and B, \\( P(A \\cup B) = P(A) + P(B) - P(A \\cap B) \\)\n\nThis means that \\( P(B) - P(A \\cap B) = 0.4 \\)\n\nAs \\( P(B) - P(A \\cap B) = P(B \\cap \\bar{A}) \\), then \\( P(B \\cap \\bar{A}) = 0.4 \\)\n\nHence, \\( P(A \\cap B) = 0.4 \\) and \\( P(B) = 0.8 \\)", "id": "./materials/247.pdf" }, { "contents": "Example 3 (Conditional probability, intersection, tree diagram)\n\nA cautious individual usually carries his umbrella around, approximately in 80% of his daily errands. Curiously, when he carries his umbrella, he estimates that the probability of raining is about 50%, whereas when he forgets his umbrella at home, it rains in approximately 60% of those occasions.\n\nIt started pouring a while ago. Calculate the probability the individual left the house carrying his umbrella.\n\nAnswer:\nConsider the following events:\nU: the individual carries his umbrella; R: it is raining\nThe situation mentioned above can be represented with the help of the tree diagram shown.\n\nWe need to calculate the probability the individual brought the umbrella, given the fact we already know it is raining, \\( P(U|R) = \\frac{P(U \\cap R)}{P(R)} \\)\n\nAs \\( P(U \\cap R) = P(R \\cap U) \\) and \\( P(R \\cap U) = P(R|U) \\times P(U) = 50\\% \\times 80\\% = 40\\% \\)\n\n\\( P(R) = P(R|U) \\times P(U) + P(R|\\bar{U}) \\times P(\\bar{U}) = 50\\% \\times 80\\% + 60\\% \\times 20\\% = 52\\% \\)\n\nSo, \\( P(U|R) = \\frac{P(U \\cap R)}{P(R)} = \\frac{40\\%}{52\\%} = 76.92\\% \\)", "id": "./materials/248.pdf" }, { "contents": "Example 4 (Conditional probability, two-way table)\n\nA retail store made recently available for its customers the online purchase service (as well as traditional shopping in the store). Regarding delivery, customers can choose between picking up their purchases directly at the store, arranging a delivery at a pick-up point or at home delivery. The following probabilities have been calculated based on information provided by the store:\n\n| PURCHASE | DELIVERY | | |\n|----------|----------------|-------|-------|\n| | Store | Pick-up point | Home |\n| Store | 0.15 | 0.05 | 0.20 |\n| Online | 0.05 | 0.25 | 0.30 |\n\nA customer placed and online (verified) purchase. Calculate the probability the customer asked to pick up the items at the store.\n\n**Answer:**\n\nConsider the following events:\n\nDS: delivery at the store; PO: online purchase\n\nWe need to calculate the probability the asked to pick up his purchase at the store, given the fact we already it was placed online,\n\n\\[\nP(DS|PO) = \\frac{P(DS \\cap PO)}{P(PO)} = \\frac{0.05}{0.05 + 0.25 + 0.30} = 8.33\\%\n\\]", "id": "./materials/249.pdf" }, { "contents": "Conditional Probability\n\nConsider A and B, two events in a sample space S. The conditional probability of A given B is defined as:\n\n\\[ P(A|B) = \\frac{P(A \\cap B)}{P(B)} \\]\n\nwhen \\( P(B) > 0 \\).\n\nKnowing that B has occurred, every outcome that is outside B should be discarded, so our sample space is reduced to the set B.\n\nThis way, A can only happen is when the outcome belongs to the set \\( A \\cap B \\).\n\nWe divide \\( P(A \\cap B) \\) by \\( P(B) \\), so that the conditional probability of the new sample space becomes 1.", "id": "./materials/250.pdf" }, { "contents": "Events\n\nAn event is an outcome from a random experiment.\n\n**Example:** Consider rolling a six sided die. The set of possible outcomes is $S = \\{1, 2, 3, 4, 5, 6\\}$.\n\nSimple events\n\nA simple event has got a single outcome.\n\n**Example:** Consider rolling a six sided die. The probability of rolling a 6 matches a single outcome in the set shown above ($S$). Therefore, rolling a 6 is a simple event.\n\nCompound event\n\nA compound event has got two or more outcomes.\n\n**Example:** Consider rolling a six sided die. The probability of rolling an even number matches several outcomes ($2, 4, 6$) in the set shown above ($S$). Therefore, rolling an even number is a compound event.", "id": "./materials/251.pdf" }, { "contents": "Example 1 (Independent events, reunion, intersection)\n\nConsider A and B, two events from a sample space S.\n\nA and B are independent and equally likely.\n\nWith $P(A \\cap B) = 0.25$, calculate $P(\\bar{A} \\cap \\bar{B})$.\n\n**Answer:**\n\nIf A and B are independent, $P(A \\cap B) = P(A) \\times P(B)$, so $P(A) \\times P(B) = 0.25$\n\nConsidering that $P(A) = P(B) = \\sqrt{0.25} = 0.5$,\n\n$P(A \\cup B) = P(A) + P(B) - P(A \\cap B) = 0.5 + 0.5 - 0.25 = 0.75$\n\nAs $P(\\bar{A} \\cap \\bar{B}) = 1 - P(A \\cup B) = 1 - 0.75 = 0.25$", "id": "./materials/252.pdf" }, { "contents": "Independent Events\n\nIn probability, events are independent if the outcome of one event does not affect the outcome of another. Given two events, A and B, independent, then the probability of A happening AND the probability of B happening is \\( P(A) \\times P(B) \\), therefore:\n\n\\[\nP(A \\cap B) = P(A) \\times P(B)\n\\]\n\nExample:\nAn inquiry was made regarding students’ preferred payment methods. 50% of students use credit card, 40% resort to bank transfers and 20% use both payment methods. Determine if paying with credit card or bank transfer are independent events.\n\nConsidering event\n\nA: paying with credit card and\n\nB: paying though bank transfer,\n\nwe know that \\( P(A) = 0.5, P(B) = 0.4 \\) and \\( P(A \\cap B) = 0.2 \\)\n\nIf A and B are independent, then \\( P(A \\cap B) = P(A) \\times P(B) = 0.5 \\times 0.4 = 0.2 \\), which is true.\n\nSo, we can confirm that A and B are independent events.", "id": "./materials/253.pdf" }, { "contents": "Mutually Exclusive Events\n\nTwo events (A and B) are said to be mutually exclusive if they cannot happen at the same time.\n\nFrom the Venn diagram below, we can see that $A \\cap B = \\emptyset$. Therefore, $P(A \\cap B) = 0$\n\nIf A and B are mutually exclusive events, then the probability of A happening OR the probability of B happening is $P(A) + P(B)$. So, when considering mutually exclusive events,\n\n$$P(A \\cup B) = P(A) + P(B)$$\nExample:\nConsider rolling a six sided die.\n\nEvent A: rolling a 1. \\( A = \\{1\\} \\)\n\nEvent B: rolling a 6. \\( B = \\{6\\} \\)\n\nThe set of possible outcomes is \\( S = \\{1, 2, 3, 4, 5, 6\\} \\)\n\n\\[\nP(A) = P(B) = \\frac{1}{6}\n\\]\n\nGiven 2 generic events, A and B, \\( P(A \\cup B) = P(A) + P(B) - P(A \\cap B) \\)\n\nAs A and B are mutually exclusive events, \\( P(A \\cap B) = 0 \\).\n\nSo, \\( P(A \\cup B) = P(A) + P(B) = \\frac{1}{6} + \\frac{1}{6} = \\frac{1}{3} \\)", "id": "./materials/254.pdf" }, { "contents": "Odds\n\nOdds are ratios of probabilities.\n\nWe can define odds in favour of an event (odds on) or odds against an event (odds against).\n\nOdds can be expressed as a ratio of the probability an event will happen divided by the probability an event won't happen:\n\n\\[\n\\text{Odds in favour of } A = \\frac{p(A)}{1-p(A)}\n\\]\n\nExample:\n\nThe occurrence probability of a certain event is 0.80.\n\nThe odds in favour are \\(0.80/0.20 = 4/1 = 4\\) to 1 = 4 : 1.\n\nThe odds against are \\(1/4 = 1 : 4\\).\n\nBetting odds are written in the form 4 : 1 (\"4 to 1\").", "id": "./materials/255.pdf" }, { "contents": "MathE project\n\nContinuity for real functions of several variables\n\nExample 1.1. Study the continuity of the function $f : \\mathbb{R}^2 \\to \\mathbb{R}$\n\n$$f(x, y) = \\begin{cases} \\sqrt{1 - x^2 - y^2}, & x^2 + y^2 \\leq 1 \\\\ \\lambda, & x^2 + y^2 > 1, \\quad \\lambda \\in \\mathbb{R}. \\end{cases}$$\n\nSolution. On the set $\\{(x, y) \\in \\mathbb{R}^2 \\mid x^2 + y^2 < 1\\}$ the function $f$ is a composition of elementary continuous functions, so $f$ is continuous. On the set $\\{(x, y) \\in \\mathbb{R}^2 \\mid x^2 + y^2 > 1\\}$ the function $f$ is continuous being a constant. We study the continuity at the points from the circle $x^2 + y^2 = 1$.\n\nLet $(x_0, y_0) \\in \\mathbb{R}^2$ such that $x_0^2 + y_0^2 = 1$. Then $f(x_0, y_0) = \\sqrt{1 - x_0^2 - y_0^2} = 1$. Obviously we have\n\n$$\\lim_{(x, y) \\to (x_0, y_0)} f(x, y) = \\lim_{x^2 + y^2 < 1} \\sqrt{1 - x^2 - y^2} = 0,$$\n\nand\n\n$$\\lim_{(x, y) \\to (x_0, y_0)} f(x, y) = \\lim_{x^2 + y^2 > 1} \\lambda = \\lambda.$$\n\nThe function $f$ is continuous at $(x_0, y_0)$, so on $\\mathbb{R}^2$, if and only if $\\lambda = 0$. If $\\lambda \\neq 0$ the function $f$ is continuous only on $\\mathbb{R}^2 \\setminus \\{(x, y) \\in \\mathbb{R}^2 \\mid x^2 + y^2 = 1\\}$.\n\nExample 1.2. Study the continuity of the function $f : \\mathbb{R}^2 \\to \\mathbb{R}$\n\n$$f(x, y) = \\begin{cases} \\frac{(x^4 - y^2)^2}{x^6}, & y^2 < x^4 \\text{ and } x \\neq 0 \\\\ 0, & y^2 \\geq x^4 \\text{ or } x = 0. \\end{cases}$$\n\nSolution. Let us denote the sets\n\n$$D_1 = \\{(x, y) \\in \\mathbb{R}^2 \\mid y^2 < x^4 \\text{ and } x \\neq 0\\} = \\{(x, y) \\in \\mathbb{R}^2 \\mid -x^2 < y < x^2 \\text{ and } x \\neq 0\\}$$\n\nand\n\n$$D_2 = \\{(x, y) \\in \\mathbb{R}^2 \\mid y^2 \\geq x^4 \\text{ or } x \\neq 0\\} = \\{(x, y) \\in \\mathbb{R}^2 \\mid y \\leq -x^2 \\text{ or } y \\geq x^2 \\text{ or } x = 0\\}.$$\n\nObviously we have $\\mathbb{R}^2 = D_1 \\cup D_2$ and\n\n$$f(x, y) = \\begin{cases} \\frac{(x^4 - y^2)^2}{x^6}, & (x, y) \\in D_1 \\\\ 0, & (x, y) \\in D_2. \\end{cases}$$\nFor the points \\((x_0, y_0)\\) with \\(x_0^4 = y_0^2\\) and \\(x_0 \\neq 0\\) we have \\(f(x_0, y_0) = 0\\) and\n\\[\n\\lim_{(x,y) \\to (0,0)} f(x, y) = \\lim_{(x,y) \\to (0,0)} \\frac{(x^4 - y^2)^2}{x^6} = \\lim_{(x,y) \\to (x_0,y_0)} f(x, y) = 0 \\quad \\text{and} \\quad \\lim_{(x,y) \\to (x_0,y_0)} f(x, y) = 0.\n\\]\n\nFor \\((x_0, y_0) = (0, 0)\\) we have \\(f(0, 0) = 0\\) and\n\\[\n\\lim_{(x,y) \\to (0,0)} f(x, y) = \\lim_{(x,y) \\to (0,0)} \\frac{(x^4 - y^2)^2}{x^6} = \\lim_{(x,y) \\to (0,0)} \\left( x - \\frac{y^2}{x^3} \\right)^2 = 0,\n\\]\nbecause for \\((x, y) \\in D_1\\) we can write \\(0 \\leq \\frac{y^2}{x^3} \\leq \\frac{x^4}{|x|^3} = |x|\\) so, \\(\\lim_{(x,y) \\to (0,0)} \\frac{y^2}{x^3} = 0\\). It results that \\(f\\) is continuous at \\((0, 0)\\). At the other points \\(f\\) is an elementary continuous function. Finally \\(f\\) is continuous on \\(\\mathbb{R}^2\\).\n\n**Example 1.3.** Find the real constant \\(\\lambda\\) such that the function \\(f : D \\to \\mathbb{R}, D = \\{(x, y) \\in \\mathbb{R}^2 \\mid x^2 + y^2 < \\pi/2\\}\\) given by\n\\[\nf(x, y) = \\begin{cases} \n1 - \\cos \\sqrt{x^2 + y^2} & \\text{if } (x, y) \\neq (0, 0) \\\\\n\\lambda & \\text{if } (x, y) = (0, 0)\n\\end{cases}\n\\]\nbe continuous on \\(D\\).\n\n**Solution.** On the set \\(D \\setminus \\{(0, 0)\\}\\) the function \\(f\\) is a composition of elementary continuous functions, so \\(f\\) is continuous. We calculate the limit at the point \\((0, 0)\\). If we denote \\(\\sqrt{x^2 + y^2} = t\\) and use \\(\\lim_{t \\to 0} \\frac{\\sin t}{t} = \\lim_{t \\to 0} \\frac{\\tan t}{t} = 1\\), the limit can be calculate as\n\\[\n\\lim_{(x,y) \\to (0,0)} f(x, y) = \\lim_{(x,y) \\to (0,0)} \\frac{1 - \\cos \\sqrt{x^2 + y^2}}{\\tan (x^2 + y^2)} = \\lim_{t \\to 0} \\frac{1 - \\cos t}{\\tan (t^2)} = \\lim_{t \\to 0} \\frac{2\\sin^2 \\left(\\frac{t}{2}\\right)}{(t^2)} \\cdot \\frac{t^2}{\\tan (t^2)} \\cdot \\frac{1}{4} = \\frac{1}{2}.\n\\]\nIt results that \\(f\\) is continuous at \\((0, 0)\\), and so on \\(D\\), if and only if \\(\\lambda = \\frac{1}{2}\\).", "id": "./materials/256.pdf" }, { "contents": "Diagonals of a polygon\n\nHow to calculate the number of diagonals of a polygon?\n\nRemember:\n\nA diagonal of a polygon is a line segment joining two non-consecutive vertices of that polygon.\n\nCheck the number of diagonals drawn on the polygons shown in figures 1, 2, 3, 4, and 5, respectively, a triangle, a quadrilateral, a pentagon, a hexagon, and a heptagon.\n\nThe following table shows the number of diagonals drawn as a function of the number of sides of each polygon.\n\n| Number of polygon sides | 3 | 4 | 5 | 6 | 7 |\n|-------------------------|----|----|----|----|----|\n| Number of diagonals | 0 | 2 | 5 | 9 | 14 |\n\n+2 +3 +4 +5\nLet's understand what is happening from a vertex.\n\nRegarding the number of diagonals, for example in figure 3 (pentagon), we observe 2 diagonals from each vertex. So, apparently, we would have 10 diagonals \\((5 \\times 2)\\).\n\nHowever, we only observe 5, because the line segments \\([IL]\\) and \\([LI]\\) represent the same diagonal, as well as \\([HJ]\\) and \\([JH]\\), and the same is true for the remaining diagonals.\n\nThus:\n\n| Number of polygon sides | 3 | 4 | 5 | 6 | 7 |\n|-------------------------|---|---|---|---|---|\n| Number of diagonals from a vertex | 0 | 1 | 2 | 3 | 4 |\n\nAnd how do we determine the number of diagonals of a polygon with 12 sides without using the number of sides of the polygon with 11 sides and so on?\n\nThe number of diagonals from a vertex is equal 9, i.e., \\(12 - 3\\). Apparently there seem to be 108 diagonals \\((12 \\times 9)\\). But we find that there are 54, i.e., half of 108.\n\nIn other words:\n\n\\[\n12 \\times (12 - 3) = 12 \\times 9 = 108 \\\\\n108 : 2 = 54\n\\]\n\nGeneralizing, to calculate the number of diagonals of a polygon we use the formula:\n\n\\[\nd = \\left[ n \\times (n - 3) \\right] : 2\n\\]\n\nwhere:\n- \\(d\\) – total number of diagonals\n- \\(n\\) – number of sides of the polygon\n- \\((n - 3)\\) – number of diagonals from a vertex", "id": "./materials/257.pdf" }, { "contents": "Geometric Transformation\n\nA geometric transformation $T$ is a correspondence that associates with each point $P$ of the plane one and a single point $P'$ of the plane, under the following conditions:\n\na) if $P$ and $Q$ are two distinct points, then the corresponding points $P'$ and $Q'$ are also distinct,\n\nb) if $R$ is any point of the plane, then there is a point $S$ in the plane such that its correspondent by geometric transformation $T$ is $R$.\n\nSimilarity\n\nA geometric transformation $S$ is a similarity if it preserves the ratios between lengths of segments, that is, given any three points $A$, $B$, and $C$, the equality $\\frac{AB}{BC} = \\frac{A'B'}{B'C'}$ is verified in which $A'=S(A)$, $B'=S(B)$, $C'=S(C)$.\n\nIn a similarity, the distances between each two points are multiplied by a constant $(r = \\frac{A'B'}{AB} = \\frac{B'C'}{BC})$, called scale factor, usually represented by $r$.\n\nDilation of centre $O$ and scale factor $k$\n\nDilation $D$ of centre $O$ and scale factor $k$ does correspond to each point $P$ of the plane the point $P'=D(P)$ of the plane, under the following conditions:\n\na) $D(O)=O$, that is, the centre $O$ is a fixed point for dilation,\n\nb) If $P\\neq O$, $P'$ is on the OP line and\n\n- $\\frac{OP'}{OP} = |k|$, \n- $P$ and $P'$ are on the same side or on opposite sides, relative to $O$, as $k$ is positive or negative.\n\n$[PQ']$ is the image of $[PQ]$\n\n- $|k|>1$, dilation image is larger than original figure – enlargement\n- $|k|=1$, dilation image is the same size than original figure – (isometry)\n- $0<|k|<1$, dilation image is smaller than original figure – reduction\n\nDilation involves “resizing” the figure, resulting in an enlargement or a reduction.\n**Isometry (Isometric Transformation)**\n\nA geometric transformation $I$ is an isometry (isometric transformation) if, for any two points $P$ and $Q$, it has $\\text{dist}(P',Q')=\\text{dist}(P,Q)$, where $P'=I(P)$ and $Q'=I(Q)$.\n\nAn isometry preserves the distances, and the figures are transformed into congruent figures.\n\n**Translation associated to vector $\\vec{u}$**\n\nTranslation associated to vector $\\vec{u}$ is a geometric transformation in which each $P$ point of the plane is transformed into another point $P'$ (image of $P$), with $P'=P+\\vec{u}$.\n\nTranslation involves “sliding” the figure from one position to another.\n\n**Rotation of centre $O$ and amplitude $\\alpha$**\n\nRotation of centre $O$ and amplitude $\\alpha$ is a geometric transformation such that:\n\na) whatever the point $P$ of the plane, the distance from $O$ to $P$ is equal to the distance from $O$ to the image of $P$ ($P'$),\n\nb) the amplitude of the oriented angle defined by $P$, $O$ and $P'$ is equal to $\\alpha$.\n\nRotation involves “turning” the figure around a point (centre of rotation).\n\n**Reflection associated to line $s$**\n\nReflection associated to line $s$ is the geometric transformation that does correspond to each point $P$ of the plane the point $P'$ (image of $P$), in such a way that:\n\na) the line $s$ is perpendicular to $[PP']$ and passes through the midpoint of $[PP']$ (or $s$ is the mediator of $[PP']$),\n\nb) if $P$ belongs to $s$, its image coincides with $P$.\n\nReflection involves “flipping” the figure over a line (line of reflection).\n\n**Glide Reflection associated to line $s$ and vector $\\vec{u}$**\n\nGlide reflection is the geometric transformation that results from the sequence of a reflection associated to line $s$ with a translation whose vector $\\vec{u}$ has the same direction as $s$.\n\nGlide reflection involves “flipping” the figure over a line and “sliding” it, maintaining the direction of the line.", "id": "./materials/258.pdf" }, { "contents": "**Symmetry**\n\nSymmetry of a figure $F$ is a particular characteristic of that figure. It means that there is an isometry $I$ of the plane that leaves the figure globally invariant, that is, such that $I(F) = F$.\n\nTo speak of symmetry is to speak of symmetry of a figure. Analysing the symmetry of a figure leads to investigate if there are isometries (different from the identity) that leave it globally invariant.\n\nThere is a symmetry for each of the four types of isometries: translational symmetry, rotational symmetry, reflectional symmetry, glide reflexional symmetry.\n\n| Reflectional Symmetry | Translational Symmetry |\n|-----------------------|------------------------|\n| There is, at least, a reflection that leaves the figure globally invariant. The figure admits, at least, one line (or axis) of symmetry. | There is, at least, a translation that leaves the figure globally invariant. It is only possible if the figure is infinite. |\n| This square has four reflectional symmetries associated to the four lines of symmetry signed (green, blue, orange, red). | This infinite figure admits, for example, a translational symmetry associated to vector $\\vec{u}$. |\n\n| Rotational Symmetry | Glide Reflectional Symmetry |\n|---------------------|-----------------------------|\n| There is, at least, one rotation with an amplitude greater than 0º and less than 360º that leaves the figure globally invariant. Only in this case, a rotational symmetry associated with an angle of 360º is also allowed. | There is, at least, a glide reflection that leaves the figure globally invariant. It is only possible if the figure is infinite. |\n| This square has four rotational symmetries centred in $E$ and associated to positive amplitudes. For example, 90º (B: image of $A$…), 180º (C: image of $A$…), 270º (D: image of $A$…), 360º (A: image of $A$…). | This infinite figure admits a glide reflectional symmetry associated to both the line $s$ and vector $\\vec{u}$. |", "id": "./materials/259.pdf" }, { "contents": "Classification of quadrilaterals\n\nWhat criteria can we use to classify quadrilaterals?\n\nTo improve your knowledge about the classification of quadrilaterals, try to find out all the possible quadrilaterals on the 3 x 3 geoboard or the 3 x 3 dotted paper (note that it is not possible to find out all the quadrilaterals that exist on the 3 x 3 geoboard or the 3 x 3 dotted paper).\nCheck to see if you have discovered the following quadrilaterals:\n\nLet's look at possible ranking criteria.\n\n**Attending to convexity**: a quadrilateral is convex if for any two of its points the line segment joining them is contained in the polygon.\n• Focusing on convex quadrilaterals…\n\n**Given the criterion:** parallelism of its sides.\n\nWe find the quadrilaterals that have at least two parallel sides (trapezoids) and those that do not have parallel sides (non-trapezoids).\n\n---\n\n**Trapezoids**\n\n![Trapezoids](image1)\n\n**Non-trapezoids**\n\n![Non-trapezoids](image2)\n\n---\n\n○ **Now the focus is on trapezoids**\n\n**Also considering the parallelism of their sides,** we will find the quadrilaterals that have two pairs of parallel sides (parallelogram trapezoids or simply parallelograms) and those that have only two parallel sides (non-parallelogram trapezoids).\n\n---\n\n**Parallelograms**\n\n![Parallelograms](image3)\n\n**Non-parallelogram trapezoids**\n\n![Non-parallelogram trapezoids](image4)\nNow the focus is on parallelograms\n\nGiven the amplitude of the angles, we find the parallelograms that have all right angles (right-angled parallelograms or simply rectangles) and those that have no right angles (non-rectangles parallelograms).\n\n| Rectangle parallelograms | Non-rectangles parallelograms |\n|---------------------------|-------------------------------|\n| ![Rectangle parallelograms](image1) | ![Non-rectangles parallelograms](image2) |\n\nCan we also define some classification criteria for rectangles?\n\nTaking into account the length of the sides, we can consider rectangles with all sides of equal length (squares) or sides with different lengths (non-squares).\n\n| Squares | Non-squares |\n|---------|-------------|\n| ![Squares](image3) | ![Non-squares](image4) |\n\nTo go further:\n\nRemember that it is not possible to find out all the quadrilaterals that exist on the 3 x 3 geoboard or the 3 x 3 dotted paper.\n\nTry finding quadrilaterals on the 5 x 5 geoboard or the 5 x 5 dotted paper. Can you find a scalene trapezoid? Can you find a rhombus?", "id": "./materials/260.pdf" }, { "contents": "Sum of the amplitudes of the exterior angles of a triangle\n\nHow to calculate the sum of the exterior angles of a triangle?\n\nWhat can you conclude about the amplitude of an exterior (or external) angle? Justify your answer.\n\n![Figure 1](image1.png)\n\nWe know that the sum of the amplitudes of the interior (or internal) angles of a triangle is equal to 180°.\n\nAngle d is an external angle, because the sum of the amplitudes of angles d and a is equal to 180°.\n\nThus, the amplitude of the external angle d is equal to the sum of the amplitudes of angles b and c.\n\nWhat can you conclude about the sum of the amplitudes of the exterior angles, from 3 different vertices, of a triangle. Justify your answer.\n\n![Figure 2](image2.png)\n\nThe sum of the amplitudes of the interior angles of a triangle is equal to 180°.\n\nThe sum of the amplitudes of the interior and exterior angles of 3 different vertices of a triangle is equal to 3 x 180°, or 540°.\n\nThus, the sum of the exterior angles is equal to the difference between 540° and 180°, that is, 360°.\n\nTo go further:\n\nHow do you calculate the sum of the amplitudes of the exterior angles of other polygons?", "id": "./materials/261.pdf" }, { "contents": "Sum of the amplitudes of the internal angles of a triangle\n\n- **How to calculate the sum of the amplitudes of internal angles of a triangle?**\n\nConsidering figure 1, let's answer the questions to, step by step, reach the conclusion.\n\n![Figure 1](image)\n\n- **Justify that the amplitude of angle BAD is equal to the amplitude of angle ABC.**\n\nBecause they are two internal alternate angles.\n\n- **Is the amplitude of angle EAC equal to the amplitude of angle ACB? Why?**\n\nYes, the amplitude of angle EAC is equal to the amplitude of angle ACB. Because they are two internal alternate angles.\n\n- **What can you conclude about the sum of the internal angles of a triangle?**\n\nThe sum of the amplitudes of angles BAD, BAC, and EAC is equal to 180°.\n\nSince angles BAD and ABC have the same amplitude and so do angles EAC and ACB. Then the sum of the amplitudes of angles BAC, ACB and CBA (internal angles of the triangle) is equal to 180°.\n\n**To go further:**\n\nHow do you calculate the sum of the amplitudes of the internal angles of other polygons?", "id": "./materials/262.pdf" }, { "contents": "It is possible to construct a triangle?\n\n- Can you draw a triangle whose sides have lengths of 6cm, 7cm and 8cm?\n\nRemember **Triangular Inequality**\n\nThe sum of the lengths of any two sides of a triangle is greater than the length of the third side.\n\nWe have,\n\n- The sum of 6 and 7 is 13 and 13 is greater than 8.\n- The sum of 7 and 8 is 15 and 12 is greater than 6.\n- The sum of 6 and 8 is 14 and 14 is greater than 7.\n\nThe set \\{6,7,8\\} of side lengths satisfies the Triangle Inequality, so it is possible to construct a triangle.\n\n- And with the same lengths, it is possible to construct a right triangle?\n\nIn that case, we must verify the **Pythagoras' Theorem**.\n\nThe area of the square whose side is the hypotenuse (opposite the right angle) is equal to the sum of the areas of the squares on the other two sides (see the picture).\n\nIf we can build a right triangle the hypotenuse would be the side with length 8cm. So, we must have \\(8^2 = 6^2 + 7^2\\), that is, \\(64 = 85\\). This is a false proposition.\n\nConcluding, it is not possible to construct a right triangle with that side lengths.\n\nNote: We can have a right triangle if the length of hypotenuse is \\(\\sqrt{85}\\) cm.\n\nTo think: Investigate what happens if you consider sides with other lengths!", "id": "./materials/263.pdf" }, { "contents": "Classifying triangles\n\nWhat criteria can we use to classify the triangles of the figure?\n\nConsidering the length of the sides\n\n**Isosceles triangle** – The triangle has at least two sides with the same length.\n\n- Triangles: C, D, F and G\n\n- Particular case: **Equilateral triangle** – The triangle has all sides with the same length.\n - Triangle: C\n\n**Scalene triangle** – The triangle has all sides with different length.\n\n- Triangles: A, B and E\n\nConsidering the amplitude of the angles\n\n**Acute triangle** – The internal angles of the triangle are all acute.\n\n- Triangles: B, C and F\n\n**Right triangle** – The triangle has one right internal angle.\n\n- Triangles: A and D\n\n**Obtuse triangle** – The triangle has one obtuse internal angle.\n\n- Triangles: E and G", "id": "./materials/264.pdf" }, { "contents": "Pyramids, Prisms and polygon of the bases – What relation?\n\nIn the pyramids, what relationship exists between the number of sides of the polygon of the base and the number of faces, vertices, and edges?\n\nCount the number of faces, vertices, and edges of each of these pyramids. What do you observe?\n\nConfirm that your results match those in the table. What happens for a pyramid with a base polygon with n-sides?\n\n| Pyramid | N.º of sides of the base polygon (n) | N.º of faces (F) | N.º of vertices (V) | N.º of edges (E) |\n|------------|-------------------------------------|------------------|---------------------|------------------|\n| Triangular | 3 | 4 | 4 | 6 |\n| Rectangular| 4 | 5 | 5 | 8 |\n| Pentagonal | 5 | 6 | 6 | 10 |\n| Hexagonal | 6 | 7 | 7 | 12 |\n| Heptagonal | 7 | 8 | 8 | 14 |\n| Octagonal | 8 | 9 | 9 | 16 |\n| ... | ... | ... | ... | ... |\n| ... | n | n + 1 | n + 1 | 2n |\n\nTherefore, we have:\n\n\\[ V = F = n + 1 \\quad E = 2n \\]\nAnd what about prisms? What relationship exists between the number of sides of the polygon of the base and the number of faces, vertices, and edges?\n\nCount the number of faces, vertices, and edges of each of these prisms. What do you observe?\n\nConfirm that your results match those in the table. What happens for a prism with a base polygon with n-sides?\n\n| Prism | N.º of sides of the base polygon (n) | N.º of faces (F) | N.º of vertices (V) | N.º of edges (E) |\n|-----------|-------------------------------------|------------------|---------------------|------------------|\n| Triangular| 3 | 5 | 6 | 9 |\n| Rectangular| 4 | 6 | 8 | 12 |\n| Pentagonal| 5 | 7 | 10 | 15 |\n| Hexagonal | 6 | 8 | 12 | 18 |\n| Heptagonal| 7 | 9 | 14 | 21 |\n| Octagonal | 8 | 10 | 16 | 24 |\n| … | … | … | … | … |\n| … | n | n + 2 | 2n | 3n |\n\nTherefore, we have:\n\n\\[ F = n + 2 \\quad V = 2n \\quad E = 3n \\]", "id": "./materials/265.pdf" }, { "contents": "Classifying quadrilaterals\n\nHow can we classify the quadrilaterals of the picture?\n\nWhich quadrilaterals are convex?\n\nAttend the\n\nDefinition: A quadrilateral is **convex** if all line segment joining two vertices are contained in the quadrilateral.\n\nwe can say that the quadrilaterals A, B, C, D, E, F, G, H and J are convex. Quadrilaterals I and K are non-convex.\n\nWhich quadrilaterals are kites?\n\nAttend the\n\nDefinition: A **kite** is a quadrilateral with two pairs of consecutive sides congruent.\n\nwe can say that the quadrilaterals F, J and K are kites.\nWhich quadrilaterals are trapezoids?\n\nAttend the\n\n**Definition:** A trapezoid is a quadrilateral with at least one pair of parallel sides.\n\nwe can say that the quadrilaterals B, C, D, E, G, H and J are trapezoids.\n\n**Observation 1:** Some authors consider one slightly different definition, this is, a Trapezoid is a quadrilateral with exactly one pair of parallel sides (exclusive definition). In this case, only the quadrilaterals C, E and H are trapezoids.\n\n**Observation 2:** Within the scope of MathE we consider the first definition (inclusive definition) of trapezoid and the others indicated below.\n\n**Particular cases of trapezoids:**\n\n- If the non-parallel sides are congruent – **Isosceles trapezoids**\n - Quadrilateral E\n- If the non-parallel sides are not congruent – **Scalene trapezoids**\n - Quadrilateral H\n- If one of the opposing non-parallel sides is perpendicular to the bases – **Right trapezoids**\n - Quadrilateral C\n\nAttend to the number of the pairs of parallel sides, we have:\n\n- Trapezoids **with exactly one pair** of parallel sides – **Non-parallelograms**\n - Quadrilaterals C, E and H\n- Trapezoids **with two pairs** of parallel sides – **Parallelograms**\n - Quadrilaterals B, D, G and J\nHow can we classify the parallelograms?\n\n**Considering the congruence of the internal angles:**\n- Parallelograms with all congruent angles – **Rectangles**\n - Quadrilaterals B and G\n- Parallelograms with exactly two pairs of congruent angles – **Non-rectangles (oblique parallelograms)**\n - Quadrilateral D and J\n\n**Considering the congruence of the sides:**\n- Parallelograms with all congruent sides – **Rhombus**\n - Quadrilaterals B and J\n- Parallelograms with exactly two pairs of congruent sides – **Non-rhombus**\n - Quadrilaterals D and G\n\n**Considering the congruence of the angles and the sides:**\n- Parallelograms with all angles congruent and all sides congruent – **Squares** *(regular quadrilaterals)*\n - Quadrilateral B\n\n**Note:** A **regular polygon** is a polygon with all internal angles congruent and all sides congruent.\n\n**To think:** What other criteria can we use to classify the quadrilaterals?", "id": "./materials/266.pdf" }, { "contents": "Euler’s formula\n\nWhat relationship exists between the number of faces, vertices, and edges of a convex polyhedron?\n\nConsider the models of the geometric solids represented in Figure 1.\n\n![Figure 1](image)\n\nCount the number of faces, vertices, and edges of each of these solids.\n\nConfirm that your results match those in the Table 1.\n\n| Geometric solid | Number of faces ($F$) | Number of vertices ($V$) | Number of edges ($E$) |\n|-----------------|-----------------------|--------------------------|-----------------------|\n| A | 4 | 4 | 6 |\n| B | 5 | 6 | 9 |\n| C | 6 | 8 | 12 |\n| D | 6 | 6 | 10 |\n| E | 8 | 12 | 18 |\n| F | 12 | 20 | 30 |\n\nTable 1\nSum the number of faces with the number of vertices, $F + V$, and record again the number of edges.\n\nCheck if you get the same results as those recorded in Table 2.\n\n| Geometric solid | $F + V$ | Number of edges ($E$) |\n|-----------------|---------|-----------------------|\n| A | 8 | 6 |\n| B | 11 | 9 |\n| C | 14 | 12 |\n| D | 12 | 10 |\n| E | 20 | 18 |\n| F | 32 | 30 |\n\nTable 2\n\n**What can you observe?**\n\nThe number of edges (last column) differs by 2 values from the sum of the number of faces and the number of vertices, that is, from $F + V$ (middle column).\n\nThis happens because the **Euler’s formula** is valid for any convex polyhedron.\n\n**Euler’s formula**\n\n$$F + V = E + 2$$\n\nWhere:\n\n- $F$ – number of faces\n- $V$ – number of vertices\n- $E$ – number of edges", "id": "./materials/268.pdf" }, { "contents": "Congruence of Triangles Criteria\n\n**SSS criterion – Side, Side, Side**\nTwo triangles are congruent if they have, from one to the other, the three sides congruent.\n\nThe triangles \\([ABC]\\) and \\([DEF]\\) are congruent because \\(\\overline{AB} = \\overline{EF}\\); \\(\\overline{BC} = \\overline{FD}\\) and \\(\\overline{CA} = \\overline{DE}\\).\n\n**ASA criterion – Angle, Side, Angle**\nTwo triangles are congruent if they have, from one to the other, a congruent side and the two adjacent angles to this side are also congruent.\n\nThe triangles \\([ABC]\\) and \\([DEF]\\) are congruent because \\(\\overline{AB} = \\overline{DE}\\); \\(\\alpha = \\delta\\) and \\(\\beta = \\gamma\\).\n\n**SAS criterion – Side, Angle, Side**\nTwo triangles are congruent if they have, from one to the other, two congruent sides and the angle formed by them is also congruent.\n\nThe triangles \\([ABC]\\) and \\([DEF]\\) are congruent because \\(\\overline{AB} = \\overline{EF}\\); \\(\\overline{CA} = \\overline{DE}\\) and \\(\\alpha = \\beta\\).\nSimilarity of Triangles Criteria\n\n**SSS criterion – Side, Side, Side**\n\nTwo triangles are similar if they have, from one to the other, the three sides proportional.\n\nThe triangles $[ABC]$ and $[DEF]$ are similar because $\\frac{DE}{AB} = \\frac{EF}{BC} = \\frac{FD}{CA} = 2$.\n\n**AA criterion – Angle, Angle**\n\nTwo triangles are similar if they have, from one to the other, two corresponding angles congruent.\n\nThe triangles $[ABC]$ and $[DEF]$ are similar because $\\beta = \\gamma$ and $\\alpha = \\delta$.\n\n**SAS criterion – Side, Angle, Side**\n\nTwo triangles are similar if they have, from the one to the other, two sides proportional and the angle formed by them is congruent.\n\nThe triangles $[ABC]$ and $[DEF]$ are similar because $\\frac{EF}{BC} = \\frac{FD}{CA} = 1.5$ and $\\alpha = \\beta$. ", "id": "./materials/270.pdf" }, { "contents": "1.1 Limite de funcții\n\nFie \\( k, p \\geq 1, A \\subset \\mathbb{R}^k \\) o mulțime nevidă și \\( f : A \\rightarrow \\mathbb{R}^p \\) o funcție. În cazul în care vom considera norme pe spații \\( \\mathbb{R}^k, \\mathbb{R}^p \\), le vom nota \\( \\| \\cdot \\|_{\\mathbb{R}^k} \\) și \\( \\| \\cdot \\|_{\\mathbb{R}^p} \\), respectiv. Uneori vom nota ambele norme prin \\( \\| \\cdot \\| \\), contextul permitându-ne să deducem pe care dintre spații este considerată norma respectivă. Cu \\( A' \\) se notează mulțimea punctelor de acumulare ale mulțimii \\( A \\), și fie \\( a \\in A' \\), \\( \\ell \\in \\mathbb{R}^p \\).\n\nDefiniția 1.1 Spunem că funcția \\( f \\) are limită \\( \\ell \\) în punctul \\( a \\), și notăm\n\\[\n\\lim_{x \\to a} f(x) = \\ell, \\text{ sau } f(x) \\to \\ell \\text{ pentru } x \\to a,\n\\]\ndacă\n\\[\n\\forall \\varepsilon > 0, \\exists \\delta > 0, \\forall x \\in A \\setminus \\{a\\}, \\|x - a\\|_{\\mathbb{R}^k} < \\delta : \\|f(x) - \\ell\\|_{\\mathbb{R}^p} < \\varepsilon. \\tag{1.1}\n\\]\n\nObservația 1.2 Punctul \\( a \\) nu trebuie să aparțină neapărat mulțimii \\( A \\), însă trebuie să existe puncte în mulțimea \\( A \\) oricât de apropiate de \\( a \\), notiunea de limită exprimând intuitiv faptul că, atunci când punctele din domeniul funcției se apropie de punctul \\( a \\), atunci valorile funcției \\( f \\) în aceste puncte se apropie oricât de mult de punctul limită \\( \\ell \\).\n\nTeorema 1.3 (Caracterizarea cu șiruri a limitelor) Fie \\( f : A \\subset \\mathbb{R}^k \\rightarrow \\mathbb{R}^p \\) și \\( a \\in A' \\), \\( \\ell \\in \\mathbb{R}^p \\). Atunci \\( f \\) are limită \\( \\ell \\) în punctul \\( a \\) dacă și numai dacă\n\\[\n\\forall (x_n) \\subset A \\setminus \\{a\\}, x_n \\to a \\text{ implică } f(x_n) \\to \\ell. \\tag{1.2}\n\\]\n\nCorolarul 1.4 Dacă există \\((x_n), (u_n) \\subset A \\setminus \\{a\\} \\) astfel încât \\( x_n \\to a, u_n \\to a \\), iar \\( f(x_n) \\to \\ell_1, f(u_n) \\to \\ell_2 \\), cu \\( \\ell_1 \\neq \\ell_2 \\), atunci nu există limită funcției \\( f \\) în punctul \\( a \\).\n\nExercițiul 1.5 Să se arate că \\( \\lim_{x \\to 3} \\sqrt{x+1} = 2 \\).\n\nSoluție. Să observăm că \\( |\\sqrt{x+1} - 2| < \\varepsilon \\iff 2 - \\varepsilon < \\sqrt{x+1} < 2 + \\varepsilon \\). Dacă \\( \\varepsilon \\in (0, 2) \\), alegem \\( \\delta := \\varepsilon(4 - \\varepsilon) > 0 \\). Atunci\n\\[\n|x - 3| < \\delta \\Rightarrow -\\varepsilon(4 - \\varepsilon) < x - 3 \\Rightarrow (2 - \\varepsilon)^2 < x + 1 \\Rightarrow 2 - \\varepsilon < \\sqrt{x + 1},\n\\]\n\\[\n|x - 3| < \\delta \\Rightarrow x - 3 < \\varepsilon(4 - \\varepsilon) < \\varepsilon(4 + \\varepsilon) \\Rightarrow x + 1 < (2 + \\varepsilon)^2\n\\]\n\\[\n\\Rightarrow 0 < x + 1 < 2 + \\varepsilon.\n\\]\nPentru $\\varepsilon > 2$, alegem $\\delta := 4 > 0$. Atunci\n\n$$|x - 3| < \\delta \\Rightarrow -4 < x - 3 \\Rightarrow 0 < x + 1 \\Rightarrow 2 - \\varepsilon < 0 < \\sqrt{x + 1},$$\n\n$$|x - 3| < \\delta \\Rightarrow x - 3 < 4 < \\varepsilon(4 + \\varepsilon) \\Rightarrow x + 1 < (2 + \\varepsilon)^2 \\Rightarrow 0 < \\frac{\\sqrt{x + 1}}{2} < 2 + \\varepsilon.$$\n\nCu alte cuvinte, pentru orice $\\varepsilon > 0$, am găsit $\\delta > 0$ astfel încât, dacă $x \\in \\mathbb{R} \\setminus \\{3\\}$, cu $|x - 3| < \\delta$, avem $|\\sqrt{x + 1} - 2| < \\varepsilon$. Folosind caracterizarea $\\varepsilon - \\delta$, rezultă afirmația dorită.\n\n**Exercițiul 1.6** Să se arate că funcția $f : \\mathbb{R}^2 \\setminus \\{(0, 0)\\} \\to \\mathbb{R}$, $f(x, y) := \\frac{x^2y^2}{x^2 + y^2}$ are limita 0 în punctul $(0, 0)$.\n\n**Soluție.** Observăm mai întâi că $\\left| \\frac{xy}{x^2 + y^2} \\right| \\leq \\frac{1}{2}$ pentru orice $(x, y) \\neq (0, 0)$.\n\nÎntr-adevăr, pentru orice $(x, y) \\neq (0, 0)$,\n\n$$\\left| \\frac{xy}{x^2 + y^2} \\right| \\leq \\frac{1}{2} \\iff -(x^2 + y^2) \\leq 2xy \\leq (x^2 + y^2) \\iff \\begin{cases} (x - y)^2 \\geq 0 \\\\ (x + y)^2 \\geq 0. \\end{cases}$$\n\nDeducem de aici că\n\n$$0 \\leq |f(x, y)| = |xy| \\cdot \\left| \\frac{xy}{x^2 + y^2} \\right| \\leq \\frac{|xy|}{2}.$$\n\nConsiderăm $\\varepsilon > 0$ arbitrar și definim $\\delta := \\sqrt{\\varepsilon}$. Dacă $(x, y) \\in \\mathbb{R}^2 \\setminus \\{(0, 0)\\}$ are proprietatea că $||(x, y) - (0, 0)||_2 < \\delta$, rezultă $\\max\\{|x|, |y|\\} \\leq \\delta = \\sqrt{\\varepsilon}$, deci\n\n$$|f(x, y) - 0| \\leq \\frac{|xy|}{2} < \\frac{\\varepsilon}{2} < \\varepsilon.$$\n\nFolosind caracterizarea $\\varepsilon - \\delta$, rezultă concluzia.\n\n**Exercițiul 1.7** Să se arate că nu există limita funcției $f : \\mathbb{R} \\setminus \\{0\\} \\to \\mathbb{R}$, $f(x) = \\sin \\frac{1}{x}$ în punctul 0.\n\n**Soluție.** Să considerăm șirurile $(x_n), (u_n)$ date prin $x_n := \\frac{1}{n\\pi}$, $u_n := \\frac{2}{(4n + 1)\\pi}$ pentru orice $n \\in \\mathbb{N}^*$ și să observăm că $x_n \\to 0, u_n \\to 0$.\n\nÎnsă $f(x_n) = \\sin (n\\pi) = 0 \\to 0$, iar $f(u_n) = \\sin \\left(2n\\pi + \\frac{\\pi}{2}\\right) = 1 \\to 1$.\n\nAplicând Corolarul 1.4, obținem că nu există limita funcției $f$ în punctul 0.\n\n**Exercițiul 1.8** Să se arate că nu există limita funcției $f : \\mathbb{R} \\setminus \\{(0, 0)\\} \\to \\mathbb{R}$, $f(x, y) = \\frac{xy}{x^2 + y^2}$ în punctul $(0, 0)$.\nSoluție. Considerăm șirurile \\((u_n), (v_n) \\subset \\mathbb{R} \\setminus \\{(0,0)\\}, ambele convergente la \\((0,0)\\), date prin \\(u_n := \\left(\\frac{1}{n}, \\frac{1}{n}\\right), v_n := \\left(\\frac{1}{n}, \\frac{2}{n}\\right), \\forall n \\in \\mathbb{N}^*\\). Atunci\n\n\\[\n\\begin{align*}\n f(u_n) &= \\frac{1}{n^2} + \\frac{1}{n^2} = \\frac{2}{2} \\to \\frac{1}{2} \\\\\n f(v_n) &= \\frac{1}{n^2} + \\frac{2}{n^2} = \\frac{2}{5} \\to \\frac{2}{5}.\n\\end{align*}\n\\]\n\nAplicăm Corolarul 1.4 și deducem că nu există limita funcției \\(f\\) în punctul \\((0,0)\\).\n\nSă observăm în continuare că funcția \\(f : A \\subset \\mathbb{R}^k \\to \\mathbb{R}^p, k \\geq 1, p > 1\\) poate fi gândită ca fiind echivalentă cu \\(p\\) funcții cu valori reale. Într-adevăr, având date funcția \\(f\\) și \\(x \\in A\\), dacă notăm\n\n\\[\nf(x) = y = (y_1, y_2, ..., y_p) \\in \\mathbb{R}^p,\n\\]\n\nputem defini în punctul \\(x\\) funcțiile \\(f_i, i \\in \\overline{1,p}\\), prin \\(f_i(x) := y_i\\). Construim așadar funcțiile \\(f_i : A \\subset \\mathbb{R}^k \\to \\mathbb{R}, i \\in \\overline{1,p}\\) astfel încât\n\n\\[\nf(x) = (f_1(x), f_2(x), ..., f_p(x)), \\quad \\forall x \\in A. \\quad (1.3)\n\\]\n\nInvers, considerând un sistem format din \\(p\\) funcții cu valori reale \\(f_i : A \\subset \\mathbb{R}^k \\to \\mathbb{R}, i \\in \\overline{1,p}\\), putem defini funcția \\(f : A \\subset \\mathbb{R}^k \\to \\mathbb{R}^p\\) prin relația (1.3).\n\nDacă avem \\(k, p > 1\\), funcția \\(f\\) se numește funcție vectorială de argument vectorial, iar funcțiile \\(f_i\\) sunt numite funcțiile componente, sau funcțiile coordonate ale funcției \\(f\\), și scriem \\(f = (f_1, f_2, ..., f_p)\\). În cazul \\(k = 1, p > 1\\), funcția \\(f\\) se numește funcție vectorială de argument real, iar dacă \\(k > 1, p = 1\\), funcția \\(f\\) se numește funcție reală de argument vectorial. Dacă \\(k = p = 1\\), funcția \\(f\\) se numește funcție reală de argument real.\n\nAre loc rezultatul:\n\n**Teorema 1.9** Fie funcția \\(f = (f_1, f_2, ..., f_p) : A \\subset \\mathbb{R}^k \\to \\mathbb{R}^p\\) și \\(a \\in A'\\). Atunci \\(f\\) are limita \\(\\ell = (\\ell_1, \\ell_2, ..., \\ell_p) \\in \\mathbb{R}^p\\) în punctul \\(a\\) dacă și numai dacă există simultan \\(\\lim_{x \\to a} f_i(x) = \\ell_i, i = \\overline{1,p}\\).\n\nTeorema de mai sus permite reducerea studiului limitelor funcțiilor vectoriale de argument vectorial \\(f : A \\subset \\mathbb{R}^k \\to \\mathbb{R}^p\\) la studiul limitelor funcțiilor componente \\(f : A \\subset \\mathbb{R}^k \\to \\mathbb{R}\\).\n1.1.1 Limită după o direcție. Limită parțială\n\nFie o funcție \\( f : A \\subset \\mathbb{R}^k \\to \\mathbb{R}^p, a = (a_1, a_2, ..., a_k), v = (v_1, v_2, ..., v_k) \\in \\mathbb{R}^k \\).\n\nDefiniția 1.10 Spunem că funcția \\( f \\) are limită în direcția \\( v \\) în punctul \\( a \\) dacă mulțimea \\( B := \\{ t \\in \\mathbb{R} \\mid a + tv \\in A \\} \\) este nevidă, \\( 0 \\in B' \\) și există \\( \\ell \\in \\mathbb{R}^p \\) astfel încât\n\n\\[\n\\lim_{t \\to 0} f(a + tv) = \\lim_{t \\to 0} f(a_1 + tv_1, a_2 + tv_2, ..., a_k + tv_k) = \\ell.\n\\]\n\nÎn acest caz, elementul \\( \\ell \\) se numește limită în direcția \\( v \\) a funcției \\( f \\) în punctul \\( a \\).\n\nÎn cazul particular \\( v := e_i = (0, ..., 0, 1, 0, ..., 0) \\), vom spune că funcția \\( f \\) are limită parțială în raport cu variabila \\( x_i \\) în punctul \\( a \\), iar \\( \\ell \\) se numește limită parțială în raport cu variabila \\( x_i \\) a funcției \\( f \\) în punctul \\( a \\). Așadar,\n\n\\[\n\\lim_{t \\to 0} f(a + te_i) = \\lim_{t \\to 0} f(a_1, ..., a_{i-1}, a_i + t, a_{i+1}, ..., a_k) = \\ell.\n\\]\n\nLegătura între limita unei funcții și limita după o direcție este dată în următoarea teoremă.\n\nTeorema 1.11 Dacă există \\( \\lim_{x \\to a} f(x) = \\ell \\in \\mathbb{R}^p \\), iar \\( v \\in \\mathbb{R}^k \\) este un vector astfel încât mulțimea \\( D = A \\cap \\{ a + tv \\mid t \\in \\mathbb{R} \\} \\) este nevidă și \\( a \\in D' \\), atunci există limita în direcția \\( v \\) a funcției \\( f \\) în punctul \\( a \\) și este egală cu \\( \\ell \\).\n\nObservația 1.12 Din teorema anterioară rezultă că, dacă obținem pentru o direcție \\( v \\in \\mathbb{R}^k \\) că limita direcțională \\( \\lim_{t \\to 0} f(a + tv) \\) nu există, sau depinde de direcția \\( v \\), atunci nu va exista \\( \\lim_{x \\to a} f(x) \\). În ultimul caz, această metodă de a arăta că nu există limita se mai numește metoda direcțiilor variabile.\n\nReciproca teoremei anterioare nu are loc, după cum o arată următorul exemplu.\n\nExercițiul 1.13 Fie \\( f : \\mathbb{R}^2 \\setminus \\{(0,0)\\} \\to \\mathbb{R}, f(x, y) = \\frac{xy}{x^2 + y^2} \\). Să se arate că \\( f \\) are limită în orice direcție în punctul \\( (0,0) \\), dar limita funcției \\( f \\) (în raport cu ansamblul variabilelor) nu există.\n\nSoluție. Fie \\( v = (\\cos \\alpha, \\sin \\alpha), \\alpha \\in [0, 2\\pi) \\). Atunci\n\n\\[\n\\lim_{t \\to 0} f((0,0) + t(\\cos \\alpha, \\sin \\alpha)) = \\lim_{t \\to 0} f(t \\cos \\alpha, t \\sin \\alpha)\n\\]\n\n\\[\n= \\lim_{t \\to 0} \\frac{t^2 \\sin \\alpha \\cos \\alpha}{t^2 \\cos^2 \\alpha + t^2 \\sin^2 \\alpha} = \\sin \\alpha \\cos \\alpha.\n\\]\n\nAșadar, funcția \\( f \\) are limită în punctul \\( (0,0) \\) în orice direcție \\((\\cos \\alpha, \\sin \\alpha)\\). De asemenea, pentru \\( \\alpha = 0 \\) și \\( \\alpha = \\frac{\\pi}{2} \\), ne va rezulta că există limitele pațiale în raport cu \\( x \\) și cu \\( y \\) în \\( (0,0) \\), ambele egale cu \\( 0 \\). După cum am văzut însă într-un exercițiu anterior, nu există limita funcției \\( f \\) în punctul \\( (0,0) \\).\nLimita direcțională poate oferi uneori informații suplimentare utile despre limita unei funcții într-un punct, după cum se va vedea din exercițiile următoare.\n\n**Exercițiul 1.14** Să se arate funcția \\( f : \\mathbb{R}^2 \\setminus \\{(0,0)\\} \\to \\mathbb{R}, \\)\n\\[\nf(x, y) = \\frac{x^3y^2 \\sin y + x^2y^3 \\sin x}{x^4 + y^4}\n\\]\nare limita nulă în punctul \\((0,0)\\).\n\n**Soluție.** Să observăm că, pentru orice direcție \\((h_1, h_2) \\in \\mathbb{R}^2, \\) avem\n\\[\n\\lim_{t \\to 0} f((0,0) + t(h_1, h_2)) = \\lim_{t \\to 0} \\frac{h_1^3h_2^2 \\sin(th_2) + h_1^2h_2^3 \\sin(th_1)}{h_1^4 + h_2^4} = 0.\n\\]\nRezultă că, dacă există, limita funcției în punctul \\((0,0)\\) va fi egală cu 0.\nFolosind că\n\\[\n(x^2 - y^2)^2 \\geq 0 \\iff x^4 + y^4 \\geq 2x^2y^2,\n\\]\nobținem\n\\[\n|f(x, y) - 0| \\leq \\frac{|x|^3y^2 + x^2|y|^3}{x^4 + y^4} \\leq \\frac{1}{2} (|x| + |y|).\n\\]\nCum limita în punctul \\((0,0)\\) a funcției din dreapta este 0 în mod evident, ne va rezulta (folosind, de exemplu, criteriul de tip \\(\\varepsilon - \\delta\\)), concluzia.\n\n**Exercițiul 1.15** Să se arate că nu există limita funcției\n\\[\nf : \\mathbb{R}^3 \\setminus \\{(x, y, z) \\mid x \\neq y\\} \\to \\mathbb{R}, \\quad f(x, y, z) = \\frac{y^2 - z^2}{x - y},\n\\]\nin punctul \\((0,0,0)\\).\n\n**Soluție.** Cum\n\\[\n\\lim_{t \\to 0} f((0,0,0) + t(h_1, h_2, h_3)) = \\lim_{t \\to 0} \\frac{h_2^2 - h_3^2}{h_1 - h_2} = 0,\n\\]\nrezultă că, dacă există, limita funcției în punctul \\((0,0,0)\\) va fi egală cu 0.\nConsiderând însă sirul\n\\[\n\\left(\\frac{1}{n} + \\frac{1}{\\sqrt{n}}, \\frac{1}{\\sqrt{n}}, 0\\right)_{n \\in \\mathbb{N}}\n\\]\nconvergent la \\((0,0,0)\\) în \\(\\mathbb{R}^3, \\) vom avea că\n\\[\n\\lim_{n \\to \\infty} f\\left(\\frac{1}{n} + \\frac{1}{\\sqrt{n}}, \\frac{1}{\\sqrt{n}}, 0\\right) = 1,\n\\]\nceea ce arată că nu există limita funcției în punctul \\((0,0,0)\\).\n1.2 Continuitate\n\nDefiniția 1.16 Spunem că funcția \\( f : A \\subset \\mathbb{R}^k \\rightarrow \\mathbb{R}^p \\) este continuă în punctul \\( a \\in A \\) dacă\n\n\\[\n\\forall \\varepsilon > 0, \\exists \\delta > 0, \\forall x \\in A, \\|x - a\\|_{\\mathbb{R}^k} < \\delta : \\|f(x) - f(a)\\|_{\\mathbb{R}^p} < \\varepsilon. \\tag{1.4}\n\\]\n\nDacă funcția \\( f \\) nu este continuă în punctul \\( a \\in A \\), vom spune că \\( f \\) este discontinuu în punctul \\( a \\), sau că \\( a \\) este un punct de discontinuitate pentru funcția \\( f \\).\n\nVom spune că funcția \\( f \\) este continuă pe o mulțime \\( B \\subset A \\) dacă \\( f \\) este continuă în orice punct \\( x \\in B \\).\n\nObservația 1.17 Remarcăm mai întâi faptul că noțiunea de continuitate, spre deosebire de cea de limită, nu are sens decât pentru punctele mulțimii \\( A \\), domeniul de definiție al funcției \\( f \\).\n\nDe asemenea, să observăm că, dacă \\( a \\) este un punct izolat al mulțimii \\( A \\), atunci funcția \\( f \\) este continuă în \\( a \\).\n\nAșadar, problema continuității se va pune doar în punctele de acumulare ale mulțimii \\( A \\). Examinând definiția de mai sus, observăm similitudinea cu definiția limitei unei funcții într-un punct, ceea ce ne permite enunțarea următoarei teoreme de caracterizare, a cărei demonstrație evidentă o omitem.\n\nTeorema 1.18 (Caracterizare a continuității cu limită) Fie \\( f : A \\subset \\mathbb{R}^k \\rightarrow \\mathbb{R}^p \\) și \\( a \\in A' \\cap A \\). Atunci \\( f \\) este continuă în \\( a \\) dacă și numai dacă există \\( \\lim_{x \\to a} f(x) = f(a) \\).\n\nTeorema 1.19 (Caracterizarea cu șiruri a continuității) Fie \\( f : A \\subset \\mathbb{R}^k \\rightarrow \\mathbb{R}^p \\) și \\( a \\in A \\). Atunci \\( f \\) este continuă în punctul \\( a \\) dacă și numai dacă\n\n\\[\n\\forall (x_n) \\subset A, x_n \\to a \\text{ implică } f(x_n) \\to f(a). \\tag{1.5}\n\\]\n\nExercițiul 1.20 Să se arate că funcția:\n\n\\[\nf(x, y) = \\begin{cases} \n x^3 + y^3, & \\text{dacă } (x, y) \\neq (0, 0) \\\\\n x^2 + y^2, & \\text{dacă } (x, y) = (0, 0)\n\\end{cases}\n\\]\n\neste continuă pe \\( \\mathbb{R}^2 \\).\n\nSoluție. În orice punct \\( (a, b) \\neq (0, 0) \\), funcția este continuă, așa cum rezultă cu ușurință din definiția cu șiruri.\n\nPentru a dovedi continuitatea în \\( (0, 0) \\), observăm că:\n\n\\[\n|x^3| \\leq (x^2 + y^2)^{\\frac{3}{2}} \\text{ și } |y^3| \\leq (x^2 + y^2)^{\\frac{3}{2}},\n\\]\n\nde unde rezultă:\n\n\\[\n|f(x, y) - f(0, 0)| \\leq \\frac{2(x^2 + y^2)^{\\frac{3}{2}}}{x^2 + y^2} = 2\\sqrt{x^2 + y^2}, \\quad \\forall (x, y) \\in \\mathbb{R}^2 \\setminus \\{(0, 0)\\}.\n\\]\nDacă \\((x_n, y_n) \\to (0, 0)\\), atunci \\(\\sqrt{x_n^2 + y_n^2} \\to 0\\), deci:\n\n\\[\n\\lim_{(x,y) \\to (0,0)} f(x, y) = 0 = f(0,0).\n\\]\n\nAșadar, \\(f\\) este continuă și în \\((0,0)\\).\n\nUrmătoarea teoremă permite caracterizarea continuății unei funcții vectoriale de variabilă vectorială prin intermediul proprietății similare a funcțiilor componente.\n\n**Teorema 1.21** Fie \\(f = (f_1, f_2, \\ldots, f_p) : A \\subset \\mathbb{R}^k \\to \\mathbb{R}^p, k \\geq 1, p > 1\\) și \\(a \\in A\\). Atunci \\(f\\) este continuă în punctul \\(a\\) dacă și numai dacă funcțiile \\(f_1, f_2, \\ldots, f_p : A \\subset \\mathbb{R}^k \\to \\mathbb{R}\\) sunt continue în \\(a\\).", "id": "./materials/271.pdf" }, { "contents": "Evaluate $\\iiint_E (x + y) \\, dV$ with\n\n$$E = \\{(x, y, z) \\in \\mathbb{R}^3 : x \\geq 0 \\land y \\geq 0 \\land x^2 + y^2 \\leq 1 \\land 0 \\leq z \\leq 4 - \\sqrt{x^2 + y^2}\\}.$$\n\n- Let’s first sketch $E$ over $xyz$-coordinate planes:\n\n![3D sketch of E](image)\n\nFigure 1: 3D sketch of $E$\n\n- Tip: change to cylindrical coordinates\n\n$$\\begin{cases} x = r \\cos(\\theta) \\\\ y = r \\sin(\\theta) \\\\ z = z \\end{cases}$$\n• This means that we can define\n\n\\[ E = \\{(r, \\theta, z) | 0 \\leq r \\leq 1 \\land 0 \\leq \\theta \\leq \\frac{\\pi}{2} \\land 0 \\leq z \\leq 4 - r\\} \\]\n\n• Thus, we are able to write the triple integral as\n\n\\[\n\\int_0^{\\frac{\\pi}{2}} \\int_0^1 \\int_0^{4-r} (r \\cos(\\theta) + r \\sin(\\theta)) r \\, dz \\, dr \\, d\\theta \\\\\n= \\int_0^{\\frac{\\pi}{2}} \\int_0^1 r^2 (\\cos(\\theta) + \\sin(\\theta)) \\left[ z \\right]_{z=0}^{z=4-r} \\, dr \\, d\\theta \\\\\n= \\int_0^{\\frac{\\pi}{2}} \\int_0^1 r^2 (4 - r)(\\cos(\\theta) + \\sin(\\theta)) \\, dr \\, d\\theta \\\\\n= \\int_0^{\\frac{\\pi}{2}} (\\cos(\\theta) + \\sin(\\theta)) \\left[ \\frac{4r^3}{3} - \\frac{r^4}{4} \\right]_{r=0}^{r=1} \\, d\\theta \\\\\n= \\int_0^{\\frac{\\pi}{2}} (\\cos(\\theta) + \\sin(\\theta)) \\left( \\frac{4}{3} - \\frac{1}{4} \\right) \\, d\\theta \\\\\n= \\frac{13}{12} \\left[ \\sin(\\theta) - \\cos(\\theta) \\right]_{\\theta=0}^{\\theta=\\frac{\\pi}{2}} \\\\\n= \\frac{13}{12} (1 - 0 - 0 + 1) \\\\\n= \\frac{13}{6}\n\\]", "id": "./materials/280.pdf" }, { "contents": "Evaluate $\\iiint_E 1 \\, dV$ where $E$ is the region from the first octant between $y + z = 4$ and $y = x^2$.\n\n- Let’s first sketch the solid $E$ over $xyz$-coordinate planes:\n\n![3D sketch of E](image)\n\nFigure 1: 3D sketch of $E$\n\n- Now, let’s sketch the projection of $E$ over the $xz$-plane:\n\n$$y = x^2 \\quad \\land \\quad y + z = 4 \\quad \\Rightarrow \\quad z = 4 - x^2$$\nYou could have chosen to use the projection over the xy-plane instead, for instance.\n\nSo, using the projection over xz-plane, we can define\n\n\\[ E = \\{(x, y, z) \\in \\mathbb{R}^3 : 0 \\leq x \\leq 2 \\land 0 \\leq z \\leq 4 - x^2 \\land x^2 \\leq y \\leq 4 - z\\} \\]\n\nand, therefore, write the triple integral as:\n\n\\[\n\\iiint_E 1 \\, dV = \\int_0^2 \\int_0^{4-x^2} \\int_{x^2}^{4-z} 1 \\, dy \\, dz \\, dx\n\\]\n\nNow is just to solve it.\n\nAt the end you should get: \\( \\frac{128}{15} \\)", "id": "./materials/281.pdf" }, { "contents": "Evaluate $\\iiint_E z \\, dV$ where $E$ is the region from the first octant bounded by $x + y = 2$, $x + 2y = 2$ and $y^2 + z^2 = 4$.\n\n- Let’s first sketch the solid $E$ over $xyz$-coordinate planes:\n\n![3D sketch of E](image)\n\nFigure 1: 3D sketch of $E$\n\n- Now, let’s sketch the projection of $E$ over the $xy$-plane:\n\n$$x + y = 2 \\quad \\land \\quad x + 2y = 2$$\nSo, using the projection over xz-plane, we can define\n\n\\[ E = \\{(x, y, z) \\in \\mathbb{R}^3 : 0 \\leq x \\leq 2 \\land 1 - \\frac{x}{2} \\leq y \\leq 2 - x \\land 0 \\leq z \\leq \\sqrt{4 - y^2}\\} \\]\n\nand, therefore, write the triple integral as:\n\n\\[\n\\iiint_E z \\, dV = \\int_0^2 \\int_{1 - \\frac{x}{2}}^{2-x} \\int_0^{\\sqrt{4-y^2}} z \\, dz \\, dy \\, dx\n\\]\n\nNow is just to solve it.\n\nAt the end you should get: \\(\\frac{17}{12}\\)", "id": "./materials/282.pdf" }, { "contents": "Evaluate \\( \\iiint_E \\frac{1}{(1 - x^2 - y^2)^{\\frac{3}{2}}} \\, dV \\) with\n\\[\nE = \\{(x, y, z) \\in \\mathbb{R}^3 : x^2 + y^2 \\leq z \\leq 2 - x^2 - y^2\\}.\n\\]\n\n- Let’s first sketch the solid \\( E \\) over xyz-coordinate planes:\n\n![3D sketch of E](image)\n\nFigure 1: 3D sketch of E\n\n- The projection (D) of \\( E \\) over the xy-plane:\n\n\\[\nx^2 + y^2 = 2 - x^2 - y^2 \\quad \\Rightarrow \\quad x^2 + y^2 = 1\n\\]\n• So, using the projection over xy-plane, we can write the triple integral\n\n\\[\n\\iiint_E \\frac{1}{(1 - x^2 - y^2)^{\\frac{3}{2}}} \\, dV\n= \\iiint_D \\left[ \\int_{x^2+y^2}^{2-x^2-y^2} \\frac{1}{(1 - x^2 - y^2)^{\\frac{3}{2}}} \\, dz \\right] \\, dA\n= \\iiint_D \\frac{1}{(1 - x^2 - y^2)^{\\frac{3}{2}}} \\left[ z \\right]_{z=x^2+y^2}^{z=2-x^2-y^2} \\, dA\n= \\iiint_D \\frac{1}{(1 - x^2 - y^2)^{\\frac{3}{2}}} (2 - 2x^2 - 2y^2) \\, dA\n\\]\n\n• Tip: change to cylindrical coordinates:\n\n\\[\n\\begin{align*}\nx &= r \\cos(\\theta) \\\\\ny &= r \\sin(\\theta) \\\\\nz &= z\n\\end{align*}\n\\]\nThus, by substitution, we have\n\n\\[\n\\int_0^{2\\pi} \\int_0^1 \\frac{1}{(1 - r^2)^{\\frac{3}{2}}} (2 - 2r^2 \\cos^2(\\theta) - 2r^2 \\sin^2(\\theta)) r \\, dr \\, d\\theta\n\\]\n\n\\[\n= \\int_0^{2\\pi} \\int_0^1 r \\frac{2 - 2r^2}{(1 - r^2)^{\\frac{3}{2}}} \\, dr \\, d\\theta\n\\]\n\n\\[\n= 2 \\int_0^{2\\pi} \\int_0^1 r \\frac{1 - r^2}{(1 - r^2)^{\\frac{3}{2}}} \\, dr \\, d\\theta\n\\]\n\n\\[\n= 2 \\int_0^{2\\pi} \\int_0^1 r (1 - r^2)^{-\\frac{1}{2}} \\, dr \\, d\\theta\n\\]\n\n\\[\n= -2 \\int_0^{2\\pi} \\int_0^1 -\\frac{2}{2} r (1 - r^2)^{-\\frac{1}{2}} \\, dr \\, d\\theta\n\\]\n\n\\[\n= -2 \\int_0^{2\\pi} \\left[ (1 - r^2)^{\\frac{1}{2}} \\right]_{r=0}^{r=1} \\, d\\theta\n\\]\n\n\\[\n= -2 \\int_0^{2\\pi} (0 - 1) \\, d\\theta\n\\]\n\n\\[\n= 2 \\left[ \\theta \\right]_{\\theta=0}^{\\theta=2\\pi}\n\\]\n\n\\[\n= 4\\pi\n\\]", "id": "./materials/283.pdf" }, { "contents": "Evaluate \\( \\iiint_E \\frac{1}{(x^2 + y^2 + z^2)^{\\frac{3}{2}}} \\, dV \\) with\n\\[\nE = \\{(x, y, z) \\in \\mathbb{R}^3 : 4 \\leq x^2 + y^2 + z^2 \\leq 9\\}.\n\\]\n\n- Let’s first sketch the solid \\( E \\) over xyz-coordinate planes:\n\n![3D sketch of E](image)\n\nFigure 1: 3D sketch of E\n\n- The projection (D) over the xy-plane can be defined as\n\\[\n4 \\leq x^2 + y^2 \\leq 9\n\\]\n\n- Tip: change to spherical coordinates when defining the triple integral\n\\[\n\\begin{align*}\nx &= r \\cos(\\theta) \\sin(\\phi) \\\\\ny &= r \\sin(\\theta) \\sin(\\phi) \\\\\nz &= r \\cos(\\phi)\n\\end{align*}\n\\]\nSo, using the projection over xy-plane, we can write the triple integral\n\n\\[\n\\iiint_E \\frac{1}{(x^2 + y^2 + z^2)^{\\frac{3}{2}}} \\, dV\n\\]\n\n\\[\n= \\int_2^3 \\int_0^{2\\pi} \\int_0^\\pi \\frac{1}{r^2 \\sin(\\phi)} \\, d\\phi \\, d\\theta \\, dr\n\\]\n\n\\[\n= \\int_2^3 \\int_0^{2\\pi} \\int_0^\\pi \\frac{1}{r^2 \\sin(\\phi)} \\, d\\phi \\, d\\theta \\, dr\n\\]\n\n\\[\n= \\int_2^3 \\int_0^{2\\pi} \\int_0^\\pi \\frac{\\sin(\\phi)}{r^2 \\sin(\\phi)} \\, d\\phi \\, d\\theta \\, dr\n\\]\n\n\\[\n= \\int_2^3 \\int_0^{2\\pi} \\int_0^\\pi \\frac{\\sin(\\phi)}{r^2 \\sin(\\phi)} \\, d\\phi \\, d\\theta \\, dr\n\\]\n\n\\[\n= \\int_2^3 \\int_0^{2\\pi} \\int_0^\\pi \\frac{1}{r} \\left[ -\\cos(\\phi) \\right]_{\\phi=0}^{\\phi=\\pi} \\, d\\theta \\, dr\n\\]\n\n\\[\n= \\int_2^3 \\int_0^{2\\pi} \\frac{(1 + 1)}{r} \\, d\\theta \\, dr\n\\]\n\n\\[\n= 2 \\int_2^3 \\frac{1}{r} \\left[ \\theta \\right]_{\\theta=0}^{\\theta=2\\pi} \\, dr\n\\]\n\n\\[\n= 2(2\\pi - 0) \\left[ \\ln(r) \\right]_{r=2}^{r=3}\n\\]\n\n\\[\n= 4\\pi (\\ln(3) - \\ln(2))\n\\]\n\n\\[\n= 4\\pi \\ln \\left( \\frac{3}{2} \\right)\n\\]", "id": "./materials/284.pdf" }, { "contents": "Evaluate $\\iiint_E z \\, dV$ with\n\n$$E = \\{(x, y, z) \\in \\mathbb{R}^3 : x^2 + y^2 \\geq z^2 \\land x^2 + y^2 + z^2 \\leq -2z\\}.$$\n\n- Let’s first sketch the solid $E$ over $xyz$-coordinate planes:\n\n![3D sketch of E](image)\n\nFigure 1: 3D sketch of $E$\n\n- We can evaluate the $z$-coordinates for the intersection of the two solids:\n\n$$x^2 + y^2 + z^2 = x^2 + y^2 + z^2 + 2z$$\n\n$$\\Rightarrow 0 = 2z^2 + 2z$$\n\n$$\\Rightarrow 0 = z^2 + z$$\n\n$$\\Rightarrow z = 0 \\lor z = -1$$\n• Tip: Use spherical coordinates when defining the triple integral\n\n\\[\n\\begin{align*}\n x &= r \\cos(\\theta) \\sin(\\phi) \\\\\n y &= r \\sin(\\theta) \\sin(\\phi) \\\\\n z &= r \\cos(\\phi)\n\\end{align*}\n\\]\n\n• Let’s study the behaviour of \\( \\phi \\):\n\n\\[\n(x = 0 \\quad \\land \\quad z = -1) \\quad \\Rightarrow \\quad y^2 = 1 \\quad \\Rightarrow \\quad (y = \\pm 1)\n\\]\n\n\\[\n\\tan(\\pi - \\phi) = \\frac{1}{1} \\quad \\Rightarrow \\quad \\pi - \\phi = \\arctan(1) \\quad \\Rightarrow \\quad \\phi = \\frac{3\\pi}{4}\n\\]\n\n• Let’s study the behaviour of \\( r \\):\n\n\\[\nx^2 + y^2 + z^2 \\leq -2z\n\\]\n\n\\[\n\\Rightarrow x^2 + y^2 + (z + 1)^2 \\leq 1\n\\]\n\n\\[\n\\Rightarrow (r \\cos(\\theta) \\sin(\\phi))^2 + (r \\sin(\\theta) \\sin(\\phi))^2 + (r \\cos(\\phi) + 1)^2 \\leq 1\n\\]\n\n\\[\n\\Rightarrow r^2 \\cos^2(\\theta) \\sin^2(\\phi) + r^2 \\sin^2(\\theta) \\sin^2(\\phi) + r^2 \\cos^2(\\phi) + 2r \\cos(\\phi) + 1 \\leq 1\n\\]\n\n\\[\n\\Rightarrow r \\leq -2 \\cos(\\phi)\n\\]\n\n• So we can write the triple integral as\n\n\\[\n\\int \\int \\int_E z \\, dV\n\\]\n\n\\[\n= \\int_0^{2\\pi} \\int_{\\frac{3\\pi}{4}}^{\\frac{\\pi}{2}} \\int_0^{-2 \\cos(\\phi)} r \\cos(\\phi) r^2 \\sin(\\phi) \\, dr \\, d\\phi \\, d\\theta\n\\]\n\n\\[\n= \\int_0^{2\\pi} \\int_{\\frac{3\\pi}{4}}^{\\frac{\\pi}{2}} \\int_0^{-2 \\cos(\\phi)} r^3 \\cos(\\phi) \\sin(\\phi) \\, dr \\, d\\phi \\, d\\theta\n\\]\n\n• Now is just to solve the triple integral.\n\n• In the end, the result should be: \\(-\\frac{\\pi}{6}\\)", "id": "./materials/285.pdf" }, { "contents": "Evaluate $\\iiint_E y \\, dV$ with\n\n$$E = \\{(x, y, z) \\in \\mathbb{R}^3 : x^2 + y^2 + z^2 \\leq 1 \\land z \\geq -\\sqrt{x^2 + y^2}\\}$$\n\n- Let’s first sketch the solid $E$ over $xyz$-coordinate planes:\n\n![3D sketch of E](image)\n\nFigure 1: 3D sketch of $E$\n\n- We can evaluate the $z$-coordinates for the intersection of the two solids:\n\n$$x^2 + y^2 + z^2 - 1 = x^2 + y^2 - z^2$$\n\n$$\\Rightarrow z^2 = \\frac{1}{2}$$\n\n$$\\Rightarrow z = -\\frac{\\sqrt{2}}{2}$$\n• Tip: Use spherical coordinates when defining the triple integral\n\n\\[\n\\begin{align*}\n x &= r \\cos(\\theta) \\sin(\\phi) \\\\\n y &= r \\sin(\\theta) \\sin(\\phi) \\\\\n z &= r \\cos(\\phi)\n\\end{align*}\n\\]\n\n• Let’s study the behaviour of \\( \\phi \\):\n\n\\[\n\\tan(\\pi - \\phi) = \\frac{\\sqrt{2}}{\\sqrt{2}} \\quad \\Rightarrow \\quad \\pi - \\phi = \\arctan(1) \\quad \\Rightarrow \\quad \\phi = \\frac{3\\pi}{4}\n\\]\n\n• So we can write the triple integral as\n\n\\[\n\\iiint_E y \\, dV = \\int_0^{\\frac{3\\pi}{4}} \\int_0^{2\\pi} \\int_0^1 r \\sin(\\theta) \\sin(\\phi) r^2 \\sin(\\phi) \\, dr \\, d\\theta \\, d\\phi\n\\]\n\n\\[\n= \\int_0^{\\frac{3\\pi}{4}} \\int_0^{2\\pi} \\int_0^1 r^3 \\sin(\\theta) \\sin(\\phi) \\sin(\\phi) \\, dr \\, d\\theta \\, d\\phi\n\\]\n\n• Now is just to solve the triple integral.\n\n• In the end, the result should be: 0", "id": "./materials/286.pdf" }, { "contents": "Evaluate $\\iiint_E \\sqrt{x^2 + y^2 + z^2} \\, dV$ with\n\n$E = \\{(x, y, z) \\in \\mathbb{R}^3 : x^2 + y^2 + z^2 \\leq 1 \\land x^2 + y^2 + (z - 1)^2 \\leq 1\\}$\n\n- Let’s first sketch the solid $E$ over $xyz$-coordinate planes:\n\n![3D sketch of E](image)\n\nFigure 1: 3D sketch of $E$\n\n- Tip: Use spherical coordinates when defining the triple integral\n\n$$\\begin{align*}\n x &= r \\cos(\\theta) \\sin(\\phi) \\\\\n y &= r \\sin(\\theta) \\sin(\\phi) \\\\\n z &= r \\cos(\\phi)\n\\end{align*}$$\n\nIn this case, we can easily define\n\n$$0 \\leq \\theta \\leq 2\\pi \\quad \\land \\quad 0 \\leq \\phi \\leq \\frac{\\pi}{2}$$\n• We can evaluate the z-coordinates for the intersection of the two solids:\n\n\\[ x^2 + y^2 + z^2 - 1 = x^2 + y^2 + z^2 - 2z \\]\n\n\\[ \\Rightarrow z = \\frac{1}{2} \\]\n\n• Let’s study the behaviour of r:\n\nWe can also assess that r does not behave the same way along \\( \\phi \\)'s interval. So, we are going to divide into two sections\n\n\\[ \\left( x = 0 \\land z = \\frac{1}{2} \\right) \\Rightarrow y = \\pm \\frac{\\sqrt{3}}{2} \\]\n\n\\[ \\tan(\\phi) = \\frac{\\sqrt{3}}{2} \\Rightarrow \\phi = \\arctan(\\sqrt{3}) \\]\n\n\\[ \\Rightarrow \\phi = \\frac{\\pi}{3} \\]\n\nWhen\n\n\\[ 0 \\leq \\phi \\leq \\frac{\\pi}{3} \\Rightarrow 0 \\leq r \\leq 1 \\]\n\nBut when\n\n\\[ \\frac{\\pi}{3} \\leq \\phi \\leq \\frac{\\pi}{2} \\Rightarrow x^2 + y^2 + z^2 - 2z + 1 \\leq 1 \\]\n\n\\[ \\Rightarrow r^2((\\cos(\\theta) \\sin(\\phi))^2 + (\\sin(\\theta) \\sin(\\phi))^2) + (r \\cos(\\phi) - 1)^2 \\leq 1 \\]\n\n\\[ \\Rightarrow r^2((\\cos(\\theta) \\sin(\\phi))^2 + (\\sin(\\theta) \\sin(\\phi))^2) + r^2 \\cos^2(\\phi) - 2r \\cos(\\phi) + 1 \\leq 1 \\]\n\n\\[ \\Rightarrow r^2 - 2r \\cos(\\phi) \\leq 0 \\]\n\n\\[ \\Rightarrow r \\leq 2 \\cos(\\phi) \\]\nSo we can write the triple integral as\n\n\\[\n\\iiint_E \\sqrt{x^2 + y^2 + z^2} \\, dV\n\\]\n\n\\[\n= \\int_0^{2\\pi} \\int_0^{\\frac{\\pi}{2}} \\int_0^2 r^3 \\sin(\\phi) \\, dr \\, d\\phi \\, d\\theta + \\int_0^{2\\pi} \\int_0^{\\frac{\\pi}{2}} \\int_0^1 r^3 \\sin(\\phi) \\, dr \\, d\\phi \\, d\\theta\n\\]\n\n\\[\n= \\int_0^{2\\pi} \\int_0^{\\frac{\\pi}{2}} \\sin(\\phi) \\left[ \\frac{r^4}{4} \\right]_{r=0}^{r=2 \\cos(\\phi)} \\, d\\phi \\, d\\theta + \\int_0^{2\\pi} \\int_0^{\\frac{\\pi}{2}} \\sin(\\phi) \\left[ \\frac{r^4}{4} \\right]_{r=0}^{r=1} \\, d\\phi \\, d\\theta\n\\]\n\n\\[\n= 4 \\int_0^{2\\pi} \\int_0^{\\frac{\\pi}{2}} \\sin(\\phi) \\cos^4(\\phi) \\, d\\phi \\, d\\theta + \\frac{1}{4} \\int_0^{2\\pi} \\int_0^{\\frac{\\pi}{2}} \\sin(\\phi) \\, d\\phi \\, d\\theta\n\\]\n\n\\[\n= -\\frac{4}{5} \\int_0^{2\\pi} \\left[ \\cos^5(\\phi) \\right]_{\\phi=0}^{\\phi=\\frac{\\pi}{2}} \\, d\\theta + \\frac{1}{4} \\int_0^{2\\pi} \\left[ -\\cos(\\phi) \\right]_{\\phi=0}^{\\phi=\\frac{\\pi}{2}} \\, d\\theta\n\\]\n\n\\[\n= -\\frac{4}{5} \\int_0^{2\\pi} \\left( 0 - \\left( \\frac{1}{2} \\right)^5 \\right) \\, d\\theta + \\frac{1}{4} \\int_0^{2\\pi} \\left( - \\frac{1}{2} + 1 \\right) \\, d\\theta\n\\]\n\n\\[\n= \\frac{1}{40} \\left[ \\theta \\right]_{\\theta=0}^{\\theta=2\\pi} + \\frac{1}{8} \\left[ \\theta \\right]_{\\theta=0}^{\\theta=2\\pi}\n\\]\n\n\\[\n= \\frac{1}{40} (2\\pi) + \\frac{1}{8} (2\\pi)\n\\]\n\n\\[\n= \\frac{3\\pi}{10}\n\\]", "id": "./materials/287.pdf" }, { "contents": "Evaluate \\[ \\int_0^{2\\pi} \\int_0^1 \\int_{r^2-1}^{2-\\cos(\\theta)} 4r^2 \\sin(\\theta) \\, dz \\, dr \\, d\\theta \\]\n\n- Let’s evaluate first:\n\n\\[\n\\int_{r^2-1}^{2-\\cos(\\theta)} 4r^2 \\sin(\\theta) \\, dz\n\\]\n\n\\[\n= 4r^2 \\sin(\\theta) \\left[ z \\right]_{z=r^2-1}^{z=2-\\cos(\\theta)}\n\\]\n\n\\[\n= 4r^2 \\sin(\\theta) \\left[ 2 - \\cos(\\theta) - r^2 + 1 \\right]\n\\]\n\n\\[\n= 8r^2 \\sin(\\theta) - 4r^2 \\sin(\\theta) \\cos(\\theta) - 4r^4 \\sin(\\theta) + 4r^2 \\sin(\\theta)\n\\]\n\n\\[\n= 12r^2 \\sin(\\theta) - 4r^2 \\sin(\\theta) \\cos(\\theta) - 4r^4 \\sin(\\theta)\n\\]\n\n- The idea is now to do exactly the same for the next iterations, first in respect to \\( r \\) and finally in respect to \\( \\theta \\).\n\n- At the end you should get: 0", "id": "./materials/288.pdf" }, { "contents": "Use a triple integral to determine the volume defined by $x^2 + y^2 + z^2 \\leq r_1^2$, $r_1 > 0$.\n\n- According to the expression given, we can assess that we have a sphere of radius $r_1$, centered at (0,0,0).\n\n- This means that we can evaluate the volume of E through\n\n$$\n\\int_0^{2\\pi} \\int_0^\\pi \\int_0^{r_1} 1 \\times r^2 \\sin(\\phi) \\, dr \\, d\\phi \\, d\\theta\n$$\n\n- At the end you should get: $\\frac{4\\pi(r_1)^3}{3}$\n\n- Note that the result is the expression of the volume of a sphere, commonly used!", "id": "./materials/289.pdf" }, { "contents": "Use a triple integral to determine the volume defined of the solid bounded by $x^2 + y^2 + z^2 \\leq 8$ and $z^2 \\geq x^2 + y^2$\n\n- Let’s first sketch the solid E over xyz-coordinate planes:\n\n![3D sketch of E](image)\n\nFigure 1: 3D sketch of E\n\n- Tip: Use spherical coordinates when defining the triple integral\n\n$$\n\\begin{align*}\n x &= r \\cos(\\theta) \\sin(\\phi) \\\\\n y &= r \\sin(\\theta) \\sin(\\phi) \\\\\n z &= r \\cos(\\phi)\n\\end{align*}\n$$\n\nIn this case, we can easily define\n\n$$\n0 \\leq \\theta \\leq 2\\pi \\quad \\land \\quad 0 \\leq r \\leq \\sqrt{8}\n$$\n• We can evaluate the z-coordinates for the intersection of the two solids:\n\n\\[ x^2 + y^2 + z^2 - 8 = x^2 + y^2 - z^2 \\]\n\n\\[ \\Rightarrow 2z^2 = 8 \\]\n\n\\[ \\Rightarrow z = 2 \\quad \\lor \\quad z = -2 \\]\n\n• We can see from figure 1 that the total volume can be calculated by two times the volume of one of the equal parts. So, let’s evaluate the value of \\( \\phi \\), related to the part of the solid with positive z-coordinates.\n\n\\[ (x = 0 \\quad \\land \\quad z = 2) \\quad \\Rightarrow \\quad y = \\pm 2 \\]\n\n\\[ \\tan(\\phi) = \\frac{2}{2} \\quad \\Rightarrow \\quad \\phi = \\frac{\\pi}{4} \\]\n\n• So we can write evaluate the volume of the solid through\n\n\\[\n\\iiint_E 1 \\, dV = 2 \\int_0^{2\\pi} \\int_0^{\\sqrt{8}} \\int_0^{\\frac{\\pi}{4}} r^2 \\sin(\\phi) \\, d\\phi \\, dr \\, d\\theta\n\\]\n\n• Now is just to solve the triple integral.\n\n• At the end you should get: \\( \\frac{64\\pi(\\sqrt{2} - 1)}{3} \\)", "id": "./materials/290.pdf" }, { "contents": "Use a triple integral to determine the volume defined of the solid bounded by \n\\[ x^2 + y^2 \\leq z \\leq \\sqrt{x^2 + y^2}. \\]\n\n- Let’s first sketch the solid E over xyz-coordinate planes:\n\n![3D sketch of E](image)\n\nFigure 1: 3D sketch of E\n\n- Tip: change to cylindrical coordinates\n\n\\[\n\\begin{align*}\n x &= r \\cos(\\theta) \\\\\n y &= r \\sin(\\theta) \\\\\n z &= z\n\\end{align*}\n\\]\n\n- The projection (D) of the solid E over the xy-plane can be defined as\n\n\\[ x^2 + y^2 \\leq 1 \\]\n• Let’s assess the behavior of $z$:\n\n\\[\nx^2 + y^2 \\leq z \\land z \\leq \\sqrt{x^2 + y^2}\n\\]\n\n\\[\n\\Rightarrow (r \\cos(\\theta))^2 + (r \\sin(\\theta))^2 \\leq z \\land z \\leq \\sqrt{r^2}\n\\]\n\n\\[\n\\Rightarrow r^2 \\leq z \\land z \\leq r\n\\]\n\n\\[\n\\Rightarrow r^2 \\leq z \\leq r\n\\]\n\n• So we can write evaluate the volume of the solid through\n\n\\[\n\\iiint_E 1 \\, dV = \\int_0^{2\\pi} \\int_0^1 \\int_{r^2}^r r \\, dz \\, dr \\, d\\theta\n\\]\n\n• Now is just to solve the triple integral.\n\n• At the end you should get: $\\frac{\\pi}{6}$", "id": "./materials/291.pdf" }, { "contents": "Use a triple integral to determine the volume defined by Evaluate\n\\[ x^2 + y^2 + z^2 \\leq 4 \\land x^2 + y^2 + (z - 2)^2 \\leq 4 \\}\n\n- Let’s first sketch the solid E over xyz-coordinate planes:\n\n![3D sketch of E](image)\n\nFigure 1: 3D sketch of E\n\n- Tip: Use spherical coordinates when defining the triple integral\n\n\\[\n\\begin{align*}\n x &= r \\cos(\\theta) \\sin(\\phi) \\\\\n y &= r \\sin(\\theta) \\sin(\\phi) \\\\\n z &= r \\cos(\\phi)\n\\end{align*}\n\\]\n\nIn this case, we can easily define\n\n\\[ 0 \\leq \\theta \\leq 2\\pi \\land 0 \\leq \\phi \\leq \\frac{\\pi}{2} \\]\n• We can evaluate the z-coordinates for the intersection of the two solids:\n\n\\[ x^2 + y^2 + z^2 - 4 = x^2 + y^2 + z^2 - 4z + 4 - 4 \\]\n\n\\[ \\Rightarrow z = 1 \\]\n\n• Let’s study the behaviour of r:\n\nWe can also assess that r does not behave the same way along \\( \\phi \\)'s interval. So, we are going to divide into two sections\n\n\\[ (x = 0 \\land z = 1) \\Rightarrow y = \\pm \\sqrt{3} \\]\n\n\\[ \\tan(\\phi) = \\frac{\\sqrt{3}}{1} \\Rightarrow \\phi = \\arctan(\\sqrt{3}) \\]\n\n\\[ \\Rightarrow \\phi = \\frac{\\pi}{3} \\]\n\nWhen\n\n\\[ 0 \\leq \\phi \\leq \\frac{\\pi}{3} \\Rightarrow 0 \\leq r \\leq 2 \\]\n\nBut when\n\n\\[ \\frac{\\pi}{3} \\leq \\phi \\leq \\frac{\\pi}{2} \\Rightarrow x^2 + y^2 + z^2 - 4z + 4 \\leq 4 \\]\n\n\\[ \\Rightarrow r^2((\\cos(\\theta) \\sin(\\phi))^2 + (\\sin(\\theta) \\sin(\\phi))^2) + (r \\cos(\\phi) - 2)^2 \\leq 4 \\]\n\n\\[ \\Rightarrow r^2((\\cos(\\theta) \\sin(\\phi))^2 + (\\sin(\\theta) \\sin(\\phi))^2) + r^2 \\cos^2(\\phi) - 4r \\cos(\\phi) \\leq 0 \\]\n\n\\[ \\Rightarrow r^2 - 4r \\cos(\\phi) \\leq 0 \\]\n\n\\[ \\Rightarrow r \\leq 4 \\cos(\\phi) \\]\n\n• So we can evaluate the volume through\n\n\\[\n\\iiint_E 1 \\, dV = \\int_0^{2\\pi} \\int_0^{\\frac{\\pi}{3}} \\int_0^{4 \\cos(\\phi)} r^2 \\sin(\\phi) \\, dr \\, d\\phi \\, d\\theta + \\int_0^{2\\pi} \\int_0^{\\frac{\\pi}{3}} \\int_0^{2} r^2 \\sin(\\phi) \\, dr \\, d\\phi \\, d\\theta\n\\]\n\n• Now is just to solve the integrals.\n\n• In the end you should get: \\( \\frac{10\\pi}{3} \\)", "id": "./materials/292.pdf" }, { "contents": "Use a triple integral to determine the volume defined by\n\\[ E = \\{(x, y, z) \\in \\mathbb{R}^3 : 1 \\leq z \\leq 2 \\land x^2 + y^2 \\leq z\\} . \\]\n\n- Let’s first sketch the solid E over xyz-coordinate planes:\n\n![3D sketch of E](image)\n\nFigure 1: 3D sketch of E\n\n- The projection (D) of the solid E over the xy-plane can be defined as\n \\[ x^2 + y^2 \\leq 2 \\]\n\n- Tip: change to cylindrical coordinates\n \\[\n \\begin{cases}\n x = r \\cos(\\theta) \\\\\n y = r \\sin(\\theta) \\\\\n z = z\n \\end{cases}\n \\]\n• So, substituting, we have\n\n\\[ x^2 + y^2 \\leq z \\]\n\\[ \\Rightarrow r^2 \\leq z \\]\n\\[ \\Rightarrow r \\leq \\sqrt{z} \\quad , \\quad r > 0 \\]\n\n• So we can evaluate the volume through\n\n\\[\n\\iiint_E 1 \\, dV = \\int_0^{2\\pi} \\int_0^2 \\int_0^{\\sqrt{z}} r \\, dr \\, dz \\, d\\theta\n\\]\n\n• Now is just to solve the integrals.\n\n• In the end you should get: \\( \\frac{3\\pi}{2} \\)", "id": "./materials/293.pdf" }, { "contents": "Use a triple integral to determine the volume defined by\n\\[ E = \\{(x, y, z) \\in \\mathbb{R}^3 : 0 \\leq z \\leq 2 \\land x^2 + y^2 - z^2 \\leq 1\\} . \\]\n\n- Let’s first sketch the solid E over xyz-coordinate planes:\n\n![3D sketch of E](image)\n\nFigure 1: 3D sketch of E\n\n- The projection (D) of the solid E over the xy-plane can be defined as\n \\[ x^2 + y^2 \\leq 5 \\]\n\n- Tip: change to cylindrical coordinates\n \\[\n \\begin{cases}\n x = r \\cos(\\theta) \\\\\n y = r \\sin(\\theta) \\\\\n z = z\n \\end{cases}\n \\]\n• So, substituting, we have\n\n\\[ x^2 + y^2 - z^2 \\leq 1 \\]\n\\[ \\Rightarrow r^2 \\leq z^2 + 1 \\]\n\\[ \\Rightarrow r \\leq \\sqrt{z^2 + 1}, \\quad r > 0 \\]\n\n• So we can evaluate the volume through\n\n\\[\n\\iiint_E 1 \\, dV = \\int_0^{2\\pi} \\int_0^2 \\int_0^{\\sqrt{z^2 + 1}} r \\, dr \\, dz \\, d\\theta\n\\]\n\\[\n= \\int_0^{2\\pi} \\int_0^2 \\left[ \\frac{r^2}{2} \\right]_{r=0}^{r=\\sqrt{z^2 + 1}} \\, dz \\, d\\theta\n\\]\n\\[\n= \\int_0^{2\\pi} \\int_0^2 \\frac{z^2 + 1}{2} \\, dz \\, d\\theta\n\\]\n\\[\n= \\frac{1}{2} \\int_0^{2\\pi} \\left[ \\frac{z^3}{3} + z \\right]_{z=0}^{z=2} \\, d\\theta\n\\]\n\\[\n= \\frac{7}{3} \\left[ \\theta \\right]_{\\theta=0}^{\\theta=2\\pi}\n\\]\n\\[\n= \\frac{14\\pi}{3}\n\\]", "id": "./materials/294.pdf" }, { "contents": "Use a triple integral to determine the volume defined by\n\\[ E = \\{(x, y, z) \\in \\mathbb{R}^3 : x^2 + y^2 + z^2 \\leq 2z \\land x^2 + y^2 + z^2 \\leq 3 \\land y \\geq x\\} \\]\n\n- Let’s first sketch the solid E over xyz-coordinate planes:\n\n![3D sketch of E](image)\n\nFigure 1: 3D sketch of E\n\n- Tip: Use spherical coordinates when defining the triple integral\n\n\\[\n\\begin{align*}\n x &= r \\cos(\\theta) \\sin(\\phi) \\\\\n y &= r \\sin(\\theta) \\sin(\\phi) \\\\\n z &= r \\cos(\\phi)\n\\end{align*}\n\\]\n\nIn this case, we can easily define\n\\[\n\\frac{\\pi}{4} \\leq \\theta \\leq \\frac{5\\pi}{4} \\land 0 \\leq \\phi \\leq \\frac{\\pi}{2}\n\\]\n• We can evaluate the z-coordinates for the intersection of the two solids:\n\n\\[ x^2 + y^2 + z^2 - 2z = x^2 + y^2 + z^2 - 3 \\]\n\n\\[ \\Rightarrow -2z = -3 \\]\n\n\\[ \\Rightarrow z = \\frac{3}{2} \\]\n\n• Let’s study the behaviour of r:\n\nWe can also assess that r does not behave the same way along \\( \\phi \\)'s interval. So, we are going to divide into two sections\n\n\\[ (x = 0 \\land z = \\frac{3}{2}) \\Rightarrow y = \\pm \\frac{\\sqrt{3}}{2} \\]\n\n\\[ \\tan(\\phi) = \\frac{\\sqrt{3}}{\\frac{3}{2}} \\Rightarrow \\phi = \\arctan \\left( \\frac{1}{\\sqrt{3}} \\right) \\]\n\n\\[ \\Rightarrow \\phi = \\frac{\\pi}{6} \\]\n\nWhen\n\n\\[ 0 \\leq \\phi \\leq \\frac{\\pi}{6} \\Rightarrow 0 \\leq r \\leq \\sqrt{3} \\]\n\nBut when\n\n\\[ \\frac{\\pi}{6} \\leq \\phi \\leq \\frac{\\pi}{2} \\Rightarrow x^2 + y^2 + z^2 \\leq 2z \\]\n\n\\[ \\Rightarrow r^2((\\cos(\\theta) \\sin(\\phi))^2 + (\\sin(\\theta) \\sin(\\phi))^2) + (r \\cos(\\phi) - 1)^2 \\leq 1 \\]\n\n\\[ \\Rightarrow r^2((\\cos(\\theta) \\sin(\\phi))^2 + (\\sin(\\theta) \\sin(\\phi))^2) + r^2 \\cos^2(\\phi) - 2r \\cos(\\phi) + 1 \\leq 1 \\]\n\n\\[ \\Rightarrow r^2 - 2r \\cos(\\phi) \\leq 0 \\]\n\n\\[ \\Rightarrow r \\leq 2 \\cos(\\phi) \\]\n\n• So we can write the triple integral as\n\n\\[\n\\iiint_E 1 \\, dV = \\int_{\\frac{\\pi}{4}}^{\\frac{5\\pi}{4}} \\int_{\\frac{\\pi}{4}}^{\\frac{\\pi}{2}} \\int_0^{2 \\cos(\\phi)} r^2 \\sin(\\phi) \\, dr \\, d\\phi \\, d\\theta + \\int_{\\frac{\\pi}{4}}^{\\frac{5\\pi}{4}} \\int_{\\frac{\\pi}{6}}^{\\frac{\\pi}{2}} \\int_0^{\\sqrt{3}} r^2 \\sin(\\phi) \\, dr \\, d\\phi \\, d\\theta\n\\]\n\n• Now, is just to evaluate the triple integral.\n\n• At the end, you should get: \\( \\pi \\left( \\sqrt{3} - \\frac{9}{8} \\right) \\)", "id": "./materials/295.pdf" }, { "contents": "Use a triple integral to determine the volume defined by\n\\[ E = \\{(x, y, z) \\in \\mathbb{R}^3 : x^2 + z \\leq 4 \\land z + y \\leq 4 \\land y \\geq 0 \\land z \\geq 0\\}. \\]\n\n- Let’s first sketch the solid E over xyz-coordinate planes:\n\n![3D sketch of E](image)\n\nFigure 1: 3D sketch of E\n\n- Now, let’s sketch the projection of E over the yz-plane:\n\n\\[ y + z = 4 \\]\nSo, using the projection over yz-plane, we can write\n\n\\[\n\\iiint_E 1 \\, dV = \\int_0^4 \\int_0^{4-y} \\int_{-\\sqrt{4-z^2}}^{\\sqrt{4-z^2}} 1 \\, dx \\, dz \\, dy\n\\]\n\n\\[\n= \\int_0^4 \\int_0^{4-y} \\left[ x \\right]_{x=-\\sqrt{4-z^2}}^{x=\\sqrt{4-z^2}} \\, dz \\, dy\n\\]\n\n\\[\n= 2 \\int_0^4 \\int_0^{4-y} \\sqrt{4-z^2} \\, dz \\, dy\n\\]\n\n\\[\n= -\\frac{4}{3} \\int_0^4 \\int_0^{4-y} -\\frac{3}{2} \\sqrt{4-z^2} \\, dz \\, dy\n\\]\n\n\\[\n= -\\frac{4}{3} \\int_0^4 \\left[ (4-z^2)^{\\frac{3}{2}} \\right]_{z=0}^{z=4-y} \\, dy\n\\]\n\n\\[\n= -\\frac{4}{3} \\int_0^4 \\left( y^\\frac{3}{2} - 4^\\frac{3}{2} \\right) \\, dy\n\\]\n\n\\[\n= -\\frac{4}{3} \\left[ \\frac{y^{\\frac{5}{2}}}{\\frac{5}{2}} - 8y \\right]_{y=0}^{y=4}\n\\]\n\n\\[\n= -\\frac{4}{3} \\left( \\frac{4^{\\frac{5}{2}}}{\\frac{5}{2}} - 32 \\right)\n\\]\n\n\\[\n= -\\frac{4}{3} \\left( \\frac{64}{5} - 32 \\right)\n\\]\n\n\\[\n= -\\frac{128}{3} \\left( -\\frac{3}{5} \\right)\n\\]\n\n\\[\n= \\frac{128}{5}\n\\]", "id": "./materials/296.pdf" }, { "contents": "Use a triple integral to determine the volume defined by\n\\[ E = \\{(x, y, z) \\in \\mathbb{R}^3 : -1 + \\sqrt{x^2 + y^2} \\leq z \\leq 1 - \\sqrt{x^2 + y^2}\\}. \\]\n\n- Let’s first sketch the solid E over xyz-coordinate planes:\n\n![3D sketch of E](image)\n\nFigure 1: 3D sketch of E\n\n- The projection (D) of the solid E over the xy-plane can be defined as\n \\[ x^2 + y^2 = 1 \\]\n\n- Tip: change to cylindrical coordinates\n \\[\n \\begin{cases}\n x = r \\cos(\\theta) \\\\\n y = r \\sin(\\theta) \\\\\n z = z\n \\end{cases}\n \\]\n• So, substituting, we have\n\n\\[-1 + \\sqrt{x^2 + y^2} \\leq z \\leq 1 - \\sqrt{x^2 + y^2}\\]\n\n\\[\\Rightarrow r - 1 \\leq z \\leq 1 - r\\]\n\n• So we can evaluate the volume through\n\n\\[\n\\iiint_E 1 \\, dV = \\int_0^{2\\pi} \\int_0^1 \\int_{r-1}^{1-r} r \\, dz \\, dr \\, d\\theta\n\\]\n\n\\[\n= \\int_0^{2\\pi} \\int_0^1 \\left[ r \\right]_{z=r-1}^{z=1-r} \\, dr \\, d\\theta\n\\]\n\n\\[\n= \\int_0^{2\\pi} \\int_0^1 2r - 2r^2 \\, dr \\, d\\theta\n\\]\n\n\\[\n= \\int_0^{2\\pi} \\left[ r^2 - \\frac{2r^3}{3} \\right]_{r=0}^{r=1} \\, d\\theta\n\\]\n\n\\[\n= \\frac{1}{3} (2\\pi - 0)\n\\]\n\n\\[\n= \\frac{2\\pi}{3}\n\\]", "id": "./materials/297.pdf" }, { "contents": "Use a triple integral to determine the volume defined by\n\\[ E = \\{(x, y, z) \\in \\mathbb{R}^3 : x^2 + y^2 + z^2 \\leq 4 \\land x^2 + y^2 \\leq z^2 \\land y \\geq 0\\}. \\]\n\n- Let’s first sketch the solid E over xyz-coordinate planes:\n\n![3D sketch of E](image)\n\nFigure 1: 3D sketch of E\n\n- Tip: Use spherical coordinates when defining the triple integral\n\n\\[\n\\begin{align*}\n x &= r \\cos(\\theta) \\sin(\\phi) \\\\\n y &= r \\sin(\\theta) \\sin(\\phi) \\\\\n z &= r \\cos(\\phi)\n\\end{align*}\n\\]\n\nIn this case, we can easily define\n\n\\[ 0 \\leq \\theta \\leq \\pi \\land 0 \\leq r \\leq 2 \\]\n• We can evaluate the z-coordinates for the intersection of the two solids:\n\n\\[ x^2 + y^2 + z^2 - 4 = x^2 + y^2 - z^2 \\]\n\n\\[ \\Rightarrow 2z^2 = 2 \\]\n\n\\[ \\Rightarrow z = \\sqrt{2} \\quad \\lor \\quad z = -\\sqrt{2} \\]\n\n• We can see from figure 1 that the total volume can be calculated by two times the volume of one of the equal parts. So, let’s evaluate the value of \\( \\phi \\), related to the part of the solid with positive z-coordinates.\n\n\\[ \\left( x = 0 \\quad \\land \\quad z = \\sqrt{2} \\right) \\quad \\Rightarrow \\quad y = \\pm \\sqrt{2} \\]\n\n\\[ \\tan(\\phi) = \\frac{\\sqrt{2}}{\\sqrt{2}} \\quad \\Rightarrow \\quad \\phi = \\frac{\\pi}{4} \\]\n\n• So we can write evaluate the volume of the solid through\n\n\\[\n\\iiint_E 1 \\, dV = 2 \\int_0^\\pi \\int_0^2 \\int_0^{\\frac{\\pi}{4}} r^2 \\sin(\\phi) \\, d\\phi \\, dr \\, d\\theta\n\\]\n\n• Now is just to solve the triple integral.\n\n• At the end you should get: \\( \\frac{8\\pi(2 - \\sqrt{2})}{3} \\)", "id": "./materials/298.pdf" }, { "contents": "Evaluate \\( \\int_{0}^{2} \\int_{-1}^{y^2} \\int_{-1}^{z} yz \\, dx \\, dz \\, dy \\)\n\n- Let’s evaluate first:\n\n\\[\n\\int_{-1}^{z} yz \\, dx = yz \\left[ x \\right]_{x=-1}^{x=z} = yz(z + 1) = yz^2 + yz\n\\]\n\n- The idea is now to do exactly the same for the next iterations, first in respect to \\( z \\) and finally in respect to \\( y \\).\n\n- At the end you should get: \\( \\frac{47}{3} \\)", "id": "./materials/299.pdf" }, { "contents": "Evaluate \\( \\int_0^1 \\int_0^1 \\int_0^{x^2} x \\cos(y) \\, dz \\, dx \\, dy \\)\n\n- Let’s evaluate first:\n\n\\[\n\\int_1^{x^2} x \\cos(y) \\, dz = x \\cos(y) \\left[ z \\right]_{z=1}^{z=x^2} = x \\cos(y)(x^2 - 1) = x^3 \\cos(y) - x \\cos(y)\n\\]\n\n- The idea is now to do exactly the same for the next iterations, first in respect to \\( x \\) and finally in respect to \\( y \\).\n\n- At the end you should get: \\( \\frac{\\sqrt{2}}{8} \\)", "id": "./materials/300.pdf" }, { "contents": "Evaluate \\( \\int_0^3 \\int_0^{\\sqrt{9-z^2}} \\int_0^x xy \\, dy \\, dx \\, dz \\)\n\n- Let’s evaluate first:\n\n\\[\n\\int_0^x xy \\, dy = x \\left[ \\frac{y^2}{2} \\right]_{y=0}^{y=x} = \\frac{x^3}{2}\n\\]\n\n- The idea is now to do exactly the same for the next iterations, first in respect to \\( x \\) and finally in respect to \\( z \\).\n\n- At the end you should get: \\( \\frac{81}{5} \\)", "id": "./materials/301.pdf" }, { "contents": "Evaluate \\( \\int_1^3 \\int_x^{x^2} \\int_0^{\\ln(z)} x e^y \\, dy \\, dz \\, dx \\)\n\n- Let’s evaluate first:\n\n\\[\n\\int_0^{\\ln(z)} x e^y \\, dy = x \\left[ e^y \\right]_{y=0}^{y=\\ln(z)} = x(e^{\\ln(z)} - e^0) = x(z - 1)\n\\]\n\n- The idea is now to do exactly the same for the next iterations, first in respect to \\( z \\) and finally in respect to \\( x \\).\n\n- At the end you should get: \\( \\frac{118}{3} \\)", "id": "./materials/302.pdf" }, { "contents": "Evaluate \\( \\iiint_E 2x \\, dV \\) where \\( E \\) is the solid from the first octant bounded by \n\\[ z^2 = x^2 + y^2, \\quad x = 0, \\quad y = 0 \\quad \\text{and} \\quad z = 4. \\]\n\n- Let’s first sketch \\( E \\) over xyz-coordinate planes:\n\n![3D sketch of E](image)\n\nFigure 1: 3D sketch of \\( E \\)\n\n- There is more than one way to approach this problem – that difference lies in the projection we choose to use when defining the triple integral.\n\n- In one hand, we can choose to define the triple integral using the projection over the xy-plane.\nIn this case, the projection D is defined as\n\\[ x^2 + y^2 \\leq 16 \\quad \\text{with} \\quad x, y \\geq 0 \\]\nThus, we can define\n\\[ E = \\{(x, y, z) \\in \\mathbb{R}^3 : (x, y) \\in D, \\sqrt{x^2 + y^2} \\leq z \\leq 4\\} \\]\nTherefore, the triple integral may be defined as:\n\\[\n\\iiint_D \\left[ \\int_0^4 \\frac{2x}{\\sqrt{x^2+y^2}} \\, dz \\right] \\, dA\n\\]\nLet’s evaluate:\n\\[\n\\int_0^4 \\frac{2x}{\\sqrt{x^2+y^2}} \\, dz = 2x \\left[ z \\right]_{z=\\sqrt{x^2+y^2}}^{z=4} = 2x \\left( 4 - \\sqrt{x^2+y^2} \\right)\n\\]\nLet’s now switch to cylindrical coordinates, where\n\\[\n\\begin{align*}\nx &= r \\cos \\theta \\\\\ny &= r \\sin \\theta \\\\\nr &\\geq 0, \\quad 0 \\leq \\theta \\leq \\frac{\\pi}{2}\n\\end{align*}\n\\]\nSubstituting, we get:\n\n\\[\n\\int_0^4 \\int_0^{\\pi/2} 2(r \\cos(\\theta)) \\left( 4 - \\sqrt{(r \\cos(\\theta))^2 + (r \\sin(\\theta))^2} \\right) r \\, d\\theta \\, dr\n\\]\n\n\\[\n= \\int_0^4 \\int_0^{\\pi/2} 2(r^2 \\cos(\\theta)) \\left( 4 - \\sqrt{r^2 \\cos^2(\\theta) + r^2 \\sin^2(\\theta)} \\right) d\\theta \\, dr\n\\]\n\n\\[\n= \\int_0^4 \\int_0^{\\pi/2} 2(r^2 \\cos(\\theta))(4 - r) \\, d\\theta \\, dr\n\\]\n\n\\[\n= \\int_0^4 \\left[ 8r^2 \\sin(\\theta) - 2r^3 \\sin(\\theta) \\right]_{\\theta=0}^{\\theta=\\pi/2} \\, dr\n\\]\n\n\\[\n= \\int_0^4 8r^2 - 2r^3 \\, dr\n\\]\n\n\\[\n= \\left[ \\frac{8}{3} r^3 - \\frac{1}{2} r^4 \\right]_{r=0}^{r=4}\n\\]\n\n\\[\n= \\frac{512}{3} - 128\n\\]\n\n\\[\n= \\frac{128}{3}\n\\]\n\nIn the next page we are going to solve the same problem, but using a different projection.\nYou will be able to verify that the answer is the same!\n• In the other hand, we can choose to define the triple integral using the projection over the yz-plane.\n\n![Figure 3: Projection (R) of E over the yz-plane](image)\n\n• In this case, the projection R is defined as\n\n\\[ 0 \\leq y \\leq 4 \\quad \\text{and} \\quad y \\leq z \\leq 4 \\]\n\nThus, we can define\n\n\\[ E = \\{(x, y, z) \\in \\mathbb{R}^3 : (y, z) \\in R, \\ 0 \\leq x \\leq \\sqrt{z^2 - y^2}\\} \\]\n\n• Therefore, the triple integral may be defined as:\n\n\\[\n\\iiint_R \\left[ \\int_0^{\\sqrt{z^2 - y^2}} 2x \\, dx \\right] \\, dA\n\\]\n\n• Let’s evaluate:\n\n\\[\n\\int_0^{\\sqrt{z^2 - y^2}} 2x \\, dx\n\\]\n\n\\[\n= \\left[ x^2 \\right]_{x=0}^{x=\\sqrt{z^2 - y^2}}\n\\]\n\n\\[\n= z^2 - y^2\n\\]\nLet’s now focus on the double integral:\n\n\\[\n\\iint_R z^2 - y^2 \\, dA\n\\]\n\n\\[\n= \\int_0^4 \\int_y^4 z^2 - y^2 \\, dz \\, dy\n\\]\n\n\\[\n= \\int_0^4 \\left[ \\frac{z^3}{3} - zy^2 \\right]_{z=y}^{z=4} \\, dy\n\\]\n\n\\[\n= \\int_0^4 \\frac{64}{3} - 4y^2 - \\frac{y^3}{3} + y^3 \\, dy\n\\]\n\n\\[\n= \\left[ \\frac{64y}{3} - \\frac{4y^3}{3} - \\frac{y^4}{12} + \\frac{y^4}{4} \\right]_{y=0}^{y=4}\n\\]\n\n\\[\n= \\frac{256}{3} - \\frac{256}{3} - \\frac{64}{3} + 64\n\\]\n\n\\[\n= \\frac{128}{3}\n\\]\n\nAs we can see, both paths lead to the same value!", "id": "./materials/303.pdf" }, { "contents": "Evaluate \\( \\iiint_E (x^2 + y^2)z \\, dV \\) where \\( E \\) is bounded by \\( x^2 + y^2 = 1 \\) and \\( x^2 + y^2 + z^2 = 4 \\).\n\n- Let’s first sketch \\( E \\) over xyz-coordinate planes:\n\n![3D sketch of E](image)\n\nFigure 1: 3D sketch of \\( E \\)\n\n- Tip: change to cylindrical coordinates\n\n\\[\n\\begin{align*}\nx &= r \\cos(\\theta) \\\\\ny &= r \\sin(\\theta) \\\\\nz &= z\n\\end{align*}\n\\]\nSo, in this case, we have\n\n\\[ x^2 + y^2 \\leq 1 \\implies r \\leq 1 \\]\n\nand\n\n\\[ x^2 + y^2 + z^2 \\leq 4 \\implies -\\sqrt{4 - r^2} \\leq z \\leq \\sqrt{4 - r^2} \\]\n\n- This means that we can define\n\n\\[ E = \\{(r, \\theta, z) | 0 \\leq r \\leq 1 \\land 0 \\leq \\theta \\leq 2\\pi \\land -\\sqrt{4 - r^2} \\leq z \\leq \\sqrt{4 - r^2}\\} \\]\n\n- Thus, we are able to write the triple integral as\n\n\\[\n\\int_0^{2\\pi} \\int_0^1 \\int_{-\\sqrt{4-r^2}}^{\\sqrt{4-r^2}} r^2 z r \\, dz \\, dr \\, d\\theta\n\\]\n\n- Now is just to solve it.\n\n- You should get as the result: 0", "id": "./materials/304.pdf" }, { "contents": "Evaluate \\( \\iiint_E (x^2 + y^2 + z^2)^{-2} \\, dV \\) where\n\\[\nE = \\{(x, y, z) \\in \\mathbb{R}^3 : 1 \\leq x^2 + y^2 + z^2 \\leq 4\\}\n\\]\n\n- Let’s first sketch \\( E \\) over \\( xyz \\)-coordinate planes:\n\n![3D sketch of E](image)\n\nFigure 1: 3D sketch of \\( E \\)\n\n- Tip: change to spherical coordinates\n\n\\[\n\\begin{align*}\nx &= r \\cos(\\theta) \\sin(\\phi) \\\\\ny &= r \\sin(\\theta) \\sin(\\phi) \\\\\nz &= \\cos(\\phi)\n\\end{align*}\n\\]\nSo, in this case, we have\n\n\\[ E^* = [1, 2] \\times [0, 2\\pi] \\times [0, \\pi] \\]\n\n- Thus, we are able to rewrite the triple integral:\n\n\\[\n\\iiint_E (x^2 + y^2 + z^2)^{-2} \\, dV = \\iiint_{E^*} (r^2)^{-2} r^2 \\sin(\\phi) \\, dV = \\int_1^2 \\int_0^{2\\pi} \\int_0^\\pi r^{-2} \\sin(\\phi) \\, d\\phi d\\theta dr\n\\]\n\n- Now is just to solve it.\n\n- You should get as the result: \\(2\\pi\\)", "id": "./materials/305.pdf" }, { "contents": "Evaluate $\\iiint_E 1 \\, dV$ where\n\n$E = \\{(x, y, z) \\in \\mathbb{R}^3 : \\sqrt{x^2 + y^2} \\leq z \\leq 1 - 2\\sqrt{x^2 + y^2}\\}$\n\n- Tip: change to cylindrical coordinates\n\n\\[\n\\begin{align*}\n x &= r \\cos(\\theta) \\\\\n y &= r \\sin(\\theta) \\\\\n z &= z\n\\end{align*}\n\\]\n\nSo, in this case, we have\n\n$r \\leq z \\leq 1 - 2r$\n\n- We can also assess that\n\n$0 \\leq r \\leq \\frac{1}{3}, \\quad 0 \\leq \\theta \\leq 2\\pi$\n\n- Thus, we are able to rewrite the triple integral:\n\n\\[\n\\iiint_E 1 \\, dV = \\int_0^{2\\pi} \\int_0^{\\frac{1}{3}} \\int_r^{1-2r} r \\, dz \\, dr \\, d\\theta\n\\]\n\n- Now is just to solve it.\n\n- You should get as the result: $\\frac{\\pi}{27}$", "id": "./materials/306.pdf" }, { "contents": "Evaluate $\\iiint_E z \\, dV$ where $E$ is the solid tetrahedron bounded by $x = 0$, $y = 0$, $z = 0$, and $x + y + z = 1$.\n\n- Let’s first sketch $E$ over $xyz$-coordinate planes:\n\n![3D sketch of E](image)\n\nFigure 1: 3D sketch of $E$\n\n- We can define\n\n$$E = \\{(x, y, z) \\in \\mathbb{R}^3 : 0 \\leq x \\leq 1 \\land 0 \\leq y \\leq 1 - x \\land 0 \\leq z \\leq 1 - x - y\\}$$\n\n- This means that we can write the triple integral as:\n\n$$\\int_0^1 \\int_0^{1-x} \\int_0^{1-x-y} z \\, dz \\, dy \\, dx$$\n• Now it is just to solve the triple integral\n\n• At the end you should get: \\( \\frac{1}{24} \\)", "id": "./materials/307.pdf" }, { "contents": "Evaluate \\( \\iiint_E \\sqrt{x^2 + z^2} \\, dV \\) where \\( E \\) is the solid bounded by \\( y = x^2 + z^2 \\), and \\( y = 4 \\).\n\n- Let’s first sketch \\( E \\) over xyz-coordinate planes:\n\n![3D sketch of E](image)\n\nFigure 1: 3D sketch of \\( E \\)\n\n- The projection (D) of \\( E \\) over the xz-plane can be defined as\n \\[ x^2 + z^2 \\leq 4 \\]\n\n- Tip: switch to cylindrical coordinates\n \\[\n \\begin{align*}\n x &= r \\cos(\\theta) \\\\\n z &= r \\sin(\\theta) \\\\\n y &= y\n \\end{align*}\n \\]\n• This means that we can write the triple integral as:\n\n\\[\n\\iiint_D \\left[ \\int_0^4 \\sqrt{x^2 + z^2} \\, dy \\right] \\, dA\n\\]\n\n\\[\n= \\iiint_D (4 - x^2 - z^2) \\sqrt{x^2 + z^2} \\, dA\n\\]\n\n\\[\n= \\int_0^{2\\pi} \\int_0^2 (4 - r^2) r \\, dr \\, d\\theta\n\\]\n\n• Now it is just to solve the double integral.\n\n• At the end you should get: \\( \\frac{128\\pi}{15} \\)", "id": "./materials/308.pdf" }, { "contents": "Use a triple integral to determine the volume of the tetrahedron bounded by \n\\( x + 2y + z = 2 \\), \\( x = 2y \\), \\( x = 0 \\) and \\( z = 0 \\).\n\n- Let’s first sketch \\( E \\) over \\( xyz \\)-coordinate planes:\n\n![3D sketch of E](image)\n\nFigure 1: 3D sketch of \\( E \\)\n\n- We can write the triple integral as:\n\n\\[\n\\int_0^1 \\int_{\\frac{x}{2}}^{1-\\frac{x}{2}} \\int_0^{2-x-2y} 1 \\, dz \\, dy \\, dx\n\\]\n• Let’s evaluate first:\n\n\\[\n\\int_0^{2-x-2y} 1 \\, dz\n\\]\n\n\\[\n= \\left[ z \\right]_{z=0}^{z=2-x-2y}\n\\]\n\n\\[\n= 2 - x - 2y\n\\]\n\n• The idea is now to do exactly the same for the next iterations, first in respect to y and finally in respect to x.\n\n• At the end you should get: \\( \\frac{1}{3} \\)", "id": "./materials/309.pdf" }, { "contents": "Discussing a linear system depending on a parameter\n\nConsider the following linear system in the variables $x, y, z, t$ and depending on the parameter $k \\in \\mathbb{R}$:\n\n\\[\n\\begin{align*}\n x - ky + z + (k-1)t &= 1 \\\\\n -x + ky - kz &= -1 \\\\\n (k+1)x - 2y + 2z &= 2\n\\end{align*}\n\\]\n\n1. Discuss the solutions of the linear system with respect to $k$.\n2. Find the solution of the linear system for $k = 1$.\n3. Add an equation so that the linear system has no solutions for every $k$.\n\nSolution.\n\n1. By reducing the complete matrix of linear the system (using $R_2 \\rightarrow R_2 + R_1; R_3 \\rightarrow R_3 - (k+1)R_1; R_2 \\leftrightarrow R_3$) one finds\n\n\\[\n\\begin{pmatrix}\n 1 & -k & 1 & k-1 & 1 \\\\\n -1 & k & -k & 0 & -1 \\\\\n k+1 & -2 & 2 & 0 & 2\n\\end{pmatrix}\n\\rightarrow\n\\begin{pmatrix}\n 1 & -k & 1 & k-1 & 1 \\\\\n 0 & (k+2)(k-1) & 1-k & 1-k^2 & 1-k \\\\\n 0 & 0 & 1-k & k-1 & 0\n\\end{pmatrix}\n\\]\n\nfrom which one deduces that the ranks of the incomplete matrix and of the complete matrix are equal for every value of $k$, so the system has a solution for every $k \\in \\mathbb{R}$. More precisely: if $k \\neq -2 \\land k \\neq 1$ the rank of both matrices is 3 and the system has $\\infty^{4-3} = \\infty^1$ solutions (depending on $t$); if $k = -2$, the rank is 3 and the system has $\\infty^{4-3} = \\infty^1$ solutions (depending on $y$); if $k = -1$, the rank of both matrices is 1 and the system has $\\infty^{4-1} = \\infty^3$ solutions (depending on $y$, $z$ and $t$).\n\n2. If $k = 1$ we have\n\n\\[\n\\begin{pmatrix}\n 1 & -1 & 1 & 0 & 1 \\\\\n -1 & 1 & -1 & 0 & -1 \\\\\n 2 & -2 & 2 & 0 & 2\n\\end{pmatrix}\n\\rightarrow\n\\begin{pmatrix}\n 1 & -1 & 1 & 0 & 1 \\\\\n 0 & 0 & 0 & 0 & 0 \\\\\n 0 & 0 & 0 & 0 & 0\n\\end{pmatrix}\n\\]\n\nso $x = y - z + 1$; the solutions of the system are then\n\n\\[\n(x; y; z; t) = (y - z + 1; y; z; t) = (1; 0; 0; 0) + y(1; 1; 0; 0) + z(-1; 0; 1; 0) + t(0; 0; 0; 1),\n\\]\n\nfor every $y, z, t \\in \\mathbb{R}$.\n\n3. It is enough to add to the linear system an impossible equation which does not depend on $k$ (like $0 = 1$), or an equation which is incompatible with the previous ones. As instance, the sum of the left hand sides equals a value which is different from the some of the right hand sides: $(k+1)x - 2y + (3-k)z + (k-1)t = A$ for any $A \\neq 2$, or $x - ky + z + (k-1)t = 0$. \n", "id": "./materials/31.pdf" }, { "contents": "Evaluate \\( \\int_0^3 \\int_0^1 \\int_0^{\\sqrt{1-z^2}} z e^y \\, dx \\, dz \\, dy \\)\n\n- Let’s evaluate first:\n\n\\[\n\\int_0^{\\sqrt{1-z^2}} z e^y \\, dx = z e^y \\left[ x \\right]_{x=0}^{x=\\sqrt{1-z^2}} = z e^y \\sqrt{1-z^2}\n\\]\n\n- The idea is now to do exactly the same for the next iterations, first in respect to \\( z \\) and finally in respect to \\( y \\).\n\n- At the end you should get: \\( \\frac{e^3 - 1}{3} \\)", "id": "./materials/310.pdf" }, { "contents": "Evaluate \\( \\int_0^\\pi \\int_0^y \\int_0^x \\cos(x + y + z) \\, dz \\, dx \\, dy \\)\n\n- Let’s evaluate first:\n\n\\[\n\\int_0^x \\cos(x + y + z) \\, dz = \\left[ \\sin(x + y + z) \\right]_{z=0}^{z=x} = \\sin(2x + y) - \\sin(x + y)\n\\]\n\n- The idea is now to do exactly the same for the next iterations, first in respect to \\( x \\) and finally in respect to \\( y \\).\n\n- At the end you should get: \\(-\\frac{1}{3}\\)", "id": "./materials/311.pdf" }, { "contents": "Evaluate $\\iiint_E 6xy \\, dV$ where $E$ lies under $z = 1 + x + y$ and above the region in the xy-plane bounded by $y = \\sqrt{x}$, $y = 0$ and $x = 1$.\n\n- Let’s first sketch the projection of $E$ over xy-plane:\n\n![Figure 1: Projection over xy-plane](image)\n\n- In this case:\n\n$$E = \\{(x, y, z) \\in \\mathbb{R}^3 : 0 \\leq x \\leq 1 \\land 0 \\leq y \\leq \\sqrt{x} \\land 0 \\leq z \\leq 1 + x + y\\}$$\n• This means that we can write the triple integral as:\n\n\\[\n\\int_0^1 \\int_0^{\\sqrt{x}} \\int_0^{1+x+y} 6xy \\, dz \\, dy \\, dx\n\\]\n\n• At the end you should get: \\( \\frac{65}{28} \\)", "id": "./materials/312.pdf" }, { "contents": "Evaluate \\( \\iiint_E x^2 e^y \\, dV \\) where \\( E \\) is bounded by \\( z = 1 - y^2 \\), \\( z = 0 \\), \\( x = 1 \\) and \\( x = -1 \\).\n\n- Let’s first sketch \\( E \\) over xyz-coordinate planes:\n\n![Figure 1: 3D sketch of the solid E](image)\n\n- \\( E \\) is the region below the parabolic cylinder\n\n\\[\nz = 1 - y^2\n\\]\n\nand above the square \\([-1, 1] \\times [-1, 1]\\) in the xy-plane.\n• This means that we can write the triple integral as:\n\n\\[\n\\int_{-1}^{1} \\int_{-1}^{1} \\int_{0}^{1-y^2} x^2 e^y \\, dz \\, dy \\, dx\n\\]\n\n• Integrate by parts twice – After the first integration you must get:\n\n\\[\n\\int_{-1}^{1} \\int_{-1}^{1} x^2 e^y (1 - y^2) \\, dy \\, dx\n\\]\n\n• At the end you should get: \\( \\frac{8}{3e} \\)", "id": "./materials/313.pdf" }, { "contents": "Evaluate $\\iiint_Q x^2 \\, dV$ where $Q$ is the solid tetrahedron with vertices $(0, 0, 0)$, $(1, 0, 0)$, $(0, 1, 0)$ and $(0, 0, 1)$.\n\n- Let’s first sketch $Q$ over $xyz$-coordinate planes:\n\n![Figure 1: 3D sketch of the solid $Q$](image)\n\n- If we think of the projection of $Q$ over $xy$-plane, we can define\n\n$$Q = \\{(x, y, z) \\in \\mathbb{R}^3 : 0 \\leq x \\leq 1 \\land 0 \\leq y \\leq 1 - x \\land 0 \\leq z \\leq 1 - x - y\\}$$\n\n- This means that we can write the triple integral as:\n\n$$\\int_0^1 \\int_0^{1-x} \\int_0^{1-x-y} x^2 \\, dz \\, dy \\, dx$$\n• At the end you should get: \\( \\frac{1}{60} \\)", "id": "./materials/314.pdf" }, { "contents": "Evaluate \\( \\iiint_E x \\, dV \\) where \\( E \\) is bounded by \\( x = 4y^2 + 4z^2 \\) and \\( x = 4 \\).\n\n- Let’s first sketch \\( E \\) over \\( xyz \\)-coordinate planes:\n\n![3D sketch of the solid E](image)\n\nFigure 1: 3D sketch of the solid \\( E \\)\n\n- The projection of \\( E \\) over \\( yz \\)-plane is the disk \\( D \\) defined as\n\n\\[\ny^2 + z^2 \\leq 1\n\\]\n\n- This means that we can write the triple integral as:\n\n\\[\n\\iint_D \\left[ \\int_{4y^2+4z^2}^{4} x \\, dx \\right] \\, dA\n\\]\n• Tip: switch to cylindrical coordinates after the first iteration:\n\n\\[\n\\begin{align*}\n y &= r \\cos(\\theta) \\\\\n z &= r \\sin(\\theta) \\\\\n x &= x\n\\end{align*}\n\\]\n\n• At the end you should get: \\( \\frac{16\\pi}{3} \\)", "id": "./materials/315.pdf" }, { "contents": "Use a triple integral to determine the volume of a tetrahedron $E$ enclosed by the coordinate planes and $2x + y + z = 4$\n\n- Let’s first sketch $E$ over $xyz$-coordinate planes:\n\n![3D sketch of E](image)\n\nFigure 1: 3D sketch of $E$\n\n- Taking into consideration the projection of the tetrahedron over $xy$-plane, we can define\n\n$$E = \\{(x, y, z) \\in \\mathbb{R}^3 : 0 \\leq x \\leq 2 \\land 0 \\leq y \\leq 4-2x \\land 0 \\leq z \\leq 4-2x-y\\}$$\n\n- Since we want to determine the volume of the solid, we can write the triple integral as:\n\n$$\\int_0^2 \\int_0^{4-2x} \\int_0^{4-2x-y} 1 \\, dz \\, dy \\, dx$$\n• At the end you should get: \\( \\frac{16}{3} \\)", "id": "./materials/316.pdf" }, { "contents": "Use a triple integral to determine the volume of the solid bounded by \n\\( x^2 + y^2 = 9 \\), \\( y + z = 5 \\) and \\( z = 1 \\)\n\n- Let’s first sketch the solid \\( E \\) over \\( xyz \\)-coordinate planes:\n\n![3D sketch of E](image)\n\nFigure 1: 3D sketch of \\( E \\)\n\n- Since we want to determine the volume of the solid, we can write the triple integral as:\n\n\\[\n\\int_{-3}^{3} \\int_{-\\sqrt{9-x^2}}^{\\sqrt{9-x^2}} \\int_{1}^{5-y} 1 \\, dz \\, dy \\, dx\n\\]\n\n- At the end you should get: \\( 36\\pi \\)", "id": "./materials/317.pdf" }, { "contents": "Evaluate \\( \\int_0^1 \\int_x^{2x} \\int_0^y 2xyz \\, dz \\, dy \\, dx \\)\n\n- Let’s first check how to determine a similar example:\n \\[\n \\int_0^2 \\int_{\\frac{z}{2}}^{2z} \\int_1^{2y} (x + yz) \\, dx \\, dy \\, dz\n \\]\n\n- Let’s evaluate first:\n \\[\n \\int_1^{2y} (x + yz) \\, dx\n = \\left[ \\frac{x^2}{2} + xyz \\right]_{x=1}^{x=2y}\n = 2y^2 - \\frac{1}{2} + 2y^2z - yz\n \\]\n\n- Now, let’s evaluate:\n \\[\n \\int_{\\frac{z}{2}}^{2z} 2y^2 - \\frac{1}{2} + 2y^2z - yz \\, dy\n = \\left[ \\frac{2y^3}{3} - \\frac{y}{2} + \\frac{2y^3z}{3} - \\frac{y^2z}{2} \\right]_{y=\\frac{z}{2}}^{y=2z}\n = -z + \\frac{z}{4} + \\frac{21z^3}{4} - 2z^3 + \\frac{z^3}{8} + \\frac{21z^4}{4}\n \\]\n\n- Finally, let’s evaluate:\n \\[\n \\int_0^2 -z + \\frac{z}{4} + \\frac{21z^3}{4} - 2z^3 + \\frac{z^3}{8} + \\frac{21z^4}{4} \\, dz\n = \\left[ -\\frac{z^2}{2} + \\frac{z^2}{8} + \\frac{21z^4}{16} - \\frac{z^4}{2} + \\frac{z^4}{32} + \\frac{21z^5}{20} \\right]_{z=0}^{z=2}\n = \\frac{228}{5}\n \\]\n• The idea is now to do exactly the same for the triple integral you were asked to evaluate:\n\n\\[ \\int_0^1 \\int_x^{2x} \\int_0^y 2xyz \\, dz \\, dy \\, dx \\]\n\n• At the end you should get: \\( \\frac{5}{8} \\)", "id": "./materials/318.pdf" }, { "contents": "Evaluate \\( \\int_1^2 \\int_0^{2z} \\int_0^{\\ln(x)} x e^{-y} \\, dy \\, dx \\, dz \\)\n\n- Let’s evaluate first:\n\n\\[\n\\int_0^{\\ln(x)} x e^{-y} \\, dy = \\left[ -x e^{-y} \\right]_{y=0}^{x=\\ln(x)} = -x e^{-\\ln(x)} + x e^0 = -1 + x\n\\]\n\n- The idea is now to do exactly the same for the next iterations, first in respect to \\( x \\) and finally in respect to \\( z \\).\n\n- At the end you should get: \\( \\frac{5}{3} \\)", "id": "./materials/319.pdf" }, { "contents": "Systems of linear equations\n\nTest. Let $Ax = b$ be a system of linear equations, where $A$ is a square matrix of order $n$ with coefficients in a field $K$.\n\nDecide whether the following sentences are true or false. Provide full explanation of your answers.\n\n(i) If $\\det A \\neq 0$, then the given system is equivalent to one with same variables but with the identity matrix as coefficient matrix.\n\n(ii) If $\\det A = 0$, then the system is not consistent.\n\n(iii) If $\\det A \\neq 0$, then the system has a unique solution.\n\n(iv) If $\\det A \\neq 0$, then $x = A^{-1}b$ is a solution of the system.\n\nSolution\n\n(i) The sentence is true. As $\\det A \\neq 0$, then $A$ has matrix inverse $A^{-1}$ and by left multiplication by it we get the equivalent system\n\n$$A^{-1}(Ax) = A^{-1}b \\iff (A^{-1}A)x = A^{-1}b \\iff Ix = A^{-1}b.$$\n\n(ii) The sentence is false in general. If $\\det A = 0$ we can only say that the rank of $A$ is not full. In this case the system can either have infinitely many solutions, as for\n\n$$\\begin{pmatrix} 1 & -1 \\\\ 0 & 0 \\end{pmatrix} \\begin{pmatrix} x_1 \\\\ x_2 \\end{pmatrix} = \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} \\iff \\begin{pmatrix} x_1 \\\\ x_2 \\end{pmatrix} = \\begin{pmatrix} 1 + h \\\\ h \\end{pmatrix} \\text{ for each } h \\in K,$$\n\nor no solution at all, as for\n\n$$\\begin{pmatrix} 1 & -1 \\\\ 0 & 0 \\end{pmatrix} \\begin{pmatrix} x_1 \\\\ x_2 \\end{pmatrix} = \\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix} \\text{ (no solution)}.$$\n\n(iii) The sentence is true. If $\\det A \\neq 0$, then the rank of the square matrix $A$ is equal to the number of its (rows and) columns, which in turns is equal to the number of variables; therefore the system has a unique solution by the Rouché-Capelli Theorem.\n\nOne can also argue, directly, that if $x^*$ is any solution of the given system, that is $Ax^* = b$, then by left multiplication by $A^{-1}$ one finds\n\n$$A^{-1}(Ax^*) = A^{-1}b \\iff (A^{-1}A)x^* = A^{-1}b \\iff Ix^* = A^{-1}b \\iff x^* = A^{-1}b,$$\n\nhence the system, if it has solution, has a unique solution, on the other hand one can easily check that $A^{-1}b$ is a solution.\n\n(iv) The sentence is true. As $\\det A \\neq 0$, then $A$ has matrix inverse $A^{-1}$ and by substitution we find\n\n$$A(A^{-1}b) = (AA^{-1})b = Ib = b,$$\n\nhence $x := A^{-1}b$ is a solution of the system.", "id": "./materials/32.pdf" }, { "contents": "Evaluate \\( \\int_0^1 \\int_0^1 \\int_0^{\\sqrt{1-z^2}} \\frac{z}{y+1} \\, dx \\, dz \\, dy \\)\n\n- Let’s evaluate first:\n\n\\[\n\\int_0^{\\sqrt{1-z^2}} \\frac{z}{y+1} \\, dx = \\left[ \\frac{zx}{y+1} \\right]_{x=0}^{x=\\sqrt{1-z^2}} = \\frac{z\\sqrt{1-z^2}}{y+1}\n\\]\n\n- The idea is now to do exactly the same for the next iterations, first in respect to \\( z \\) and finally in respect to \\( y \\).\n\n- At the end you should get: \\( \\frac{\\ln(2)}{3} \\)", "id": "./materials/320.pdf" }, { "contents": "Evaluate \\( \\iiint_E e^z \\, dV \\) where\n\n\\[ E = \\{(x, y, z) \\in \\mathbb{R}^3 : 1 \\leq y \\leq 2 \\land y \\leq x \\leq 1 \\land 0 \\leq z \\leq xy\\} \\]\n\n- We can write the triple integral as:\n\n\\[\n\\int_1^2 \\int_y^1 \\int_0^{xy} e^z \\, dz \\, dy \\, dx\n\\]\n\n- Let’s evaluate first:\n\n\\[\n\\int_0^{xy} e^z \\, dz = \\left[ ye^z \\right]_{z=0}^{z=xy} = ye^x - y\n\\]\n\n- The idea is now to do exactly the same for the next iterations, first in respect to \\( y \\) and finally in respect to \\( x \\).\n\n- At the end you should get: \\( \\frac{3e}{2} + \\frac{5}{6} - e^2 \\)", "id": "./materials/321.pdf" }, { "contents": "Evaluate \\( \\iiint_E \\frac{z}{x^2 + z^2} \\, dV \\) where\n\n\\[ E = \\{(x, y, z) \\in \\mathbb{R}^3 : 1 \\leq y \\leq 4 \\land y \\leq z \\leq 4 \\land 0 \\leq x \\leq z\\} \\]\n\n- We can write the triple integral as:\n\n\\[\n\\int_1^4 \\int_y^4 \\int_0^z \\frac{z}{x^2 + z^2} \\, dx \\, dz \\, dy\n\\]\n\n- Let’s evaluate first:\n\n\\[\n\\int_0^z \\frac{z}{x^2 + z^2} \\, dx = \\left[ \\frac{z \\arctan \\left( \\frac{x}{z} \\right)}{z} \\right]_{x=0}^{x=z} = \\arctan(1) - \\arctan(0) = \\frac{\\pi}{4}\n\\]\n\n- The idea is now to do exactly the same for the next iterations, first in respect to \\( z \\) and finally in respect to \\( y \\).\n\n- At the end you should get: \\( \\frac{9\\pi}{8} \\)", "id": "./materials/322.pdf" }, { "contents": "Evaluate $\\iiint_E \\sin(y) \\, dV$ where $E$ lies between $z = x$ and the triangular region with vertices $(0, 0, 0)$, $(\\pi, 0, 0)$ and $(0, \\pi, 0)$.\n\n- Let’s first sketch the solid $E$ over $xyz$-coordinate planes:\n\n![3D sketch of E](image)\n\nFigure 1: 3D sketch of $E$\n\n- This means that we can define\n\n$$E = \\{(x, y, z) \\in \\mathbb{R}^3 : 0 \\leq x \\leq \\pi \\land 0 \\leq y \\leq \\pi - x \\land 0 \\leq z \\leq x\\}$$\n\n- We can write the triple integral as:\n\n$$\\int_0^\\pi \\int_0^{\\pi-x} \\int_0^x \\sin(y) \\, dz \\, dy \\, dx$$\n• At the end you should get: $\\frac{\\pi^2}{2} - 2$", "id": "./materials/323.pdf" }, { "contents": "Evaluate $\\iiint_E xy \\, dV$ where $E$ is bounded by $y = x^2$, $x = y^2$, $z = 0$ and $z = x + y$.\n\n- Let’s first sketch the solid $E$ over $xyz$-coordinate planes:\n\nFigure 1: 3D sketch of $E$\nThus, we can write the triple integral as:\n\n\\[\n\\int_0^1 \\int_{x^2}^{\\sqrt{x}} \\int_0^{x+y} xy \\, dz \\, dy \\, dx\n\\]\n\n\\[\n= \\int_0^1 \\int_{x^2}^{\\sqrt{x}} xy \\left[ z \\right]_{z=0}^{z=x+y} \\, dy \\, dx\n\\]\n\n\\[\n= \\int_0^1 \\int_{x^2}^{\\sqrt{x}} x^2 y + xy^2 \\, dy \\, dx\n\\]\n\n\\[\n= \\int_0^1 \\left[ \\frac{x^2 y^2}{2} + \\frac{xy^3}{3} \\right]_{y=x^2}^{y=\\sqrt{x}} \\, dx\n\\]\n\n\\[\n= \\int_0^1 \\left( \\frac{x^3}{2} + \\frac{x^{5/2}}{3} - \\frac{x^6}{2} - \\frac{x^{7/3}}{3} \\right) \\, dx\n\\]\n\n\\[\n= \\left[ \\frac{x^4}{8} + \\frac{2x^{7/2}}{21} - \\frac{x^7}{14} - \\frac{x^{8/3}}{24} \\right]_{x=0}^{x=1}\n\\]\n\n\\[\n= \\frac{1}{8} + \\frac{2}{21} - \\frac{1}{14} - \\frac{1}{24}\n\\]\n\n\\[\n= \\frac{3}{28}\n\\]", "id": "./materials/324.pdf" }, { "contents": "Evaluate $\\iiint_E xyz \\, dV$ where $E$ is the solid tetrahedron with vertices \n$(0, 0, 0), (1, 0, 0), (1, 1, 0)$ and $(1, 0, 1)$.\n\n- Let’s first sketch the solid $E$ over $xyz$-coordinate planes:\n\n![3D sketch of E](image)\n\nFigure 1: 3D sketch of $E$\n\n- This means that we can define\n\n$$E = \\{(x, y, z) \\in \\mathbb{R}^3 : 0 \\leq x \\leq 1 \\land 0 \\leq y \\leq x \\land 0 \\leq z \\leq x - y\\}$$\n\n- Thus, we can write the triple integral as:\n\n$$\\int_0^1 \\int_0^x \\int_0^{x-y} xyz \\, dz \\, dy \\, dx$$\n• At the end you should get: \\( \\frac{1}{144} \\)", "id": "./materials/325.pdf" }, { "contents": "Evaluate $\\iiint_E z \\, dV$ where $E$ is bounded by $y^2 + z^2 = 9$, $x = 0$, $y = 3x$ and $z = 0$ in the first octant.\n\n- Let’s first sketch the solid $E$ over $xyz$-coordinate planes:\n\n![3D sketch of E](image)\n\nFigure 1: 3D sketch of $E$\n\n- Thus, we can write the triple integral as:\n\n$$\\int_0^1 \\int_{3x}^3 \\int_0^{\\sqrt{9-y^2}} z \\, dz \\, dy \\, dx$$\n\n- At the end you should get: $\\frac{27}{8}$", "id": "./materials/326.pdf" }, { "contents": "Use a triple integral to determine the volume of a solid bounded by \n\\( y = x^2 + z^2 \\) and \\( y = 8 - x^2 - z^2 \\).\n\n- Let’s first sketch \\( E \\) over \\( xyz \\)-coordinate planes:\n\n![3D sketch of E](image)\n\nFigure 1: 3D sketch of \\( E \\)\n\n- Taking into consideration the projection of the tetrahedron over \n \\( xz \\)-plane, which is the disk \\( D \\) defined by \n \\[ x^2 + z^2 \\leq 4 \\]\n\nwe can define\n\n\\[\nE = \\{(x, y, z) \\in \\mathbb{R}^3 : x^2 + z^2 \\leq y \\leq 8 - x^2 - z^2 \\land x^2 + z^2 \\leq 4\\}\n\\]\n• Since we want to determine the volume of the solid, we can write the triple integral as:\n\n\\[\n\\iiint_D \\left[ \\int_{x^2+z^2}^{8-x^2-z^2} 1 \\, dy \\right] \\, dA\n\\]\n\n• Tip: After the first iteration, switch to cylindrical coordinates:\n\n\\[\n\\begin{align*}\n x &= r \\cos(\\theta) \\\\\n z &= r \\sin(\\theta) \\\\\n z &= z\n\\end{align*}\n\\]\n\n• At the end you should get: \\(16\\pi\\)", "id": "./materials/327.pdf" }, { "contents": "Use a triple integral to determine the volume of a solid bounded by \\( y = x^2 \\), \\( z = 0 \\) and \\( y + z = 1 \\).\n\n- Let’s first sketch \\( E \\) over xyz-coordinate planes:\n\n![3D sketch of E](image)\n\nFigure 1: 3D sketch of \\( E \\)\n\n- The plane \\( y + z = 1 \\) intersects the xy-plane in the line \\( y = 1 \\), so we can define\n\n\\[\nE = \\{(x, y, z) \\in \\mathbb{R}^3 : -1 \\leq x \\leq 1 \\land x^2 \\leq y \\leq 1 \\land 0 \\leq z \\leq 1 - y\\}\n\\]\n• Since we want to determine the volume of the solid, we can write the triple integral as:\n\n\\[ \\int_{-1}^{1} \\int_{x^2}^{1} \\int_{0}^{1-y} dz \\, dy \\, dx \\]\n\n• At the end you should get: \\( \\frac{8}{15} \\)", "id": "./materials/328.pdf" }, { "contents": "Use a triple integral to determine the volume of a solid bounded by \n\\( x^2 + z^2 = 4 \\), \\( y = -1 \\) and \\( y + z = 4 \\).\n\n- Let’s first sketch \\( E \\) over \\( xyz \\)-coordinate planes:\n\n![3D sketch of E](image)\n\n**Figure 1: 3D sketch of E**\n\n- We can define\n\n\\[\nE = \\{(x, y, z) \\in \\mathbb{R}^3 : -1 \\leq y \\leq 4 - z \\land x^2 + z^2 \\leq 4\\}\n\\]\n\n- Since we want to determine the volume of the solid, we can write the triple integral as:\n\n\\[\n\\int_{-2}^{2} \\int_{-\\sqrt{4-x^2}}^{\\sqrt{4-x^2}} \\int_{-1}^{4-z} 1 \\, dy \\, dz \\, dx\n\\]\n\n- At the end you should get: \\( 20\\pi \\)", "id": "./materials/329.pdf" }, { "contents": "Test. Consider the following two systems of linear equations:\n\n\\[\n(I) \\begin{cases}\n x + y + z + t = 1 \\\\\n x + y + 5z + 7t = 5\n\\end{cases}\n\\quad\n(II) \\begin{cases}\n x + y + z + t = 1 \\\\\n 2z + 3t = 2 \\\\\n x + y - z - 2t = -1\n\\end{cases}\n\\]\n\nDecide whether the following sentences are true or false. Provide full explanation of your answers.\n\n(i) The two systems can’t be equivalent as they do not have the same number of equations.\n\n(ii) Every solution of the first system is a solution of the second, but not the converse.\n\n(iii) The two systems are equivalent.\n\nSolution\n\n(i) The sentence is false. In general, two equivalent systems can have, each one, any number of equations. What is true is that equivalent consistent linear systems must have the same number of maximum (linearly) independent equations, that is to say that their matrices must have the same rank. We will see that the two given systems are actually equivalent, albeit they do not have the same number of equations.\n\n(ii) The sentence is false. We will show that the two given systems are equivalent, hence every solution of the first system is a solution of the second, and the converse is also true.\n\n(iii) The sentence is true. We will show the two systems are equivalent in two ways: working with the corresponding (augmented) matrices and also by direct inspection of the equations.\n\nUsing the matrices, we get\n\n\\[\n\\begin{pmatrix}\n 1 & 1 & 1 & 1 & 1 \\\\\n 1 & 1 & 5 & 7 & 5\n\\end{pmatrix}\n\\quad\n\\begin{pmatrix}\n 1 & 1 & 1 & 1 & 1 \\\\\n 0 & 0 & 2 & 3 & 3 \\\\\n 1 & 1 & -1 & -2 & -1\n\\end{pmatrix}\n\\]\n\nby row elementary operations the two matrices reduce to\n\n\\[\n\\begin{pmatrix}\n 1 & 1 & 1 & 1 & 1 \\\\\n 0 & 0 & 4 & 6 & 4\n\\end{pmatrix}\n\\quad\n\\begin{pmatrix}\n 1 & 1 & 1 & 1 & 1 \\\\\n 0 & 0 & 2 & 3 & 2 \\\\\n 0 & 0 & -2 & -3 & -2\n\\end{pmatrix}\n\\]\n\nand eventually to\n\n\\[\n\\begin{pmatrix}\n 1 & 1 & 1 & 1 & 1 \\\\\n 0 & 0 & 2 & 3 & 2\n\\end{pmatrix}\n\\quad\n\\begin{pmatrix}\n 1 & 1 & 1 & 1 & 1 \\\\\n 0 & 0 & 2 & 3 & 2 \\\\\n 0 & 0 & 0 & 0 & 0\n\\end{pmatrix}\n\\]\n\nThe last two matrices represent the same system of linear equations, hence the given two systems are equivalent.\n\nThe same result can be found as follows. Let us call \\( E_1, E_2 \\) the two linear equations of the first system, and \\( E_3, E_4, E_5 \\) those of the last one. Then we can observe: \\( E_1 = E_3, E_2 = E_3 + 2E_4, E_5 = E_3 - E_4 \\) (hence the equation \\( E_5 \\) is actually redundant in the second system and therefore can be discarded), and also \\( E_3 = E_1 \\) and \\( E_4 = \\frac{1}{2}(E_2 - E_1) \\). Thus every equation of the first system is a linear combination of equations from the second and viceversa, hence the two systems are equivalent.", "id": "./materials/33.pdf" }, { "contents": "Evaluate \\( \\iiint_E \\sqrt{x^2 + y^2} \\, dV \\) where \\( E \\) is the region bounded by\n\\[ x^2 + y^2 = 16, \\quad z = -5 \\text{ and } z = 4. \\]\n\n- Let’s first sketch \\( E \\) over \\( xyz \\)-coordinate planes:\n\n![Figure 1: 3D sketch of \\( E \\)](image)\n\n- Using cylindrical coordinates, we can define:\n\\[ E = \\{(r, \\theta, z) \\mid 0 \\leq \\theta \\leq 2\\pi \\land 0 \\leq r \\leq 4 \\land -5 \\leq z \\leq 4\\} \\]\n\n- We can write the triple integral as:\n\\[ \\int_0^{2\\pi} \\int_0^4 \\int_{-5}^4 \\sqrt{r^2} \\, dz \\, dr \\, d\\theta \\]\n\n- At the end you should get: 384\\( \\pi \\)", "id": "./materials/330.pdf" }, { "contents": "Evaluate $\\iiint_E z \\, dV$ where $E$ is the region bounded by $z = x^2 + y^2$ and $z = 4$.\n\n- Let’s first sketch $E$ over $xyz$-coordinate planes:\n\n![3D sketch of E](image)\n\nFigure 1: 3D sketch of $E$\n\n- The paraboloid $z = x^2 + y^2$ intersects the plane $z = 4$ in the circle $x^2 + y^2 = 4$ or $r^2 = 4 \\Rightarrow r = 2$.\nso in cylindrical coordinates we can define\n\n\\[ E = \\{(r, \\theta, z) \\mid 0 \\leq \\theta \\leq 2\\pi \\land 0 \\leq r \\leq 2 \\land r^2 \\leq z \\leq 4\\} \\]\n\n- We can write the triple integral as:\n\n\\[\n\\int_0^{2\\pi} \\int_0^2 \\int_{r^2}^4 zr \\, dz \\, dr \\, d\\theta\n\\]\n\n- At the end you should get: \\( \\frac{64\\pi}{3} \\)", "id": "./materials/331.pdf" }, { "contents": "Evaluate $\\iiint_E (x + y + z) \\, dV$ where $E$ is the solid in the first octant that lies under $z = 4 - x^2 - y^2$.\n\n- Let’s first sketch $E$ over $xyz$-coordinate planes:\n\n![3D sketch of E](image)\n\nFigure 1: 3D sketch of $E$\n\n- The paraboloid\n\n$$z = 4 - x^2 - y^2 \\quad \\text{or} \\quad z = 4 - r^2$$\n\nintersects the $xy$-plane in the circle\n\n$$x^2 + y^2 = 4 \\quad \\text{or} \\quad r^2 = 4 \\Rightarrow r = 2$$\nso in cylindrical coordinates, we can define\n\n\\[ E = \\{(r, \\theta, z) \\mid 0 \\leq \\theta \\leq \\frac{\\pi}{2} \\land 0 \\leq r \\leq 2 \\land 0 \\leq z \\leq 4 - r^2\\} \\]\n\n- We can write the triple integral as:\n\n\\[\n\\int_0^{\\frac{\\pi}{2}} \\int_0^2 \\int_0^{4-r^2} (r \\cos \\theta + r \\sin \\theta + z) r \\, dz \\, dr \\, d\\theta\n\\]\n\n- At the end you should get: \\( \\frac{8\\pi}{3} + \\frac{128}{15} \\)", "id": "./materials/332.pdf" }, { "contents": "Evaluate $\\iiint_E x \\, dV$ where $E$ is the solid bounded by $z = 0$, $z = x + y + 5$, $x^2 + y^2 = 4$ and $x^2 + y^2 = 9$.\n\n- Let’s first sketch $E$ over $xyz$-coordinate planes:\n\n![Figure 1: 3D sketch of $E$](image)\n\n- In cylindrical coordinates $E$ is bounded by the planes\n\n$$z = 0 \\quad \\text{and} \\quad z = r \\cos(\\theta) + r \\sin(\\theta) + 5,$$\n\nand the cylinders\n\n$$r = 2 \\quad \\text{and} \\quad r = 3,$$\n\nso we can define\n\n$$E = \\{(r, \\theta, z) \\mid 0 \\leq \\theta \\leq 2\\pi \\land 2 \\leq r \\leq 3 \\land 0 \\leq z \\leq r \\cos(\\theta) + r \\sin(\\theta) + 5\\}$$\n• We can write the triple integral as:\n\n\\[\n\\int_0^{2\\pi} \\int_0^3 \\int_0^r (r \\cos(\\theta) + r \\sin(\\theta) + 5) r \\, dz \\, dr \\, d\\theta\n\\]\n\n• At the end you should get: \\( \\frac{65\\pi}{4} \\)", "id": "./materials/333.pdf" }, { "contents": "Find the volume of the solid that lies within both $x^2 + y^2 = 1$ and $x^2 + y^2 + z^2 = 4$.\n\n- Let’s first sketch $E$ over $xyz$-coordinate planes:\n\n![3D sketch of E](image)\n\nFigure 1: 3D sketch of $E$\n\n- In cylindrical coordinates $E$ is the solid region within the cylinder\n\n$$r = 1$$\n\nand bounded above and below by the sphere\n\n$$r^2 + z^2 = 4,$$\n\nso we can define\n\n$$E = \\{(r, \\theta, z) \\mid 0 \\leq \\theta \\leq 2\\pi \\land 0 \\leq r \\leq 1 \\land -\\sqrt{4-r^2} \\leq z \\leq \\sqrt{4-r^2}\\}$$\n• We can write the triple integral as:\n\n\\[ \\int_0^{2\\pi} \\int_0^1 \\int_{-\\sqrt{4-r^2}}^{\\sqrt{4-r^2}} r \\, dz \\, dr \\, d\\theta \\]\n\n• At the end you should get: \\( \\frac{4\\pi(8 - \\sqrt{27})}{3} \\)", "id": "./materials/334.pdf" }, { "contents": "Find the volume of the solid enclosed by \\( z = \\sqrt{x^2 + y^2} \\) and \\( x^2 + y^2 + z^2 = 2 \\).\n\n- Let’s first sketch \\( E \\) over \\( xyz \\)-coordinate planes:\n\n![3D sketch of \\( E \\)](image)\n\nFigure 1: 3D sketch of \\( E \\)\n\n- In cylindrical coordinates \\( E \\) is bounded below the cone\n\n\\[\nz = r\n\\]\n\nand above the sphere\n\n\\[\nr^2 + z^2 = 2.\n\\]\n\nThe cone and the sphere intersect when\n\n\\[\n2r^2 = 2 \\quad \\Rightarrow \\quad r = 1,\n\\]\n\nso we can define\n\n\\[\nE = \\{(r, \\theta, z) \\mid 0 \\leq \\theta \\leq 2\\pi \\land 0 \\leq r \\leq 1 \\land r \\leq z \\leq \\sqrt{2 - r^2}\\}\n\\]\n• We can write the triple integral as:\n\n\\[ \\int_0^{2\\pi} \\int_0^1 \\int_r^{\\sqrt{2-r^2}} r \\, dz \\, dr \\, d\\theta \\]\n\n• At the end you should get: \\( \\frac{4\\pi(\\sqrt{2} - 1)}{3} \\)", "id": "./materials/335.pdf" }, { "contents": "Find the volume of the solid enclosed by \\( z = x^2 + y^2 \\) and \\( x^2 + y^2 + z^2 = 2 \\).\n\n- Let’s first sketch \\( E \\) over \\( xyz \\)-coordinate planes:\n\n![3D sketch of E](image)\n\nFigure 1: 3D sketch of \\( E \\)\n\n- In cylindrical coordinates \\( E \\) is bounded below\n\n\\[\nz = r^2\n\\]\n\nand above\n\n\\[\nr^2 + z^2 = 2.\n\\]\n\nThe cone and the sphere intersect when\n\n\\[\nr^2 + r^4 = 2 \\quad \\Rightarrow \\quad (r^2 + 2)(r^2 - 1) = 0 \\quad \\Rightarrow \\quad r = 1,\n\\]\n\nso we can define\n\n\\[\nE = \\{(r, \\theta, z) \\mid 0 \\leq \\theta \\leq 2\\pi \\land 0 \\leq r \\leq 1 \\land r^2 \\leq z \\leq \\sqrt{2 - r^2}\\}\n\\]\n• We can write the triple integral as:\n\n\\[ \\int_0^{2\\pi} \\int_0^1 \\int_{r^2}^{\\sqrt{2-r^2}} r \\, dz \\, dr \\, d\\theta \\]\n\n• At the end you should get: \\( \\pi \\left( -\\frac{7}{6} + \\frac{4\\sqrt{2}}{3} \\right) \\)", "id": "./materials/336.pdf" }, { "contents": "Find the volume of the solid enclosed by \\( z = x^2 + y^2 \\) and \\( z = 36 - 3x^2 - 3y^2 \\).\n\n- Let’s first sketch \\( E \\) over \\( xyz \\)-coordinate planes:\n\n![3D sketch of E](image)\n\n**Figure 1: 3D sketch of \\( E \\)**\n\n- The paraboloids intersect when\n\n\\[\nx^2 + y^2 = 36 - 3x^2 - 3y^2 \\quad \\Rightarrow \\quad x^2 + y^2 = 9,\n\\]\n\nso the region of integration is\n\n\\[\nD = \\{(x, y) \\mid x^2 + y^2 \\leq 9\\}.\n\\]\n\nThus, in cylindrical coordinates, we can define\n\n\\[\nE = \\{(r, \\theta, z) \\mid r^2 \\leq z \\leq 36 - 3r^2 \\land 0 \\leq r \\leq 3 \\land 0 \\leq \\theta \\leq 2\\pi\\}\n\\]\n• We can write the triple integral as:\n\n\\[\n\\int_0^{2\\pi} \\int_0^3 \\int_{r^2}^{36-3r^2} r \\, dz \\, dr \\, d\\theta\n\\]\n\n• At the end you should get: 162\\pi", "id": "./materials/337.pdf" }, { "contents": "Evaluate \\[ \\int_{-2}^{2} \\int_{-\\sqrt{4-y^2}}^{\\sqrt{4-y^2}} \\int_{-\\sqrt{4-y^2}}^{\\sqrt{4-y^2}} xz \\, dz \\, dx \\, dy \\]\n\n- Tip: switch to cylindrical coordinates after the first iteration:\n \\[\n \\begin{align*}\n y &= r \\cos(\\theta) \\\\\n z &= r \\sin(\\theta) \\\\\n x &= x\n \\end{align*}\n \\]\n\n- The region of integration is the region above the cone\n \\[ z = \\sqrt{x^2 + y^2} \\quad \\text{or} \\quad z = r \\]\n and below the plane\n \\[ z = 2 \\]\n Also, we have\n \\[-2 \\leq y \\leq 2 \\quad \\text{and} \\quad -\\sqrt{4-y^2} \\leq x \\leq \\sqrt{4-y^2} \\]\n which describes a circle of radius 2 in the xy-plane centered at (0, 0).\n\n- This means that we can rewrite the triple integral as:\n \\[\n \\int_{0}^{2\\pi} \\int_{0}^{2} \\int_{r}^{2} (r \\cos(\\theta))zr \\, dz \\, dr \\, d\\theta\n \\]\n\n- Now, it is just to determine this triple integral with simpler bounds.\n\n- At the end you should get: 0", "id": "./materials/338.pdf" }, { "contents": "Evaluate \\( \\int_{-3}^{3} \\int_{0}^{\\sqrt{9-y^2}} \\int_{0}^{9-x^2-y^2} \\sqrt{x^2 + y^2} \\, dz \\, dy \\, dx \\)\n\n- The region of integration is the region above the plane \n \\( z = 0 \\)\n and below the paraboloid \n \\( z = 9 - x^2 - y^2 \\).\n\n- Also, we have \n \\(-3 \\leq x \\leq 3 \\quad \\land \\quad 0 \\leq y \\leq \\sqrt{9-x^2}\\)\n which describes the upper half of a circle of radius 3 in the xy-plane, centered at (0,0).\n\n- So we can rewrite the triple integral, using cylindrical coordinates:\n \\[\n \\int_{-3}^{3} \\int_{0}^{\\sqrt{9-y^2}} \\int_{0}^{9-x^2-y^2} \\sqrt{x^2 + y^2} \\, dz \\, dy \\, dx \\\\\n = \\int_{0}^{\\pi} \\int_{0}^{3} \\int_{0}^{9-r^2} \\sqrt{r^2} \\, r \\, dz \\, dr \\, d\\theta \\\\\n = \\int_{0}^{\\pi} \\int_{0}^{3} \\int_{0}^{9-r^2} r^2 \\, dz \\, dr \\, d\\theta\n \\]\n\n- Now is just to solve the triple integral.\n\n- At the end you should get: \\( \\frac{162\\pi}{5} \\)", "id": "./materials/339.pdf" }, { "contents": "Systems of linear equations\n\nTest. Consider a system of \\( n \\) linear equations. Decide whether the following sentences are true or false. Provide full explanation of your answers.\n\n(i) Multiplying by a non zero scalar its first equation, one gets an equivalent system.\n(ii) If it is homogeneous, then it is consistent.\n(iii) By substitution of one of its equation by a linear combination of the others, one gets an equivalent system.\n(iv) If it is homogeneous, then it has a unique solution.\n\nSolution\n\n(i) The sentence is true, as the new equation one gets in this way has exactly the same solutions of the starting one.\n\n(ii) The sentence is true. As the system is homogeneous (i.e. its constant term is zero), then the zero vector whose size equals to the number of the variables is certainly a solution. Hence the system is necessarily consistent.\n\nFrom a different perspective: if the system is homogeneous, then its augmented matrix and coefficient matrix have the same rank, hence the system is necessarily consistent by the Rouché-Capelli Theorem.\n\n(iii) The sentence is false in general. It would be true only if the discarded equation is itself a linear combination of the others and it is replaced with a non-zero linear combination of the others. Let us consider the following examples:\n\n\\[\n\\begin{align*}\n(a) & \\quad \\begin{cases} x = 0 \\\\ y = 0 \\\\ z = 0 \\end{cases} & (b) & \\quad \\begin{cases} x + y = 0 \\\\ y = 0 \\\\ z = 0 \\end{cases}\n\\end{align*}\n\\]\n\nIn (a), one cannot replace \\( x = 0 \\) with \\( y + z = 0 \\): the new system is not equivalent to system (a):\n\n\\[\n\\begin{align*}\n\\begin{cases} y + z = 0 \\\\ y = 0 \\\\ z = 0 \\end{cases} & \\iff \\begin{cases} y = 0 \\\\ z = 0 \\end{cases}\n\\end{align*}\n\\]\n\nhas solutions \\( \\{(x; 0; 0) \\mid x \\in \\mathbb{R}\\} \\) while (a) has solution \\( \\{(0; 0; 0)\\} \\), hence the two systems are not equivalent.\n\nIn (b), one cannot replace \\( x + y = 0 \\) with \\( 0y + 0z = 0 \\) (trivial linear combination): the new system is not equivalent to system (b):\n\n\\[\n\\begin{align*}\n\\begin{cases} 0 = 0 \\\\ y = 0 \\\\ z = 0 \\end{cases} & \\iff \\begin{cases} y = 0 \\\\ z = 0 \\end{cases}\n\\end{align*}\n\\]\n\nhas solutions \\( \\{(x; 0; 0) \\mid x \\in \\mathbb{R}\\} \\) while (b) has solution \\( \\{(0; 0; 0)\\} \\).\n\n(iv) The sentence is false in general. It would be true only if its coefficient matrix is a nondegenerate square matrix (that is a square matrix with non zero determinant).\n\nTo show that the sentence is false in general consider the system given by a unique homogeneous equation \\( x + y = 0 \\), which has the infinitely many solutions \\( \\{(t; -t) \\mid t \\in \\mathbb{R}\\} \\).\n\nAlso the homogeneous linear system consisting of the single equation \\( x = 0 \\) in two variables \\( x \\) and \\( y \\) has the infinitely many solutions \\( \\{(0; t) \\mid t \\in \\mathbb{R}\\} \\).", "id": "./materials/34.pdf" }, { "contents": "Evaluate \\( \\iiint_E (9 - x^2 - y^2) \\, dV \\) where \\( E \\) is the solid hemisphere \\( x^2 + y^2 + z^2 \\leq 9 \\) with \\( z \\geq 0 \\).\n\n- Let’s first sketch \\( E \\) over \\( xyz \\)-coordinate planes:\n\n![Figure 1: 3D sketch of \\( E \\)](image)\n\n- Tip: change to spherical coordinates\n\n\\[\n\\begin{align*}\nx &= r \\cos(\\theta) \\sin(\\phi) \\\\\ny &= r \\sin(\\theta) \\sin(\\phi) \\\\\nz &= r \\cos(\\phi)\n\\end{align*}\n\\]\n\nSo, in this case, can define\n\n\\[\nE = \\{(r, \\theta, \\phi) | 0 \\leq r \\leq 3 \\land 0 \\leq \\theta \\leq 2\\pi \\land 0 \\leq \\phi \\leq \\frac{\\pi}{2}\\}\n\\]\nThus, we are able to write the triple integral:\n\n\\[\n\\iiint_E (9 - x^2 - y^2) \\, dV\n\\]\n\n\\[\n= \\int_0^{\\frac{\\pi}{2}} \\int_0^{2\\pi} \\int_0^3 \\left[ 9 - (r^2 \\sin^2(\\phi) \\cos^2(\\theta) + r^2 \\sin^2(\\phi) \\sin^2(\\theta)) \\right] r^2 \\sin(\\phi) \\, dr \\, d\\theta \\, d\\phi\n\\]\n\nNow is just to solve it.\n\nYou should get as the result: \\( \\frac{486\\pi}{5} \\)", "id": "./materials/340.pdf" }, { "contents": "Evaluate \\( \\iiint_E (x^2 + y^2) \\, dV \\) where \\( E \\) lies between \\( x^2 + y^2 + z^2 = 4 \\) and \\( x^2 + y^2 + z^2 = 9 \\).\n\n- Let’s first sketch \\( E \\) over xyz-coordinate planes:\n\n![3D sketch of E](image)\n\nFigure 1: 3D sketch of \\( E \\)\n\n- Tip: change to spherical coordinates\n\n\\[\n\\begin{align*}\n x &= r \\cos(\\theta) \\sin(\\phi) \\\\\n y &= r \\sin(\\theta) \\sin(\\phi) \\\\\n z &= r \\cos(\\phi)\n\\end{align*}\n\\]\nSo, in this case, can define\n\n\\[ E = \\{(r, \\theta, \\phi) | 2 \\leq r \\leq 3 \\land 0 \\leq \\theta \\leq 2\\pi \\land 0 \\leq \\phi \\leq \\pi \\} \\]\n\n- We can also assess\n\n\\[\nx^2 + y^2 = r^2 \\sin^2(\\phi) \\cos^2(\\theta) + r^2 \\sin^2(\\phi) \\sin^2(\\theta) = r^2 \\sin^2(\\phi)(\\cos^2(\\theta) + \\sin^2(\\theta)) = r^2 \\sin^2(\\phi)\n\\]\n\n- Thus, we are able to write the triple integral:\n\n\\[\n\\iiint_E (x^2 + y^2) \\, dV = \\int_0^{2\\pi} \\int_0^\\pi \\int_2^3 (r^2 \\sin^2(\\phi))(r^2 \\sin(\\phi)) \\, dr \\, d\\theta \\, d\\phi\n\\]\n\n- Now is just to solve it.\n\n- You should get as the result: \\( \\frac{1688\\pi}{15} \\)", "id": "./materials/341.pdf" }, { "contents": "Evaluate $\\iiint_E y^2 \\, dV$ where $E$ is the solid hemisphere $x^2 + y^2 + z^2 \\leq 9$ with $y \\geq 0$.\n\n- Let’s first sketch $E$ over $xyz$-coordinate planes:\n\n![3D sketch of E](image)\n\nFigure 1: 3D sketch of $E$\n\n- Tip: change to spherical coordinates\n\n$$\n\\begin{align*}\n x &= r \\cos(\\theta) \\sin(\\phi) \\\\\n y &= r \\sin(\\theta) \\sin(\\phi) \\\\\n z &= r \\cos(\\phi)\n\\end{align*}\n$$\n\nSo, in this case, can define\n\n$$\nE = \\{(r, \\theta, \\phi) | 0 \\leq r \\leq 3 \\land 0 \\leq \\theta \\leq \\pi \\land 0 \\leq \\phi \\leq \\pi\\}\n$$\n• Thus, we are able to write the triple integral:\n\n\\[\n\\iiint_E y^2 \\, dV = \\int_0^\\pi \\int_0^\\pi \\int_0^3 (r \\sin(\\theta) \\sin(\\phi))^2 (r^2 \\sin(\\phi)) \\, dr \\, d\\theta \\, d\\phi\n\\]\n\n• Now is just to solve it.\n\n• You should get as the result: \\( \\frac{162\\pi}{5} \\)", "id": "./materials/342.pdf" }, { "contents": "Evaluate \\( \\iiint_E x e^{x^2+y^2+z^2} \\, dV \\) where \\( E \\) is the solid constrained by \\( x^2 + y^2 + z^2 = 1 \\), that lies in the first octant.\n\n- Let’s first sketch \\( E \\) over \\( xyz \\)-coordinate planes:\n\n![3D sketch of E](image)\n\nFigure 1: 3D sketch of \\( E \\)\n\n- Tip: change to spherical coordinates\n\n\\[\n\\begin{align*}\n x &= r \\cos(\\theta) \\sin(\\phi) \\\\\n y &= r \\sin(\\theta) \\sin(\\phi) \\\\\n z &= r \\cos(\\phi)\n\\end{align*}\n\\]\n\nSo, in this case, can define\n\n\\[\nE = \\{(r, \\theta, \\phi) | 0 \\leq r \\leq 1 \\land 0 \\leq \\theta \\leq \\frac{\\pi}{2} \\land 0 \\leq \\phi \\leq \\frac{\\pi}{2}\\}\n\\]\n• Thus, we are able to write the triple integral:\n\n\\[\n\\iiint_E x e^{x^2+y^2+z^2} \\, dV\n\\]\n\n\\[\n= \\int_0^{\\frac{\\pi}{2}} \\int_0^{\\frac{\\pi}{2}} \\int_0^1 (r \\sin(\\phi) \\cos(\\theta)) e^{r^2} r^2 \\sin(\\phi) \\, dr \\, d\\theta \\, d\\phi\n\\]\n\n• Now is just to solve it.\n\n• You should get as the result: \\( \\frac{\\pi}{8} \\)", "id": "./materials/343.pdf" }, { "contents": "Evaluate $\\iiint_E xyz \\, dV$ where $E$ is the solid that lies between the spheres $r = 2$ and $r = 4$, and above the cone $\\phi = \\frac{\\pi}{3}$.\n\n- Let’s first sketch $E$ over $xyz$-coordinate planes:\n\n![3D sketch of E](image)\n\nFigure 1: 3D sketch of $E$\n\n- Tip: change to spherical coordinates\n\n$$\\begin{align*}\nx &= r \\cos(\\theta) \\sin(\\phi) \\\\\ny &= r \\sin(\\theta) \\sin(\\phi) \\\\\nz &= r \\cos(\\phi)\n\\end{align*}$$\nThus, we are able to write the triple integral:\n\n\\[\n\\iiint_E xyz \\, dV = \\int_0^\\pi \\int_0^{2\\pi} \\int_2^4 (r \\sin(\\phi) \\cos(\\theta))(r \\sin(\\theta) \\sin(\\phi))(r \\cos(\\phi))r^2 \\sin(\\phi) \\, dr \\, d\\theta \\, d\\phi\n\\]\n\n\\[\n= \\int_0^\\pi \\sin^3(\\phi) \\cos(\\phi) \\, d\\phi \\left[ \\int_0^{2\\pi} \\sin(\\theta) \\cos(\\theta) \\, d\\theta \\left[ \\int_2^4 r^5 \\, dr \\right] \\right]\n\\]\n\n\\[\n= \\left[ \\frac{1}{4} \\sin^4(\\phi) \\right]_{\\phi=0}^{\\phi=\\frac{\\pi}{3}} \\left[ \\frac{1}{2} \\sin^2(\\theta) \\right]_{\\theta=0}^{\\theta=2\\pi} \\left[ \\frac{1}{6} r^6 \\right]_{r=2}^{r=4}\n\\]\n\n\\[\n= \\frac{1}{48} \\left[ \\left( \\frac{\\sqrt{3}}{2} \\right)^4 - 0 \\right] \\left[ 0 - 0 \\right] \\left[ 4^6 - 2^6 \\right]\n\\]\n\n\\[\n= 0\n\\]", "id": "./materials/344.pdf" }, { "contents": "Use a triple integral to determine the volume defined by $r \\leq a$ and $\\frac{\\pi}{3} \\leq \\phi \\leq \\frac{\\pi}{6}$.\n\n- We can define\n \n $$E = \\{(r, \\theta, \\phi) | 0 \\leq r \\leq a \\land 0 \\leq \\theta \\leq 2\\pi \\land \\frac{\\pi}{6} \\leq \\phi \\leq \\frac{\\pi}{3}\\}$$\n\n- Thus, we are able to write the triple integral:\n\n $$\\iiint_E 1 \\, dV = \\int_{\\frac{\\pi}{3}}^{\\frac{\\pi}{6}} \\int_{0}^{2\\pi} \\int_{0}^{a} r^2 \\sin(\\phi) \\, dr \\, d\\theta \\, d\\phi$$\n\n $$= \\int_{\\frac{\\pi}{3}}^{\\frac{\\pi}{6}} \\int_{0}^{2\\pi} \\sin(\\phi) \\left[ \\frac{r^3}{3} \\right]_{r=0}^{r=a} \\, d\\theta \\, d\\phi$$\n\n $$= \\int_{\\frac{\\pi}{3}}^{\\frac{\\pi}{6}} \\int_{0}^{2\\pi} \\frac{a^3}{3} \\sin(\\phi) \\, d\\theta \\, d\\phi$$\n\n- Now is just to solve the last two iterations.\n\n- You should get as the result: $\\frac{a^3\\pi(\\sqrt{3} - 1)}{3}$", "id": "./materials/345.pdf" }, { "contents": "Use a triple integral to determine the volume that lies above $\\phi = \\frac{\\pi}{3}$ and below $r = 4 \\cos(\\phi)$.\n\n- Since\n \n $$r = 4 \\cos(\\phi) \\quad \\Rightarrow \\quad r^2 = 4r \\cos(\\phi)$$\n\n which is the equation of a sphere of radius 2, centered at (0,0,2).\n\n- Thus, we are able to write the triple integral:\n\n $$\\iiint_E 1 \\, dV = \\int_0^{2\\pi} \\int_0^{\\frac{\\pi}{3}} \\int_0^{4 \\cos(\\phi)} r^2 \\sin(\\phi) \\, dr \\, d\\phi \\, d\\theta$$\n\n $$= \\int_0^{2\\pi} \\int_0^{\\frac{\\pi}{3}} \\sin(\\phi) \\left[ \\frac{r^3}{3} \\right]_{r=0}^{r=4 \\cos(\\phi)} \\, d\\phi \\, d\\theta$$\n\n $$= \\int_0^{2\\pi} \\int_0^{\\frac{\\pi}{3}} \\sin(\\phi) \\frac{(4 \\cos(\\phi))^3}{3} \\, d\\phi \\, d\\theta$$\n\n- Now is just to solve the last two iterations.\n\n- You should get as the result: $10\\pi$", "id": "./materials/346.pdf" }, { "contents": "Use a triple integral to determine the volume of the solid that lies within $x^2 + y^2 + z^2 = 4$, above the xy-plane and below $z = \\sqrt{x^2 + y^2}$.\n\n- Let’s first sketch E over xyz-coordinate planes:\n\n![Figure 1: 3D sketch of E](image)\n\n- Tip: change to spherical coordinates\n\n$$\\begin{align*}\n x &= r \\cos(\\theta) \\sin(\\phi) \\\\\n y &= r \\sin(\\theta) \\sin(\\phi) \\\\\n z &= r \\cos(\\phi)\n\\end{align*}$$\n\nSubstituting we have the sphere\n\n$$x^2 + y^2 + z^2 = 4 \\quad \\Rightarrow \\quad r = 2$$\nand the cone\n\\[ z = \\sqrt{x^2 + y^2} \\Rightarrow \\phi = \\frac{\\pi}{4} \\]\n\n- Therefore, we can define\n\\[ E = \\{(r, \\theta, \\phi) | 0 \\leq r \\leq 2 \\land 0 \\leq \\theta \\leq 2\\pi \\land \\frac{\\pi}{4} \\leq \\phi \\leq \\frac{\\pi}{2}\\} \\]\n\n- Thus, we are able to write the triple integral:\n\\[\n\\iiint_E 1 \\, dV = \\int_{\\frac{\\pi}{4}}^{\\frac{\\pi}{2}} \\int_0^{2\\pi} \\int_0^2 r^2 \\sin(\\phi) \\, dr \\, d\\theta \\, d\\phi\n\\]\n\n- Now is just to solve it.\n\n- You should get as the result: \\( \\frac{8\\pi \\sqrt{2}}{3} \\)", "id": "./materials/347.pdf" }, { "contents": "Use a triple integral to determine the volume of the solid above \n\\( z = \\sqrt{x^2 + y^2} \\) and below \\( x^2 + y^2 + z^2 = 1 \\).\n\n- Let’s first sketch \\( E \\) over \\( xyz \\)-coordinate planes:\n\n![3D sketch of E](image)\n\nFigure 1: 3D sketch of \\( E \\)\n\n- Tip: change to spherical coordinates\n\n\\[\n\\begin{align*}\n x &= r \\cos(\\theta) \\sin(\\phi) \\\\\n y &= r \\sin(\\theta) \\sin(\\phi) \\\\\n z &= r \\cos(\\phi)\n\\end{align*}\n\\]\n\nSubstituting we have\n\n\\[\nz = \\sqrt{x^2 + y^2} \\quad \\Rightarrow \\quad \\cos(\\phi) = \\sin(\\phi) \\quad \\Rightarrow \\quad \\phi = \\frac{\\pi}{4}\n\\]\n• Thus, we are able to write the triple integral:\n\n\\[\n\\iiint_E 1 \\, dV = \\int_0^{2\\pi} \\int_0^{\\frac{\\pi}{4}} \\int_0^1 r^2 \\sin(\\phi) \\, dr \\, d\\phi \\, d\\theta\n\\]\n\n\\[\n= \\int_0^{2\\pi} \\int_0^{\\frac{\\pi}{4}} \\left[ \\frac{r^3}{3} \\right]_{r=0}^{r=1} \\, d\\phi \\, d\\theta\n\\]\n\n\\[\n= \\int_0^{2\\pi} \\int_0^{\\frac{\\pi}{4}} \\frac{1}{3} \\, d\\phi \\, d\\theta\n\\]\n\n• Now is just to solve the last two iterations.\n\n• You should get as the result: \\( \\frac{\\pi(2 - \\sqrt{2})}{3} \\)", "id": "./materials/348.pdf" }, { "contents": "Evaluate \\[ \\int_0^1 \\int_0^{1-x^2} \\int_0^{\\sqrt{2-x^2-y^2}} xy \\, dz \\, dy \\, dx \\]\n\n- The region of integration is the region above the cone \n \\[ z = \\sqrt{x^2 + y^2} \\]\n and below the sphere \n \\[ x^2 + y^2 + z^2 = 2 \\]\n in the first octant.\n\n- This means that (switching to spherical coordinates):\n \\[ 0 \\leq \\theta \\leq \\frac{\\pi}{2} \\quad \\land \\quad 0 \\leq \\phi \\leq \\frac{\\pi}{4} \\quad \\land \\quad 0 \\leq r \\leq \\sqrt{2} \\]\n\n- So, rewriting the triple integral, we have:\n \\[\n \\int_0^1 \\int_0^{1-x^2} \\int_0^{\\sqrt{2-x^2-y^2}} xy \\, dz \\, dy \\, dx \\\\\n = \\int_0^{\\frac{\\pi}{4}} \\int_0^{\\frac{\\pi}{2}} \\int_0^{\\sqrt{2}} (r \\cos(\\theta) \\sin(\\phi))(r \\sin(\\theta) \\sin(\\phi)) r^2 \\sin(\\phi) \\, dr \\, d\\theta \\, d\\phi\n \\]\n\n- Let’s evaluate first:\n \\[\n \\int_0^{\\sqrt{2}} (r \\cos(\\theta) \\sin(\\phi))(r \\sin(\\theta) \\sin(\\phi)) r^2 \\sin(\\phi) \\, dr \\\\\n = \\int_0^{\\sqrt{2}} \\sin^3(\\phi) \\sin(\\theta) \\cos(\\theta) r^4 \\, dr \\\\\n = \\sin^3(\\phi) \\sin(\\theta) \\cos(\\theta) \\left[ \\frac{r^5}{5} \\right]_{r=0}^{r=\\sqrt{2}} \\\\\n = \\frac{2\\sqrt{2}}{5} \\sin^3(\\phi) \\sin(\\theta) \\cos(\\theta)\n \\]\n• The idea is now to do exactly the same for the next iterations, first in respect to $\\theta$ and finally in respect to $\\phi$.\n\n• At the end you should get: $\\frac{4\\sqrt{2} - 5}{15}$", "id": "./materials/349.pdf" }, { "contents": "Systems of linear equations\n\nExercise. Solve the following system of linear equations:\n\n\\[\n\\begin{align*}\n2x_1 - x_2 - x_3 - 4x_4 &= 9 \\\\\n4x_1 - 3x_3 - x_4 &= 0 \\\\\n8x_1 - 2x_2 - 5x_3 - 9x_4 &= 18\n\\end{align*}\n\\]\n\nSolution The system has augmented matrix \\( \\overline{A} \\), where \\( A \\) is the coefficient matrix, and \\( b \\) is the constant column term:\n\n\\[\n\\overline{A} := (A|b) = \\begin{pmatrix}\n2 & -1 & -1 & -4 & 9 \\\\\n4 & 0 & -3 & -1 & 0 \\\\\n8 & -2 & -5 & -9 & 18\n\\end{pmatrix}\n\\]\n\nperforming row operations \\( (R_2 \\rightarrow R_2 - 2R_1, R_3 \\rightarrow R_3 - 4R_1) \\) we get\n\n\\[\n\\overline{A} \\rightarrow \\begin{pmatrix}\n2 & -1 & -1 & -4 & 9 \\\\\n0 & 2 & -1 & 7 & -18 \\\\\n0 & 2 & -1 & 7 & -18\n\\end{pmatrix}\n\\]\n\nEventually, performing \\( R_3 \\rightarrow R_3 - R_1 \\) we find the row echelon form\n\n\\[\n\\overline{A} \\rightarrow \\begin{pmatrix}\n2 & -1 & -1 & -4 & 9 \\\\\n0 & 2 & -1 & 7 & -18 \\\\\n0 & 0 & 0 & 0 & 0\n\\end{pmatrix}\n\\]\n\nfrom which, by the Rouché-Capelli Theorem, we deduce that the system is consistent, as \\( \\overline{A} \\) and \\( A \\) both have rank 2 (number of non zero lines of the row echelon form), and that the number of its solutions is \\( \\infty^{4-2} = \\infty^2 \\): the solution set will depends on 2 reals parameters, which correspond to the free (i.e. non-pivotal) variables \\( x_3, x_4 \\).\n\nTo write down the solution set we can go ahead to a complete reduction performing the following row operations:\n\n\\[\nR_2 \\rightarrow (1/2)R_2 =: R'_2, R_1 \\rightarrow R_1 - 1 + R'_2 =: R'_1:\n\\]\n\n\\[\n\\begin{pmatrix}\n2 & -1 & -1 & -4 & 9 \\\\\n0 & 2 & -1 & 7 & -18 \\\\\n0 & 0 & 0 & 0 & 0\n\\end{pmatrix} \\rightarrow \\begin{pmatrix}\n2 & -1 & -1 & -4 & 9 \\\\\n0 & 1 & -1/2 & 7/2 & -9 \\\\\n0 & 0 & 0 & 0 & 0\n\\end{pmatrix} \\rightarrow \\begin{pmatrix}\n2 & 0 & -3/2 & -1/2 & 0 \\\\\n0 & 1 & -1/2 & 7/2 & -9 \\\\\n0 & 0 & 0 & 0 & 0\n\\end{pmatrix}\n\\]\n\nand finally \\( R'_1 \\rightarrow (1/2)R'_1 \\):\n\n\\[\n\\overline{A} \\rightarrow \\begin{pmatrix}\n1 & 0 & -3/4 & -1/4 & 0 \\\\\n0 & 1 & -1/2 & 7/2 & -9 \\\\\n0 & 0 & 0 & 0 & 0\n\\end{pmatrix}\n\\]\n\nHence the starting system is equivalent to the reduced system\n\n\\[\n\\begin{align*}\nx_1 - \\frac{3}{4}x_3 - \\frac{1}{2}x_4 &= 0 \\\\\nx_2 - \\frac{1}{2}x_3 + \\frac{7}{2}x_4 &= -9\n\\end{align*}\n\\]\n\n\\[\n\\iff \\begin{align*}\nx_1 &= \\frac{3}{2}h + \\frac{1}{2}k \\\\\nx_2 &= -9 + \\frac{1}{2}h - \\frac{7}{2}k \\\\\nx_3 &= h \\\\\nx_4 &= k\n\\end{align*}\n\\]\n\nTherefore, its solution set is\n\n\\[\nS = \\left\\{ \\begin{pmatrix} 0 \\\\ -9 \\\\ 0 \\\\ 0 \\end{pmatrix} + h \\begin{pmatrix} \\frac{3}{2} \\\\ \\frac{1}{2} \\\\ 1 \\\\ 0 \\end{pmatrix} + k \\begin{pmatrix} \\frac{1}{2} \\\\ -\\frac{7}{2} \\\\ 0 \\\\ 1 \\end{pmatrix} \\in \\mathbb{R}^4 \\mid h, k \\in \\mathbb{R} \\right\\}\n\\]", "id": "./materials/35.pdf" }, { "contents": "Evaluate \\[ \\int_{-a}^{a} \\int_{-\\sqrt{a^2-y^2}}^{\\sqrt{a^2-y^2}} \\int_{-\\sqrt{a^2-y^2-x^2}}^{\\sqrt{a^2-y^2-x^2}} (x^2 z + y^2 z + z^3) \\, dz \\, dx \\, dy \\]\n\n- The region of integration is the solid sphere\n \\[ x^2 + y^2 + z^2 \\leq a^2 \\]\n so, in spherical coordinates, we have:\n \\[ 0 \\leq \\theta \\leq 2\\pi \\quad \\land \\quad 0 \\leq \\phi \\leq \\pi \\quad \\land \\quad 0 \\leq r \\leq a \\]\n\n- We can also transform:\n \\[ x^2 z + y^2 z + z^3 = (x^2 + y^2 + z^2)z = r^2 z = r^3 \\cos(\\phi) \\]\n\n- So, rewriting the triple integral, we have:\n \\[\n \\int_{-a}^{a} \\int_{-\\sqrt{a^2-y^2}}^{\\sqrt{a^2-y^2}} \\int_{-\\sqrt{a^2-y^2-x^2}}^{\\sqrt{a^2-y^2-x^2}} (x^2 z + y^2 z + z^3) \\, dz \\, dx \\, dy \\\\\n = \\int_{0}^{\\pi} \\int_{0}^{2\\pi} \\int_{0}^{a} r^3 \\cos(\\phi) r^2 \\sin(\\phi) \\, dr \\, d\\theta \\, d\\phi \\\\\n = \\int_{0}^{\\pi} \\int_{0}^{2\\pi} \\int_{0}^{a} r^5 \\cos(\\phi) \\sin(\\phi) \\, dr \\, d\\theta \\, d\\phi \\\\\n \\]\n\n- Let’s evaluate first:\n \\[\n \\int_{0}^{a} r^5 \\cos(\\phi) \\sin(\\phi) \\, dr \\\\\n = \\cos(\\phi) \\sin(\\phi) \\left[ \\frac{r^6}{6} \\right]_{r=0}^{r=a} \\\\\n = \\frac{a^6}{6} \\cos(\\phi) \\sin(\\phi) \n \\]\n• The idea is now to do exactly the same for the next iterations, first in respect to $\\theta$ and finally in respect to $\\phi$.\n\n• At the end you should get: 0", "id": "./materials/350.pdf" }, { "contents": "Evaluate \\[ \\int_{-2}^{2} \\int_{-\\sqrt{4-x^2}}^{\\sqrt{4-x^2}} \\int_{2-\\sqrt{4-y^2-x^2}}^{2+\\sqrt{4-y^2-x^2}} (x^2 + y^2 + z^2)^{\\frac{3}{2}} \\, dz \\, dy \\, dx \\]\n\n- The region of integration is the solid sphere\n \\[ x^2 + y^2 + (z - 2)^2 \\leq 4 \\]\n so, in spherical coordinates, we have:\n \\[ r^2 \\sin^2(\\phi) + (r \\cos(\\phi) - 2)^2 \\leq 4 \\]\n \\[ \\Rightarrow r^2 - 4r \\cos(\\phi) + 4 \\leq 4 \\]\n \\[ \\Rightarrow r \\leq 4 \\cos(\\phi) \\]\n and this means that\n \\[ 0 \\leq \\theta \\leq 2\\pi \\quad \\land \\quad 0 \\leq \\phi \\leq \\frac{\\pi}{2} \\quad \\land \\quad 0 \\leq r \\leq 4 \\cos(\\phi) \\]\n\n- We can also transform:\n \\[ (x^2 + y^2 + z^2)^{\\frac{3}{2}} = (r^2)^{\\frac{3}{2}} = (r^3) \\]\n\n- So, rewriting the triple integral, we have:\n \\[ \\int_{-2}^{2} \\int_{-\\sqrt{4-x^2}}^{\\sqrt{4-x^2}} \\int_{2-\\sqrt{4-y^2-x^2}}^{2+\\sqrt{4-y^2-x^2}} (x^2 + y^2 + z^2)^{\\frac{3}{2}} \\, dz \\, dy \\, dx \\]\n \\[ = \\int_{0}^{\\frac{\\pi}{2}} \\int_{0}^{2\\pi} \\int_{0}^{4 \\cos(\\phi)} r^3 r^2 \\sin(\\phi) \\, dr \\, d\\theta \\, d\\phi \\]\n \\[ = \\int_{0}^{\\frac{\\pi}{2}} \\int_{0}^{2\\pi} \\int_{0}^{4 \\cos(\\phi)} r^5 \\sin(\\phi) \\, dr \\, d\\theta \\, d\\phi \\]\n\n- Now is just to solve it.\n\n- At the end you should get: \\[ \\frac{4096\\pi}{21} \\]", "id": "./materials/351.pdf" }, { "contents": "Arithmetic Sequences\n\nA simple way to generate a sequence is to start with a number $a$, and add to it a fixed constant $d$, over and over again. This type of sequence is called an arithmetic sequence.\n\n**Definition:** An arithmetic sequence is a sequence of the form\n\n$$a, a + d, a + 2d, a + 3d, a + 4d, \\ldots$$\n\nThe number $a$ is the first term, and $d$ is the common difference of the sequence. The $n$th term of an arithmetic sequence is given by\n\n$$a_n = a + (n - 1)d$$\n\nThe number $d$ is called the common difference because any two consecutive terms of an arithmetic sequence differ by $d$, and it is found by subtracting any pair of terms $a_n$ and $a_{n+1}$. That is\n\n$$d = a_{n+1} - a_n$$\n\n**Is the Sequence Arithmetic?**\n\n**Example 1:** Determine whether or not the sequence is arithmetic. If it is arithmetic, find the common difference.\n\n(a) 2, 5, 8, 11, ...\n(b) 1, 2, 3, 5, 8, ...\n\n**Solution (a):** In order for a sequence to be arithmetic, the differences between each pair of adjacent terms should be the same. If the differences are all the same, then $d$, the common difference, is that value.\n\n**Step 1:** First, calculate the difference between each pair of adjacent terms.\n\n$$5 - 2 = 3$$\n$$8 - 5 = 3$$\n$$11 - 8 = 3$$\n\n**Step 2:** Now, compare the differences. Since each pair of adjacent terms has the same difference 3, the sequence is arithmetic and the common difference $d = 3$. \n\nBy: Crystal Hull\nExample 1 (Continued):\n\nSolution (b):\n\nStep 1: Calculate the difference between each pair of adjacent terms.\n\n\\[\n\\begin{align*}\n2 - 1 &= 1 \\\\\n3 - 2 &= 1 \\\\\n5 - 3 &= 2 \\\\\n8 - 5 &= 3\n\\end{align*}\n\\]\n\nStep 2: Compare the differences. Since the differences between each pair of adjacent terms are not all the same, the sequence is not arithmetic.\n\nAn arithmetic sequence is determined completely by the first term \\(a\\), and the common difference \\(d\\). Thus, if we know the first two terms of an arithmetic sequence, then we can find the equation for the \\(n\\)th term.\n\nFinding the Terms of an Arithmetic Sequence:\n\nExample 2: Find the \\(n\\)th term, the fifth term, and the 100th term, of the arithmetic sequence determined by \\(a = 2\\) and \\(d = 3\\).\n\nSolution: To find a specific term of an arithmetic sequence, we use the formula for finding the \\(n\\)th term.\n\nStep 1: The \\(n\\)th term of an arithmetic sequence is given by\n\n\\[\na_n = a + (n - 1)d\n\\]\n\nSo, to find the \\(n\\)th term, substitute the given values \\(a = 2\\) and \\(d = 3\\) into the formula.\n\n\\[\na_n = 2 + (n - 1)3\n\\]\n\nStep 2: Now, to find the fifth term, substitute \\(n = 5\\) into the equation for the \\(n\\)th term.\n\n\\[\na_5 = 2 + (5 - 1)3\n\\]\n\n\\[\n= 14\n\\]\n\nStep 3: Finally, find the 100th term in the same way as the fifth term.\n\n\\[\na_{100} = 2 + (100 - 1)3\n\\]\n\n\\[\n= 299\n\\]\nExample 3: Find the common difference, the fifth term, the \\( n \\)th term, and the 100th term of the arithmetic sequence.\n\n(a) 4, 14, 24, 34, ...\n\n(b) \\( t + 3, t + \\frac{15}{4}, t + \\frac{9}{2}, t + \\frac{21}{4}, ... \\)\n\nSolution (a): In order to find the \\( n \\)th and 100th terms, we will first have to determine what \\( a \\) and \\( d \\) are. We will then use the formula for finding the \\( n \\)th term.\n\nStep 1: First, we will determine what \\( a \\) and \\( d \\) are. The number \\( a \\) is always the first term of the sequence, so\n\n\\[ a = 4 \\]\n\nThe difference between any pair of adjacent terms should be the same because the sequence is arithmetic, so we can choose any one pair to find the common difference \\( d \\). If we choose the first two terms then\n\n\\[ d = 14 - 4 \\]\n\\[ = 10 \\]\n\nStep 2: Since we are given the fourth term, we can add the common difference \\( d = 10 \\) to it to get the fifth term.\n\n\\[ a_5 = 34 + 10 \\]\n\\[ = 44 \\]\n\nStep 3: Now to find the \\( n \\)th term, substitute \\( a = 4 \\) and \\( d = 10 \\) into the formula for the \\( n \\)th term.\n\n\\[ a_n = 4 + (n - 1)10 \\]\n\nStep 4: Finally, substitute \\( n = 100 \\) into the equation for the \\( n \\)th term to get the 100th term.\n\n\\[ a_{100} = 4 + (100 - 1)10 \\]\n\\[ = 994 \\]\n\nBy: Crystal Hull\nExample 3 (Continued):\n\nSolution (b):\n\nStep 1: Calculate $a$ and $d$.\n\n\\[ a = t + 3 \\]\n\n\\[ d = \\left( t + \\frac{15}{4} \\right) - (t + 3) \\]\n\n\\[ = t + \\frac{15}{4} - t - 3 \\]\n\n\\[ = \\frac{15}{4} - 3 \\]\n\n\\[ = \\frac{3}{2} \\]\n\nStep 2: The fifth term is the fourth term plus the common difference. Therefore,\n\n\\[ a_5 = \\left( t + \\frac{21}{4} \\right) + \\frac{3}{2} \\]\n\n\\[ = t + \\frac{24}{4} \\]\n\n\\[ = t + 6 \\]\n\nStep 3: Now, substitute $a = t + 3$, $d = \\frac{3}{2}$ into the formula for the $n$th term.\n\n\\[ a_n = (t + 3) + (n - 1) \\frac{3}{2} \\]\n\nStep 4: Finally, substitute $n = 100$ into the equation for the $n$th term that we just found.\n\n\\[ a_n = (t + 3) + (100 - 1) \\frac{3}{2} \\]\n\n\\[ = t + 3 + (99) \\frac{3}{2} \\]\n\n\\[ = t + \\frac{303}{2} \\]\n\nBy: Crystal Hull\nPartial Sums of an Arithmetic Sequence:\n\nTo find a formula for the sum, $S_n$, of the first $n$ terms of an arithmetic sequence, we can write out the terms as\n\n$$S_n = a + (a + d) + (a + 2d) + ... + [a + (n-1)d].$$\n\nThis same sum can be written in reverse as\n\n$$S_n = a_n + (a_n - d) + (a_n - 2d) + ... + [a_n - (n-1)d]$$\n\nNow, add the corresponding terms of these two expressions for $S_n$ to get\n\n$$S_n = a + (a + d) + (a + 2d) + ... + [a + (n-1)d]$$\n$$S_n = a_n + (a_n - d) + (a_n - 2d) + ... + [a_n - (n-1)d]$$\n$$\\frac{2S_n}{} = (a + a_n) + (a + a_n) + (a + a_n) + ... + (a + a_n)$$\n\nThe right hand side of this expression contains $n$ terms, each equal to $a + a_n$, so\n\n$$2S_n = n(a + a_n)$$\n$$S_n = \\frac{n}{2}(a + a_n).$$\n\n**Definition:** For the arithmetic sequence $a_n = a + (n-1)d$, the **nth partial sum**\n\n$$S_n = a + (a + d) + (a + 2d) + (a + 3d) + ... + [a + (n-1)d]$$\n\nis given by either of the following formulas.\n\n1. $S_n = \\frac{n}{2}[2a + (n-1)d]$\n\n2. $S_n = n\\left(\\frac{a + a_n}{2}\\right)$\n\nBy: Crystal Hull\nThe $n$th partial sum of an arithmetic sequence can also be written using summation notation.\n\n$$\\sum_{i=1}^{n} ki - c$$\n\nrepresents the sum of the first $n$ terms of an arithmetic sequence having the first term $a = k(1) + c = k + c$ and the $n$th term $a_n = k(n) + c = kn + c$. We can find this sum with the second formula for $S_n$ given above.\n\n**Example 4:** Find the partial sum $S_n$ of the arithmetic sequence that satisfies the given conditions.\n\n(a) $a = 6$, $d = 3$, and $n = 7$\n\n(b) $\\sum_{i=1}^{14} 2i - 7$\n\n**Solution (a):** To find the $n$th partial sum of an arithmetic sequence, we can use either of the formulas\n\n$$S_n = \\frac{n}{2} [2a + (n-1)d] \\quad \\text{or} \\quad S_n = n \\left( \\frac{a + a_n}{2} \\right)$$\n\n**Step 1:** To use the first formula for the $n$th partial sum, we only need to substitute the given values $a = 6$, $d = 3$, and $n = 7$ into the equation.\n\n$$S_n = \\frac{7}{2} [2(6) + (7-1)3]$$\n\n$$= \\frac{7}{2} [12 + 18]$$\n\n$$= 105$$\nExample 4 (Continued):\n\nSolution (b): This is the sum of the first fourteen terms of the arithmetic sequence having $a_n = 2n - 7$.\n\nStep 1: Since the partial sum is given in summation notation, we must first find $a$ and $a_n$. From the given information we know $k = 2$, $c = -7$, and $n = 14$, so\n\n$$a = k + c$$\n$$= 2 + (-7)$$\n$$= -5$$\n\n$$a_n = kn + c$$\n$$a_{14} = 2(14) + (-7)$$\n$$= 21$$\n\nStep 2: Now that we know $a = -5$, $n = 14$, and $a_{14} = 21$, we can substitute these values into the second formula for the $n$th partial sum to find the fourteenth partial sum.\n\n$$S_n = n\\left(\\frac{a + a_n}{2}\\right)$$\n$$= 14\\left(\\frac{-5 + 21}{2}\\right)$$\n$$= 112$$\n\nExample 5: Find the sum of the first 37 even numbers.\n\nSolution:\n\nStep 1: First, we must find the values $a$, $d$, and $n$. Since the first even number is zero, $a = 0$. The next even number is 2, so $d = 2 - 0 = 2$. Since we are told to find the sum of the first 37 even numbers, $n = 37$. \n\nBy: Crystal Hull\nExample 5 (Continued):\n\n**Step 2:** Now that we know \\( a = 0, \\ d = 2, \\) and \\( n = 37 \\) we can solve this problem the same way as in the previous example. First find \\( a_{37} \\), and then substitute the values for \\( a, \\ d, \\) and \\( a_{37} \\) into the equation for the \\( n \\)th partial sum. Thus,\n\n\\[\na_{37} = 0 + (37 - 1)2 = 18\n\\]\n\n\\[\nS_{37} = 37 \\left( \\frac{0 + 18}{2} \\right) = 363\n\\]\n\nExample 6: A partial sum of an arithmetic sequence is given. Find the sum.\n\n\\[1 + 8 + 15 + \\ldots + 78\\]\n\n**Solution:**\n\n**Step 1:** As in the previous example, we must first find \\( a, \\ d, \\) and \\( n \\). The values \\( a \\) and \\( d \\) are easy to find.\n\n\\[a = 1\\]\n\n\\[d = 8 - 1 = 7\\]\n\nNow, finding \\( n \\) is a bit more work because we are not explicitly told how many numbers we will be summing. We know \\( a \\) and \\( d \\), and we know the \\( n \\)th term, so we will substitute these values into the formula for the \\( n \\)th term of a sequence.\n\n\\[a_n = a + (n-1)d\\]\n\n\\[78 = 1 + (n-1)7\\]\n\nNow solve for \\( n \\)\n\n\\[77 = (n-1)7\\]\n\n\\[11 = n - 1\\]\n\n\\[12 = n\\]\n\nTherefore, we will be summing twelve terms and \\( 78 = a_{12} \\).\nExample 6 (Continued):\n\n**Step 2:** Now that we know $a = 1$, $n = 12$, and $a_{12} = 78$ we can solve this problem the same way as in example 4. Substitute the values for $a$, $d$, and $a_{12}$ into the formula for the $n$th partial sum.\n\n$$S_{12} = 12 \\left( \\frac{1 + 78}{2} \\right)$$\n\n$$= 474$$", "id": "./materials/353.pdf" }, { "contents": "Arithmetic and geometric progressions\n\nThis unit introduces sequences and series, and gives some simple examples of each. It also explores particular types of sequence known as arithmetic progressions (APs) and geometric progressions (GPs), and the corresponding series.\n\nIn order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature.\n\nAfter reading this text, and/or viewing the video tutorial on this topic, you should be able to:\n\n- recognise the difference between a sequence and a series;\n- recognise an arithmetic progression;\n- find the $n$-th term of an arithmetic progression;\n- find the sum of an arithmetic series;\n- recognise a geometric progression;\n- find the $n$-th term of a geometric progression;\n- find the sum of a geometric series;\n- find the sum to infinity of a geometric series with common ratio $|r| < 1$.\n\nContents\n\n1. Sequences 2\n2. Series 3\n3. Arithmetic progressions 4\n4. The sum of an arithmetic series 5\n5. Geometric progressions 8\n6. The sum of a geometric series 9\n7. Convergence of geometric series 12\n1. Sequences\n\nWhat is a sequence? It is a set of numbers which are written in some particular order. For example, take the numbers\n\n\\[ 1, \\ 3, \\ 5, \\ 7, \\ 9, \\ \\ldots. \\]\n\nHere, we seem to have a rule. We have a sequence of odd numbers. To put this another way, we start with the number 1, which is an odd number, and then each successive number is obtained by adding 2 to give the next odd number.\n\nHere is another sequence:\n\n\\[ 1, \\ 4, \\ 9, \\ 16, \\ 25, \\ \\ldots. \\]\n\nThis is the sequence of square numbers. And this sequence,\n\n\\[ 1, \\ -1, \\ 1, \\ -1, \\ 1, \\ -1, \\ \\ldots, \\]\n\nis a sequence of numbers alternating between 1 and \\(-1\\). In each case, the dots written at the end indicate that we must consider the sequence as an infinite sequence, so that it goes on for ever.\n\nOn the other hand, we can also have finite sequences. The numbers\n\n\\[ 1, \\ 3, \\ 5, \\ 9 \\]\n\nform a finite sequence containing just four numbers. The numbers\n\n\\[ 1, \\ 4, \\ 9, \\ 16 \\]\n\nalso form a finite sequence. And so do these, the numbers\n\n\\[ 1, \\ 2, \\ 3, \\ 4, \\ 5, \\ 6, \\ \\ldots, \\ n. \\]\n\nThese are the numbers we use for counting, and we have included \\(n\\) of them. Here, the dots indicate that we have not written all the numbers down explicitly. The \\(n\\) after the dots tells us that this is a finite sequence, and that the last number is \\(n\\).\n\nHere is a sequence that you might recognise:\n\n\\[ 1, \\ 1, \\ 2, \\ 3, \\ 5, \\ 8, \\ \\ldots. \\]\n\nThis is an infinite sequence where each term (from the third term onwards) is obtained by adding together the two previous terms. This is called the Fibonacci sequence.\n\nWe often use an algebraic notation for sequences. We might call the first term in a sequence \\(u_1\\), the second term \\(u_2\\), and so on. With this same notation, we would write \\(u_n\\) to represent the \\(n\\)-th term in the sequence. So\n\n\\[ u_1, \\ u_2, \\ u_3, \\ \\ldots, \\ u_n \\]\n\nwould represent a finite sequence containing \\(n\\) terms. As another example, we could use this notation to represent the rule for the Fibonacci sequence. We would write\n\n\\[ u_n = u_{n-1} + u_{n-2} \\]\n\nto say that each term was the sum of the two preceding terms.\nA sequence is a set of numbers written in a particular order. We sometimes write $u_1$ for the first term of the sequence, $u_2$ for the second term, and so on. We write the $n$-th term as $u_n$.\n\n**Exercise 1**\n\n(a) A sequence is given by the formula $u_n = 3n + 5$, for $n = 1, 2, 3, \\ldots$. Write down the first five terms of this sequence.\n\n(b) A sequence is given by $u_n = 1/n^2$, for $n = 1, 2, 3, \\ldots$. Write down the first four terms of this sequence. What is the 10th term?\n\n(c) Write down the first eight terms of the Fibonacci sequence defined by $u_n = u_{n-1} + u_{n-2}$, when $u_1 = 1$, and $u_2 = 1$.\n\n(d) Write down the first five terms of the sequence given by $u_n = (-1)^{n+1}/n$.\n\n**2. Series**\n\nA series is something we obtain from a sequence by adding all the terms together.\n\nFor example, suppose we have the sequence\n\n$$u_1, \\ u_2, \\ u_3, \\ \\ldots, \\ u_n.$$ \n\nThe series we obtain from this is\n\n$$u_1 + u_2 + u_3 + \\ldots + u_n,$$\n\nand we write $S_n$ for the sum of these $n$ terms. So although the ideas of a ‘sequence’ and a ‘series’ are related, there is an important distinction between them.\n\nFor example, let us consider the sequence of numbers\n\n$$1, \\ 2, \\ 3, \\ 4, \\ 5, \\ 6, \\ \\ldots, \\ n.$$ \n\nThen $S_1 = 1$, as it is the sum of just the first term on its own. The sum of the first two terms is $S_2 = 1 + 2 = 3$. Continuing, we get\n\n$$S_3 = 1 + 2 + 3 = 6,$$\n\n$$S_4 = 1 + 2 + 3 + 4 = 10,$$\n\nand so on.\nA series is a sum of the terms in a sequence. If there are \\( n \\) terms in the sequence and we evaluate the sum then we often write \\( S_n \\) for the result, so that\n\n\\[\nS_n = u_1 + u_2 + u_3 + \\ldots + u_n.\n\\]\n\n**Exercise 2**\n\nWrite down \\( S_1, S_2, \\ldots, S_n \\) for the sequences\n\n(a) \\( 1, 3, 5, 7, 9, 11; \\)\n(b) \\( 4, 2, 0, -2, -4. \\)\n\n### 3. Arithmetic progressions\n\nConsider these two common sequences\n\n\\[\n1, \\ 3, \\ 5, \\ 7, \\ \\ldots\n\\]\n\nand\n\n\\[\n0, \\ 10, \\ 20, \\ 30, \\ 40, \\ \\ldots.\n\\]\n\nIt is easy to see how these sequences are formed. They each start with a particular first term, and then to get successive terms we just add a fixed value to the previous term. In the first sequence we add 2 to get the next term, and in the second sequence we add 10. So the difference between consecutive terms in each sequence is a constant. We could also subtract a constant instead, because that is just the same as adding a negative constant. For example, in the sequence\n\n\\[\n8, \\ 5, \\ 2, \\ -1, \\ -4, \\ \\ldots\n\\]\n\nthe difference between consecutive terms is \\(-3\\). Any sequence with this property is called an arithmetic progression, or AP for short.\n\nWe can use algebraic notation to represent an arithmetic progression. We shall let \\( a \\) stand for the first term of the sequence, and let \\( d \\) stand for the common difference between successive terms. For example, our first sequence could be written as\n\n\\[\n1, \\ 3, \\ 5, \\ 7, \\ 9, \\ \\ldots\n\\]\n\nand this can be written as\n\n\\[\na, \\ a + d, \\ a + 2d, \\ a + 3d, \\ a + 4d, \\ \\ldots\n\\]\n\nwhere \\( a = 1 \\) is the first term, and \\( d = 2 \\) is the common difference. If we wanted to write down the \\( n \\)-th term, we would have\n\n\\[\na + (n - 1)d,\n\\]\nbecause if there are \\( n \\) terms in the sequence there must be \\((n - 1)\\) common differences between successive terms, so that we must add on \\((n - 1)d\\) to the starting value \\(a\\). We also sometimes write \\(\\ell\\) for the last term of a finite sequence, and so in this case we would have\n\n\\[\n\\ell = a + (n - 1)d.\n\\]\n\n**Key Point**\n\nAn arithmetic progression, or AP, is a sequence where each new term after the first is obtained by adding a constant \\(d\\), called the *common difference*, to the preceding term. If the first term of the sequence is \\(a\\) then the arithmetic progression is\n\n\\[\na, \\ a + d, \\ a + 2d, \\ a + 3d, \\ldots\n\\]\n\nwhere the \\(n\\)-th term is \\(a + (n - 1)d\\).\n\n**Exercise 3**\n\n(a) Write down the first five terms of the AP with first term 8 and common difference 7.\n\n(b) Write down the first five terms of the AP with first term 2 and common difference \\(-5\\).\n\n(c) What is the common difference of the AP \\(11, -1, -13, -25, \\ldots\\) ?\n\n(d) Find the 17th term of the arithmetic progression with first term 5 and common difference 2.\n\n(e) Write down the 10th and 19th terms of the APs\n (i) \\(8, 11, 14, \\ldots\\),\n (ii) \\(8, 5, 2, \\ldots\\)\n\n(f) An AP is given by \\(k, 2k/3, k/3, 0, \\ldots\\)\n (i) Find the sixth term.\n (ii) Find the \\(n\\)th term.\n (iii) If the 20th term is equal to 15, find \\(k\\).\n\n**4. The sum of an arithmetic series**\n\nSometimes we want to add the terms of a sequence. What would we get if we wanted to add the first \\(n\\) terms of an arithmetic progression? We would get\n\n\\[\nS_n = a + (a + d) + (a + 2d) + \\ldots + (\\ell - 2d) + (\\ell - d) + \\ell.\n\\]\n\nNow this is now a series, as we have added together the \\(n\\) terms of a sequence. This is an *arithmetic series*, and we can find its sum by using a trick. Let us write the series down again, but this time we shall write it down with the terms in reverse order. We get\n\n\\[\nS_n = \\ell + (\\ell - d) + (\\ell - 2d) + \\ldots + (a + 2d) + (a + d) + a.\n\\]\nWe are now going to add these two series together. On the left-hand side, we just get $2S_n$. But on the right-hand side, we are going to add the terms in the two series so that each term in the first series will be added to the term vertically below it in the second series. We get\n\n$$2S_n = (a + \\ell) + (a + \\ell) + (a + \\ell) + \\ldots + (a + \\ell) + (a + \\ell),$$\n\nand on the right-hand side there are $n$ copies of $(a + \\ell)$ so we get\n\n$$2S_n = n(a + \\ell).$$\n\nBut of course we want $S_n$ rather than $2S_n$, and so we divide by 2 to get\n\n$$S_n = \\frac{1}{2}n(a + \\ell).$$\n\nWe have found the sum of an arithmetic progression in terms of its first and last terms, $a$ and $\\ell$, and the number of terms $n$.\n\nWe can also find an expression for the sum in terms of the $a$, $n$ and the common difference $d$. To do this, we just substitute our formula for $\\ell$ into our formula for $S_n$. From\n\n$$\\ell = a + (n - 1)d, \\quad S_n = \\frac{1}{2}n(a + \\ell)$$\n\nwe obtain\n\n$$S_n = \\frac{1}{2}n(a + a + (n - 1)d)$$\n\n$$= \\frac{1}{2}n(2a + (n - 1)d).$$\n\n**Key Point**\n\nThe sum of the terms of an arithmetic progression gives an arithmetic series. If the starting value is $a$ and the common difference is $d$ then the sum of the first $n$ terms is\n\n$$S_n = \\frac{1}{2}n(2a + (n - 1)d).$$\n\nIf we know the value of the last term $\\ell$ instead of the common difference $d$ then we can write the sum as\n\n$$S_n = \\frac{1}{2}n(a + \\ell).$$\n\n**Example**\n\nFind the sum of the first 50 terms of the sequence\n\n$$1, \\ 3, \\ 5, \\ 7, \\ 9, \\ \\ldots.$$\nSolution\n\nThis is an arithmetic progression, and we can write down\n\n\\[ a = 1, \\quad d = 2, \\quad n = 50. \\]\n\nWe now use the formula, so that\n\n\\[\nS_n = \\frac{1}{2}n(2a + (n - 1)d)\n\\]\n\n\\[\nS_{50} = \\frac{1}{2} \\times 50 \\times (2 \\times 1 + (50 - 1) \\times 2)\n\\]\n\n\\[\n= 25 \\times (2 + 49 \\times 2)\n\\]\n\n\\[\n= 25 \\times (2 + 98)\n\\]\n\n\\[\n= 2500.\n\\]\n\nExample\n\nFind the sum of the series\n\n\\[ 1 + 3 \\cdot 5 + 6 + 8 \\cdot 5 + \\ldots + 101. \\]\n\nSolution\n\nThis is an arithmetic series, because the difference between the terms is a constant value, 2·5. We also know that the first term is 1, and the last term is 101. But we do not know how many terms are in the series. So we will need to use the formula for the last term of an arithmetic progression,\n\n\\[ \\ell = a + (n - 1)d \\]\n\nto give us\n\n\\[ 101 = 1 + (n - 1) \\times 2.5. \\]\n\nNow this is just an equation for \\( n \\), the number of terms in the series, and we can solve it. If we subtract 1 from each side we get\n\n\\[ 100 = (n - 1) \\times 2.5 \\]\n\nand then dividing both sides by 2·5 gives us\n\n\\[ 40 = n - 1 \\]\n\nso that \\( n = 41 \\). Now we can use the formula for the sum of an arithmetic progression, in the version using \\( \\ell \\), to give us\n\n\\[\nS_n = \\frac{1}{2}n(a + \\ell)\n\\]\n\n\\[\nS_{41} = \\frac{1}{2} \\times 41 \\times (1 + 101)\n\\]\n\n\\[\n= \\frac{1}{2} \\times 41 \\times 102\n\\]\n\n\\[\n= 41 \\times 51\n\\]\n\n\\[\n= 2091.\n\\]\nExample\nAn arithmetic progression has 3 as its first term. Also, the sum of the first 8 terms is twice the sum of the first 5 terms. Find the common difference.\n\nSolution\nWe are given that \\( a = 3 \\). We are also given some information about the sums \\( S_8 \\) and \\( S_5 \\), and we want to find the common difference. So we shall use the formula\n\n\\[\nS_n = \\frac{1}{2}n(2a + (n - 1)d)\n\\]\n\nfor the sum of the first \\( n \\) terms. This tells us that\n\n\\[\nS_8 = \\frac{1}{2} \\times 8 \\times (6 + 7d)\n\\]\n\nand that\n\n\\[\nS_5 = \\frac{1}{2} \\times 5 \\times (6 + 4d)\n\\]\n\nSo, using the given fact that \\( S_8 = 2S_5 \\), we see that\n\n\\[\n\\frac{1}{2} \\times 8 \\times (6 + 7d) = 2 \\times \\frac{1}{2} \\times 5 \\times (6 + 4d)\n\\]\n\n\\[\n4 \\times (6 + 7d) = 5 \\times (6 + 4d)\n\\]\n\n\\[\n24 + 28d = 30 + 20d\n\\]\n\n\\[\n8d = 6\n\\]\n\n\\[\nd = \\frac{3}{4}\n\\]\n\nExercise 4\n(a) Find the sum of the first 23 terms of the AP \\( 4, -3, -10, \\ldots \\).\n(b) An arithmetic series has first term 4 and common difference \\( \\frac{1}{2} \\). Find\n (i) the sum of the first 20 terms,\n (ii) the sum of the first 100 terms.\n(c) Find the sum of the arithmetic series with first term 1, common difference 3, and last term 100.\n(d) The sum of the first 20 terms of an arithmetic series is identical to the sum of the first 22 terms. If the common difference is \\( -2 \\), find the first term.\n\n5. Geometric progressions\nWe shall now move on to the other type of sequence we want to explore.\nConsider the sequence\n\n\\[\n2, \\ 6, \\ 18, \\ 54, \\ \\ldots\n\\]\n\nHere, each term in the sequence is 3 times the previous term. And in the sequence\n\n\\[\n1, \\ -2, \\ 4, \\ -8, \\ \\ldots\n\\]\n\neach term is \\( -2 \\) times the previous term. Sequences such as these are called geometric progressions, or GPs for short.\nLet us write down a general geometric progression, using algebra. We shall take \\( a \\) to be the first term, as we did with arithmetic progressions. But here, there is no common difference. Instead there is a common ratio, as the ratio of successive terms is always constant. So we shall let \\( r \\) be this common ratio. With this notation, the general geometric progression can be expressed as\n\n\\[\na, \\ ar, \\ ar^2, \\ ar^3, \\ldots.\n\\]\n\nSo the \\( n \\)-th can be calculated quite easily. It is \\( ar^{n-1} \\), where the power \\((n - 1)\\) is always one less than the position \\( n \\) of the term in the sequence. In our first example, we had \\( a = 2 \\) and \\( r = 3 \\), so we could write the first sequence as\n\n\\[\n2, \\ 2 \\times 3, \\ 2 \\times 3^2, \\ 2 \\times 3^3, \\ldots.\n\\]\n\nIn our second example, \\( a = 1 \\) and \\( r = -2 \\), so that we could write it as\n\n\\[\n1, \\ 1 \\times (-2), \\ 1 \\times (-2)^2, \\ 1 \\times (-2)^3, \\ldots.\n\\]\n\n**Key Point**\n\nA geometric progression, or GP, is a sequence where each new term after the first is obtained by multiplying the preceding term by a constant \\( r \\), called the *common ratio*. If the first term of the sequence is \\( a \\) then the geometric progression is\n\n\\[\na, ar, ar^2, ar^3, \\ldots\n\\]\n\nwhere the \\( n \\)-th term is \\( ar^{n-1} \\).\n\n**Exercise 5**\n\n(a) Write down the first five terms of the geometric progression which has first term 1 and common ratio \\( \\frac{1}{2} \\).\n\n(b) Find the 10th and 20th terms of the GP with first term 3 and common ratio 2.\n\n(c) Find the 7th term of the GP \\( 2, -6, 18, \\ldots \\).\n\n**6. The sum of a geometric series**\n\nSuppose that we want to find the sum of the first \\( n \\) terms of a geometric progression. What we get is\n\n\\[\nS_n = a + ar + ar^2 + ar^3 + \\ldots + ar^{n-1},\n\\]\n\nand this is called a *geometric series*. Now the trick here to find the sum is to multiply by \\( r \\) and then subtract:\n\n\\[\nS_n = a + ar + ar^2 + ar^3 + \\ldots + ar^{n-1}\n\\]\n\n\\[\nrS_n = ar + ar^2 + ar^3 + \\ldots + ar^{n-1} + ar^n\n\\]\n\n\\[\nS_n - rS_n = a - ar^n\n\\]\nso that\n\n\\[ S_n(1 - r) = a(1 - r^n). \\]\n\nNow divide by \\(1 - r\\) (as long as \\(r \\neq 1\\)) to give\n\n\\[ S_n = \\frac{a(1 - r^n)}{1 - r}. \\]\n\n**Key Point**\n\nThe sum of the terms of a geometric progression gives a geometric series. If the starting value is \\(a\\) and the common ratio is \\(r\\) then the sum of the first \\(n\\) terms is\n\n\\[ S_n = \\frac{a(1 - r^n)}{1 - r} \\]\n\nprovided that \\(r \\neq 1\\).\n\n**Example**\n\nFind the sum of the geometric series\n\n\\[ 2 + 6 + 18 + 54 + \\ldots \\]\n\nwhere there are 6 terms in the series.\n\n**Solution**\n\nFor this series, we have \\(a = 2\\), \\(r = 3\\) and \\(n = 6\\). So\n\n\\[ S_n = \\frac{a(1 - r^n)}{1 - r} \\]\n\n\\[ S_6 = \\frac{2(1 - 3^6)}{1 - 3} \\]\n\n\\[ = \\frac{2(1 - 729)}{-2} \\]\n\n\\[ = -(-728) \\]\n\n\\[ = 728. \\]\n\n**Example**\n\nFind the sum of the geometric series\n\n\\[ 8 - 4 + 2 - 1 + \\ldots \\]\n\nwhere there are 5 terms in the series.\nSolution\n\nFor this series, we have \\( a = 8 \\), \\( r = -\\frac{1}{2} \\) and \\( n = 5 \\). So\n\n\\[\nS_n = \\frac{a(1 - r^n)}{1 - r}\n\\]\n\n\\[\nS_5 = \\frac{8(1 - (-\\frac{1}{2})^5)}{1 - (-\\frac{1}{2})}\n\\]\n\n\\[\n= \\frac{8(1 - (-\\frac{1}{32}))}{\\frac{3}{2}}\n\\]\n\n\\[\n= \\frac{2 \\times 8 \\times \\frac{33}{32}}{3}\n\\]\n\n\\[\n= \\frac{11}{2}\n\\]\n\n\\[\n= 5\\frac{1}{2}.\n\\]\n\nExample\n\nHow many terms are there in the geometric progression\n\n\\[2, 4, 8, \\ldots, 128?\\]\n\nSolution\n\nIn this sequence \\( a = 2 \\) and \\( r = 2 \\). We also know that the \\( n \\)-th term is 128. But the formula for the \\( n \\)-th term is \\( ar^{n-1} \\). So\n\n\\[\n128 = 2 \\times 2^{n-1}\n\\]\n\n\\[\n64 = 2^{n-1}\n\\]\n\n\\[\n2^6 = 2^{n-1}\n\\]\n\n\\[\n6 = n - 1\n\\]\n\n\\[\nn = 7.\n\\]\n\nSo there are 7 terms in this geometric progression.\n\nExample\n\nHow many terms in the geometric progression\n\n\\[1, 1\\cdot1, 1\\cdot21, 1\\cdot331, \\ldots\\]\n\nwill be needed so that the sum of the first \\( n \\) terms is greater than 20?\n\nSolution\n\nThe sequence is a geometric progression with \\( a = 1 \\) and \\( r = 1\\cdot1 \\). We want to find the smallest value of \\( n \\) such that \\( S_n > 20 \\). Now\n\n\\[\nS_n = \\frac{a(1 - r^n)}{1 - r},\n\\]\nso\n\n\\[\n\\frac{1 \\times (1 - 1 \\cdot 1^n)}{1 - 1 \\cdot 1} > 20\n\\]\n\n\\[\n\\frac{1 - 1 \\cdot 1^n}{-0.1} > 20\n\\]\n\n\\[\n(1 \\cdot 1^n - 1) \\times 10 > 20\n\\]\n\n\\[\n1 \\cdot 1^n - 1 > 2\n\\]\n\n\\[\n1 \\cdot 1^n > 3.\n\\]\n\nIf we now take logarithms of both sides, we get\n\n\\[\nn \\ln 1 \\cdot 1 > \\ln 3\n\\]\n\nand as \\( \\ln 1 \\cdot 1 > 0 \\) we obtain\n\n\\[\nn > \\frac{\\ln 3}{\\ln 1 \\cdot 1} = 11.5267 \\ldots\n\\]\n\nand therefore the smallest whole number value of \\( n \\) is 12.\n\n**Exercise 6**\n\n(a) Find the sum of the first five terms of the GP with first term 3 and common ratio 2.\n\n(b) Find the sum of the first 20 terms of the GP with first term 3 and common ratio 1.5.\n\n(c) The sum of the first 3 terms of a geometric series is \\( \\frac{37}{8} \\). The sum of the first six terms is \\( \\frac{3367}{512} \\). Find the first term and common ratio.\n\n(d) How many terms in the GP 4, 3.6, 3.24, \\ldots are needed so that the sum exceeds 35?\n\n### 7. Convergence of geometric series\n\nConsider the geometric progression\n\n\\[\n1, \\ \\frac{1}{2}, \\ \\frac{1}{4}, \\ \\frac{1}{8}, \\ \\frac{1}{16}, \\ \\ldots\n\\]\n\nWe have \\( a = 1 \\) and \\( r = \\frac{1}{2} \\), and so we can calculate some sums. We get\n\n\\[\nS_1 = 1\n\\]\n\n\\[\nS_2 = 1 + \\frac{1}{2} = \\frac{3}{2}\n\\]\n\n\\[\nS_3 = 1 + \\frac{1}{2} + \\frac{1}{4} = \\frac{7}{4}\n\\]\n\n\\[\nS_4 = 1 + \\frac{1}{2} + \\frac{1}{4} + \\frac{1}{8} = \\frac{15}{8}\n\\]\n\n\\[\n\\vdots\n\\]\n\nand there seems to be a pattern because\n\n\\[\n1 = 2 - 1\n\\]\n\n\\[\n\\frac{3}{2} = 2 - \\frac{1}{2}\n\\]\n\n\\[\n\\frac{7}{4} = 2 - \\frac{1}{4}\n\\]\n\n\\[\n\\frac{15}{8} = 2 - \\frac{1}{8}.\n\\]\nIn each case, we subtract a small quantity from 2, and as we take successive sums the quantity gets smaller and smaller. If we were able to add ‘infinitely many’ terms, then the answer ‘ought to be’ 2 — or as near as we want to get to 2.\n\nLet us see if we can explain this by using some algebra. We know that\n\n\\[ S_n = \\frac{a(1 - r^n)}{1 - r}, \\]\n\nand we want to examine this formula in the case of our particular example where \\( r = \\frac{1}{2} \\). Now the formula contains the term \\( r^n \\) and, as \\(-1 < r < 1\\), this term will get closer and closer to zero as \\( n \\) gets larger and larger. So, if \\(-1 < r < 1\\), we can say that the ‘sum to infinity’ of a geometric series is\n\n\\[ S_\\infty = \\frac{a}{1 - r}, \\]\n\nwhere we have omitted the term \\( r^n \\). We say that this is the limit of the sums \\( S_n \\) as \\( n \\) ‘tends to infinity’. You will find more details of this concept in another unit.\n\n**Example** Find the sum to infinity of the geometric progression\n\n\\[ 1, \\ \\frac{1}{3}, \\ \\frac{1}{9}, \\ \\frac{1}{27}, \\ldots. \\]\n\n**Solution** For this geometric progression we have \\( a = 1 \\) and \\( r = \\frac{1}{3} \\). As \\(-1 < r < 1\\) we can use the formula, so that\n\n\\[ S_\\infty = \\frac{1}{1 - \\frac{1}{3}} = \\frac{1}{\\frac{2}{3}} = \\frac{3}{2}. \\]\n\n**Key Point**\n\nThe sum to infinity of a geometric progression with starting value \\( a \\) and common ratio \\( r \\) is given by\n\n\\[ S_\\infty = \\frac{a}{1 - r} \\]\n\nwhere \\(-1 < r < 1\\).\n\n**Exercise 7**\n\n(a) Find the sum to infinity of the GP with first term 3 and common ratio \\( \\frac{1}{2} \\).\n\n(b) The sum to infinity of a GP is four times the first term. Find the common ratio.\n\n(c) The sum to infinity of a GP is twice the sum of the first two terms. Find possible values of the common ratio.\nAnswers\n\n1. \n (a) 8, 11, 14, 17, 20\n (b) 1, $\\frac{1}{4}$, $\\frac{1}{9}$, $\\frac{1}{16}$; tenth term is $\\frac{1}{100}$\n (c) 1, 1, 2, 3, 5, 8, 13, 21\n (d) 1, $-\\frac{1}{2}$, $\\frac{1}{3}$, $-\\frac{1}{4}$, $\\frac{1}{5}$\n\n2. \n (a) 1, 4, 9, 16, 25, 36\n (b) 4, 6, 6, 4, 0\n\n3. \n (a) 8, 15, 22, 29, 36\n (b) 2, $-3$, $-8$, $-13$, $-18$\n (c) $-12$\n (d) 37\n (e) (i) 35, 62 (ii) $-19$, $-46$\n (f) (i) $-2k/3$ (ii) $k(4 - n)/3$ (iii) $-\\frac{45}{16}$\n\n4. \n (a) $-1679$ (b) (i) 175, (ii) 2875 (c) 1717 (d) 41\n\n5. \n (a) 1, $\\frac{1}{2}$, $\\frac{1}{4}$, $\\frac{1}{8}$, $\\frac{1}{16}$ (b) 1536, 1,572,864 (c) 1458\n\n6. \n (a) 93 (b) 19,946 (c) 2, $\\frac{3}{4}$ (d) 20 terms\n\n7. \n (a) 6 (b) $\\frac{3}{4}$ (c) $\\pm 1/\\sqrt{2}$", "id": "./materials/354.pdf" }, { "contents": "LINEAR RECURRENCE RELATIONS: HANDOUT 2\n\nExercise 1. For each recurrence relation, find the unique root, \\( r \\), of the characteristic equation, and find a closed form for the sequence with \\( a_0 = 0 \\) and \\( a_1 = r \\).\n\n1. \\( a_n = -2a_{n-1} - a_{n-2} \\)\n2. \\( a_n = 4a_{n-1} - 4a_{n-2} \\)\n\nExercise 2. Find a closed form for each of the following recurrence relations:\n\n1. \\( a_n = 6a_{n-1} - 9a_{n-2}; a_0 = 2, a_1 = 3 \\)\n2. \\( a_n = 2a_{n-1} - a_{n-2}; a_0 = 5, a_1 = 3 \\)\n3. \\( a_n = -4a_{n-1} - 4a_{n-2}; a_0 = 1, a_1 = -1 \\)\n\nExercise 3. Have fun!\n\n1. Consider the recurrence relation \\( a_n = \\frac{3}{2}a_{n-1} - \\frac{1}{2}a_{n-2} \\). Choose any starting values you like, \\( a_0 \\) and \\( a_1 \\). Before computing the sequence, compute \\( 2a_1 - a_0 \\). Now find the first ten or so terms of the sequence. You may want to use a calculator. Explain what you notice.\n\n2. The number \\( 3 + 2\\sqrt{2} \\) is approximately 5.82842712474619009. It is irrational, so it never repeats or terminates. Using your calculator, look at the decimal expansion of \\((3 + 2\\sqrt{2})^n\\) for \\( n = 2, 3, 4, 5 \\), and maybe a few more if you like. You should notice that they get very close to taking on integer values. Try to explain why this happens. Here are a few hints:\n - Find a recurrence relation whose closed form is \\( a_n = (3 + 2\\sqrt{2})^n + (3 - 2\\sqrt{2})^n \\).\n - Look at the value of \\((3 - 2\\sqrt{2})^n\\) for a few \\( n \\). What do you notice?\n\n3. Recall the recurrence relation defined by \\( a_n = 2a_{n-1} + a_{n-2} \\) with \\( a_0 = 1 \\) and \\( a_1 = 1 \\). Compute the first few terms. Now compute the first few terms of the sequence \\( a_n = 6a_{n-1} - a_{n-2} \\) with \\( a_0 = 1 \\) and \\( a_1 = 3 \\). Look closely at the two sequences and explain the connection. (Hint: Solve both recurrence relations, trying to find a connection between the roots.)\n\n4. One of the first examples we did was the recurrence relation \\( a_n = -a_{n-1} - a_{n-2} \\). Pick any \\( a_0 \\) and \\( a_1 \\) you like, and compute the first few terms of the sequence. You should notice a pattern. Explain it! (Hint: Look at powers of the roots of the characteristic equation.)\n(5) In the very first lecture, we saw that if \\( f_n \\) is the \\( n \\)th term of the Fibonacci sequence, that \\( f_{n+1}/f_n \\) gets very close to the value \\( \\frac{1+\\sqrt{5}}{2} \\). By similar ideas from (2), explain why this happens.\n\n(6) Compute the first few terms of \\( a_n = 3a_{n-1} - 2a_{n-1} \\) with \\( a_0 \\) and \\( a_1 \\) chosen to your liking. Find a recurrence relation that is every third term of the one you just created. That is, the sequence generated is \\( a_0, a_3, a_6, a_9, \\ldots \\).\n\n(7) Consider the third order linear recurrence relation \\( a_n = 2a_{n-1} + a_{n-2} - 2a_{n-3} \\) with \\( a_0 = 3, a_1 = 2, \\) and \\( a_2 = 6 \\). Try to find a closed form for this sequence using the same type of technique we have already seen.\n\n(8) In this exercise, you will find rational approximations to \\( \\sqrt{3} \\). That is, rational numbers that are very close to \\( \\sqrt{3} \\). To do this, find a recurrence relation with closed form \\( a_n = (1 + \\sqrt{3})^n + (1 - \\sqrt{3})^n \\). Now figure out what happens to the \\( (1 - \\sqrt{3})^n \\) part as \\( n \\) gets large. What do you expect to happen to \\( a_n/a_{n-1} \\) for large \\( n \\)? Now compute \\( (a_n/a_{n-1} - 1)^2 \\) for a few \\( n \\) and look at the decimal expansion.\n\n(9) BONUS Pick any integer, \\( k \\), that isn’t a square. Find a recurrence relation and an integer \\( c \\) such that the sequence has the property that \\( a_n/a_{n-1} - c \\) is a very good approximation of \\( \\sqrt{k} \\) for large \\( n \\). (This is hard)", "id": "./materials/355.pdf" }, { "contents": "6 Generating functions\n\nThe (ordinary) **generating function** of an infinite sequence\n\n\\[ a_0, a_1, a_2, \\ldots, a_n, \\ldots \\]\n\nis the infinite series\n\n\\[ A(x) = a_0 + a_1 x + a_2 x^2 + \\cdots + a_n x^n + \\cdots. \\]\n\nA finite sequence\n\n\\[ a_0, a_1, a_2, \\ldots, a_n \\]\n\ncan be regarded as the infinite sequence\n\n\\[ a_0, a_1, a_2, \\ldots, a_n, 0, 0, \\ldots \\]\n\nand its generating function\n\n\\[ A(x) = a_0 + a_1 x + a_2 x^2 + \\cdots + a_n x^n \\]\n\nis a polynomial.\n\n**Example 6.1.** The generating function of the constant infinite sequence\n\n\\[ 1, 1, \\ldots, 1, \\ldots \\]\n\nis the function\n\n\\[ A(x) = 1 + x + x^2 + \\cdots + x^n + \\cdots = \\frac{1}{1 - x}. \\]\n\n**Example 6.2.** For any positive integer \\( n \\), the generating function for the binomial coefficients\n\n\\[ \\binom{n}{0}, \\binom{n}{1}, \\binom{n}{2}, \\ldots, \\binom{n}{n}, 0, \\ldots \\]\n\nis the function\n\n\\[ \\sum_{k=0}^{n} \\binom{n}{k} x^k = (1 + x)^n. \\]\nExample 6.3. For any real number $\\alpha$, the generating function for the infinite sequence of binomial coefficients\n\n\\[\n\\binom{\\alpha}{0}, \\binom{\\alpha}{1}, \\binom{\\alpha}{2}, \\ldots, \\binom{\\alpha}{n}, \\ldots\n\\]\n\nis the function\n\n\\[\n\\sum_{n=0}^{\\infty} \\binom{\\alpha}{n} x^n = (1 + x)^\\alpha.\n\\]\n\nExample 6.4. Let $k$ be a positive integer and let\n\n\\[a_0, a_1, a_2, \\ldots, a_n, \\ldots\\]\n\nbe the infinite sequence whose general term $a_n$ is the number of nonnegative integral solutions of the equation\n\n\\[x_1 + x_2 + \\cdots + x_k = n.\\]\n\nThen the generating function of the sequence $(a_n)$ is\n\n\\[\nA(x) = \\sum_{n=0}^{\\infty} \\left( \\sum_{i_1+\\cdots+i_k=n} 1 \\right) x^n = \\sum_{n=0}^{\\infty} \\sum_{i_1+\\cdots+i_k=n} x^{i_1+\\cdots+i_k}\n\\]\n\n\\[\n= \\left( \\sum_{i_1=0}^{\\infty} x^{i_1} \\right) \\cdots \\left( \\sum_{i_k=0}^{\\infty} x^{i_k} \\right) = \\frac{1}{(1-x)^k}\n\\]\n\n\\[\n= \\sum_{n=0}^{\\infty} (-1)^n \\binom{-k}{n} x^n = \\sum_{n=0}^{\\infty} \\binom{n+k-1}{n} x^n.\n\\]\n\nExample 6.5. Let $a_n$ be the number of integral solutions of the equation\n\n\\[x_1 + x_2 + x_3 + x_4 = n,\\]\n\nwhere $0 \\leq x_1 \\leq 3$, $0 \\leq x_2 \\leq 2$, $x_3 \\geq 2$, and $3 \\leq x_4 \\leq 5$. The generating function of the sequence $(a_n)$ is\n\n\\[\nA(x) = (1 + x + x^2 + x^3) (1 + x + x^2) (x^2 + x^3 + \\cdots) (x^3 + x^4 + x^5)\n\\]\n\n\\[\n= \\frac{x^5 (1 + x + x^2 + x^3) (1 + x + x^2)^2}{1 - x}.\n\\]\nExample 6.6. Determine the generating function for the number of $n$-combinations of apples, bananas, oranges, and pears where in each $n$-combination the number of apples is even, the number of bananas is odd, the number of oranges is between 0 and 4, and the number of pears is at least two.\n\nThe required generating function is\n\n$$A(x) = \\left( \\sum_{i=0}^{\\infty} x^{2i} \\right) \\left( \\sum_{i=0}^{\\infty} x^{2i+1} \\right) \\left( \\sum_{i=0}^{4} x^{i} \\right) \\left( \\sum_{i=2}^{\\infty} x^{i} \\right)$$\n\n$$= \\frac{x^3(1 - x^5)}{(1 - x^2)^2(1 - x)^2}.$$\n\nExample 6.7. Determine the number $a_n$ of bags with $n$ pieces of fruit (apples, bananas, oranges, and pears) such that the number of apples is even, the number bananas is a multiple of 5, the number oranges is at most 4, and the number of pears is either one or zero.\n\nThe generating function of the sequence $(a_n)$ is\n\n$$A(x) = \\left( \\sum_{i=0}^{\\infty} x^{2i} \\right) \\left( \\sum_{i=0}^{\\infty} x^{5i} \\right) \\left( \\sum_{i=0}^{4} x^{i} \\right) \\left( \\sum_{i=0}^{1} x^{i} \\right)$$\n\n$$= \\frac{(1 + x + x^2 + x^3 + x^4)(1 + x)}{(1 - x^2)(1 - x^5)}$$\n\n$$= \\frac{(1 + x)(1 - x^5)/(1 - x)}{(1 + x)(1 - x)(1 - x^5)}$$\n\n$$= \\frac{1}{(1 - x)^2} = \\sum_{n=0}^{\\infty} (-1)^n \\binom{-2}{n} x^n$$\n\n$$= \\sum_{n=0}^{\\infty} \\binom{n+1}{n} x^n = \\sum_{n=0}^{\\infty} (n+1)x^n.$$\n\nThus $a_n = n + 1$.\n\nExample 6.8. Find a formula for the number $a_{n,k}$ of integer solutions $(i_1, i_2, \\ldots, i_k)$ of the equation\n\n$$x_1 + x_2 + \\cdots + x_k = n$$\n\nsuch that $i_1, i_2, \\ldots, i_k$ are nonnegative odd numbers.\nThe generating function of the sequence \\((a_n)\\) is\n\n\\[\nA(x) = \\left( \\sum_{i=0}^{\\infty} x^{2i+1} \\right) \\cdots \\left( \\sum_{i=0}^{\\infty} x^{2i+1} \\right) = \\frac{x^k}{(1 - x^2)^k}\n\\]\n\n\\[\n= x^k \\sum_{i=0}^{\\infty} \\binom{i + k - 1}{i} x^{2i} = \\sum_{i=0}^{\\infty} \\binom{i + k - 1}{i} x^{2i+k}\n\\]\n\n\\[\n= \\begin{cases} \n\\sum_{j=r}^{\\infty} \\binom{j+r-1}{j-r} x^{2j} & \\text{for } k = 2r \\\\\n\\sum_{j=r}^{\\infty} \\binom{j+r}{j-r} x^{2j+1} & \\text{for } k = 2r + 1.\n\\end{cases}\n\\]\n\nWe then conclude that \\(a_{2s,2r} = \\binom{s+r-1}{s-r}\\), \\(a_{2s+1,2r+1} = \\binom{s+r}{s-r}\\), and \\(a_{n,k} = 0\\) otherwise. We may combine the three cases into\n\n\\[\na_{n,k} = \\begin{cases} \n\\left( \\left\\lfloor \\frac{n}{2} \\right\\rfloor + \\left\\lceil \\frac{k}{2} \\right\\rceil - 1 \\right) & \\text{if } n - k = \\text{even}, \\\\\n0 & \\text{if } n - k = \\text{odd}.\n\\end{cases}\n\\]\n\n**Example 6.9.** Let \\(a_n\\) denote the number of nonnegative integral solutions of the equation\n\n\\[\n2x_1 + 3x_2 + 4x_3 + 5x_4 = n.\n\\]\n\nThen the generating function of the sequence \\((a_n)\\) is\n\n\\[\nA(x) = \\sum_{n=0}^{\\infty} \\left( \\sum_{i,j,k,l \\geq 0} 1 \\right) x^n\n\\]\n\n\\[\n= \\left( \\sum_{i=0}^{\\infty} x^{2i} \\right) \\left( \\sum_{j=0}^{\\infty} x^{3j} \\right) \\left( \\sum_{k=0}^{\\infty} x^{4k} \\right) \\left( \\sum_{l=0}^{\\infty} x^{5l} \\right)\n\\]\n\n\\[\n= \\frac{1}{(1 - x^2)(1 - x^3)(1 - x^4)(1 - x^5)}.\n\\]\n\n**Theorem 6.1.** Let \\(s_n\\) be the number of nonnegative integral solutions of the equation\n\n\\[\na_1x_1 + a_2x_2 + \\cdots + a_kx_k = n.\n\\]\nThen the generating function of the sequence \\((s_n)\\) is\n\n\\[\nA(x) = \\frac{1}{(1 - x^{a_1})(1 - x^{a_2}) \\cdots (1 - x^{a_k})}.\n\\]\n\n7 Recurrence and generating functions\n\nSince\n\n\\[\n\\frac{1}{(1 - x)^n} = \\sum_{k=0}^{\\infty} \\binom{-n}{k} (-x)^k = \\sum_{k=0}^{\\infty} \\binom{n+k-1}{k} x^k, \\quad |x| < 1;\n\\]\n\nthen\n\n\\[\n\\frac{1}{(1 - ax)^n} = \\sum_{k=0}^{\\infty} \\binom{-n}{k} (-ax)^k = \\sum_{k=0}^{\\infty} \\binom{n+k-1}{k} a^k x^k, \\quad |x| < \\frac{1}{|a|}.\n\\]\n\nExample 7.1. Determine the generating function of the sequence\n\n\\[0, 1, 2^2, \\ldots, n^2, \\ldots.\\]\n\nSince \\(\\frac{1}{1-x} = \\sum_{k=0}^{\\infty} x^k\\), then\n\n\\[\n\\frac{1}{(1 - x)^2} = \\frac{d}{dx} \\left( \\frac{1}{1-x} \\right) = \\sum_{k=0}^{\\infty} \\frac{d}{dx} (x^k) = \\sum_{k=0}^{\\infty} kx^{k-1}.\n\\]\n\nThus \\(\\frac{x}{(1-x)^2} = \\sum_{k=0}^{\\infty} kx^k\\). Taking the derivative with respect to \\(x\\) we have\n\n\\[\n\\frac{1 + x}{(1 - x)^3} = \\sum_{k=0}^{\\infty} k^2 x^{k-1}.\n\\]\n\nTherefore the desired generating function is\n\n\\[\nA(x) = \\frac{x(1 + x)}{(1 - x)^3}.\n\\]\n\nExample 7.2. Solve the recurrence relation\n\n\\[\n\\begin{cases}\n a_n = 5a_{n-1} - 6a_{n-2}, & n \\geq 2 \\\\\n a_0 = 1 \\\\\n a_1 = -2\n\\end{cases}\n\\]\nLet \\( A(x) = \\sum_{n=0}^{\\infty} a_n x^n \\). Applying the recurrence relation, we have\n\n\\[\nA(x) = a_0 + a_1 x + \\sum_{n=2}^{\\infty} (5a_{n-1} - 6a_{n-2}) x^n\n\\]\n\n\\[\n= a_0 + a_1 x - 5xa_0 + 5xA(x) - 6x^2A(x).\n\\]\n\nApplying the initial values and collecting the coefficient functions of \\( A(x) \\), we further have\n\n\\[\n(1 - 5x + 6x^2) A(x) = 1 - 7x.\n\\]\n\nThus the function \\( g(x) \\) is solved as\n\n\\[\nA(x) = \\frac{1 - 7x}{1 - 5x + 6x^2}.\n\\]\n\nObserving that \\( 1 - 5x + 6x^2 = (1 - 2x)(1 - 3x) \\) and applying partial fraction,\n\n\\[\n\\frac{1 - 7x}{1 - 5x + 6x^2} = \\frac{A}{1 - 2x} + \\frac{B}{1 - 3x}.\n\\]\n\nThe constants \\( A \\) and \\( B \\) can be determined by\n\n\\[\nA(1 - 3x) + B(1 - 2x) = 1 - 7x.\n\\]\n\nThen\n\n\\[\n\\begin{cases}\nA + B = 1 \\\\\n-3A - 2B = -7\n\\end{cases}\n\\]\n\nThus \\( A = 5, B = -4 \\). Hence\n\n\\[\nA(x) = \\frac{1 - 7x}{1 - 5x + 6x^2} = \\frac{5}{1 - 2x} - \\frac{4}{1 - 3x}.\n\\]\n\nSince\n\n\\[\n\\frac{1}{1 - 2x} = \\sum_{n=0}^{\\infty} 2^n x^n \\quad \\text{and} \\quad \\frac{1}{1 - 3x} = \\sum_{n=0}^{\\infty} 3^n x^n\n\\]\n\nWe obtain the sequence\n\n\\[\na_n = 5 \\times 2^n - 4 \\times 3^n, \\quad n \\geq 0.\n\\]\nTheorem 7.1. Let \\((a_n; n \\geq 0)\\) be a sequence satisfying the homogeneous linear recurrence relation of order \\(k\\) with constant coefficients, i.e.,\n\n\\[\na_n = \\alpha_1 a_{n-1} + \\alpha_2 a_{n-2} + \\cdots + \\alpha_k a_{n-k},\n\\]\n\nwhere \\(\\alpha_k \\neq 0\\), \\(n \\geq k\\). Then its generating function \\(A(x) = \\sum_{n=0}^{\\infty} a_n x^n\\) is a rational function of the form\n\n\\[\nA(x) = \\frac{P(x)}{Q(x)},\n\\]\n\nwhere \\(Q(x)\\) is a polynomial of degree \\(k\\) with a nonzero constant term and \\(P(x)\\) is a polynomial of degree strictly less than \\(k\\).\n\nConversely, given such polynomials \\(P(x)\\) and \\(Q(x)\\), there exists a unique sequence \\((a_n)\\) satisfying the linear homogeneous recurrence relation (18), and its generating function is the rational function in (19).\n\nProof. The generating function \\(A(x)\\) of the sequence \\((a_n)\\) can be written as\n\n\\[\nA(x) = \\sum_{i=0}^{k-1} a_i x^i + \\sum_{n=k}^{\\infty} a_n x^n = \\sum_{i=0}^{k-1} a_i x^i + \\sum_{n=k}^{\\infty} \\left( \\sum_{i=1}^{k} \\alpha_i a_{n-i} \\right) x^n\n\\]\n\n\\[\n= \\sum_{i=0}^{k-1} a_i x^i + \\sum_{i=1}^{k} \\alpha_i \\sum_{n=k}^{\\infty} a_{n-i} x^n = \\sum_{i=0}^{k-1} a_i x^i + \\sum_{i=1}^{k} \\alpha_i \\sum_{n=k-i}^{\\infty} a_n x^{n+i}\n\\]\n\n\\[\n= \\sum_{i=0}^{k-1} a_i x^i + \\alpha_k x^k \\sum_{n=0}^{\\infty} a_n x^n + \\sum_{i=1}^{k} \\alpha_i x^i \\left( \\sum_{n=0}^{\\infty} a_n x^n - \\sum_{j=0}^{k-1} a_j x^j \\right)\n\\]\n\n\\[\n= \\sum_{i=0}^{k-1} a_i x^i + A(x) \\sum_{i=1}^{k} \\alpha_i x^i - \\sum_{i=1}^{k-1} \\alpha_i x^i \\sum_{j=0}^{k-1} a_j x^j\n\\]\n\n\\[\n= A(x) \\sum_{i=1}^{k} \\alpha_i x^i + \\sum_{i=0}^{k-1} a_i x^i - \\sum_{l=1}^{k-1} x^l \\sum_{i=1}^{l} \\alpha_i a_{l-i}.\n\\]\nThen\n\\[\nA(x) \\left( 1 - \\sum_{i=1}^{k} \\alpha_i x^i \\right) = \\sum_{i=0}^{k-1} a_i x^i - \\sum_{l=1}^{k-1} x^l \\sum_{i=1}^{l} \\alpha_i a_{l-i}\n\\]\n\\[\n= a_0 + \\sum_{l=1}^{k-1} \\left( a_l - \\sum_{i=1}^{l} \\alpha_i a_{l-i} \\right) x^l.\n\\]\nThus\n\\[\nP(x) = a_0 + \\sum_{l=1}^{k-1} \\left( a_l - \\sum_{i=1}^{l} \\alpha_i a_{l-i} \\right) x^l,\n\\]\n\\[\nQ(x) = 1 - \\sum_{i=1}^{k} \\alpha_i x^i.\n\\]\n\nConversely, let \\((a_n)\\) be the sequence whose generating function is \\(A(x) = P(x)/Q(x)\\). Write\n\\[\nA(x) = \\sum_{n=0}^{\\infty} a_n x^n, \\quad P(x) = \\sum_{i=0}^{k} b_i x^i, \\quad Q(x) = 1 - \\sum_{i=1}^{k} \\alpha_i x^i.\n\\]\nThen \\(A(x) = \\frac{P(x)}{Q(x)}\\) is equivalent to\n\\[\n\\left( 1 - \\sum_{i=1}^{k} \\alpha_i x^i \\right) \\left( \\sum_{n=0}^{\\infty} a_n x^n \\right) = \\sum_{i=0}^{k} b_i x^i.\n\\]\nThe polynomial \\(Q(x)\\) can be viewed as an infinite series with \\(\\alpha_i = 0\\) for \\(i > k\\). Thus\n\\[\n\\sum_{n=0}^{\\infty} a_n x^n - \\sum_{n=0}^{\\infty} \\left( \\sum_{i=1}^{n} \\alpha_i a_{n-i} \\right) x^n = \\sum_{i=0}^{k} b_i x^i.\n\\]\nEquating the coefficients of \\(x^n\\), we have the recurrence relation\n\\[\na_n = \\sum_{i=1}^{k} \\alpha_i a_{n-i}, \\quad n \\geq k.\n\\]\nProposition 7.2 (Partial Fractions). (a) If $P(x)$ is a polynomial of degree less than $k$, then\n\n$$\n\\frac{P(x)}{(1-ax)^k} = \\frac{A_1}{1-ax} + \\frac{A_2}{(1-ax)^2} + \\cdots + \\frac{A_k}{(1-ax)^k},\n$$\n\nwhere $A_1, A_2, \\ldots, A_k$ are constants to be determined.\n\n(b) If $P(x)$ is a polynomial of degree less than $p+q+r$, then\n\n$$\n\\frac{P(x)}{(1-ax)^p(1-bx)^q(1-cx)^r} = \\frac{A_1(x)}{(1-ax)^p} + \\frac{A_2(x)}{(1-bx)^q} + \\frac{A_3(x)}{(1-cx)^r},\n$$\n\nwhere $A_1(x), A_2(x),$ and $A_3(x)$ are polynomials of degree $q+r$, $p+r$, and $p+q$, respectively.\n\n8 A geometry example\n\nA polygon $P$ in $\\mathbb{R}^2$ is called convex if the segment joining any two points in $P$ is also contained in $P$. Let $C_n$ denote the number of ways to divide a labeled convex polygon with $n+2$ sides into triangles. The first a few such numbers are $C_1 = 1$, $C_2 = 2$, $C_3 = 5$.\n\nWe first establish a recurrence relation for $C_{n+1}$ in terms of $C_0$, $C_1$, $\\ldots$, $C_n$. Let $P_{v_1v_2\\ldots v_{n+3}}$ denote a convex $(n+3)$-polygon with vertices $v_1, v_2, \\ldots, v_{n+3}$. In each triangular decomposition of $P_{v_1v_2\\ldots v_{n+3}}$ into triangles, the segment $v_1v_{n+3}$ is one side of a triangle $\\Delta$ in the decomposition; the third vertex of the triangle $\\Delta$ is one of the vertices $v_2, v_3, \\ldots, v_{n+2}$. Let $v_{k+2}$ be the third vertex of $\\Delta$ other than $v_1$ and $v_{n+3}$ ($0 \\leq k \\leq n$); see Figure 1 below. Then we have a convex $(k+2)$-polygon $P_{v_1v_2\\ldots v_{k+2}}$ and another convex $(n-k+2)$-polygon $P_{v_{k+2}v_{k+3}\\ldots v_{n+3}}$. Then\n\n![Figure 1: $v_{k+2}$ is the third vertex of the triangle with the side $v_1v_{n+3}$](image-url)\nby induction there are $C_k$ ways to divide $P_{v_1v_2...v_{k+2}}$ into triangles, and $C_{n-k}$ ways to divide $P_{v_{k+2}v_{k+3}...v_{n+3}}$ into triangles. We thus have the recurrence relation\n\n$$C_{n+1} = \\sum_{k=0}^{n} C_k C_{n-k} \\quad \\text{with} \\quad C_0 = 1.$$ \n\nConsider the generating function $F(x) = \\sum_{n=0}^{\\infty} C_n x^n$. Then\n\n$$F(x)F(x) = \\left( \\sum_{n=0}^{\\infty} C_n x^n \\right) \\left( \\sum_{n=0}^{\\infty} C_n x^n \\right)$$\n\n$$= \\sum_{n=0}^{\\infty} \\left( \\sum_{k=0}^{n} C_k C_{n-k} \\right) x^n$$\n\n$$= \\sum_{n=0}^{\\infty} C_{n+1} x^n = \\frac{1}{x} \\sum_{n=1}^{\\infty} C_n x^n$$\n\n$$= \\frac{F(x)}{x} - \\frac{1}{x}.$$ \n\nWe thus obtain the equation\n\n$$xF(x)^2 - F(x) + 1 = 0.$$ \n\nSolving for $F(x)$, we obtain\n\n$$F(x) = \\frac{1 \\pm \\sqrt{1 - 4x}}{2x}.$$ \n\nNote that\n\n$$\\sqrt{1 - 4x} = 1 + \\sum_{n=1}^{\\infty} (-1)^n \\left( \\frac{1}{2} \\right)^n 4^n x^n$$\n\n$$= 1 + \\sum_{n=1}^{\\infty} a_n x^n,$$\nwhere\n\\[\na_n = (-1)^n \\left[ \\frac{1}{2} \\left( \\frac{1}{2} - 1 \\right) \\cdots \\left( \\frac{1}{2} - n + 1 \\right) / n! \\right] \\cdot 2^n \\cdot 2^n\n\\]\n\\[\n= (-1)^n \\cdot \\frac{(-1)(-3)(-5) \\cdots (-2n-1+1)}{n!} \\cdot 2^n\n\\]\n\\[\n= -\\frac{1 \\cdot 3 \\cdot 5 \\cdots (2n-1-1)}{n!} \\cdot 2^n\n\\]\n\\[\n= -2 \\cdot \\frac{(2(n-1))!}{n!(n-1)!}.\n\\]\n\nThen\n\\[\n\\sqrt{1 - 4x} = 1 - 2 \\sum_{n=1}^{\\infty} \\frac{(2(n-1))!}{n!(n-1)!} x^n\n\\]\n\\[\n= 1 - 2 \\sum_{n=0}^{\\infty} \\frac{(2n)!}{n!(n+1)!} x^{n+1}.\n\\]\n\nWe conclude that\n\\[\nF(x) = \\frac{1 - \\sqrt{1 - 4x}}{2x}\n\\]\n\\[\n= \\sum_{n=0}^{\\infty} \\frac{(2n)!}{n!(n+1)!} x^n\n\\]\n\\[\n= \\sum_{n=0}^{\\infty} \\frac{1}{n+1} \\binom{2n}{n} x^n.\n\\]\n\nHence the sequence \\((C_n)\\) is given by the binomial coefficients:\n\\[\nC_n = \\frac{1}{n+1} \\binom{2n}{n}, \\quad n \\geq 0.\n\\]\n\nThe sequence \\((C_n)\\) is known as the **Catalan sequence** and the numbers \\(C_n\\) as the **Catalan numbers**.\n\n**Example 8.1.** Let \\(C_n\\) be the number of ways to evaluate a matrix product\n\\[\nA_1 A_2 \\cdots A_{n+1}, \\quad n \\geq 0\n\\]\nby adding various parentheses. For instance, $C_0 = 1$, $C_1 = 1$, $C_2 = 2$, and $C_3 = 5$. In general the formula is given by\n\n$$C_n = \\frac{1}{n+1} \\binom{2n}{n}.$$ \n\nNote that each way of evaluating the matrix product $A_1A_2 \\cdots A_{n+2}$ will be finished by multiplying of two matrices at the end. There are exactly $n+1$ ways of multiplying the two matrices at the end:\n\n$$A_1A_2 \\cdots A_{n+2} = (A_1 \\cdots A_{k+1})(A_{k+2} \\cdots A_{n+2}), \\quad 0 \\leq k \\leq n.$$ \n\nThis yields the recurrence relation\n\n$$C_{n+1} = \\sum_{k=0}^{n} C_k C_{n-k}.$$ \n\nThus $C_n = \\frac{1}{n+1} \\binom{2n}{n}$, $n \\geq 0$.\n\n9 Exponential generating functions\n\nThe ordinary generating function method is a powerful algebraic tool for finding unknown sequences, especially when the sequences are certain binomial coefficients or the order is immaterial. However, when the sequences are not binomial type or the order is material in defining the sequences, we may need to consider a different type of generating functions. For example, the sequence $a_n = n!$ is the counting of the number of permutations of $n$ distinct objects; its ordinary generating function\n\n$$\\sum_{n=0}^{\\infty} a_n x^n = \\sum_{n=0}^{\\infty} n! x^n$$\n\ncannot be easily figure out as a closed known expression. However, the generating function\n\n$$\\sum_{n=0}^{\\infty} \\frac{a_n}{n!} x^n = \\sum_{n=0}^{\\infty} x^n = \\frac{1}{1-x}$$\n\nis obvious.\nThe **exponential generating function** of a sequence \\((a_n; n \\geq 0)\\) is the infinite series\n\n\\[\nE(x) = \\sum_{n=0}^{\\infty} \\frac{a_n}{n!} x^n.\n\\]\n\n**Example 9.1.** The exponential generating function of the sequence\n\n\\[P(n, 0), P(n, 1), \\ldots, P(n, n), 0, \\ldots\\]\n\nis given by\n\n\\[\nE(x) = \\sum_{k=0}^{n} \\frac{P(n, k)}{k!} x^k\n= \\sum_{k=0}^{n} \\binom{n}{k} x^k\n= (1 + x)^n.\n\\]\n\n**Example 9.2.** The exponential generating function of the constant sequence \\((a_n = 1; n \\geq 0)\\) is\n\n\\[\nE(x) = \\sum_{n=0}^{\\infty} \\frac{x^n}{n!} = e^x.\n\\]\n\nThe exponential generating function of the geometric sequence \\((a_n = a^n; n \\geq 0)\\) is\n\n\\[\nE(x) = \\sum_{n=0}^{\\infty} \\frac{a^n x^n}{n!} = e^{ax}.\n\\]\n\n**Theorem 9.1.** Let \\(M = \\{n_1 \\alpha_1, n_2 \\alpha_2, \\ldots, n_k \\alpha_k\\}\\) be a multiset over the set \\(S = \\{\\alpha_1, \\alpha_2, \\ldots, \\alpha_k\\}\\) with \\(n_1\\) many \\(\\alpha_1\\)'s, \\(n_2\\) many \\(\\alpha_2\\)'s, \\ldots, \\(n_k\\) many \\(\\alpha_k\\)'s. Let \\(a_n\\) be the number of \\(n\\)-permutations of the multiset \\(M\\). Then the exponential generating function of the sequence \\((a_n; n \\geq 0)\\) is given by\n\n\\[\nE(x) = \\left(\\sum_{i=0}^{n_1} \\frac{x^i}{i!}\\right) \\left(\\sum_{i=0}^{n_2} \\frac{x^i}{i!}\\right) \\cdots \\left(\\sum_{i=0}^{n_k} \\frac{x^i}{i!}\\right).\n\\]\nProof. Note that $a_n = 0$ for $n > n_1 + \\cdots + n_k$. Thus $E(x)$ is a polynomial. The right side of (20) can be expanded to the form\n\n$$\n\\sum_{i_1, i_2, \\ldots, i_k = 0}^{n_1, n_2, \\ldots, n_k} \\frac{x^{i_1+i_2+\\cdots+i_k}}{i_1!i_2!\\cdots i_k!} = \\sum_{n=0}^{n_1+n_2+\\cdots+n_k} \\frac{x^n}{n!} \\sum_{0 \\leq i_1 \\leq n_1, \\ldots, 0 \\leq i_k \\leq n_k} \\frac{n!}{i_1!i_2!\\cdots i_k!}.\n$$\n\nNote that the number of permutation of $M$ with exactly $i_1 \\alpha_1$’s, $i_2 \\alpha_2$’s, $\\ldots$, and $i_k \\alpha_k$’s such that\n\n$$i_1 + i_2 + \\cdots + i_k = n$$\n\nis the multinomial coefficient\n\n$$\\binom{n}{i_1, i_2, \\ldots, i_k} = \\frac{n!}{i_1!i_2!\\cdots i_k!}.$$\n\nIt turns out that the sequence $(a_n)$ is given by\n\n$$a_n = \\sum_{0 \\leq i_1 \\leq n_1, \\ldots, 0 \\leq i_k \\leq n_k} \\binom{n}{i_1, i_2, \\ldots, i_k}, \\quad n \\geq 0.$$\n\nWe proved the closed form of the required exponential generating function. \\qed\n\nExample 9.3. Determine the number of ways to color the squares of a 1-by-$n$ chessboard using the colors, red, white, and blue, if an even number of squares are colored red.\n\nLet $a_n$ denote the number of ways of such colorings and set $a_0 = 1$. Each such coloring can be considered as a permutation of three objects $r$ (for red), $w$ (for white), and $b$ (for blue) with repetition allowed, and the element $r$ appears even\nnumber of times. The exponential generating function of the sequence \\((a_n)\\) is\n\n\\[\nE(x) = \\left( \\sum_{n=0}^{\\infty} \\frac{x^{2n}}{(2n)!} \\right) \\left( \\sum_{n=0}^{\\infty} \\frac{x^n}{n!} \\right)^2\n\\]\n\n\\[\n= \\frac{e^x + e^{-x}}{2} e^{2x} = \\frac{1}{2} (e^{3x} + e^x)\n\\]\n\n\\[\n= \\frac{1}{2} \\left( \\sum_{n=0}^{\\infty} \\frac{3^n x^n}{n!} + \\sum_{n=0}^{\\infty} \\frac{x^n}{n!} \\right)\n\\]\n\n\\[\n= \\frac{1}{2} \\sum_{n=0}^{\\infty} (3^n + 1) \\cdot \\frac{x^n}{n!}.\n\\]\n\nThus the sequence is given by\n\n\\[\na_n = \\frac{3^n + 1}{2}, \\quad n \\geq 0.\n\\]\n\n**Example 9.4.** Determine the number \\(a_n\\) of \\(n\\) digit (under base 10) numbers with each digit odd where the digit 1 and 3 occur an even number of times.\n\nAssume \\(a_0 = 1\\). The number \\(a_n\\) equals the number of \\(n\\)-permutations of the multiset \\(M = \\{\\infty 1, \\infty 3, \\infty 5, \\infty 7, \\infty 9\\}\\), in which 1 and 3 occur an even number of times. The exponential generating function of the sequence \\(a_n\\) is\n\n\\[\nE(x) = \\left( \\sum_{n=0}^{\\infty} \\frac{x^{2n}}{(2n)!} \\right) \\left( \\sum_{n=0}^{\\infty} \\frac{x^n}{n!} \\right)^3\n\\]\n\n\\[\n= \\left( \\frac{e^x + e^{-x}}{2} \\right)^2 e^{3x}\n\\]\n\n\\[\n= \\frac{1}{4} \\left( e^{5x} + 2e^{3x} + e^x \\right)\n\\]\n\n\\[\n= \\frac{1}{4} \\left( \\sum_{n=0}^{\\infty} \\frac{5^n x^n}{n!} + \\sum_{n=0}^{\\infty} \\frac{3^n x^n}{n!} + \\sum_{n=0}^{\\infty} \\frac{x^n}{n!} \\right)\n\\]\n\n\\[\n= \\sum_{n=0}^{\\infty} \\left( \\frac{5^n + 2 \\times 3^n + 1}{4} \\right) \\frac{x^n}{n!}.\n\\]\n\nThus\n\n\\[\na_n = \\frac{5^n + 2 \\times 3^n + 1}{4}, \\quad n \\geq 0.\n\\]\nExample 9.5. Determine the number of ways to color the squares of a 1-by-$n$ board with the colors, red, blue, and white, where the number of red squares is even and there is at least one blue square.\n\nThe exponential generating function for the sequence is\n\n$$E(x) = \\left( \\sum_{i=0}^{\\infty} \\frac{x^{2i}}{(2i)!} \\right) \\left( \\sum_{i=0}^{\\infty} \\frac{x^i}{i!} \\right) \\left( \\sum_{i=1}^{\\infty} \\frac{x^i}{i!} \\right)$$\n\n$$= \\frac{e^x + e^{-x}}{2} e^x (e^x - 1)$$\n\n$$= \\frac{1}{2} (e^{3x} - e^{2x} + e^x - 1)$$\n\n$$= -\\frac{1}{2} + \\sum_{n=0}^{\\infty} \\frac{3^n - 2^n + 1}{2} \\cdot \\frac{x^n}{n!}$$\n\nThus\n\n$$a_n = \\frac{3^n - 2^n + 1}{2}, \\quad n \\geq 1$$\n\nand\n\n$$a_0 = 0.$$\n\n10 Combinatorial interpretations\n\nTheorem 10.1. The combinatorial interpretations of ordinary generating functions:\n\n(a) The number of ways of placing $n$ indistinguishable balls into $m$ distinguishable boxes is the coefficient of $x^n$ in\n\n$$(1 + x + x^2 + \\cdots)^m = \\left( \\sum_{k=0}^{\\infty} x^k \\right)^m = \\frac{1}{(1 - x)^m}.$$\n\n(b) The number of ways of placing $n$ indistinguishable balls into $m$ distinguishable boxes with at most $r_k$ balls in the $k$th box is the coefficient of $x^n$ in the expression\n\n$$\\prod_{k=1}^{m} (1 + x + x^2 + \\cdots + x^{r_k}).$$\n(c) The number of ways of placing \\( n \\) indistinguishable balls into \\( m \\) distinguishable boxes with at least \\( s_k \\) balls in the \\( k \\)th box is the coefficient of \\( x^n \\) in the expression\n\n\\[\n\\prod_{k=1}^{m} x^{s_k} (1 + x + x^2 + \\cdots) = \\frac{x^{s_1+\\cdots+s_m}}{(1 - x)^m}.\n\\]\n\n(d) The number of ways of placing \\( n \\) indistinguishable balls into \\( m \\) distinguishable boxes, such that the number of balls held in the \\( k \\)th box is allowed in a subset \\( C_k \\subseteq \\mathbb{Z}_{\\geq 0} \\) \\((1 \\leq k \\leq m)\\), is the coefficient of \\( x^n \\) in the expression\n\n\\[\n\\left( \\sum_{k \\in C_1} x^k \\right) \\left( \\sum_{k \\in C_2} x^k \\right) \\cdots \\left( \\sum_{k \\in C_m} x^k \\right).\n\\]\n\n**Proof.** Let us consider the situation of placing infinitely many indistinguishable balls into \\( m \\) distinguishable boxes. We use the symbol \\( x \\) to represent a ball and \\( x^k \\) to represent indistinguishable \\( k \\) balls. In the course of placing the balls into \\( m \\) boxes, each box contains either none of balls \\((x^0 = 1)\\), or one ball \\((x)\\), or two balls \\((x^2)\\), or three balls \\((x^3)\\), etc.; that is, each box contains\n\n\\[\n1 + x + x^2 + x^3 + \\cdots\n\\]\n\nballs, where the plus sign “+” means “or”.\n\n| Box 1 | Box 2 | Box 3 | \\( \\cdots \\) | Box \\( m \\) |\n|-------|-------|-------|-------------|-------------|\n| 1 | 1 | 1 | 1 | 1 |\n| \\( x \\) | \\( x \\) | \\( x \\) | \\( x \\) | \\( x \\) |\n| \\( x^2 \\) | \\( x^2 \\) | \\( x^2 \\) | \\( x^2 \\) | \\( x^2 \\) |\n| \\( x^3 \\) | \\( x^3 \\) | \\( x^3 \\) | \\( x^3 \\) | \\( x^3 \\) |\n| \\( \\vdots \\) | \\( \\vdots \\) | \\( \\vdots \\) | \\( \\vdots \\) | \\( \\vdots \\) |\n\nIf we use multiplication to represent “and”, then all possible distributions of balls in the \\( m \\) boxes in the course of placing are contained in the expression\n\n\\[\n(1 + x + x^2 + \\cdots)^m.\n\\]\nThus, when there are \\( n \\) distinguishable balls to be placed into the \\( m \\) distinguishable boxes, the number of distributions of the \\( n \\) balls in the \\( m \\) boxes will be the coefficient of \\( x^n \\) in the expression (21). Indeed, to compute the coefficient of \\( x^n \\) in (21), we select \\( x^{n_1} \\) from column 1, \\( x^{n_2} \\) from column 2, \\ldots, \\( x^{n_m} \\) from column \\( m \\) in the above table such that \\( x^{n_1}x^{n_2}\\cdots x^{n_m} = x^n \\), and do this in all possible ways. The number of such ways is clearly the number of nonnegative integer solutions of the equation\n\n\\[\ny_1 + y_2 + \\cdots + y_m = n,\n\\]\n\nand is given by \\( \\binom{m+n-1}{n} \\).\n\nThe situation for other cases are similar. \\( \\square \\)\n\n**Theorem 10.2.** The combinatorial interpretation of exponential generating functions:\n\n(a) The number of ways of selecting an ordered \\( n \\) objects with repetition allowed from an \\( m \\)-set is the coefficient of \\( \\frac{x^n}{n!} \\) in the expression\n\n\\[\n\\left( 1 + \\frac{x}{1!} + \\frac{x^2}{2!} + \\frac{x^3}{3!} + \\cdots \\right)^m = \\left( \\sum_{k=0}^{\\infty} \\frac{x^k}{k!} \\right)^m = e^{mx}.\n\\] \n\n(b) The number of ways of selecting an ordered \\( n \\) objects from an \\( m \\)-set such that the number of \\( k \\)th objects is allowed in a subset \\( C_k \\subseteq \\mathbb{Z}_{\\geq 0} \\), is the coefficient of \\( \\frac{x^n}{n!} \\) in the expression\n\n\\[\n\\left( \\sum_{k \\in C_1} \\frac{x^k}{k!} \\right) \\left( \\sum_{k \\in C_2} \\frac{x^k}{k!} \\right) \\cdots \\left( \\sum_{k \\in C_m} \\frac{x^k}{k!} \\right).\n\\]\n\n**Example 10.1.** Find the number of 4-permutations of the multiset \\( M = \\{a, a, b, b, b, c\\} \\).\nThere are 5 sub-multisets of $M$ having cardinality 4.\n\n| Sub-multiset | 4!/2!2!0! | 4!/2!1!1! | 4!/1!3!0! | 4!/1!2!1! | 4!/0!3!1! |\n|--------------|-----------|-----------|-----------|-----------|-----------|\n\nThen the answer is\n\n$$\\frac{4!}{2!2!0!} + \\frac{4!}{2!1!1!} + \\frac{4!}{1!3!0!} + \\frac{4!}{1!2!1!} + \\frac{4!}{0!3!1!} = 38,$$\n\nwhich is the coefficient of $\\frac{x^4}{4!}$ in the expansion of\n\n$$E(x) = \\left(1 + \\frac{x}{1!} + \\frac{x^2}{2!}\\right) \\left(1 + \\frac{x}{1!} + \\frac{x^2}{2!} + \\frac{x^3}{3!}\\right) \\left(1 + \\frac{x}{1!}\\right).$$\n\nIn fact, the coefficient of $x^4/4!$ is\n\n$$a_4 = \\sum_{i+j+k=4 \\atop i \\leq 2, j \\leq 3, k \\leq 1} \\frac{4!}{i!j!k!} = \\frac{4!}{2!2!0!} + \\frac{4!}{2!1!1!} + \\frac{4!}{1!3!0!} + \\frac{4!}{1!2!1!} + \\frac{4!}{0!3!1!}.$$", "id": "./materials/356.pdf" }, { "contents": "Section 1 Arithmetic Progression\n\nAn arithmetic progression is a list of numbers where the difference between successive numbers is constant. The terms in an arithmetic progression are usually denoted as \\( u_1, u_2, u_3 \\) etc. where \\( u_1 \\) is the initial term in the progression, \\( u_2 \\) is the second term, and so on; \\( u_n \\) is the \\( n \\)th term. An example of an arithmetic progression is\n\n\\[\n2, 4, 6, 8, 10, 12, 14, \\ldots\n\\]\n\nSince the difference between successive terms is constant, we have\n\n\\[\nu_3 - u_2 = u_2 - u_1\n\\]\n\nand in general\n\n\\[\nu_{n+1} - u_n = u_2 - u_1\n\\]\n\nWe will denote the difference \\( u_2 - u_1 \\) as \\( d \\), which is a common notation.\n\n**Example 1**: Given that 3, 7 and 11 are the first three terms in an arithmetic progression, what is \\( d \\)?\n\n\\[\n7 - 3 = 11 - 7 = 4\n\\]\n\nThen \\( d = 4 \\). That is, the common difference between the terms is 4.\n\nIf we know the first term in an arithmetic progression, and the difference between terms, then we can work out the \\( n \\)th term, i.e. we can work out what any term will be. The formula which tells us what the \\( n \\)th term in an arithmetic progression is\n\n\\[\nu_n = a + (n - 1) \\times d\n\\]\n\nwhere \\( a \\) is the first term.\n\n**Example 2**: If the first 3 terms in an arithmetic progression are 3, 7, 11 then what is the 10th term? The first term is \\( a = 3 \\), and the common difference is \\( d = 4 \\).\n\n\\[\nu_n = a + (n - 1)d\n\\]\n\n\\[\nu_{10} = 3 + (10 - 1)4\n\\]\n\n\\[\n= 3 + 9 \\times 4\n\\]\n\n\\[\n= 39\n\\]\nExample 3: If the first 3 terms in an arithmetic progression are 8, 5, 2 then what is the 16th term? In this progression \\( a = 8 \\) and \\( d = -3 \\).\n\n\\[\n\\begin{align*}\n u_n &= a + (n - 1)d \\\\\n u_{16} &= 8 + (10 - 1) \\times (-3) \\\\\n &= -37\n\\end{align*}\n\\]\n\nExample 4: Given that \\( 2x, 5 \\) and \\( 6 - x \\) are the first three terms in an arithmetic progression, what is \\( d \\)?\n\n\\[\n\\begin{align*}\n 5 - 2x &= (6 - x) - 5 \\\\\n x &= 4\n\\end{align*}\n\\]\n\nSince \\( x = 4 \\), the terms are 8, 5, 2 and the difference is \\(-3\\). The next term in the arithmetic progression will be \\(-1\\).\n\nAn arithmetic series is an arithmetic progression with plus signs between the terms instead of commas. We can find the sum of the first \\( n \\) terms, which we will denote by \\( S_n \\), using another formula:\n\n\\[\nS_n = \\frac{n}{2} [2a + (n - 1)d]\n\\]\n\nExample 5: If the first 3 terms in an arithmetic progression are 3, 7, 11 then what is the sum of the first 10 terms?\n\nNote that \\( a = 3 \\), \\( d = 4 \\) and \\( n = 10 \\).\n\n\\[\n\\begin{align*}\n S_{10} &= \\frac{10}{2} (2 \\times 3 + (10 - 1) \\times 4) \\\\\n &= 5(6 + 36) \\\\\n &= 210\n\\end{align*}\n\\]\n\nAlternatively, but more tediously, we add the first 10 terms together:\n\n\\[\nS_{10} = 3 + 7 + 11 + 15 + 19 + 23 + 27 + 31 + 35 + 39 = 210\n\\]\n\nThis method would have drawbacks if we had to add 100 terms together!\n\nExample 6: If the first 3 terms in an arithmetic progression are 8, 5, 2 then what is the sum of the first 16 terms?\n\n\\[\n\\begin{align*}\n S_{16} &= \\frac{16}{2} (2 \\times 8 + (16 - 1) \\times (-3)) \\\\\n &= 8(16 - 45) \\\\\n &= -232\n\\end{align*}\n\\]\nExercises:\n\n1. For each of the following arithmetic progressions, find the values of $a$, $d$, and the $u_n$ indicated.\n\n (a) 1, 4, 7, ..., $(u_{10})$\n (b) $-8, -6, -4, ..., (u_{12})$\n (c) 8, 4, 0, ..., $(u_{20})$\n (d) $-20, -15, -10, ..., (u_6)$\n (e) 40, 30, 20, ..., $(u_{18})$\n (f) $-6, -8, -10, ..., (u_{12})$\n (g) 2, $2\\frac{1}{2}, 3, ..., (u_{19})$\n (h) 6, $5\\frac{1}{2}, 5\\frac{1}{2}, ..., (u_{10})$\n (i) $-7, -6\\frac{1}{2}, -6, ..., (u_{14})$\n (j) 0, $-5, -10, ..., (u_{15})$\n\n2. For each of the following arithmetic progressions, find the values of $a$, $d$, and the $S_n$ indicated.\n\n (a) 1, 3, 5, ..., $(S_8)$\n (b) 2, 5, 8, ..., $(S_{10})$\n (c) 10, 7, 4, ..., $(S_{20})$\n (d) $6, 6\\frac{1}{2}, 7, ..., (S_8)$\n (e) $-8, -7, -6, ..., (S_{14})$\n (f) $-2, 0, 2, ..., (S_5)$\n (g) $-20, -16, -12, ..., (S_4)$\n (h) 40, 35, 30, ..., $(S_{11})$\n (i) $12, 10\\frac{1}{2}, 9, ..., (S_9)$\n (j) $-8, -5, -2, ..., (S_{20})$\n\n---\n\nSection 2 Geometric Progressions\n\nA geometric progression is a list of terms as in an arithmetic progression but in this case the ratio of successive terms is a constant. In other words, each term is a constant times the term that immediately precedes it. Let’s write the terms in a geometric progression as $u_1, u_2, u_3, u_4$ and so on. An example of a geometric progression is\n\n$$10, 100, 1000, 10000, \\ldots$$\n\nSince the ratio of successive terms is constant, we have\n\n$$\\frac{u_3}{u_2} = \\frac{u_2}{u_1} \\quad \\text{and} \\quad \\frac{u_{n+1}}{u_n} = \\frac{u_2}{u_1}$$\n\nThe ratio of successive terms is usually denoted by $r$ and the first term again is usually written $a$. \n\n3\nExample 1: Find $r$ for the geometric progression whose first three terms are 2, 4, 8.\n\n\\[\n\\frac{4}{2} = \\frac{8}{4} = 2\n\\]\n\nThen $r = 2$.\n\nExample 2: Find $r$ for the geometric progression whose first three terms are 5, $\\frac{1}{2}$, and $\\frac{1}{20}$.\n\n\\[\n\\frac{1}{2} \\div 5 = \\frac{1}{20} \\div \\frac{1}{2} = \\frac{1}{10}\n\\]\n\nThen $r = \\frac{1}{10}$.\n\nIf we know the first term in a geometric progression and the ratio between successive terms, then we can work out the value of any term in the geometric progression. The $n$th term is given by\n\n\\[\nu_n = ar^{n-1}\\]\n\nAgain, $a$ is the first term and $r$ is the ratio. Remember that $ar^{n-1} \\neq (ar)^{n-1}$.\n\nExample 3: Given the first two terms in a geometric progression as 2 and 4, what is the 10th term?\n\n\\[\na = 2 \\quad r = \\frac{4}{2} = 2\n\\]\n\nThen $u_{10} = 2 \\times 2^9 = 1024$.\n\nExample 4: Given the first two terms in a geometric progression as 5 and $\\frac{1}{2}$, what is the 7th term?\n\n\\[\na = 5 \\quad r = \\frac{1}{10}\n\\]\n\nThen\n\n\\[\nu_7 = 5 \\times \\left(\\frac{1}{10}\\right)^{7-1}\n\\]\n\n\\[\n= \\frac{5}{1000000}\n\\]\n\n\\[\n= 0.000005\n\\]\nA geometric series is a geometric progression with plus signs between the terms instead of commas. So an example of a geometric series is\n\n\\[ 1 + \\frac{1}{10} + \\frac{1}{100} + \\frac{1}{1000} + \\cdots \\]\n\nWe can take the sum of the first \\( n \\) terms of a geometric series and this is denoted by \\( S_n \\):\n\n\\[ S_n = \\frac{a(1 - r^n)}{1 - r} \\]\n\nExample 5: Given the first two terms of a geometric progression as 2 and 4, what is the sum of the first 10 terms? We know that \\( a = 2 \\) and \\( r = 2 \\). Then\n\n\\[ S_{10} = \\frac{2(1 - 2^{10})}{1 - 2} = 2046 \\]\n\nExample 6: Given the first two terms of a geometric progression as 5 and \\( \\frac{1}{2} \\), what is the sum of the first 7 terms? We know that \\( a = 5 \\) and \\( r = \\frac{1}{10} \\). Then\n\n\\[ S_7 = \\frac{5(1 - \\frac{1}{10^7})}{1 - \\frac{1}{10}} = 5 \\frac{1 - \\frac{1}{10^7}}{\\frac{9}{10}} = 5.555555 \\]\n\nIn certain cases, the sum of the terms in a geometric progression has a limit (note that this is summing together an infinite number of terms). A series like this has a limit partly because each successive term we are adding is smaller and smaller (but this fact in itself is not enough to say that the limiting sum exists). When the sum of a geometric series has a limit we say that \\( S_\\infty \\) exists and we can find the limit of the sum. For more information on limits, see worksheet 3.7. The condition that \\( S_\\infty \\) exists is that \\( r \\) is greater than \\(-1\\) but less than 1, i.e. \\(|r| < 1\\). If this is the case, then we can use the formula for \\( S_n \\) above and let \\( n \\) grow arbitrarily big so that \\( r^n \\) becomes as close as we like to zero. Then\n\n\\[ S_\\infty = \\frac{a}{1 - r} \\]\n\nis the limit of the geometric progression so long as \\(-1 < r < 1\\).\nExample 7: The geometric progression whose first two terms are 2 and 4 does not have a $S_\\infty$ because $r = 2 \\neq 1$.\n\nExample 8: For the geometric progression whose first two terms are 5 and $\\frac{1}{2}$, find $S_\\infty$. Note that $r = \\frac{1}{10}$ so $|r| < 1$, so that $S_\\infty$ exists. Now\n\n$$S_\\infty = \\frac{a}{1 - r} = \\frac{5}{1 - \\frac{1}{10}} = \\frac{5}{\\frac{9}{10}}$$\n\nSo the sum of $5 + \\frac{1}{2} + \\frac{1}{20} + \\frac{1}{200} + \\ldots$ is $\\frac{55}{9}$.\n\nExample 9: Consider a geometric progression whose first three terms are 12, $-6$ and 3. Notice that $r = -\\frac{1}{2}$. Find both $S_8$ and $S_\\infty$.\n\n$$S_8 = \\frac{a(1 - r^n)}{1 - r} = \\frac{12(1 - (-\\frac{1}{2})^8)}{1 - (-\\frac{1}{2})} \\approx 7.967$$\n\n$$S_\\infty = \\frac{a}{1 - r} = \\frac{12}{1 - (-\\frac{1}{2})} = \\frac{12}{3/2} = 8$$\n\nExercises:\n\n1. Find the term indicated for each of the geometric progressions.\n\n (a) 1, 3, 9, ..., $(u_9)$\n (b) 4, $-8$, 16, ..., $(u_{10})$\n (c) 18, $-6$, 2, ..., $(u_{12})$\n (d) 1000, 100, 10, ..., $(u_7)$\n (e) 32, $-8$, 2, ..., $(u_{14})$\n (f) $-0.005$, $-0.05$, $-0.5$, ..., $(u_{10})$\n (g) $-6$, $-12$, $-24$, ..., $(u_6)$\n (h) 1.4, 0.7, 0.35, ..., $(u_5)$\n (i) 68, $-34$, 17, ..., $(u_9)$\n (j) 8, 2, $\\frac{1}{2}$, ..., $(u_{11})$\n2. Find the sum indicated for each of the following geometric series\n\n(a) $6 + 9 + 13.5 + \\cdots (S_{10})$\n(b) $18 - 9 + 4.5 + \\cdots (S_{12})$\n(c) $6 + 3 + \\frac{3}{2} + \\cdots (S_{10})$\n(d) $6000 + 600 + 60 + \\cdots (S_{20})$\n(e) $80 - 20 + 5 + \\cdots (S_{9})$\nExercises 3.6 Arithmetic and Geometric Progressions\n\n1. For each of the following progressions, determine whether it is arithmetic, geometric, or neither:\n (a) 5, 9, 13, 17, ...\n (b) 1, −2, 4, −8, ...\n (c) 1, 1, 2, 3, 5, 8, 13, 21, ...\n (d) 81, −9, 3, −1, ...\n (e) 512, 474, 436, 398, ...\n\n2. Find the sixth and twentieth terms, and the sum of the first 10 terms of each of the following sequences:\n (a) −15, −9, −3, ...\n (b) log 7, log 14, log 28, ...\n (c) $\\frac{1}{16}$, $\\frac{1}{8}$, $\\frac{1}{4}$, ...\n (d) 0.5, 0.45, 0.405, ...\n (e) 64, −32, 16, ...\n\n3. (a) The third and eighth terms of an AP are 470 and 380 respectively. Find the first term and the common difference. Hint: write expressions for $u_3$ and $u_8$ and solve simultaneously.\n (b) Find the sum to 5 terms of the geometric progression whose first term is 54 and fourth term is 2.\n (c) Find the second term of a geometric progression whose third term is $\\frac{9}{7}$ and sixth term is $-\\frac{16}{81}$.\n (d) Find the sum to $n$ terms of an arithmetic progression whose fourth and fifth terms are 13 and 15.\n\n4. (a) A university lecturer has an annual salary of $40,000. If this increases by 2% each year, how much will she have grossed in total after 10 years?\n (b) A bob of a pendulum swings through an arc of 50 cm on its first swing. Each successive swing is 90% of the length of the previous swing. Find the total distance the bob travels before coming to rest.\nAnswers 3.6\n\n1. (a) Arithmetic (b) Geometric (c) Neither (d) Neither (e) Arithmetic\n\n2. (a) $T_6 = 15$, $T_{20} = 99$, $S_{10} = 120$\n (b) $T_6 = \\log 7 + 5 \\log 2$, $T_{20} = \\log 7 + 19 \\log 2$, $S_{10} = \\frac{10}{2}(2 \\log 7 + 9 \\log 2)$\n (c) $T_6 = 2$, $T_{20} = 2^{15}$, $S_{10} = \\frac{1}{16}(2^{10} - 1)$\n (d) $T_6 = (0.5)(0.9)^5$, $T_{20} = (0.5)(0.9)^{19}$, $S_{10} = 5(1 - .9^{10})$\n (e) $T_6 = -2$, $T_{20} = -\\frac{1}{2^{11}}$, $S_{10} = \\frac{128}{3}(1 + 2^{-10})$\n\n3. (a) $a = 506$, $d = \\frac{81(1 - (\\frac{1}{3})^5)}{-18}$ (b) $81(1 - (\\frac{1}{3})^5)$ (c) $T_2 = (\\frac{2}{3})^3$ (d) $n^2 + 6n$\n\n4. (a) $437,988.84$ (b) 5 metres", "id": "./materials/357.pdf" }, { "contents": "Recurrence Relations\n\n1 Infinite Sequences\n\nAn infinite sequence is a function from the set of positive integers to the set of real numbers or to the set of complex numbers.\n\nExample 1.1. The game of Hanoi Tower is to play with a set of disks of graduated size with holes in their centers and a playing board having three spokes for holding the disks.\n\nThe object of the game is to transfer all the disks from spoke A to spoke C by moving one disk at a time without placing a larger disk on top of a smaller one. What is the minimal number of moves required when there are \\( n \\) disks?\n\nSolution. Let \\( a_n \\) be the minimum number of moves to transfer \\( n \\) disks from one spoke to another. In order to move \\( n \\) disks from spoke A to spoke C, one must move the first \\( n - 1 \\) disks from spoke A to spoke B by \\( a_{n-1} \\) moves, then move the last (also the largest) disk from spoke A to spoke C by one move, and then remove the \\( n - 1 \\) disks again from spoke B to spoke C by \\( a_{n-1} \\) moves. Thus the total number of moves should be\n\n\\[\na_n = a_{n-1} + 1 + a_{n-1} = 2a_{n-1} + 1.\n\\]\n\nThis means that the sequence \\( \\{a_n \\mid n \\geq 1\\} \\) satisfies the recurrence relation\n\n\\[\n\\begin{align*}\na_n &= 2a_{n-1} + 1, \\quad n \\geq 1 \\\\\na_1 &= 1.\n\\end{align*}\n\\]\nApplying the recurrence relation again and again, we have\n\n\\[\n\\begin{align*}\na_1 &= 2a_0 + 1 \\\\\na_2 &= 2a_1 + 1 = 2(2a_0 + 1) + 1 \\\\\n&= 2^2a_0 + 2 + 1 \\\\\na_3 &= 2a_2 + 1 = 2(2^2a_0 + 2 + 1) + 1 \\\\\n&= 2^3a_0 + 2^2 + 2 + 1 \\\\\na_4 &= 2a_3 + 1 = 2(2^3a_0 + 2^2 + 2 + 1) + 1 \\\\\n&= 2^4a_0 + 2^3 + 2^2 + 2 + 1 \\\\\n&\\vdots \\\\\na_n &= 2^n a_0 + 2^{n-1} + 2^{n-2} + \\cdots + 2 + 1 \\\\\n&= 2^n a_0 + 2^n - 1.\n\\end{align*}\n\\]\n\nLet \\( a_0 = 0 \\). The general term is given by\n\n\\[ a_n = 2^n - 1, \\quad n \\geq 1. \\]\n\nGiven a recurrence relation for a sequence with initial conditions. Solving the recurrence relation means to find a formula to express the general term \\( a_n \\) of the sequence.\n\n## 2 Homogeneous Recurrence Relations\n\nAny recurrence relation of the form\n\n\\[ x_n = ax_{n-1} + bx_{n-2} \\tag{2} \\]\n\nis called a **second order homogeneous linear recurrence relation**.\n\nLet \\( x_n = s_n \\) and \\( x_n = t_n \\) be two solutions, i.e.,\n\n\\[ s_n = as_{n-1} + bs_{n-2} \\quad \\text{and} \\quad t_n = at_{n-1} + bt_{n-2}. \\]\n\nThen for constants \\( c_1 \\) and \\( c_2 \\)\n\n\\[\n\\begin{align*}\nc_1s_n + c_2t_n &= c_1(as_{n-1} + bs_{n-2}) + c_2(at_{n-1} + bt_{n-2}) \\\\\n&= a(c_1s_{n-1} + c_2t_{n-1}) + b(c_1s_{n-2} + c_2t_{n-2}).\n\\end{align*}\n\\]\n\nThis means that \\( x_n = c_1s_n + c_2t_n \\) is a solution of (2).\nTheorem 2.1. Any linear combination of solutions of a homogeneous recurrence linear relation is also a solution.\n\nIn solving the first order homogeneous recurrence linear relation\n\n\\[ x_n = ax_{n-1}, \\]\n\nit is clear that the general solution is\n\n\\[ x_n = a^n x_0. \\]\n\nThis means that \\( x_n = a^n \\) is a solution. This suggests that, for the second order homogeneous recurrence linear relation (2), we may have the solutions of the form\n\n\\[ x_n = r^n. \\]\n\nIndeed, put \\( x_n = r^n \\) into (2). We have\n\n\\[ r^n = ar^{n-1} + br^{n-2} \\quad \\text{or} \\quad r^{n-2}(r^2 - ar - b) = 0. \\]\n\nThus either \\( r = 0 \\) or\n\n\\[ r^2 - ar - b = 0. \\] (3)\n\nThe equation (3) is called the characteristic equation of (2).\n\nTheorem 2.2. If the characteristic equation (3) has two distinct roots \\( r_1 \\) and \\( r_2 \\), then the general solution for (2) is given by\n\n\\[ x_n = c_1 r_1^n + c_2 r_2^n. \\]\n\nIf the characteristic equation (3) has only one root \\( r \\), then the general solution for (2) is given by\n\n\\[ x_n = c_1 r^n + c_2 nr^n. \\]\n\nProof. When the characteristic equation (3) has two distinct roots \\( r_1 \\) and \\( r_2 \\) it is clear that both\n\n\\[ x_n = r_1^n \\quad \\text{and} \\quad x_n = r_2^n \\]\n\nare solutions of (2), so are their linear combinations.\n\nRecall that \\( r = \\frac{a \\pm \\sqrt{a^2 + 4b}}{2} \\). Now assume that (2) has only one root \\( r \\). Then\n\n\\[ a^2 + 4b = 0 \\quad \\text{and} \\quad r = a/2. \\]\nThus\n\\[ b = -\\frac{a^2}{4} \\quad \\text{and} \\quad r = \\frac{a}{2}. \\]\n\nWe verify that \\( x_n = nr^n \\) is a solution of (2). In fact,\n\\[\nax_{n-1} + bx_{n-2} = a(n-1) \\left( \\frac{a}{2} \\right)^{n-1} + \\left( -\\frac{a^2}{4} \\right) (n-2) \\left( \\frac{a}{2} \\right)^{n-2}\n\\]\n\\[\n= [2(n-1) - (n-2)] \\left( \\frac{a}{2} \\right)^n = n \\left( \\frac{a}{2} \\right)^n = x_n.\n\\]\n\n\\[ \\square \\]\n\n**Remark.** There is heuristic method to explain why \\( x_n = nr^n \\) is a solution when the two roots are the same. If two roots \\( r_1 \\) and \\( r_2 \\) are distinct but very close to each other, then \\( r_1^n - r_2^n \\) is a solution. So is \\( (r_1^n - r_2^n)/(r_1 - r_2) \\). It follows that the limit\n\\[\n\\lim_{r_2 \\to r_1} \\frac{r_1^n - r_2^n}{r_1 - r_2} = nr_1^{n-1}\n\\]\nwould be a solution. Thus its multiple \\( x_n = r_1(nr_1^{n-1}) = nr_1^n \\) by the constant \\( r_1 \\) is also a solution. Please note that this is not a mathematical proof, but a mathematical idea.\n\n**Example 2.1.** Find a general formula for the **Fibonacci sequence**\n\n\\[\n\\begin{align*}\n\\begin{cases}\n f_n &= f_{n-1} + f_{n-2} \\\\\n f_0 &= 0 \\\\\n f_1 &= 1\n\\end{cases}\n\\end{align*}\n\\]\n\n**Solution.** The characteristic equation \\( r^2 = r + 1 \\) has two distinct roots\n\\[\nr_1 = \\frac{1 + \\sqrt{5}}{2} \\quad \\text{and} \\quad r_2 = \\frac{1 - \\sqrt{5}}{2}.\n\\]\n\nThe general solution is given by\n\\[\nf_n = c_1 \\left( \\frac{1 + \\sqrt{5}}{2} \\right)^n + c_2 \\left( \\frac{1 - \\sqrt{5}}{2} \\right)^n.\n\\]\nSet\n\\[\n\\begin{align*}\n0 &= c_1 + c_2 \\\\\n1 &= c_1 \\left( \\frac{1 + \\sqrt{5}}{2} \\right) + c_2 \\left( \\frac{1 - \\sqrt{5}}{2} \\right).\n\\end{align*}\n\\]\nWe have \\( c_1 = -c_2 = \\frac{1}{\\sqrt{5}} \\). Thus\n\\[\nf_n = \\frac{1}{\\sqrt{5}} \\left( \\frac{1 + \\sqrt{5}}{2} \\right)^n - \\frac{1}{\\sqrt{5}} \\left( \\frac{1 - \\sqrt{5}}{2} \\right)^n, \\quad n \\geq 0.\n\\]\n\n**Remark.** The Fibonacci sequence \\( f_n \\) is an integer sequence, but it “looks like” a sequence of irrational numbers from its general formula above.\n\n**Example 2.2.** Find the solution for the recurrence relation\n\\[\n\\begin{align*}\nx_n &= 6x_{n-1} - 9x_{n-2} \\\\\nx_0 &= 2 \\\\\nx_1 &= 3\n\\end{align*}\n\\]\n\n**Solution.** The characteristic equation\n\\[\nr^2 - 6r + 9 = 0 \\iff (r - 3)^2 = 0\n\\]\nhas only one root \\( r = 3 \\). Then the general solution is\n\\[\nx_n = c_1 3^n + c_2 n 3^n.\n\\]\nThe initial conditions \\( x_0 = 2 \\) and \\( x_1 = 3 \\) imply that \\( c_1 = 2 \\) and \\( c_2 = -1 \\). Thus the solution is\n\\[\nx_n = 2 \\cdot 3^n - n \\cdot 3^n = (2 - n)3^n, \\quad n \\geq 0.\n\\]\n\n**Example 2.3.** Find the solution for the recurrence relation\n\\[\n\\begin{align*}\nx_n &= 2x_{n-1} - 5x_{n-2}, \\quad n \\geq 2 \\\\\nx_0 &= 1 \\\\\nx_1 &= 5\n\\end{align*}\n\\]\n\n**Solution.** The characteristic equation\n\\[\nr^2 - 2r + 5 = 0 \\iff (x - 1 - 2i)(x - 1 + 2i) = 0\n\\]\nhas two distinct complex roots \\( r_1 = 1 + 2i \\) and \\( r_2 = 1 - 2i \\). The initial conditions imply that\n\\[\nc_1 + c_2 = 1 \\quad c_1(1 + 2i) + c_2(1 - 2i) = 5.\n\\]\nSo \\( c_1 = \\frac{1 - 2i}{2} \\) and \\( c_2 = \\frac{1 + 2i}{2} \\). Thus the solutions is\n\\[\nx_n = \\frac{1 - 2i}{2} \\cdot (1 + 2i)^n + \\frac{1 + 2i}{2} \\cdot (1 - 2i)^n\n\\]\n\\[\n= \\frac{5}{2} (1 + 2i)^{n+1} + \\frac{5}{2} (1 - 2i)^{n+1}, \\quad n \\geq 0.\n\\]\n\n**Remark.** The sequence is obviously a real sequence. However, its general formula involves complex numbers.\n\n**Example 2.4.** Two persons A and B gamble dollars on the toss of a fair coin. A has $70 and B has $30. In each play either A wins $1 from B or loss $1 to B. The game is played without stop until one wins all the money of the other or goes forever. Find the probabilities of the following three possibilities:\n\n(a) A wins all the money of B.\n\n(b) A loss all his money to B.\n\n(c) The game continues forever.\n\n**Solution.** Either A or B can keep track of the game simply by counting their own money. Their position \\( n \\) (number of dollars) can be one of the numbers 0, 1, 2, \\ldots, 100. Let\n\\[\np_n = \\text{probability that A reaches 100 at position } n.\n\\]\nAfter one toss, A enters into either position \\( n + 1 \\) or position \\( n - 1 \\). The new probability that A reaches 100 is either \\( p_{n+1} \\) or \\( p_{n-1} \\). Since the probability of A moving to position \\( n + 1 \\) or \\( n - 1 \\) from \\( n \\) is \\( \\frac{1}{2} \\). We obtain the recurrence relation\n\\[\n\\begin{cases}\np_n &= \\frac{1}{2} p_{n+1} + \\frac{1}{2} p_{n-1} \\\\\np_0 &= 0 \\\\\np_{100} &= 1\n\\end{cases}\n\\]\n\n**First Method:** The characteristic equation\n\\[\nr^2 - 2r + 1 = 0.\n\\]\nhas only one root \\( r = 1 \\). The general solutions is\n\n\\[ p_n = c_1 + c_2 n. \\]\n\nApplying the boundary conditions \\( p_0 = 0 \\) and \\( p_{100} = 1 \\), we have\n\n\\[ c_1 = 0 \\quad \\text{and} \\quad c_2 = \\frac{1}{100}. \\]\n\nThus\n\n\\[ p_n = \\frac{n}{100}, \\quad 0 \\leq n \\leq 100. \\]\n\nOf course, \\( p_n = \\frac{n}{100} \\) for \\( n > 100 \\) is nonsense to the original problem. The probabilities for (a), (b), and (c) are 70%, 30%, and 0, respectively.\n\n**Second Method:** The recurrence relation \\( p_n = \\frac{1}{2} p_{n+1} + \\frac{1}{2} p_{n-1} \\) can be written as\n\n\\[ p_{n+1} - p_n = p_n - p_{n-1}. \\]\n\nThen\n\n\\[ p_{n+1} - p_n = p_n - p_{n-1} = \\cdots = p_1 - p_0. \\]\n\nSince \\( p_0 = 0 \\), we have \\( p_n = p_{n-1} + p_1 \\). Applying the recurrence relation again and again, we obtain\n\n\\[ p_n = p_0 + np_1. \\]\n\nApplying the conditions \\( p_0 = 0 \\) and \\( p_{100} = 1 \\), we have \\( p_n = \\frac{n}{100} \\).\n\n### 3 Higher Order Homogeneous Recurrence Relations\n\nFor a higher order homogeneous recurrence relation\n\n\\[ x_{n+k} = a_1 x_{n+k-1} + a_2 x_{n+k-2} + \\cdots + a_{n-k} x_n, \\quad n \\geq 0 \\quad (4) \\]\n\nwe also have the characteristic equation\n\n\\[ t^k = a_1 t^{k-1} + a_2 t^{k-1} + \\cdots + a_{n-k+1} t + a_{n-k} \\quad (5) \\]\n\nor\n\n\\[ t^k - a_1 t^{k-1} - a_2 t^{k-1} - \\cdots - a_{n-k+1} t - a_{n-k} = 0. \\]\nTheorem 3.1. For the recurrence relation (4), if its characteristic equation (5) has distinct roots \\( r_1, r_2, \\ldots, r_k \\), then the general solution for (4) is\n\n\\[ x_n = c_1 r_1^n + c_2 r_2^n + \\cdots + c_k r_k^n \\]\n\nwhere \\( c_1, c_2, \\ldots, c_k \\) are arbitrary constants. If the characteristic equation has repeated roots \\( r_1, r_2, \\ldots, r_s \\) with multiplicities \\( m_1, m_2, \\ldots, m_s \\) respectively, then the general solution of (4) is a linear combination of the solutions\n\n\\[ r_1^n, \\ n r_1^n, \\ldots, \\ n^{m_1-1} r_1^n; \\]\n\\[ r_2^n, \\ n r_2^n, \\ldots, \\ n^{m_2-1} r_2^n; \\]\n\\[ \\ldots; \\]\n\\[ r_s^n, \\ n r_s^n, \\ldots, \\ n^{m_s-1} r_s^n. \\]\n\nExample 3.1. Find an explicit formula for the sequence given by the recurrence relation\n\n\\[\n\\begin{align*}\n\\begin{cases}\nx_n &= 15x_{n-2} - 10x_{n-3} - 60x_{n-4} + 72x_{n-5} \\\\\nx_0 &= 1, \\ x_1 = 6, \\ x_2 = 9, \\ x_3 = -110, \\ x_4 = -45\n\\end{cases}\n\\end{align*}\n\\]\n\nSolution. The characteristic equation\n\n\\[ r^5 = 15r^3 - 10r^2 - 60r + 72 \\]\n\ncan be simplified as\n\n\\[ (r - 2)^3(r + 3)^2 = 0. \\]\n\nThere are roots \\( r_1 = 2 \\) with multiplicity 3 and \\( r_2 = -3 \\) with multiplicity 2. The general solution is given by\n\n\\[ x_n = c_1 2^n + c_2 n 2^n + c_3 n^2 2^n + c_4 (-3)^n + c_5 n (-3)^n. \\]\n\nThe initial condition means that\n\n\\[\n\\begin{align*}\n\\begin{cases}\nc_1 &+ c_4 = 1 \\\\\n2c_1 &+ 2c_2 + 2c_3 - 3c_4 - 3c_5 = 1 \\\\\n4c_1 &+ 8c_2 + 16c_3 + 9c_4 + 18c_5 = 1 \\\\\n8c_1 &+ 24c_2 + 72c_3 - 27c_4 - 81c_5 = 1 \\\\\n16c_1 &+ 64c_2 + 256c_3 + 81c_4 + 324c_5 = 1\n\\end{cases}\n\\end{align*}\n\\]\n\nSolving the linear system we have\n\n\\[ c_1 = 2, \\ c_2 = 3, \\ c_3 = -2, \\ c_4 = -1, \\ c_5 = 1. \\]\n4 Non-homogeneous Equations\n\nA recurrence relation of the form\n\n\\[ x_n = ax_{n-1} + bx_{n-2} + f(n) \\] \\hspace{1cm} (6)\n\nis called a **non-homogeneous recurrence relation**.\n\nLet \\( x_n^{(s)} \\) be a solution of (6), called a **special solution**. Then the general solution for (6) is\n\n\\[ x_n = x_n^{(s)} + x_n^{(h)}, \\] \\hspace{1cm} (7)\n\nwhere \\( x_n^{(h)} \\) is the general solution for the corresponding homogeneous recurrence relation\n\n\\[ x_n = ax_{n-1} + bx_{n-2}. \\] \\hspace{1cm} (8)\n\n**Theorem 4.1.** Let \\( f(n) = cr^n \\) in (6). Let \\( r_1 \\) and \\( r_2 \\) be the roots of the characteristic equation\n\n\\[ t^2 = at + b. \\] \\hspace{1cm} (9)\n\n(a) If \\( r \\neq r_1, r \\neq r_2 \\), then \\( x_n^{(s)} = Ar^n \\);\n\n(b) If \\( r = r_1, r_1 \\neq r_2 \\), then \\( x_n^{(s)} = Anr^n \\);\n\n(c) If \\( r = r_1 = r_2 \\), then \\( x_n^{(s)} = An^2r^n \\);\n\nwhere \\( A \\) is a constant to be determined in all cases.\n\n**Proof.** We assume \\( r \\neq 0 \\). Otherwise the recurrence relation is homogeneous.\n\n(a) Put \\( x_n = Ar^n \\) into (6). We have\n\n\\[ Ar^n = aAr^{n-1} + bAr^{n-2} + cr^n. \\]\n\nThus\n\n\\[ A(r^2 - ar - b) = cr^2. \\]\n\nSince \\( r \\) is not a root of the characteristic equation (9), then \\( r^2 - ar - b \\neq 0 \\). Hence\n\n\\[ A = \\frac{cr^2}{r^2 - ar - b}. \\]\n(b) Since \\( r = r_1 \\neq r_2 \\), it is clear that \\( x_n = nr^n \\) is not a solution for its corresponding homogeneous equation (8), i.e.,\n\n\\[\nnr^2 - a(n-1)r - b(n-2) = n(r^2 - ar - b) + ar + 2b \\\\\n= ar + 2b \\neq 0.\n\\]\n\nPut \\( x_n = Anr^n \\) into (6). We have\n\n\\[\nAnr^n = aA(n-1)r^{n-1} + bA(n-2)r^{n-2} + cr^n,\n\\]\n\nThus \\( A(nr^2 - a(n-1)r - b(n-2)) = cr^2 \\). Therefore\n\n\\[\nA = \\frac{cr^2}{ar + 2b}.\n\\]\n\n(c) Since \\( r = r_1 = r_2 \\), then \\( a^2 + 4b = 0 \\) (discriminant of \\( r^2 - ar - b = 0 \\) must be zero), \\( r = a/2 \\), and \\( x_n = n^2r^n \\) is not a solution of the corresponding homogeneous equation (8), i.e.,\n\n\\[\nn^2r^2 - a(n-1)^2r - b(n-2)^2 \\\\\n= n^2(r^2 - ar - b) + 2n(ar + 2b) - ar - 4b \\\\\n= -ar - 4b \\neq 0.\n\\]\n\nPut \\( x_n = An^2r^n \\) into (6). We have\n\n\\[\nAr^{n-2}(n^2r^2 - a(n-1)^2r - b(n-2)^2) = cr^n.\n\\]\n\nThus\n\n\\[\nA = -\\frac{cr^2}{ar + 4b}.\n\\]\n\nExample 4.1. Consider the non-homogeneous equation\n\n\\[\n\\begin{cases}\nx_n = 3x_{n-1} + 10x_{n-2} + 7 \\cdot 5^n \\\\\nx_0 = 4 \\\\\nx_1 = 3\n\\end{cases}\n\\]\n\nSolution. The characteristic equation is\n\n\\[\nt^2 - 3t - 10 = 0 \\iff (t - 5)(t + 2) = 0.\n\\]\nWe have roots $r_1 = 5$, $r_2 = -2$. Since $r = 5$, then $r = r_1$ and $r \\neq r_2$. A special solution can be of the type $x_n = An5^n$. Put the solution into the non-homogeneous relation. We have\n\n$$An5^n = 3A(n - 1)5^{n-1} + 10A(n - 2)5^{n-2} + 7 \\cdot 5^n$$\n\nDividing both sides by $5^{n-2}$,\n\n$$An5^2 = 3A(n - 1)5 + 10A(n - 2) + 7 \\cdot 5^2.$$ \n\nThus\n\n$$-35A + 7 \\cdot 25 = 0 \\implies A = 5.$$ \n\nSo\n\n$$x_n = n5^{n+1}.$$ \n\nThe general solution is\n\n$$x_n = n5^{n+1} + c_15^n + c_2(-2)^n.$$ \n\nThe initial condition implies $c_1 = -2$ and $c_2 = 6$. Therefore\n\n$$x_n = n5^{n+1} - 2 \\cdot 5^n + 6(-2)^n.$$ \n\n**Example 4.2.** Consider the non-homogeneous equation\n\n$$\\begin{cases} \n x_n = 10x_{n-1} - 25x_{n-2} + 8 \\cdot 5^n \\\\\n x_0 = 6 \\\\\n x_1 = 10 \n\\end{cases}$$\n\n**Solution.** The characteristic equation is\n\n$$t^2 - 10t + 25 = 0 \\iff (t - 5)^2 = 0.$$ \n\nWe have roots $r_1 = r_2 = 5$, then $r = r_1 = r_2 = 5$. A special solution can be of the type $x_n = An^25^n$. Put the solution into the non-homogeneous relation. We have\n\n$$An^25^n = 10A(n - 1)^25^{n-1} - 25A(n - 2)^25^{n-2} + 8 \\cdot 5^n$$\n\nDividing both sides by $5^{n-2}$,\n\n$$An^25^2 = 10A(n - 1)^25 - 25A(n - 2)^2 + 8 \\cdot 5^2.$$\nSince $A n^2 5^2 = 10 A n^2 5 - 25 n^2$, we have\n\n$$10 A (-2n + 1) 5 - 25 A (-4n + 4) + 8 \\cdot 5^2 = 0 \\implies A = 4.$$ \n\nSo a nonhomogeneous solution is\n\n$$x_n = 4 n^2 5^n.$$ \n\nThe general solution is\n\n$$x_n = 4 n 5^n + c_1 5^n + c_2 n 5^n.$$ \n\nThe initial condition implies $c_1 = 6$ and $c_2 = -8$. Therefore\n\n$$x_n = (4 n^2 - 8n + 6) 5^n.$$", "id": "./materials/358.pdf" }, { "contents": "St Andrew’s Academy\n\nMathematics Department\n\nHigher Mathematics\n\nRECURRENCE RELATIONS\nSequences\n\nContents\n\nSequences 1\n\n1 Introduction to Sequences A 1\n2 Linear Recurrence Relations A 3\n3 Divergence and Convergence A 4\n4 The Limit of a Sequence A 5\n5 Finding a Recurrence Relation for a Sequence A 6\nSequences\n\n1 Introduction to Sequences\n\nA sequence is an ordered list of objects (usually numbers).\n\nUsually we are interested in sequences which follow a particular pattern. For example, 1, 2, 3, 4, 5, 6, … is a sequence of numbers – the “…” just indicates that the list keeps going forever.\n\nWriting a sequence in this way assumes that you can tell what pattern the numbers are following but this is not always clear, e.g.\n\n28, 22, 19, 17 1/2, …\n\nFor this reason, we prefer to have a formula or rule which explicitly defines the terms of the sequence.\n\nIt is common to use subscript numbers to label the terms, e.g.\n\n\\( u_1, u_2, u_3, u_4, \\ldots \\)\n\nso that we can use \\( u_n \\) to represent the \\( n \\)th term.\n\nWe can then define sequences with a formula for the \\( n \\)th term. For example:\n\n| Formula | List of terms |\n|------------------|---------------|\n| \\( u_n = n \\) | 1, 2, 3, 4, … |\n| \\( u_n = 2n \\) | 2, 4, 6, 8, … |\n| \\( u_n = \\frac{1}{2}n(n+1) \\) | 1, 3, 6, 10, … |\n| \\( u_n = \\cos \\left( \\frac{n\\pi}{2} \\right) \\) | 0, -1, 0, 1, … |\n\nNotice that if we have a formula for \\( u_n \\), it is possible to work out any term in the sequence. For example, you could easily find \\( u_{1000} \\) for any of the sequences above without having to list all the previous terms.\nRecurrence Relations\n\nAnother way to define a sequence is with a recurrence relation. This is a rule which defines each term of a sequence using previous terms.\n\nFor example:\n\n\\[ u_{n+1} = u_n + 2, \\quad u_0 = 4 \\]\n\nsays “the first term \\( u_0 \\) is 4, and each other term is 2 more than the previous one”, giving the sequence 4, 6, 8, 10, 12, 14, ….\n\nNotice that with a recurrence relation, we need to work out all earlier terms in the sequence before we can find a particular term. It would take a long time to find \\( u_{1000} \\).\n\nAnother example is interest on a bank account. If we deposit £100 and get 4% interest per year, the balance at the end of each year will be 104% of what it was at the start of the year.\n\n\\[\n\\begin{align*}\n u_0 &= 100 \\\\\n u_1 &= 104\\% \\text{ of } 100 = 1.04 \\times 100 = 104 \\\\\n u_2 &= 104\\% \\text{ of } 104 = 1.04 \\times 104 = 108.16 \\\\\n &\\vdots \\\\\n u_n &= 1.04u_{n-1} \\text{ with } u_0 = 100,\n\\end{align*}\n\\]\n\nThe complete sequence is given by the recurrence relation\n\n\\[ u_{n+1} = 1.04u_n \\text{ with } u_0 = 100, \\]\n\nwhere \\( u_n \\) is the amount in the bank account after \\( n \\) years.\n\n**EXAMPLE**\n\nThe value of an endowment policy increases at the rate of 5% per annum. The initial value is £7000.\n\n(a) Write down a recurrence relation for the policy’s value after \\( n \\) years.\n\n(b) Calculate the value of the policy after 4 years.\n\n(a) Let \\( u_n \\) be the value of the policy after \\( n \\) years.\n\nSo \\( u_{n+1} = 1.05u_n \\) with \\( u_0 = 7000 \\).\n\n(b) \\( u_0 = 7000 \\)\n\n\\[\n\\begin{align*}\n u_1 &= 1.05 \\times 7000 = 7350 \\\\\n u_2 &= 1.05 \\times 7350 = 7717.5 \\\\\n u_3 &= 1.05 \\times 7717.5 = 8103.375 \\\\\n u_4 &= 1.05 \\times 8103.375 = 8508.54375\n\\end{align*}\n\\]\n\nAfter 4 years, the policy is worth £8508.54.\n2 Linear Recurrence Relations\n\nIn Higher, we will deal with recurrence relations of the form\n\n\\[ u_{n+1} = au_n + b \\]\n\nwhere \\( a \\) and \\( b \\) are any real numbers and \\( u_0 \\) is specified. These are called linear recurrence relations of order one.\n\nNote\n\nTo properly define a sequence using a recurrence relation, we must specify the initial value \\( u_0 \\).\n\nEXAMPLES\n\n1. A patient is injected with 156 ml of a drug. Every 8 hours, 22% of the drug passes out of his bloodstream. To compensate, a further 25 ml dose is given every 8 hours.\n (a) Find a recurrence relation for the amount of drug in his bloodstream.\n (b) Calculate the amount of drug remaining after 24 hours.\n\n (a) Let \\( u_n \\) be the amount of drug in his bloodstream after \\( 8n \\) hours.\n \\[ u_{n+1} = 0.78u_n + 25 \\text{ with } u_0 = 156 \\]\n\n (b) \\( u_0 = 156 \\)\n \\[ u_1 = 0.78 \\times 156 + 25 = 146.68 \\]\n \\[ u_2 = 0.78 \\times 146.68 + 25 = 139.4104 \\]\n \\[ u_3 = 0.78 \\times 139.4104 + 25 = 133.7401 \\]\n\n After 24 hours, he will have 133.74 ml of drug in his bloodstream.\n\n2. A sequence is defined by the recurrence relation \\( u_{n+1} = 0.6u_n + 4 \\) with \\( u_0 = 7 \\).\n Calculate the value of \\( u_3 \\) and the smallest value of \\( n \\) for which \\( u_n > 9.7 \\).\n\n \\[ u_0 = 7 \\]\n \\[ u_1 = 0.6 \\times 7 + 4 = 8.2 \\]\n \\[ u_2 = 0.6 \\times 8.2 + 4 = 8.92 \\]\n \\[ u_3 = 0.6 \\times 8.92 + 4 = 9.352 \\]\n\n The value of \\( u_3 \\) is 9.352\n\n \\[ u_4 = 9.6112 \\]\n \\[ u_5 = 9.76672 \\]\n\n The smallest value of \\( n \\) for which \\( u_n > 9.7 \\) is 5\nUsing a Calculator\nUsing the ANS button on the calculator, we can carry out the above calculation more efficiently.\n\n\\[\n\\begin{align*}\n7 &= 0 \\cdot 6 \\times \\text{ANS} + 4 \\\\\n&= \\\\\n&= \\\\\n&= \\\\\n\\end{align*}\n\\]\n\n3 Divergence and Convergence\n\nIf we plot the graphs of some of the sequences that we have been dealing with, then some similarities will occur.\n\nDivergence\nSequences defined by recurrence relations in the form \\( u_{n+1} = au_n + b \\) where \\( a < -1 \\) or \\( a > 1 \\), will have a graph like this:\n\nSequences like this will continue to increase or decrease forever.\nThey are said to diverge.\n\nConvergence\nSequences defined by recurrence relations in the form \\( u_{n+1} = au_n + b \\) where \\( -1 < a < 1 \\), will have a graph like this:\n\nSequences like this “tend to a limit”.\nThey are said to converge.\n4 The Limit of a Sequence\n\nWe saw that sequences defined by \\( u_{n+1} = au_n + b \\) with \\(-1 < a < 1\\) “tend to a limit”. In fact, it is possible to work out this limit just from knowing \\( a \\) and \\( b \\).\n\nThe sequence defined by \\( u_{n+1} = au_n + b \\) with \\(-1 < a < 1\\) tends to a limit \\( l \\) as \\( n \\to \\infty \\) (i.e. as \\( n \\) gets larger and larger) given by\n\n\\[\nl = \\frac{b}{1-a}.\n\\]\n\nYou will need to know this formula, as it is not given in the exam.\n\nEXAMPLES\n\n1. The deer population in a forest is estimated to drop by 7.3% each year. Each year, 20 deer are introduced to the forest. The initial deer population is 200.\n\n (a) How many deer will there be in the forest after 3 years?\n (b) What is the long term effect on the population?\n\n (a) \\( u_{n+1} = 0.927u_n + 20 \\)\n\n \\( u_0 = 200 \\)\n\n \\( u_1 = 0.927 \\times 200 + 20 = 205.4 \\)\n\n \\( u_2 = 0.927 \\times 205.4 + 20 = 210.4058 \\)\n\n \\( u_3 = 0.927 \\times 210.4058 + 20 = 215.0461 \\)\n\n Therefore there are 215 deer living in the forest after 3 years.\n\n (b) A limit exists, since \\(-1 < 0.927 < 1\\).\n\n \\[\nl = \\frac{b}{1-a} \\quad \\text{where} \\quad a = 0.927 \\quad \\text{and} \\quad b = 20\n \\]\n\n \\[\n = \\frac{20}{1-0.927}\n \\]\n\n \\[\n = 273.97 \\quad \\text{(to 2 d.p.)}\n \\]\n\n Therefore the number of deer in the forest will settle around 273.\n\nNote\n\nWhenever you calculate a limit using this method, you must state that “A limit exists since \\(-1 < a < 1\\).”\n2. A sequence is defined by the recurrence relation \\( u_{n+1} = ku_n + 2k \\) and the first term is \\( u_0 \\).\n\nGiven that the limit of the sequence is 27, find the value of \\( k \\).\n\nThe limit is given by \\( \\frac{b}{1-a} = \\frac{2k}{1-k} \\), and so\n\n\\[\n\\frac{2k}{1-k} = 27\n\\]\n\n\\[\n27(1-k) = 2k\n\\]\n\n\\[\n29k = 27\n\\]\n\n\\[\nk = \\frac{27}{29}.\n\\]\n\n5 Finding a Recurrence Relation for a Sequence\n\nIf we know that a sequence is defined by a linear recurrence relation of the form \\( u_{n+1} = au_n + b \\), and we know three consecutive terms of the sequence, then we can find the values of \\( a \\) and \\( b \\).\n\nThis can be done easily by forming two equations and solving them simultaneously.\n\n**EXAMPLE**\n\nA sequence is defined by \\( u_{n+1} = au_n + b \\) with \\( u_1 = 4 \\), \\( u_2 = 3.6 \\) and \\( u_3 = 2.04 \\).\n\nFind the values of \\( a \\) and \\( b \\).\n\nForm two equations using the given terms of the sequence:\n\n\\[\nu_2 = au_1 + b \\quad \\text{and} \\quad u_3 = au_2 + b\n\\]\n\n\\[\n3.6 = 4a + b \\quad \\text{①} \\quad \\quad 2.04 = 3.6a + b \\quad \\text{②}.\n\\]\n\nEliminate \\( b \\):\n\n\\[\n\\text{①} - \\text{②}: \\quad 1.56 = 0.4a\n\\]\n\n\\[\na = \\frac{1.56}{0.4} = 3.9.\n\\]\n\nPut \\( a = 3.9 \\) into ①:\n\n\\[\n4 \\times 3.9 + b = 3.6\n\\]\n\n\\[\nb = 3.6 - 15.6\n\\]\n\n\\[\nb = -12.\n\\]\n\nSo \\( a = 3.9 \\) and \\( b = -12 \\).\nEXERCISES:\n\nExercise 1:\n\n1. Arthur contributes £100 on the first of every month to an I.S.A. The fund is growing at 0.5% per month and was worth £1000 on the first of June (after his June contribution). Show that it will be worth £1105 a month later, and calculate what it will be worth on the following 1 December (after his December contribution).\n\n2. Betty pays off a loan of £2000 by monthly instalments of £250 on the first of each month. The interest is 1.5% per month on the outstanding balance. If she took out the loan on 1 February, show that on 1 March she was still due £1780. Find also the outstanding balance at 1 October and the payment required to clear off the loan at 1 November.\n\n3. It has been estimated that when the total amount of a certain waste chemical in a particular sea loch reaches 6 tonnes, the water in the loch is sufficiently polluted to endanger marine life. The natural flushing action of the tide is known to remove 40% of whatever amount of this waste chemical is in the loch every week. A factory producing this waste chemical as a by-product seeks the permission of the local water authority to release a batch of 2 tonnes of it into the loch once a week. Should this permission be granted?\n\n4. If you invest £1000 for two years at a compound interest rate of 8% per annum, does it make any difference if the interest is added annually or quarterly?\n\nExercise 2:\n\n1. The formulae for the $n^{th}$ term of some sequences are given below. Write down the first four terms and the 20th term of each sequence.\n\n a) $u_n = 2n - 1$\n b) $u_n = \\frac{1}{2}n + 2$\n c) $u_n = 4n - 10$\n d) $u_n = 5n^2$\n e) $u_n = n^3$\n f) $u_n = 3^n$\n g) $u_n = 7 - \\frac{1}{n}$\n h) $u_n = 5\\cos(180n^\\circ)$\n\n2. Find a formula for $u_n$, the $n^{th}$ term of each sequence below.\n\n a) 6, 10, 14, 18, 22, ...\n b) 7, 15, 23, 31, 39, ...\n c) -5, -3, -1, 1, 3, ...\n d) 21, 17, 13, 9, 5, ...\n e) 1, 4, 9, 16, 25, ...\n f) 2, 4, 8, 16, 32, ...\nExercise 3:\n\n1. For each recurrence relation calculate the value of $u_5$:\n a) $u_{n+1} = 3u_n + 1$, $u_0 = \\ldots$\n b) $u_n = 0.5u_{n-1} + 2$, $u_0 = 12$\n c) $u_{n+1} = -4u_n + 3$, $u_0 = \\ldots$\n d) $u_{n+1} = \\frac{1}{3}u_n - 1$, $u_0 = 120$\n\n2. A recurrence relation is given by $u_n = 2u_{n-1} + 15$, $u_0 = 10$.\n a) Calculate the values of $u_2$ and $u_3$.\n b) Find the smallest value of $n$ such that $u_n > 200$.\n\n3. A recurrence relation is given by $u_{n+1} = 0.3u_n - 4$, $u_0 = 8$.\n a) Calculate the values of $u_2$ and $u_3$.\n b) Find the smallest value of $n$ such that $u_n < -5.7$.\n\nExercise 4:\n\n1. Ali was given £100 on his 10th birthday and £50 on each birthday thereafter. He invests this money in a savings account that pays interest at 2.5% per annum.\n a) Write down a recurrence relation to model this situation.\n b) How much will he have in his account on his 16th birthday?\n\n2. Dave, who is retired, has £150 000 invested in a pension fund. This fund earns Dave interest at the rate of 5% per annum. At the end of each year Dave takes out £15 000 for living expenses for the following year.\n a) Write down a recurrence relation to model this situation.\n b) Dave’s financial advisor tells him his pension fund will ‘easily last 15 years’. Is this sound advice? Justify your answer.\n\n3. Gemma borrows £2000 from a finance company on February 1st. On the last day of each month she is charged 1.5% interest on the outstanding balance. She makes repayments on the first of each subsequent month. Each repayment is £250 except for the smaller final amount which will pay off the loan.\n a) Write down a recurrence relation for the outstanding balance on the first of each month.\n b) Find the date and the amount of the final payment.\n\n4. Joan has a balloon which contains 1500 millilitres of air. She blows more air into it. Each puff she gives it increases the amount of air in the balloon by 15%. However, 100 millilitres of air escapes at the same time.\n a) Write down a recurrence relation to model this situation.\n b) How much air will be in the balloon after 5 puffs?\n c) The balloon will burst when it reaches 3000 millilitres. After how many puffs should Joan stop?\n\n5. Mike’s dog is ill. The vet gives the dog an injection of 100 millilitres of drug. Every 4 hours, 12% of the drug passes through the dog’s bloodstream. To compensate, a further 10 millilitre dose is given every 4 hours.\n a) Write down a recurrence relation to model this situation.\n b) How much drug will be in Mike’s dog after 24 hours!\nExercise 5:\n\n1. Write down the first four terms of the sequence defined by the recurrence relation\n a) \\( u_{n+1} = 2u_n + 3 \\) and \\( u_0 = 1 \\)\n b) \\( u_{n+1} = 3u_n - 2 \\) and \\( u_0 = 2 \\)\n c) \\( u_{n+1} = \\frac{1}{2} u_n + 12 \\) and \\( u_0 = 16 \\)\n d) \\( u_{n+1} = \\frac{1}{4} u_n - 16 \\) and \\( u_0 = 128 \\).\n\n2. A sequence is defined by the recurrence relation \\( u_{n+1} = 3u_n + 1 \\) and \\( u_0 = 2 \\).\n a) Calculate mentally the values of \\( u_1, u_2, u_3 \\) and \\( u_4 \\).\n b) Check these values on your calculator and find \\( u_5, u_6 \\) and \\( u_7 \\), starting with \\( 2 \\times 3 + 1 \\) followed by \\( \\times 3 + 1 \\) as often as necessary.\n c) This sequence can be found more easily by using the [ANS] key on the calculator.\n Key in \\( 2 \\equiv \\) (or \\( \\text{EXE} \\)); \\( 3 \\times \\text{ANS} \\equiv 1 \\equiv \\) followed by \\( \\text{EXE} \\) as often as necessary. It is essential that you can evaluate sequences quickly like this.\n d) Use the procedure described in part (c) to find which term of this sequence is the first to exceed a million.\n\n3. For the sequence defined by the recurrence relation \\( u_{n+1} = 2u_n - 1 \\) and \\( u_0 = 3 \\),\n a) find the values of \\( u_1, u_2, u_3, u_4, u_5 \\) and \\( u_6 \\)\n b) find which term of the sequence is the first to exceed 50,000.\n\n4. For the sequence defined by the recurrence relation \\( u_{n+1} = 5u_n - 3 \\) and \\( u_0 = 1 \\),\n a) find the values of \\( u_1, u_2, u_3 \\) and \\( u_4 \\)\n b) find a simple reason why 123,456,789 is not a member of this sequence.\n\n5. a) Find the next term in each sequence given one particular term and the appropriate recurrence relation\n (i) \\( u_{35} = 20, \\ u_{n+1} = 0.7u_n + 6 \\)\n (ii) \\( u_{29} = 10, \\ u_{n+1} = 0.3u_n + 7 \\)\n (iii) \\( u_{78} = 5, \\ u_{n+1} = 0.4u_n + 3 \\)\n (iv) \\( u_{41} = 2, \\ u_{n+1} = 0.5u_n + 1 \\)\n (v) \\( u_n = 4, \\ u_{n+1} = 0.25u_n + 3 \\)\n (vi) \\( u_n = 110, \\ u_{n+1} = 0.9u_n + 11 \\)\n b) Write down the limit of each sequence.\n c) Check these values for the limits by solving the equation obtained by replacing both \\( u_{n+1} \\) and \\( u_n \\) by \\( L \\) in the recurrence relation equation.\n d) Apply this process to \\( u_{n+1} = au_n + b \\) to obtain a formula for the limit.\n e) Under what circumstances is this formula valid?\n\n6B. Find the values of \\( u_1, u_2, u_3, u_4 \\) and \\( u_5 \\) for the sequence defined by:\n a) \\( u_{n+1} = -2u_n + 1; \\ u_0 = 1 \\)\n b) \\( u_{n+1} = -2u_n + 1; \\ u_0 = 4 \\)\n c) \\( u_{n+1} = -3u_n + 2; \\ u_0 = 1 \\)\n d) \\( u_{n+1} = 4u_n + 3; \\ u_0 = -3 \\)\n e) \\( u_{n+1} = 6u_n - 2; \\ u_0 = -1 \\)\n Observe the effects in \\( u_{n+1} = au_n + b \\) of \\( a \\) or \\( b \\) being negative.\n\n7B. a) Find \\( u_{15}, u_{16}, u_{17} \\), for the sequence defined by \\( u_{n+1} = -0.2u_n + 12; \\ u_0 = 5 \\).\n b) Find \\( u_{35}, u_{36}, u_{37} \\), for the sequence defined by \\( u_{n+1} = -0.5u_n + 3; \\ u_0 = 5 \\).\n c) If the sequence defined by the recurrence relation \\( u_{n+1} = au_n + b \\) approaches a limit, it does so in different ways depending on whether \\( a > 0 \\) or \\( a < 0 \\). Explain.\nExercise 6:\n\n1. Determine which of the following recurrence relations generate sequences that:\n - are convergent;\n - are divergent;\n - oscillate between two values.\n a) \\( u_{n+1} = 3u_n + 1, \\quad u_0 = 2 \\)\n b) \\( u_{n+1} = 0.5u_n + 2, \\quad u_0 = 10 \\)\n c) \\( u_{n+1} = -2u_n + 7, \\quad u_0 = 1 \\)\n d) \\( u_{n+1} = -0.6u_n + 2, \\quad u_0 = 15 \\)\n e) \\( u_n = u_{n-1} - 20, \\quad u_0 = 25 \\)\n f) \\( u_{n+1} = -u_n + 4, \\quad u_0 = 100 \\)\n g) \\( u_{n+1} = 0.8u_n - 3, \\quad u_0 = 5 \\)\n\n2. For those sequences in question 1 which are convergent, find the limit of the sequence.\n\n3. Find the limit of the sequence generated by the recurrence relation \\( u_{n+1} = 0.75u_n + 3 \\) for each of the following starting values:\n a) \\( u_0 = 4 \\)\n b) \\( u_0 = 200 \\)\n c) \\( u_0 = -30 \\)\n\n4. Find the limit of the sequence generated by the recurrence relation \\( u_{n+1} = -0.2u_n - 10 \\) for each of the following starting values:\n a) \\( u_0 = 4 \\)\n b) \\( u_0 = 200 \\)\n c) \\( u_0 = -30 \\)\n\n5. How can you tell if the sequence generated by the linear recurrence relation \\( u_{n+1} = au_n + b \\) converges to a limit? Make a conjecture based on your answers to the questions in this exercise.\n\nExercise 7:\n\n1. For each of the following recurrence relations:\n - explain why the sequence generated by it converges or diverges;\n - where the sequence converges, find the limit.\n a) \\( u_{n+1} = 0.6u_n + 12, \\quad u_0 = 4 \\)\n b) \\( u_{n+1} = 0.25u_n - 9, \\quad u_0 = 6 \\)\n c) \\( u_{n+1} = 2u_n + 5, \\quad u_0 = 0 \\)\n d) \\( u_{n+1} = -0.5u_n + 3, \\quad u_0 = 10 \\)\n e) \\( u_{n+1} = u_n + 10, \\quad u_0 = 3 \\)\n f) \\( u_{n+1} = -\\frac{2}{3}u_n - 4, \\quad u_0 = 5 \\)\n\n2. A sequence is defined by the recurrence relation \\( u_{n+1} = 0.3u_n + 2, \\quad u_0 = 2 \\).\n a) Explain why this sequence has a limit as \\( n \\to \\infty \\).\n b) Find the exact value of this limit.\n\n3. A sequence is defined by the recurrence relation \\( u_{n+1} = \\frac{4}{7}u_n + 4, \\quad u_0 = 5 \\).\n a) Explain why this sequence has a limit as \\( n \\to \\infty \\).\n b) Find the exact value of this limit.\n\n4. Two sequences are defined by the recurrence relations.\n \\( u_{n+1} = 4u_n - 0.7, \\quad u_0 = 2 \\), and \\( w_{n+1} = 0.4w_n + 3, \\quad w_0 = 2 \\).\n a) Explain why only one of these sequences has a limit as \\( n \\to \\infty \\).\n b) Find this limit.\n\n5. A sequence, defined by the recurrence relation \\( u_{n+1} = 0.84u_n + b, \\quad u_0 = 6 \\), converges to a limit of 25. Find the value of \\( b \\).\nExercise 8:\n\n1. In a pond, 35% of the existing tadpoles die off each day but during the night 800 tadpoles are hatched. There are \\( t_n \\) tadpoles at the start of a given day.\n a) Write down a recurrence relation for \\( t_{n+1} \\), the number of tadpoles at the start of the next day.\n b) Find the limit of this sequence.\n c) Explain what this limit means in the context of the question.\n\n2. A farmer has 200 chickens. Unfortunately, 27% of the chickens are killed by foxes each month. At the end of each month the farmer buys 20 chickens to replenish his stock.\n a) Set up a recurrence relation to model this situation.\n b) State why a limit exists for this sequence.\n c) Calculate this limit.\n d) Explain what this limit means in the context of the question.\n\n3. Algae grow in a pond at a rate of 275 grams per week. The pond is cleaned every week using a process which removes 55% of any algae present.\n a) Set up a recurrence relation to model this situation.\n b) What will happen, in the long run, to the mass of algae in the pond?\n\n4. A hospital patient is put on medication which is taken once per day. The dose is 35 milligrams, and each day the patient’s metabolism burns off 65% of the drug in her system. If the level of drug reaches 50 milligrams the consequences are very serious. Is it safe for the patient to take this drug indefinitely?\n\n5. A factory wishes to release waste containing 2 tonnes of pollutant chemicals annually into a loch. The factory can remove 70% of the pollutant chemicals through its filter system. The maximum amount that the authorities will allow to be in the loch in total is 3 tonnes. Would the factory meet this requirement in the long run? Justify your answer.\n\n6. Once a month, the environmental health department removes sticky gum from the pavements in Dalburgh town centre. This operation removes 85% of the gum. However, each month, the public drop another 3 kilograms of gum on the pavements.\n a) In the long run, how much gum will there be on the pavements in Dalburgh?\n b) The council runs an awareness-raising campaign which they think will reduce the monthly amount of gum dropped to 2 kilograms. How will this affect the sticky gum problem in the long run?\n\n7. In a membership drive a health club is trying to recruit new members. In any month it estimates that it loses 2.5% of its members to competitors and attracts 30 new members. It has 1250 members at the start of the recruitment drive.\n a) What would happen to the number of members in the long term if this situation continued?\n b) How many new members would the club have to recruit each month in order to maintain its membership level at 1400?\nExercise 9:\n\n1. A bird feeder is initially filled with 200 g of bird seed. The birds eat 60% of whatever seed is in the feeder every day. Every morning thereafter, the feeder is topped-up with 60 g of bird seed. If this pattern continues, how much seed is eventually in the feeder each morning after it has been topped-up?\n [Hint: Let $u_n$ be the number of grams of seed in the feeder after it has been topped-up on the $n^{th}$ morning, and show that $u_{n+1} = 0.4u_n + 60$.]\n\n2. The population of the Isle of Scrabley was 250 ten years ago. Since then a pattern has developed of 10% of the population leaving each year to seek work on the mainland, and 30 city escapees from the ‘rat-race’ retiring to the island. If this pattern continues for many years, what will the population of the island eventually be? [Obtain an appropriate recurrence relation.]\n\n3. The tea stain on the inside of the bottom of Granny’s tea pot is 150 units thick. On each wash, her dishwasher removes half of whatever thickness of stain there is. Each time Granny makes a pot of tea, the stain becomes 25 units thicker. If she only washes the tea pot once after each use, what will the thickness of the stain eventually become?\n\n4. An angina sufferer is prescribed one capsule containing 250 mg of a certain drug per day. Over any 24 hour period the human body loses half of the amount of drug it had initially. If the patient takes this medicine regularly at the same time each day over a long period of time, what is the least amount of drug present in the body? [Interpret the limit of your recurrence relation sensibly.]\n\n5. A schoolboy, whose piggy bank is empty, negotiates with his parents to receive £10 pocket money each Friday night. He decides to spend 80% of the contents of his piggy bank each week. If he maintains this arrangement over a long time, what are the greatest and least amounts of money he ever has in his piggy bank?\n If, on the other hand, he decided to save up for his holidays by only spending 20% of the bank’s contents every week, what difference would this make?\n\n6. The rear tyre on a tractor has a slow puncture which reduces the pressure in the tyre by 20% each day. The tractor driver uses a foot pump every morning to add 5 units to the tyre pressure. If the puncture gets no worse over a long period, what will the pressure in the tyre be each morning after it has been re-inflated?\n\n7. Algae is growing in a fish tank at a rate of 100 g per week. The tank is cleaned every week using a process which removes only 35% of any algae present. Is this cleaning process adequate, if 300 g of algae in the tank is thought to be too much for the fish?\n\n8. Victoria Park Gardens are combed for litter every day, resulting in 80% of all litter being removed. Every day however, 20 kg of litter is dropped in the gardens. What can be said about the mass of litter in the gardens after several weeks?\n9 The sequences generated by the recurrence relations\n\\[ u_{n+1} = au_n + 12 \\text{ and } u_0 = 1, \\text{ and } v_{n+1} = a^2v_n + 16 \\text{ and } v_0 = 1, \\]\nhave the same limit. Find this limit and the value of \\( a \\).\n\n10 Two sequences are generated by the recurrence relations\n\\[ u_{n+1} = pu_n + 2 \\text{ and } u_0 = a, \\text{ and } v_{n+1} = qv_n + 3 \\text{ and } v_0 = b. \\]\nThe limit of \\( \\{u_n\\} \\) is twice the limit of \\( \\{v_n\\} \\). Express \\( q \\) in terms of \\( p \\).\n\nExercise 10:\n\n6 A sequence is defined by the recurrence relation \\( u_{n+1} = au_n + b \\).\nIf \\( u_1 = -3, \\ u_2 = 7 \\text{ and } u_3 = 10 \\), find the values of \\( a \\) and \\( b \\).\n\n7 A sequence is defined by the recurrence relation \\( u_{n+1} = mu_{n-1} + c \\).\nIf \\( u_0 = 100, \\ u_1 = 0 \\text{ and } u_2 = 2 \\), find the values of \\( m \\) and \\( c \\).\n\n8 The amounts in a bank account at the end of three consecutive years were £2800, £3112 and £3436.48 respectively. The interest rate remained constant over this period and an extra fixed amount was also invested each year.\nWhat was the interest and the fixed amount invested each year?\n\n9 A recurrence relation is defined by \\( u_{n+1} = au_n - 4, \\ u_0 = 5 \\).\n a) Find expressions for \\( u_1 \\) and \\( u_2 \\) in terms of \\( a \\).\n b) Given \\( u_2 = 26 \\), find the values of \\( a \\).\n\n10 A recurrence relation is defined by \\( u_{n+1} = mu_{n-1} + 1, \\ u_0 = 3 \\).\n a) Find expressions for \\( u_1 \\) and \\( u_2 \\) in terms of \\( m \\).\n b) If \\( u_2 = 5 \\), find the values of \\( m \\).\nANSWERS:\n\nExercise 1:\n\n1 a) 1, 3, 5, 7; 39\n b) $2\\frac{1}{2}$, 3, $3\\frac{1}{2}$, 4; 12\n c) −6, −2, 2, 6; 70\n d) 5, 20, 45, 80; 2000\n e) 1, 8, 27, 64; 8000\n f) 3, 9, 27, 81; 3486784401\n g) $6, 6\\frac{1}{2}, 6\\frac{2}{3}, 6\\frac{3}{4}, 6\\frac{19}{20}$\n b) −5, 5, −5, 5; 5\n\n2 a) $u_n = 4n + 2$\n b) $u_n = 8n - 1$\n c) $u_n = 2n - 7$\n d) $u_n = 25 - 4n$\n e) $u_n = n^2$\n f) $u_n = 2^n$\n\nExercise 2:\n\n1. £1637.93.\n2. £144.78, £146.95.\n3. Yes, it always stays below the danger level or No, it is too close to the danger level. The level approaches 5 tonnes.\n4. annually £1164.40, quarterly £1171.66.\n\nExercise 3:\n\n1 a) 607\n b) 4.25\n c) −409\n d) −1\n\n2 a) 44.5, 90.5\n b) 5\n\n3 a) −4.48, −5.344\n b) 6\n\nExercise 4:\n\n1 a) $u_{n+1} = 1.025u_n + 50, \\ u_0 = 100$\n b) £435.36\n\n2 a) $u_{n+1} = 1.05u_n - 15000, \\ u_0 = 150000$\n b) No, after 14 years Dave would only have £3010.26 left, so after 15 years he would be £11 839.23 in the red.\n\n3 a) $u_{n+1} = 1.015u_n - 250, \\ u_0 = 2000$\n b) November 1st; £146.95\n\n4 a) $u_{n+1} = 1.15u_n - 100, \\ u_0 = 1500$\n b) 2343 ml (to the nearest ml)\n c) 7\n\n5 a) $u_{n+1} = 0.88u_n + 10, \\ u_0 = 100$\n b) 91 ml (to the nearest ml)\nExercise 5:\n\n1. a) 1, 5, 13, 29 b) 2, 4, 10, 28 \n c) 16, 20, 22, 23 d) 128, 16, −12, −19.\n\n2. a) 7, 22, 67, 202 \n b) 607, 1822, 5467 d) $u_{12}$.\n\n3. a) 5, 9, 17, 33, 65, 129 b) $u_{15}$.\n\n4. a) 2, 7, 32, ……, 488282 \n b) it doesn’t end in a 2 or a 7 or \n $u_{12} < 123456789 < u_{13}$.\n\n5. a) (i) 20 (ii) 10 (iii) 5 (iv) 2 \n (v) 4 (vi) 110 \n b) (i) 20 (ii) 10 (iii) 5 (iv) 2 \n (v) 4 (vi) 110 \n d) $L = \\frac{b}{1-a}$ e) $|a| < 1$.\n\n6. a) −1, 3, −5, 11, −21 \n b) −7, 15, −29, 59, −117 \n c) −1, 5, −13, 41, −121 \n d) −9, −33, −129, −513, −2049 \n e) −8, −50, −302, −1814, −10 886.\n\n7. a) 10, 10, 10 b) 2, 2, 2 \n c) $a > 0 \\Rightarrow$ the limit is approached \n only from above or only from below:\n\n ![Diagram](image)\n\n a < 0 $\\Rightarrow$ the limit is approached \n with the sequence oscillating \n between above and below:\n\n ![Diagram](image)\nExercise 6:\n\n1 i) b, d, g\n ii) a, c, e\n iii) f\n2 b) 4\nd) 1.25\ng) −15\n3 a) 12\n b) 12\n c) 12\n4 a) $-8\\frac{1}{3}$\n b) $-8\\frac{1}{3}$\n c) $-8\\frac{1}{3}$\n5 Sequence converges to a limit if $-1 < a < 1$.\n\nExercise 7:\n\n1 a) Converges since $-1 < 0.6 < 1$; limit = 30\n b) Converges since $-1 < 0.25 < 1$; limit = −12\n c) Diverges since $2 > 1$\n d) Converges since $-1 < -0.5 < 1$; limit = 2\n e) Diverges since $1 = 1$\n f) Converges since $-1 < -\\frac{2}{3} < 1$; limit = −2.4\n2 a) $-1 < 0.3 < 1$\n b) $\\frac{20}{7}$\n3 a) $-1 < \\frac{4}{7} < 1$\n b) $\\frac{28}{3}$\n4 a) $u$ does not have a limit since $4 > 1$; $w$ does have a limit since $-1 < 0.4 < 1$\n b) 5\n5 4\n6 0.2\n7 $p = 2q$\n8 a) $a = 1.5$, $b = 3$\n b) Limit does not exist since $1.5 > 1$\n\nExercise 8:\n\n1 a) $t_{n+1} = 0.65t_n + 800$\n b) 2286\n c) In the long run the number of tadpoles in the pond converges to 2286.\n2 a) $u_{n+1} = 0.73u_n + 20$\n b) $-1 < 0.73 < 1$\n c) 74\n d) In the long run the number of chickens converges to 74.\n3 a) $u_{n+1} = 0.45u_n + 275$\n b) In the long run the mass of algae in the pond converges to 500 g.\n4 It is unsafe, since in the long run the level of drug in the patient's system converges to $53.8$ mg. This is more than 50 mg at which level the consequences are very serious.\n5 Yes, in the long run the amount of pollutant chemicals in the loch would converge to $2.86$ tonnes. This is below the safe level of 3 tonnes.\n6 a) 3.53 kg\n b) In the long run the amount of sticky gum on the pavements will converge to 2.35 kg.\n7 a) In the long term the number of members will converge to 1200.\n b) 35\n8 a) 3.75 m\n b) 30%\nExercise 9:\n\n1. 100g.\n2. 300.\n3. 50 units after each use before washing\n4. 250 mg.\n5. a) £2.50 - £12.50 b) £40 - £50.\n6. 25.\n7. yes (limit < 300).\n8. There will be 25 kg of litter every morning before the cleaning is done.\n9. a = \\frac{1}{3} \\text{ limit = 18.}\n10. q = 3p - 2.\n\nExercise 10:\n\n6. a = 0.3, b = 7.9\n7. m = 0.4, c = -10\n8. 4%; £200\n9. a) \\ u_1 = 2a - 4, \\ u_2 = 2a^2 - 4a - 4\n b) -3.5\n10. a) \\ u_1 = 3m + 1, \\ u_2 = 3m^2 + m + 1\n b) -\\frac{4}{3}, 1\n### Higher : Recurrence Relations Revision\n\n| 2008 P1 | |\n|---|---|\n| **4.** A sequence is generated by the recurrence relation \\( u_{n+1} = 0.4u_n - 240 \\). What is the limit of this sequence as \\( n \\to \\infty \\)? | 2 |\n| A \\(-800\\) | |\n| B \\(-400\\) | |\n| C \\(200\\) | |\n| D \\(400\\) | |\n| **Ans** | B |\n\n| 2008 P1 | |\n|---|---|\n| **1.** A sequence is defined by the recurrence relation \\( u_{n+1} = 0.3u_n + 6 \\) with \\( u_{10} = 10 \\). What is the value of \\( u_{12} \\)? | 2 |\n| A \\(6.6\\) | |\n| B \\(7.8\\) | |\n| C \\(8.7\\) | |\n| D \\(9.6\\) | |\n| **Ans** | C |\n\n| 2007 P1 | |\n|---|---|\n| **7.** A sequence is defined by the recurrence relation \\( u_{n+1} = \\frac{1}{4}u_n + 16 \\), \\( u_0 = 0 \\). (a) Calculate the values of \\( u_1 \\), \\( u_2 \\) and \\( u_3 \\). | 3 |\n| Four terms of this sequence, \\( u_1 \\), \\( u_2 \\), \\( u_3 \\) and \\( u_4 \\) are plotted as shown in the graph. As \\( n \\to \\infty \\), the points on the graph approach the line \\( u_n = k \\), where \\( k \\) is the limit of this sequence. (b) (i) Give a reason why this sequence has a limit. (ii) Find the exact value of \\( k \\). | 3 |\n| **Ans** | |\n| \\( (a) u_1 = 16 \\) | |\n| \\( u_2 = 20 \\) | |\n| \\( u_3 = 21 \\) | |\n| \\( (b) (i) -1 < \\frac{1}{4} < 1 \\) | |\n| \\( (ii) \\frac{64}{3} \\) | |\n\nwww.national5maths.co.uk for all you need to pass Maths in one place\n| Year | Question | Answer |\n|------|----------|--------|\n| 2006 P1 | 4. A sequence is defined by the recurrence relation $u_{n+1} = 0.8u_n + 12$, $u_0 = 4$. \n (a) State why this sequence has a limit. \n (b) Find this limit. | Ans \n (a) sequence has limit since $-1 < 0.8 < 1$ \n (b) limit = 60 |\n| 2005 P1 | 6. (a) The terms of a sequence satisfy $u_{n+1} = ku_n + 5$. Find the value of $k$ which produces a sequence with a limit of 4. \n (b) A sequence satisfies the recurrence relation $u_{n+1} = mu_n + 5$, $u_0 = 3$. \n (i) Express $u_1$ and $u_2$ in terms of $m$. \n (ii) Given that $u_2 = 7$, find the value of $m$ which produces a sequence with no limit. | Ans \n (a) $k = -\\frac{1}{4}$ \n (b) (i) $u_1 = 3m + 5$, $u_2 = m(3m + 5) + 5$ \n (ii) $m = -2$ |\n| 2004 P2 | 4. A sequence is defined by the recurrence relation $u_{n+1} = ku_n + 3$. \n (a) Write down the condition on $k$ for this sequence to have a limit. \n (b) The sequence tends to a limit of 5 as $n \\to \\infty$. Determine the value of $k$. | Ans \n (a) $-1 < k < 1$ \n (b) $k = \\frac{2}{5}$ |\n| 2003 P1 | 4. A recurrence relation is defined by $u_{n+1} = pu_n + q$, where $-1 < p < 1$ and $u_0 = 12$. \n (a) If $u_1 = 15$ and $u_2 = 16$, find the values of $p$ and $q$. \n (b) Find the limit of this recurrence relation as $n \\to \\infty$. | Ans \n (a) $p = \\frac{1}{3}$, $q = 11$ \n (b) $16\\frac{1}{2}$ |\n| 2002 W P2 | 3. (a) Calculate the limit as $n \\to \\infty$ of the sequence defined by $u_{n+1} = 0.9u_n + 10$, $u_0 = 1$. \n (b) Determine the least value of $n$ for which $u_n$ is greater than half of this limit and the corresponding value of $u_n$. | Ans \n (a) 100 \n (b) $n = 7$, $u_7 = 52.65$ |\n4. A man decides to plant a number of fast-growing trees as a boundary between his property and the property of his next door neighbour. He has been warned, however, by the local garden centre that, during any year, the trees are expected to increase in height by 0.5 metres. In response to this warning he decides to trim 20% off the height of the trees at the start of any year.\n\n(a) If he adopts the “20% pruning policy”, to what height will he expect the trees to grow in the long run?\n\n(b) His neighbour is concerned that the trees are growing at an alarming rate and wants assurances that the trees will grow no taller than 2 metres. What is the minimum percentage that the trees will need to be trimmed each year so as to meet this condition?\n\nAns\n\n(a) \\(-1 < 0.8 < 1\\) and limit = 2.5 metres\n\n(b) trim 25%\n\n3. On the first day of March, a bank loans a man £2500 at a fixed rate of interest of 1.5% per month. This interest is added on the last day of each month and is calculated on the amount due on the first day of the month. He agrees to make repayments on the first day of each subsequent month. Each repayment is £300 except for the smaller final amount which will pay off the loan.\n\n(a) The amount that he owes at the start of each month is taken to be the amount still owing just after the monthly repayment has been made.\n\nLet \\(u_n\\) and \\(u_{n+1}\\) represent the amounts that he owes at the starts of two successive months. Write down a recurrence relation involving \\(u_{n+1}\\) and \\(u_n\\).\n\n(b) Find the date and the amount of the final payment.\n\nAns\n\n(a) \\(u_{n+1} = 1.015u_n - 300, u_0 = 2500\\)\n\n(b) Dec 1st, £290.68\n\n5. (a) Solve the equation \\(\\sin 2x - \\cos x = 0\\) in the interval \\(0 \\leq x \\leq 180\\).\n\n(b) The diagram shows parts of two trigonometric graphs, \\(y = \\sin 2x\\) and \\(y = \\cos x\\).\n\nUse your solutions in (a) to write down the coordinates of the point P.\n| Ans | \\( a = \\frac{3}{5}, \\ L = 25 \\) |\n|-----|----------------------------------|\n| Specimen 2 P2 | 4. Two sequences are defined by the recurrence relations \n\\[ u_{n+1} = 0.2u_n + p, \\quad u_0 = 1 \\quad \\text{and} \\quad v_{n+1} = 0.6v_n + q, \\quad v_0 = 1. \\] \n(a) Explain why each of these sequences has a limit. \n(b) If both sequences have the same limit, express \\( p \\) in terms of \\( q \\). |\n| Ans | \\( (a) \\ -1 < 0.2 < 1 \\) and \\( -1 < 0.6 < 1 \\) \n(b) \\( \\text{Limit} = \\frac{p}{0.8} \\) and \\( \\text{Limit} = \\frac{q}{0.4} \\Rightarrow p = 2q \\) |\n| Specimen 1 P1 | 2. A sequence is defined by the recurrence relation \\( u_{n+1} = 0.3u_n + 5 \\) with first term \\( u_1 \\). \n(a) Explain why this sequence has a limit as \\( n \\) tends to infinity. \n(b) Find the **exact** value of this limit. |\n| Ans | \\( (a) \\ -1 < 0.3 < 1 \\) \n(b) \\( \\frac{50}{7} \\) |\n| Specimen 1 P2 | 2. Trees are sprayed weekly with the pesticide, “Killpest”, whose manufacturers claim it will destroy 60% of all pests. Between the weekly sprayings, it is estimated that 300 new pests invade the trees. \nA new pesticide, “Pestkill”, comes onto the market. The manufacturers claim that it will destroy 80% of existing pests but it is estimated that 360 new pests per week will invade the trees. \nWhich pesticide will be more effective in the long term? |\n| Ans | Pestkill |\n1. A sequence is defined by the recurrence relation \\( u_n = 0.9u_{n-1} + 2, \\ u_1 = 3. \\)\n \n (a) Calculate the value of \\( u_2. \\) \n \n (b) What is the smallest value of \\( n \\) for which \\( u_n > 10? \\) \n \n (c) Find the limit of this sequence as \\( n \\to \\infty. \\) \n\n2. (a) At 12 noon a hospital patient is given a pill containing 50 units of antibiotic. \n By 1 pm the number of units in the patient’s body has dropped by 12%. \n By 2 pm a further 12% of the units remaining in the body at 1 pm is lost. \n If this fall-off rate is maintained, find the number of units of antibiotic remaining at 6 pm. \n\n (b) A doctor considers prescribing a course of treatment which involves a patient taking one of these pills every 6 hours over a long period of time. \n The doctor knows that more than 100 units of this antibiotic in the body is regarded as too dangerous. \n Should the doctor prescribe this course of treatment? \n Give reasons for your answer. \n\n3. A sequence is defined by the recurrence relation \\( u_{n+1} = 2u_n + 3 \\) and \\( u_0 = 1. \\) \n What is the value of \\( u_2? \\) \n\n4. A sequence is defined by \\( u_{n+1} = 3u_n + 4 \\) with \\( u_1 = 2. \\) \n What is the value of \\( u_3? \\)\n5. A sequence is defined by the recurrence relation \\( u_{n+1} = 0.3u_n + 5 \\) with first term \\( u_1 \\).\n\n (a) Explain why this sequence has a limit as \\( n \\) tends to infinity. \n\n (b) Find the exact value of this limit. \n\n6. Two sequences are defined by the recurrence relations\n\n \\[\n u_{n+1} = 0.2u_n + p, \\quad u_0 = 1 \\quad \\text{and} \\quad v_{n+1} = 0.6v_n + q, \\quad v_0 = 1.\n \\]\n\n If both sequences have the same limit, express \\( p \\) in terms of \\( q \\). \n\n7. Trees are sprayed weekly with the pesticide, KILLPEST, whose manufacturers claim it will destroy 65% of all pests. Between the weekly sprayings it is estimated that 500 new pests invade the trees.\n\n A new pesticide, PESTKILL, comes onto the market. The manufacturers claim that it will destroy 85% of existing pests but it is estimated that 650 new pests per week will invade the trees.\n\n Which pesticide will be more effective in the long term? \n\nTotal 31", "id": "./materials/359.pdf" }, { "contents": "Example. Consider the system\n\\[\n\\begin{align*}\nab x + y + t &= 1 \\\\\n(b + 1)y + t &= a \\\\\n(b + 1)y + a(b - 1)z + t &= b\n\\end{align*}\n\\]\nFor each \\((a; b) \\in \\mathbb{R}^2\\), decide whether the system is consistent and find its “number” of solutions. Find, if possible, an \\(a \\in \\mathbb{R}\\) such that the system is consistent for every \\(b \\in \\mathbb{R}\\), and an \\(a \\in \\mathbb{R}\\) such that the system is inconsistent for every \\(b \\in \\mathbb{R}\\).\n\nSolution The system has augmented matrix \\(\\overline{A}\\), where \\(A\\) is the coefficient matrix, and \\(v\\) is the constant column term:\n\\[\n\\overline{A} := (A|v) = \\begin{pmatrix} ab & 1 & 0 & 1 & 1 \\\\ 0 & b + 1 & 0 & 1 & a \\\\ 0 & b + 1 & a(b - 1) & 1 & b \\end{pmatrix}\n\\]\nperforming row operations \\((R_3 \\rightarrow R_3 - R_2)\\) we get a row echelon form\n\\[\n\\overline{A} \\rightarrow \\begin{pmatrix} ab & 1 & 0 & 1 & 1 \\\\ 0 & b + 1 & 0 & 1 & a \\\\ 0 & 0 & a(b - 1) & 0 & b - a \\end{pmatrix}\n\\]\nNow, if \\(a \\neq 0 \\land b \\neq -1 \\land b \\neq 0 \\land b \\neq 1\\), \\(A\\) and \\(\\overline{A}\\) have equal rank \\(\\rho(A) = \\rho(\\overline{A}) = 3\\), hence the system is consistent and it has \\(\\infty^{4-3} = \\infty^1\\) solutions, with \\(t\\) as free variable.\nIf \\(a = 0\\) we get\n\\[\n\\overline{A} \\rightarrow \\begin{pmatrix} 0 & 1 & 0 & 1 & 1 \\\\ 0 & b + 1 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 0 & b \\end{pmatrix} \\rightarrow \\begin{pmatrix} 0 & 1 & 0 & 1 & 1 \\\\ 0 & b & 0 & 0 & -1 \\\\ 0 & 0 & 0 & 0 & b \\end{pmatrix}\n\\]\nwhose corresponding system is inconsistent for \\(b \\neq 0\\) (because of the last row) as well as for \\(b = 0\\) (because of the second row). Thus for \\(a = 0\\) the system is inconsistent for every \\(b \\in \\mathbb{R}\\).\nIf \\(b = -1\\) we get\n\\[\n\\overline{A} \\rightarrow \\begin{pmatrix} -a & 1 & 0 & 1 & 1 \\\\ 0 & 0 & 0 & 1 & a \\\\ 0 & 0 & -2a & 0 & -1 - a \\end{pmatrix} \\rightarrow \\begin{pmatrix} -a & 1 & 0 & 1 & 1 \\\\ 0 & 0 & -2a & 0 & -1 - a \\\\ 0 & 0 & 0 & 1 & a \\end{pmatrix}\n\\]\nIf \\(b = -1 \\land a \\neq 0\\) the matrix is in row echelon form with \\(\\rho(A) = \\rho(\\overline{A}) = 3\\), hence the corresponding system is consistent and it has \\(\\infty^{4-3} = \\infty^1\\) solutions, with \\(y\\) as free variable.\nIf \\(b = -1 \\land a = 0\\) the matrix becomes\n\\[\n\\overline{A} \\rightarrow \\begin{pmatrix} 0 & 1 & 0 & 1 & 1 \\\\ 0 & 0 & 0 & 0 & -1 \\\\ 0 & 0 & 0 & 1 & 0 \\end{pmatrix}\n\\]\n(no longer in row echelon form) which corresponds to an inconsistent system (because of the second row), coherently with what we found above.\nIf \\(b = 0\\) we get\n\\[\n\\overline{A} \\rightarrow \\begin{pmatrix} 0 & 1 & 0 & 1 & 1 \\\\ 0 & 1 & 0 & 1 & a \\\\ 0 & 0 & -a & 0 & -a \\end{pmatrix} \\rightarrow \\begin{pmatrix} 0 & 1 & 0 & 1 & 1 \\\\ 0 & 0 & 0 & 0 & a - 1 \\\\ 0 & 0 & -a & 0 & -a \\end{pmatrix} \\rightarrow \\begin{pmatrix} 0 & 1 & 0 & 1 & 1 \\\\ 0 & 0 & -a & 0 & -a \\\\ 0 & 0 & 0 & 0 & a - 1 \\end{pmatrix}\n\\]\nIf \\(b = 0 \\land a \\neq 1\\) the corresponding system is inconsistent (because of the last row).\nIf $b = 0 \\land a = 1$ the matrix becomes\n\\[\n\\begin{pmatrix}\n0 & 1 & 0 & 1 & 1 \\\\\n0 & 0 & -1 & 0 & -1 \\\\\n0 & 0 & 0 & 0 & 0\n\\end{pmatrix}\n\\]\nwhich is in row echelon form with $\\rho(A) = \\rho(\\overline{A}) = 2$, hence the corresponding system is consistent and it has $\\infty^{4-2} = \\infty^2$ solutions, with $x$ and $t$ as free variables.\n\nIf $b = 1$ we get\n\\[\n\\begin{pmatrix}\na & 1 & 0 & 1 & 1 \\\\\n0 & 2 & 0 & 1 & a \\\\\n0 & 0 & 0 & 0 & 1 - a\n\\end{pmatrix}\n\\]\nIf $b = 1 \\land a \\neq 1$ the corresponding system is inconsistent (because of the last raw).\n\nIf $b = 1 \\land a = 1$ the matrix is in row echelon form with $\\rho(A) = \\rho(\\overline{A}) = 2$, hence the corresponding system is consistent and it has $\\infty^{4-2} = \\infty^2$ solutions, with $z$ and $t$ as free variables.\n\nWe can graphically summarize the previous discussion as follows:\n\nThus for $a = 1$ the system is consistent for every $b \\in \\mathbb{R}$. \n\n\\[\n\\begin{array}{c}\n\\text{no solutions for } a = 0 \\\\\n\\text{no solutions on } b = 1, \\text{ but for } (1:1) \\\\\n\\text{no solutions on } b = 0, \\text{ but for } (1:0) \\\\\n\\text{For } a \\neq 0 \\land b \\neq 0 \\land b \\neq 1 \\\\\n\\text{the system has } \\infty^1 \\text{ solutions}\n\\end{array}\n\\]", "id": "./materials/36.pdf" }, { "contents": "Recurrence Relations and Generating Functions\n\nNgày 8 tháng 12 năm 2010\nQuestion\n\nCertain bacteria divide into two bacteria every second. It was noticed that when one bacterium is placed in a bottle, it fills it up in 3 minutes. How long will it take to fill half the bottle?\nQuestion\n\nCertain bacteria divide into two bacteria every second. It was noticed that when one bacterium is placed in a bottle, it fills it up in 3 minutes. How long will it take to fill half the bottle?\n\nDiscussion\n\nMany processes lend themselves to recursive handling. Many sequences are determined by previous members of the sequence.\nQuestion\n\nCertain bacteria divide into two bacteria every second. It was noticed that when one bacterium is placed in a bottle, it fills it up in 3 minutes. How long will it take to fill half the bottle?\n\nDiscussion\n\nMany processes lend themselves to recursive handling. Many sequences are determined by previous members of the sequence.\nQuestion\n\nCertain bacteria divide into two bacteria every second. It was noticed that when one bacterium is placed in a bottle, it fills it up in 3 minutes. How long will it take to fill half the bottle?\n\nDiscussion\n\nMany processes lend themselves to recursive handling. Many sequences are determined by previous members of the sequence.\n\nHow many got the bacteria process right?\nQuestion\n\nCertain bacteria divide into two bacteria every second. It was noticed that when one bacterium is placed in a bottle, it fills it up in 3 minutes. How long will it take to fill half the bottle?\n\nDiscussion\n\nMany processes lend themselves to recursive handling. Many sequences are determined by previous members of the sequence.\n\nHow many got the bacteria process right?\n\nIf we denote the number of bacteria at second number $k$ by $b_k$ then we have: $b_{k+1} = 2b_k$, $b_1 = 1$.\n\nThis is a recurrence relation.\nAnother example of a problem that lends itself to a recurrence relation is a famous puzzle: The towers of Hanoi\nThis puzzle asks you to move the disks from the left tower to the right tower, one disk at a time so that a larger disk is never placed on a smaller disk. The goal is to use the smallest number of moves.\n\nClearly, before we move the large disk from the left to the right, all but the bottom disk, have to be on the middle tower. So if we denote the smallest number of moves by $h_n$ then we have:\n\n$$h_n + 1 = 2h_n + 1$$\n\nA simple technique for solving recurrence relations is called telescoping. Start from the first term and sequentially produce the next terms until a clear pattern emerges. If you want to be mathematically rigorous you may use induction.\nThis puzzle asks you to move the disks from the left tower to the right tower, one disk at a time so that a larger disk is never placed on a smaller disk. The goal is to use the smallest number of moves.\n\nClearly, before we move the large disk from the left to the right, all but the bottom disk, have to be on the middle tower. So if we denote the smallest number of moves by $h_n$ then we have:\n\n$$h_{n+1} = 2h_n + 1$$\n\nA simple technic for solving recurrence relation is called *telescoping*. \nThis puzzle asks you to move the disks from the left tower to the right tower, one disk at a time so that a larger disk is never placed on a smaller disk. The goal is to use the smallest number of moves.\n\nClearly, before we move the large disk from the left to the right, all but the bottom disk, have to be on the middle tower. So if we denote the smallest number of moves by $h_n$ then we have:\n\n$$h_{n+1} = 2h_n + 1$$\n\nA simple technic for solving recurrence relation is called *telescoping*.\n\nStart from the first term and sequentially produce the next terms until a clear pattern emerges. If you want to be mathematically rigorous you may use induction.\nExample\n\nSolving $b_{n+1} = 2b_n$, $b_1 = 1$. \n\nProof by induction:\n\n1. $h_1 = 1 = 2^1 - 1$\n2. Assume $h_n = 2^n - 1$\n3. Prove: $h_{n+1} = 2^{n+1} - 1$.\n\n4. $h_{n+1} = 2h_n + 1 = 2(2^n - 1) + 1 = 2^{n+1} - 1$.\n\n5. Solve: $a_n = 1 + a_{n-1}$, $a_1 = 1$.\n\n6. Telescoping yields:\n\n$$1, 1, 2, 3, 5, 8, 13, \\ldots$$\n\nRecurrence Relations and Generating Functions\nExample\n\nSolving $b_{n+1} = 2b_n$, $b_1 = 1$. \n\nProof by induction:\n\n1. $h_1 = 1 = 2^1 - 1$\n2. Assume $h_n = 2^n - 1$\n3. Prove: $h_{n+1} = 2^{n+1} - 1$.\n\n4. $h_{n+1} = 2h_n + 1 = 2(2^n - 1) + 1 = 2^{n+1} - 1$.\n\n5. Solve: $a_n = 1 + a_{n-1}$, $a_1 = 1$.\n\n6. Telescoping yields:\n\n$$1, 1, 2, 3, 5, 8, 13, \\ldots$$\n\nRecurrence Relations and Generating Functions\nExample\n\nSolving $b_{n+1} = 2b_n$, $b_1 = 1$.\n\n$b_1 = 1$, $b_2 = 2$, $b_3 = 4$, ... $b_n = 2^{n-1}$. \nExample\n\nSolving $b_{n+1} = 2b_n$, $b_1 = 1$.\n\n$b_1 = 1$, $b_2 = 2$, $b_3 = 4$, ... $b_n = 2^{n-1}$.\n\nSolving the Towers of Hanoi recurrence relation:\n\n$h_1 = 1$, $h_2 = 3$, $h_3 = 7$, $h_4 = 15$, ... $h_n = 2^n - 1$\n\nProof by induction:\n\n1. $h_1 = 1 = 2^1 - 1$\nExample\n\nSolving $b_{n+1} = 2b_n$, $b_1 = 1$.\n\n$b_1 = 1$, $b_2 = 2$, $b_3 = 4$, ... $b_n = 2^{n-1}$.\n\nSolving the Towers of Hanoi recurrence relation:\n\n$h_1 = 1$, $h_2 = 3$, $h_3 = 7$, $h_4 = 15$, ... $h_n = 2^n - 1$\n\nProof by induction:\n\n1. $h_1 = 1 = 2^1 - 1$\n2. Assume $h_n = 2^n - 1$\nExample\n\nSolving $b_{n+1} = 2b_n$, $b_1 = 1$.\n\n$b_1 = 1$, $b_2 = 2$, $b_3 = 4$, ... $b_n = 2^{n-1}$.\n\nSolving the Towers of Hanoi recurrence relation:\n\n$h_1 = 1$, $h_2 = 3$, $h_3 = 7$, $h_4 = 15$, ... $h_n = 2^n - 1$\n\nProof by induction:\n\n1. $h_1 = 1 = 2^1 - 1$\n2. Assume $h_n = 2^n - 1$\n3. Prove: $h_{n+1} = 2^{n+1} - 1$. \n\nRecurrence Relations and Generating Functions\nExample\n\nSolving $b_{n+1} = 2b_n$, $b_1 = 1$.\n\n$b_1 = 1$, $b_2 = 2$, $b_3 = 4$, ... $b_n = 2^{n-1}$.\n\nSolving the Towers of Hanoi recurrence relation:\n\n$h_1 = 1$, $h_2 = 3$, $h_3 = 7$, $h_4 = 15$, ... $h_n = 2^n - 1$\n\nProof by induction:\n\n1. $h_1 = 1 = 2^1 - 1$\n2. Assume $h_n = 2^n - 1$\n3. Prove: $h_{n+1} = 2^{n+1} - 1$.\n4. $h_{n+1} = 2h_n + 1 = 2(2^n - 1) + 1 = 2^{n+1} - 1$. \n\nRecurrence Relations and Generating Functions\nExample\n\nSolving $b_{n+1} = 2b_n$, $b_1 = 1$.\n\n$b_1 = 1$, $b_2 = 2$, $b_3 = 4$, ... $b_n = 2^{n-1}$.\n\nSolving the Towers of Hanoi recurrence relation:\n\n$h_1 = 1$, $h_2 = 3$, $h_3 = 7$, $h_4 = 15$, ... $h_n = 2^n - 1$\n\nProof by induction:\n\n1. $h_1 = 1 = 2^1 - 1$\n2. Assume $h_n = 2^n - 1$\n3. Prove: $h_{n+1} = 2^{n+1} - 1$.\n4. $h_{n+1} = 2h_n + 1 = 2(2^n - 1) + 1 = 2^{n+1} - 1$.\n5. Solve: $a_n = \\frac{1}{1+a_{n-1}}$, $a_1 = 1$. \n\nRecurrence Relations and Generating Functions\nExample\n\nSolving $b_{n+1} = 2b_n$, $b_1 = 1$.\n\n$b_1 = 1$, $b_2 = 2$, $b_3 = 4$, ... $b_n = 2^{n-1}$.\n\nSolving the Towers of Hanoi recurrence relation:\n\n$h_1 = 1$, $h_2 = 3$, $h_3 = 7$, $h_4 = 15$, ... $h_n = 2^n - 1$\n\nProof by induction:\n\n1. $h_1 = 1 = 2^1 - 1$\n2. Assume $h_n = 2^n - 1$\n3. Prove: $h_{n+1} = 2^{n+1} - 1$.\n4. $h_{n+1} = 2h_n + 1 = 2(2^n - 1) + 1 = 2^{n+1} - 1$.\n5. Solve: $a_n = \\frac{1}{1+a_{n-1}}$, $a_1 = 1$.\n6. Telescoping yields: $1, \\frac{1}{2}, \\frac{2}{3}, \\frac{3}{5}, \\frac{5}{8}, \\frac{8}{13}$\nDo we see a pattern?\n\nLooks like $a_n = f_{n-1}$ where $f_n$ are the Fibonacci numbers.\n\nCan we prove it?\n\nBy induction:\n\n1. $a_1 = 1 = f_0$.\n\n2. Induction hypothesis: assume $a_n = f_{n-1}$.\n\n3. $a_{n+1} = 1 + a_n = 1 + f_{n-1} = f_n$.\n\nRecurrence Relations and Generating Functions\n1, \\frac{1}{2}, \\frac{2}{3}, \\frac{3}{5}, \\frac{5}{8}, \\frac{8}{13}\n\nDo we see a pattern?\n1, \\frac{1}{2}, \\frac{2}{3}, \\frac{3}{5}, \\frac{5}{8}, \\frac{8}{13}\n\nDo we see a pattern?\n\nLooks like \\( a_n = \\frac{f_{n-1}}{f_n} \\) where \\( f_n \\) are the Fibonacci numbers.\n\nCan we prove it?\n1, \\frac{1}{2}, \\frac{2}{3}, \\frac{3}{5}, \\frac{5}{8}, \\frac{8}{13}\n\nDo we see a pattern?\n\nLooks like \\( a_n = \\frac{f_{n-1}}{f_n} \\) where \\( f_n \\) are the Fibonacci numbers.\n\nCan we prove it?\n\nChứng minh.\n1, \\frac{1}{2}, \\frac{2}{3}, \\frac{3}{5}, \\frac{5}{8}, \\frac{8}{13}\n\nDo we see a pattern?\n\nLooks like \\( a_n = \\frac{f_{n-1}}{f_n} \\) where \\( f_n \\) are the Fibonacci numbers.\n\nCan we prove it?\n\nChứng minh.\n\n1. By induction: \\( a_1 = 1 = \\frac{f_0}{f_1} \\).\n1, \\frac{1}{2}, \\frac{2}{3}, \\frac{3}{5}, \\frac{5}{8}, \\frac{8}{13}\n\nDo we see a pattern?\n\nLooks like \\( a_n = \\frac{f_{n-1}}{f_n} \\) where \\( f_n \\) are the Fibonacci numbers.\n\nCan we prove it?\n\nChứng minh.\n\n1. By induction: \\( a_1 = 1 = \\frac{f_0}{f_1} \\).\n\n2. Induction hypothesis: assume \\( a_n = \\frac{f_{n-1}}{f_n} \\).\n1, \\frac{1}{2}, \\frac{2}{3}, \\frac{3}{5}, \\frac{5}{8}, \\frac{8}{13}\n\nDo we see a pattern?\n\nLooks like \\( a_n = \\frac{f_{n-1}}{f_n} \\) where \\( f_n \\) are the Fibonacci numbers.\n\nCan we prove it?\n\nChứng minh.\n\n1. By induction: \\( a_1 = 1 = \\frac{f_0}{f_1} \\).\n2. Induction hypothesis: assume \\( a_n = \\frac{f_{n-1}}{f_n} \\).\n3. \\[\na_{n+1} = \\frac{1}{1 + a_n} = \\frac{1}{1 + \\frac{f_{n-1}}{f_n}} = \\frac{f_n}{f_n + f_{n-1}} = \\frac{f_n}{f_{n+1}}\n\\]\nA recurrence relation for a sequence $a_n$ is a relation of the form $a_{n+1} = f(a_1, a_2, \\ldots, a_n)$.\n\nWe do not expect to have a useful method to solve all recurrence relations. This definition actually applies to any sequence! We shall break down the functions for which we do have effective methods to “solve” the recurrence relation. By solving we mean obtaining an explicit expression of the form $a_n = g(n)$. To accomplish this we need some terminology.\nRecurrence Relations Terminology\n\n**Definition**\n\nA recurrence relation for a sequence $a_n$ is a relation of the form $a_{n+1} = f(a_1, a_2, \\ldots, a_n)$.\n\nWe do not expect to have a useful method to solve all recurrence relations. This definition actually applies to any sequence! We shall break down the functions for which we do have effective methods to “solve” the recurrence relation. By solving we mean obtaining an explicit expression of the form $a_n = g(n)$. To accomplish this we need some terminology.\n\n**Definition**\n\nA recurrence relation is **linear** if:\n\n$$f(a_1, a_2, \\ldots, a_n) = \\sum_{i=1}^{n} h_i \\cdot a_i + h(n) \\quad \\text{Where } h(n) \\text{ is a function of } n.$$\nA recurrence relation is:\n\n- **homogeneous** if \\( h(n) = 0 \\)\n- With constant coefficients: if all \\( h_i \\) are constants.\n- Of order \\( k \\) if \\( f(a_1, a_2, \\ldots, a_n) = \\sum_{i=n-k}^{n} h_i \\cdot a_i \\)\n\n**Example**\n\n1. \\( f_n = f_{n-1} + f_{n-2} \\) is a linear, homogeneous recurrence relation of order 2 with constant coefficients.\n2. \\( a_n = a_{n-1} + n \\) is a linear, non-homogeneous recurrence relation of order 1 and constant coefficients.\n3. \\( d_n = (n-1) d_{n-1} + (n-1) d_{n-2} \\) is a linear, homogeneous recurrence relation of order 2. It does not have constant coefficients.\n4. \\( a_n = a_{n-1} + 2a_{n-2} + 4a_{n-3} + 2n \\) is a non-homogeneous, linear recurrence relation with constant coefficients of order 5.\n5. \\( a_n = 1 + a_{n-1} \\) is a non-linear recurrence relation.\nA recurrence relation is:\n\n1. **Homogeneous** if \\( h(n) = 0 \\)\n2. With constant coefficients: if all \\( h_i \\) are constants.\n3. Of order \\( k \\) if \\( f(a_1, a_2, \\ldots, a_n) = \\sum_{i=n-k}^{n} h_i \\cdot a_i \\)\n\n**Example**\n\n1. \\( f_n = f_{n-1} + f_{n-2} \\) is a linear, homogeneous recurrence relation of order 2 with constant coefficients.\n2. \\( a_n = a_{n-1} + n \\) is a linear, non-homogeneous recurrence relation of order 1 and constant coefficients.\n3. \\( d_n = (n-1)d_{n-1} + (n-1)d_{n-2} \\) is a linear, homogeneous recurrence relation of order 2. It does not have constant coefficients.\n4. \\( a_n = a_{n-1} + 2a_{n-2} + 4a_{n-3} + 2n \\) is a non-homogeneous, linear recurrence relation with constant coefficients of order 5.\n5. \\( a_n = 1 + a_{n-1} \\) is a non-linear recurrence relation.\nA recurrence relation is:\n\n1. homogeneous if \\( h(n) = 0 \\)\nA recurrence relation is:\n\n1. **homogeneous** if \\( h(n) = 0 \\)\n2. With constant coefficients: if all \\( h_i \\) are constants.\nA recurrence relation is:\n\n1. **homogeneous** if \\( h(n) = 0 \\)\n2. With constant coefficients: if all \\( h_i \\) are constants.\n3. Of order \\( k \\) if \\( f(a_1, a_2, \\ldots, a_n) = \\sum_{i=n-k}^{n} h_i \\cdot a_i \\)\nDefinition\n\n1. A recurrence relation is:\n2. **homogeneous** if \\( h(n) = 0 \\)\n3. With constant coefficients: if all \\( h_i \\) are constants.\n4. Of order \\( k \\) if \\( f(a_1, a_2, \\ldots, a_n) = \\sum_{i=n-k}^{n} h_i \\cdot a_i \\)\n\nExample\nDefinition\n\n1. A recurrence relation is:\n2. **homogeneous** if \\( h(n) = 0 \\)\n3. With constant coefficients: if all \\( h_i \\) are constants.\n4. Of order \\( k \\) if \\( f(a_1, a_2, \\ldots, a_n) = \\sum_{i=n-k}^{n} h_i \\cdot a_i \\)\n\nExample\n\n1. \\( f_n = f_{n-1} + f_{n-2} \\) is a linear, homogeneous recurrence relation of order 2 with constant coefficients.\nDefinition\n\n1. A recurrence relation is:\n2. **homogeneous** if \\( h(n) = 0 \\)\n3. With constant coefficients: if all \\( h_i \\) are constants.\n4. Of order \\( k \\) if \\( f(a_1, a_2, \\ldots, a_n) = \\sum_{i=n-k}^{n} h_i \\cdot a_i \\)\n\nExample\n\n1. \\( f_n = f_{n-1} + f_{n-2} \\) is a linear, homogeneous recurrence relation of order 2 with constant coefficients.\n2. \\( a_n = a_{n-1} + n \\) is a linear, non-homogeneous recurrence relation of order 1 and constant coefficients.\nDefinition\n\n1. A recurrence relation is:\n2. **homogeneous** if \\( h(n) = 0 \\)\n3. With constant coefficients: if all \\( h_i \\) are constants.\n4. Of order \\( k \\) if \\( f(a_1, a_2, \\ldots, a_n) = \\sum_{i=n-k}^{n} h_i \\cdot a_i \\)\n\nExample\n\n1. \\( f_n = f_{n-1} + f_{n-2} \\) is a linear, homogeneous recurrence relation of order 2 with constant coefficients.\n2. \\( a_n = a_{n-1} + n \\) is a linear, non-homogeneous recurrence relation of order 1 and constant coefficients.\n3. \\( d_n = (n - 1)d_{n-1} + (n - 1)d_{n-2} \\) is a linear, homogeneous recurrence relation of order 2. It does not have constant coefficients.\nDefinition\n\n1. A recurrence relation is:\n2. **homogeneous** if \\( h(n) = 0 \\)\n3. With constant coefficients: if all \\( h_i \\) are constants.\n4. Of order \\( k \\) if \\( f(a_1, a_2, \\ldots, a_n) = \\sum_{i=n-k}^{n} h_i \\cdot a_i \\)\n\nExample\n\n1. \\( f_n = f_{n-1} + f_{n-2} \\) is a linear, homogeneous recurrence relation of order 2 with constant coefficients.\n2. \\( a_n = a_{n-1} + n \\) is a linear, non-homogeneous recurrence relation of order 1 and constant coefficients.\n3. \\( d_n = (n - 1)d_{n-1} + (n - 1)d_{n-2} \\) is a linear, homogeneous recurrence relation of order 2. It does not have constant coefficients.\n4. \\( a_n = a_{n-1} + 2a_{n-2} + 4a_{n-5} + 2^n \\) is a non-homogeneous, linear recurrence relation with constant coefficients of order 5.\nDefinition\n\n1. A recurrence relation is:\n2. **homogeneous** if \\( h(n) = 0 \\)\n3. With constant coefficients: if all \\( h_i \\) are constants.\n4. Of order \\( k \\) if \\( f(a_1, a_2, \\ldots, a_n) = \\sum_{i=n-k}^{n} h_i \\cdot a_i \\)\n\nExample\n\n1. \\( f_n = f_{n-1} + f_{n-2} \\) is a linear, homogeneous recurrence relation of order 2 with constant coefficients.\n2. \\( a_n = a_{n-1} + n \\) is a linear, non-homogeneous recurrence relation of order 1 and constant coefficients.\n3. \\( d_n = (n - 1)d_{n-1} + (n - 1)d_{n-2} \\) is a linear, homogeneous recurrence relation of order 2. It does not have constant coefficients.\n4. \\( a_n = a_{n-1} + 2a_{n-2} + 4a_{n-5} + 2^n \\) is a non-homogeneous, linear recurrence relation with constant coefficients of order 5.\nA few more examples coming from verbal problems.\n\nQuestion\n\n1. In how many ways can you write the integer $n$ as a sum of $k$ distinct positive integers?\n\n2. In how many ways can you write $n$ as a sum of 5 distinct positive integers?\n\nAnswer\n\n1. To answer the first question we split the set of answers into two sets:\n - First set contains all solutions that include the number 1.\n - The second is the set of solutions for which every integer is $> 1$.\n\n2. If we denote the number of solutions by $a_{n,k}$ then we get:\n $$a_{n,k} = a_{n-1,k-1} + a_{n-k,k}$$\n\n3. This is a linear, homogeneous recurrence relation with constant coefficients, but not of finite order.\nA few more examples coming from verbal problems.\n\nQuestion\n\n1. In how many ways can you write the integer $n$ as a sum of $k$ distinct positive integers?\nA few more examples coming from verbal problems.\n\n**Question**\n\n1. *In how many ways can you write the integer $n$ as a sum of $k$ distinct positive integers?*\n\n2. *In how many ways can you write $n$ as a sum of 5 distinct positive integers?*\nA few more examples coming from verbal problems.\n\n**Question**\n\n1. *In how many ways can you write the integer n as a sum of k distinct positive integers?*\n\n2. *In how many ways can you write n as a sum of 5 distinct positive integers?*\n\n**Answer**\nA few more examples coming from verbal problems.\n\n**Question**\n\n1. *In how many ways can you write the integer n as a sum of k distinct positive integers?*\n\n2. *In how many ways can you write n as a sum of 5 distinct positive integers?*\n\n**Answer**\n\n1. *To answer the first question we split the set of answers into two sets:*\n\n - First set contains all solutions that include the number 1.\n - The second is the set of solutions for which every integer is > 1.\n\n If we denote the number of solutions by $a_{n,k}$ then we get:\n\n $$a_{n,k} = a_{n-1,k-1} + a_{n-k,k}$$\n\n This is a linear, homogeneous recurrence relation with constant coefficients, but not of finite order.\nA few more examples coming from verbal problems.\n\n**Question**\n\n1. *In how many ways can you write the integer n as a sum of k distinct positive integers?*\n\n2. *In how many ways can you write n as a sum of 5 distinct positive integers?*\n\n**Answer**\n\n1. *To answer the first question we split the set of answers into two sets:*\n - *First set contains all solutions that include the number 1.*\nA few more examples coming from verbal problems.\n\n**Question**\n\n1. *In how many ways can you write the integer n as a sum of k distinct positive integers?*\n\n2. *In how many ways can you write n as a sum of 5 distinct positive integers?*\n\n**Answer**\n\n1. *To answer the first question we split the set of answers into two sets:*\n - *First set contains all solutions that include the number 1.*\n - *The second is the set of solutions for which every integer is > 1.*\nA few more examples coming from verbal problems.\n\n**Question**\n\n1. *In how many ways can you write the integer n as a sum of k distinct positive integers?*\n\n2. *In how many ways can you write n as a sum of 5 distinct positive integers?*\n\n**Answer**\n\n1. *To answer the first question we split the set of answers into two sets:*\n - First set contains all solutions that include the number 1.\n - The second is the set of solutions for which every integer is > 1.\n\n2. *If we denote the number of solutions by \\( a_{n,k} \\) then we get:*\n\n\\[\na_{n,k} = a_{n-1,k-1} + a_{n-k,k}\n\\]\nRecurrence Relations\n\nA few more examples coming from verbal problems.\n\nQuestion\n\n1. In how many ways can you write the integer $n$ as a sum of $k$ distinct positive integers?\n\n2. In how many ways can you write $n$ as a sum of 5 distinct positive integers?\n\nAnswer\n\n1. To answer the first question we split the set of answers into two sets:\n - First set contains all solutions that include the number 1.\n - The second is the set of solutions for which every integer is $> 1$.\n\n2. If we denote the number of solutions by $a_{n,k}$ then we get:\n\n$$a_{n,k} = a_{n-1,k-1} + a_{n-k,k}$$\nFor the second equation we have:\n\n\\[ b_n, 5 = b_n, 4 + b_n - 5, 5 \\]\n\nAgain, this is a linear, homogeneous recurrence relation with constant coefficients, of order 5.\n\nRemark\n\nLinear, homogeneous recurrence relations have many solutions. Indeed if \\( f(n) \\) and \\( g(n) \\) are solutions then so is \\( \\alpha f(n) + \\beta g(n) \\).\n\nIf \\( f(n) \\) and \\( g(n) \\) are solutions to a non-homogeneous recurrence relation then \\( f(n) - g(n) \\) is a solution to the associated homogeneous recurrence relation.\nFor the second equation we have:\n\n\\[ b_n, 5 = b_n, 4 + b_n - 5, 5 \\]\n\nAgain, this is a linear, homogeneous recurrence relation with constant coefficients, of order 5.\n\nRemark\n\nLinear, homogeneous recurrence relations have many solutions. Indeed if \\( f(n) \\) and \\( g(n) \\) are solutions then so is \\( \\alpha f(n) + \\beta g(n) \\).\n\nIf \\( f(n) \\) and \\( g(n) \\) are solutions to a non-homogeneous recurrence relation then \\( f(n) - g(n) \\) is a solution to the associated homogeneous recurrence relation.\nFor the second equation we have:\n\n\\[ b_{n,5} = b_{n,4} + b_{n-5,5} \\]\nFor the second equation we have:\n\n\\[ b_{n,5} = b_{n,4} + b_{n-5,5} \\]\n\nAgain, this is a linear, homogeneous recurrence relation with constant coefficients, of order ?.\nAnswer (continued)\n\n1. For the second equation we have:\n\n\\[ b_{n,5} = b_{n,4} + b_{n-5,5} \\]\n\n2. Again, this is a linear, homogeneous recurrence relation with constant coefficients, of order ?.\n\nRemark\n\nLinear, homogeneous recurrence relations have many solutions. Indeed if \\( f(n) \\) and \\( g(n) \\) are solutions then so is \\( \\alpha f(n) + \\beta g(n) \\).\nAnswer (continued)\n\n1. For the second equation we have:\n\n\\[ b_{n,5} = b_{n,4} + b_{n-5,5} \\]\n\n2. Again, this is a linear, homogeneous recurrence relation with constant coefficients, of order ?.\n\nRemark\n\nLinear, homogeneous recurrence relations have many solutions. Indeed if \\( f(n) \\) and \\( g(n) \\) are solutions then so is \\( \\alpha f(n) + \\beta g(n) \\).\nFor the second equation we have:\n\n\\[ b_{n,5} = b_{n,4} + b_{n-5,5} \\]\n\nAgain, this is a linear, homogeneous recurrence relation with constant coefficients, of order ?.\n\nRemark\n\nLinear, homogeneous recurrence relations have many solutions. Indeed if \\( f(n) \\) and \\( g(n) \\) are solutions then so is \\( \\alpha f(n) + \\beta g(n) \\).\n\nIf \\( f(n) \\) and \\( g(n) \\) are solutions to a non homogeneous recurrence relation then \\( f(n) - g(n) \\) is a solution to the associated homogeneous recurrence relation.\nRemark\n\nThis means that in order to solve a non homogeneous linear recurrence relation all we need to do is find the general solution $g(n)$ to the homogeneous part and a particular solution $p(n)$ to the non homogeneous equation.\n\nThe general solution will be: $g(n) + p(n)$.\n\nThe following example demonstrates this:\nRemark\n\nThis means that in order to solve a non homogeneous linear recurrence relation all we need to do is find the general solution $g(n)$ to the homogeneous part and a particular solution $p(n)$ to the non homogeneous equation.\n\nThe general solution will be: $g(n) + p(n)$.\n\nThe following example demonstrates this:\n\nExample\n\nSolve: $a_n = 2a_{n-1} + 3n - 1$. \nRemark\n\nThis means that in order to solve a non homogeneous linear recurrence relation all we need to do is find the general solution \\( g(n) \\) to the homogeneous part and a particular solution \\( p(n) \\) to the non homogeneous equation.\n\nThe general solution will be: \\( g(n) + p(n) \\).\n\nThe following example demonstrates this:\n\nExample\n\nSolve: \\( a_n = 2a_{n-1} + 3n - 1 \\).\n\n1. The homogeneous part is: \\( b_n = 2b_{n-1} \\).\nRemark\n\nThis means that in order to solve a non homogeneous linear recurrence relation all we need to do is find the general solution $g(n)$ to the homogeneous part and a particular solution $p(n)$ to the non homogeneous equation.\n\nThe general solution will be: $g(n) + p(n)$.\n\nThe following example demonstrates this:\n\nExample\n\nSolve: $a_n = 2a_{n-1} + 3n - 1$.\n\n1. The homogeneous part is: $b_n = 2b_{n-1}$.\n2. The general solution is: $b_n = \\alpha 2^n$. \nRemark\n\nThis means that in order to solve a non homogeneous linear recurrence relation all we need to do is find the general solution \\( g(n) \\) to the homogeneous part and a particular solution \\( p(n) \\) to the non homogeneous equation.\n\nThe general solution will be: \\( g(n) + p(n) \\).\n\nThe following example demonstrates this:\n\nExample\n\nSolve: \\( a_n = 2a_{n-1} + 3n - 1 \\).\n\n1. The homogeneous part is: \\( b_n = 2b_{n-1} \\).\n2. The general solution is: \\( b_n = \\alpha 2^n \\).\n3. To find a particular solution we try \\( p_n = cn + d \\).\nRemark\n\nThis means that in order to solve a non homogeneous linear recurrence relation all we need to do is find the general solution \\( g(n) \\) to the homogeneous part and a particular solution \\( p(n) \\) to the non homogeneous equation.\n\nThe general solution will be: \\( g(n) + p(n) \\).\n\nThe following example demonstrates this:\n\nExample\n\nSolve: \\( a_n = 2a_{n-1} + 3n - 1 \\).\n\n1. The homogeneous part is: \\( b_n = 2b_{n-1} \\).\n2. The general solution is: \\( b_n = \\alpha 2^n \\).\n3. To find a particular solution we try \\( p_n = cn + d \\).\n4. Substituting in the original recurrence relation we get: \\( cn + d = 2(c(n - 1) + d) + 3n - 1 \\).\nRemark\n\nThis means that in order to solve a non homogeneous linear recurrence relation all we need to do is find the general solution \\( g(n) \\) to the homogeneous part and a particular solution \\( p(n) \\) to the non homogeneous equation.\n\nThe general solution will be: \\( g(n) + p(n) \\).\n\nThe following example demonstrates this:\n\nExample\n\nSolve: \\( a_n = 2a_{n-1} + 3n - 1 \\).\n\n1. The homogeneous part is: \\( b_n = 2b_{n-1} \\).\n2. The general solution is: \\( b_n = \\alpha 2^n \\).\n3. To find a particular solution we try \\( p_n = cn + d \\).\n4. Substituting in the original recurrence relation we get: \\( cn + d = 2(c(n - 1) + d) + 3n - 1 \\).\n5. Solving for \\( c \\) and \\( d \\) we get: \\( a_n = \\alpha 2^n - 3n - 5 \\).\nRemark\n\nTo simplify notation we shall limit our discussion to second order recurrence relations. The extension to higher order is straightforward.\nRemark\n\nTo simplify notation we shall limit our discussion to second order recurrence relations. The extension to higher order is straightforward.\n\nTheorem (observation)\n\nLet \\( a_n = b \\cdot a_{n-1} + c \\cdot a_{n-2} + g(n) \\), \\( a_1 = \\alpha \\), \\( a_2 = \\beta \\).\n\nFor each \\( k \\geq 3 \\), \\( a_k \\) is uniquely determined.\nRemark\n\nTo simplify notation we shall limit our discussion to second order recurrence relations. The extension to higher order is straightforward.\n\nTheorem (observation)\n\nLet \\( a_n = b \\cdot a_{n-1} + c \\cdot a_{n-2} + g(n) \\), \\( a_1 = \\alpha \\), \\( a_2 = \\beta \\).\n\nFor each \\( k \\geq 3 \\), \\( a_k \\) is uniquely determined.\n\nDefinition\n\n\\( a_1 = \\alpha \\), \\( a_2 = \\beta \\) are called the initial conditions.\nRemark\n\nTo simplify notation we shall limit our discussion to second order recurrence relations. The extension to higher order is straightforward.\n\nTheorem (observation)\n\nLet \\( a_n = b \\cdot a_{n-1} + c \\cdot a_{n-2} + g(n) \\), \\( a_1 = \\alpha \\), \\( a_2 = \\beta \\).\n\nFor each \\( k \\geq 3 \\), \\( a_k \\) is uniquely determined.\n\nDefinition\n\n\\( a_1 = \\alpha \\), \\( a_2 = \\beta \\) are called the initial conditions.\n\nCorollary\n\nAny solution that satisfies the recurrence relation and initial conditions is THE ONLY solution.\nDefinition\n\nLet \\( a_n = ba_{n-1} + ca_{n-2} \\).\n\nThe quadratic equation \\( x^2 - bx - c = 0 \\) is called the characteristic equation of the recurrence relation.\nDefinition\n\nLet \\( a_n = ba_{n-1} + ca_{n-2} \\).\n\nThe quadratic equation \\( x^2 - bx - c = 0 \\) is called the characteristic equation of the recurrence relation.\n\nTheorem (Solving Linear Homogeneous RR with Constant Coefficients)\n\n1. Let \\( a_n = ba_{n-1} + ca_{n-2} \\).\n2. Let \\( r_1, r_2 \\) be the roots of the characteristic equation.\n3. Then the general solution of this recurrence relation is \\( a_n = \\alpha r_1^n + \\beta r_2^n \\).\n4. If \\( r_1 = r_2 \\) then the general solution is \\( a_n = \\alpha r^n + \\beta nr^n \\).\nDefinition\n\nLet \\( a_n = ba_{n-1} + ca_{n-2} \\).\n\nThe quadratic equation \\( x^2 - bx - c = 0 \\) is called the characteristic equation of the recurrence relation.\n\nTheorem (Solving Linear Homogeneous RR with Constant Coefficients)\n\n1. Let \\( a_n = b \\cdot a_{n-1} + c \\cdot a_{n-2} \\).\nDefinition\n\nLet \\( a_n = ba_{n-1} + ca_{n-2} \\).\n\nThe quadratic equation \\( x^2 - bx - c = 0 \\) is called the characteristic equation of the recurrence relation.\n\nTheorem (Solving Linear Homogeneous RR with Constant Coefficients)\n\n1. Let \\( a_n = b \\cdot a_{n-1} + c \\cdot a_{n-2} \\).\n2. Let \\( r_1, r_2 \\) be the roots of the characteristic equation.\nDefinition\n\nLet \\( a_n = ba_{n-1} + ca_{n-2} \\).\n\nThe quadratic equation \\( x^2 - bx - c = 0 \\) is called the characteristic equation of the recurrence relation.\n\nTheorem (Solving Linear Homogeneous RR with Constant Coefficients)\n\n1. Let \\( a_n = b \\cdot a_{n-1} + c \\cdot a_{n-2} \\).\n2. Let \\( r_1, r_2 \\) be the roots of the characteristic equation.\n3. Then the general solution of this recurrence relation is \\( a_n = \\alpha r_1^n + \\beta r_2^n \\).\nDefinition\n\nLet \\( a_n = ba_{n-1} + ca_{n-2} \\).\n\nThe quadratic equation \\( x^2 - bx - c = 0 \\) is called the characteristic equation of the recurrence relation.\n\nTheorem (Solving Linear Homogeneous RR with Constant Coefficients)\n\n1. Let \\( a_n = b \\cdot a_{n-1} + c \\cdot a_{n-2} \\).\n2. Let \\( r_1, r_2 \\) be the roots of the characteristic equation.\n3. Then the general solution of this recurrence relation is \\( a_n = \\alpha r_1^n + \\beta r_2^n \\).\n4. If \\( r_1 = r_2 \\) then the general solution is \\( a_n = \\alpha r^n + \\beta nr^n \\).\nChứng minh.\n\nWe need to show two things:\n\nWe note that since the recurrence relation is linear it is enough to prove that \\( r_i^n = br_i^{n-1} + cr_i^{n-2} \\)\nChứng minh.\n\nWe need to show two things:\n\n1. \\( a_n = br_1^n + cr_2^n \\) is a solution (or \\( a_n = br^n + cnr^n \\) is a solution in case \\( r_1 = r_2 \\)).\n\nWe note that since the recurrence relation is linear it is enough to prove that \\( r_i^n = br_i^{n-1} + cr_i^{n-2} \\).\nChứng minh.\n\nWe need to show two things:\n\n1. \\( a_n = br_1^n + cr_2^n \\) is a solution (or \\( a_n = br^n + cnr^n \\) is a solution in case \\( r_1 = r_2 \\)).\n\n2. Every other solution is of this form.\n\nWe note that since the recurrence relation is linear it is enough to prove that \\( r_i^n = br_i^{n-1} + cr_i^{n-2} \\).\nChứng minh.\n\nWe need to show two things:\n\n1. \\( a_n = br_1^n + cr_2^n \\) is a solution (or \\( a_n = br^n + cnr^n \\) is a solution in case \\( r_1 = r_2 \\)).\n\n2. Every other solution is of this form.\n\nWe note that since the recurrence relation is linear it is enough to prove that \\( r_i^n = br_i^{n-1} + cr_i^{n-2} \\)\n\n1. \\( br_i^{n-1} + cr_i^{n-2} = r_i^{n-2}(br_i + c) \\)\nChứng minh.\n\nWe need to show two things:\n\n1. \\( a_n = br_1^n + cr_2^n \\) is a solution (or \\( a_n = br^n + cnr^n \\) is a solution in case \\( r_1 = r_2 \\)).\n\n2. Every other solution is of this form.\n\nWe note that since the recurrence relation is linear it is enough to prove that \\( r_i^n = br_i^{n-1} + cr_i^{n-2} \\)\n\n1. \\( br_i^{n-1} + cr_i^{n-2} = r_i^{n-2}(br_i + c) \\)\n\n2. Since \\( r_i \\) are roots of the characteristic equation we have: \\( r_i^2 = br_i + c \\).\nChứng minh.\n\nWe need to show two things:\n\n1. \\( a_n = br_1^n + cr_2^n \\) is a solution (or \\( a_n = br^n + cnr^n \\) is a solution in case \\( r_1 = r_2 \\)).\n\n2. Every other solution is of this form.\n\nWe note that since the recurrence relation is linear it is enough to prove that \\( r_i^n = br_i^{n-1} + cr_i^{n-2} \\)\n\n1. \\( br_i^{n-1} + cr_i^{n-2} = r_i^{n-2}(br_i + c) \\)\n\n2. Since \\( r_i \\) are roots of the characteristic equation we have: \\( r_i^2 = br_i + c \\).\n\n3. Substituting we get: \\( br_i^{n-1} + cr_i^{n-2} = r_i^n \\)\nChứng minh.\n\nWe need to show two things:\n\n1. \\( a_n = br_1^n + cr_2^n \\) is a solution (or \\( a_n = br^n + cnr^n \\) is a solution in case \\( r_1 = r_2 \\)).\n\n2. Every other solution is of this form.\n\nWe note that since the recurrence relation is linear it is enough to prove that \\( r_i^n = br_i^{n-1} + cr_i^{n-2} \\)\n\n1. \\( br_i^{n-1} + cr_i^{n-2} = r_i^{n-2}(br_i + c) \\)\n\n2. Since \\( r_i \\) are roots of the characteristic equation we have: \\( r_i^2 = br_i + c \\).\n\n3. Substituting we get: \\( br_i^{n-1} + cr_i^{n-2} = r_i^n \\)\n\n4. Thus \\( \\alpha r_1^n + \\beta r_2^n \\) solves the recurrence relation.\nChứng minh.\n\nWe need to show two things:\n\n1. \\( a_n = br_1^n + cr_2^n \\) is a solution (or \\( a_n = br^n + cnr^n \\) is a solution in case \\( r_1 = r_2 \\)).\n\n2. Every other solution is of this form.\n\nWe note that since the recurrence relation is linear it is enough to prove that \\( r_i^n = br_i^{n-1} + cr_i^{n-2} \\)\n\n1. \\( br_i^{n-1} + cr_i^{n-2} = r_i^{n-2}(br_i + c) \\)\n\n2. Since \\( r_i \\) are roots of the characteristic equation we have: \\( r_i^2 = br_i + c \\).\n\n3. Substituting we get: \\( br_i^{n-1} + cr_i^{n-2} = r_i^n \\)\n\n4. Thus \\( \\alpha r_1^n + \\beta r_2^n \\) solves the recurrence relation.\n\n5. As previously proved, \\( r^n = br^{n-1} + cr^{n-2} \\). Taking the derivative we get: \\( nr^{n-1} = b(n-1)r^{n-2} + c(n-2)r^{n-3} \\) and if we multiply both sides by \\( r \\) we get:\n\n\\[\nnr^n = b(n-1)r^{n-1} + c(n-2)r^{n-2}\n\\]\ncontinued.\n\nIt remains to show that these are the general solutions.\n\nLet $a_0 = m$, $a_1 = k$. We need to show that we can choose $\\alpha$ and $\\beta$ so that\n\n$$\\alpha r_0 + \\beta r_1 = m$$\n\nand\n\n$$\\alpha r_1 + \\beta r_2 = k.$$\n\nThis is a set of two linear equations in two unknowns. Its determinant is $r_1 - r_2 \\neq 0$ hence it has a solution.\n\nIn the second case we have:\n\n$$\\alpha = m$$\n\nand\n\n$$\\alpha + \\beta = k$$\n\nwhich obviously has a solution.\ncontinued.\n\nIt remains to show that these are the general solutions.\n\nLet $a_0 = m$, $a_1 = k$. We need to show that we can choose $\\alpha$ and $\\beta$ so that\n\n$$\\alpha r_0 + \\beta r_1 = m$$\n\nand\n\n$$\\alpha r_1 + \\beta r_2 = k.$$ \n\nThis is a set of two linear equations in two unknowns. Its determinant is $r_1 - r_2 \\neq 0$ hence it has a solution.\n\nIn the second case we have:\n\n$$\\alpha = m$$\n\nand\n\n$$\\alpha + \\beta = k$$\n\nwhich obviously has a solution.\ncontinued.\n\nIt remains to show that these are the general solutions.\n\nIt is enough to show that if for any choice of $a_0, a_1$ there is a solution of these forms for which $a_0, a_1$ will be matched.\n\n1. Let $a_0 = m$, $a_1 = k$. We need to show that we can choose $\\alpha$ and $\\beta$ so that $\\alpha r_1^0 + \\beta r_2^0 = m$ and $\\alpha r_1 + \\beta r_2 = k$. \n\nThis is a set of two linear equations in two unknowns. Its determinant is $r_1 - r_2 \\neq 0$ hence it has a solution.\n\nIn the second case we have: $\\alpha = m$ and $\\alpha + \\beta = k$ which obviously has a solution.\ncontinued.\n\nIt remains to show that these are the general solutions.\n\nIt is enough to show that if for any choice of $a_0, a_1$ there is a solution of these forms for which $a_0, a_1$ will be matched.\n\n1. Let $a_0 = m, a_1 = k$. We need to show that we can choose $\\alpha$ and $\\beta$ so that $\\alpha r_1^0 + \\beta r_2^0 = m$ and $\\alpha r_1 + \\beta r_2 = k$.\n\n2. This is a set of two linear equations in two unknowns. Its determinant is $r_1 - r_2 \\neq 0$ hence it has a solution.\ncontinued.\n\nIt remains to show that these are the general solutions.\n\nIt is enough to show that if for any choice of $a_0, a_1$ there is a solution of these forms for which $a_0, a_1$ will be matched.\n\n1. Let $a_0 = m, a_1 = k$. We need to show that we can choose $\\alpha$ and $\\beta$ so that $\\alpha r_1^0 + \\beta r_2^0 = m$ and $\\alpha r_1 + \\beta r_2 = k$.\n\n2. This is a set of two linear equations in two unknowns. Its determinant is $r_1 - r_2 \\neq 0$ hence it has a solution.\n\n3. In the second case we have: $\\alpha = m$ and $\\alpha + \\beta = k$ which obviously has a solution.\nIt remains to deal with identifying particular solutions. The best approach is an “intelligent” guess.\nIt remains to deal with identifying particular solutions. The best approach is an “intelligent” guess.\n\n- If $f(n)$ is a polynomial, try a polynomial of same degree, or higher.\nIt remains to deal with identifying particular solutions. The best approach is an “intelligent” guess.\n\n- If $f(n)$ is a polynomial, try a polynomial of same degree, or higher.\n- If it is $a^n$ try an exponential function if $a$ is not a root of the characteristic equation.\nIt remains to deal with identifying particular solutions. The best approach is an “intelligent” guess.\n\n- If $f(n)$ is a polynomial, try a polynomial of same degree, or higher.\n- If it is $a^n$ try an exponential function if $a$ is not a root of the characteristic equation.\n- If it is, try $cna^n$. \nIt remains to deal with identifying particular solutions. The best approach is an “intelligent” guess.\n\n- If $f(n)$ is a polynomial, try a polynomial of same degree, or higher.\n- If it is $a^n$ try an exponential function if $a$ is not a root of the characteristic equation.\n- If it is, try $cna^n$.\n- In general, try a function “similar” to $f(n)$. The following examples will demonstrate the general approach.\nExample\n\nSolve: \\( a_n = 3a_{n-1} + 2n \\).\n\nTry: \\( p(n) = c_2n \\).\n\nSubstitute we get: \\( c_2 \\cdot 2n = 3 \\cdot c_2 \\cdot 2n - 1 + 2n \\).\n\nSolution: \\( a_n = k \\cdot 3^n - 2n + 1 \\).\n\nRecurrence Relations and Generating Functions\nExample\n\n1. Solve: \\( a_n = 3a_{n-1} + 2^n \\).\nExample\n\n1. Solve: \\( a_n = 3a_{n-1} + 2^n \\).\n - Try: \\( p(n) = c2^n \\).\nExample\n\n1. Solve: \\( a_n = 3a_{n-1} + 2^n \\).\n - Try: \\( p(n) = c2^n \\).\n - Substitute we get: \\( c \\cdot 2^n = 3 \\cdot c \\cdot 2^{n-1} + 2^n \\)\nExample\n\n1. Solve: \\( a_n = 3a_{n-1} + 2^n \\).\n - Try: \\( p(n) = c2^n \\).\n - Substitute we get: \\( c \\cdot 2^n = 3 \\cdot c \\cdot 2^{n-1} + 2^n \\)\n - Solution: \\( a_n = k \\cdot 3^n - 2^{n+1} \\).\nExample\n\n1. Solve: \\( a_n = 3a_{n-1} + 2^n \\).\n - Try: \\( p(n) = c2^n \\).\n - Substitute we get: \\( c \\cdot 2^n = 3 \\cdot c \\cdot 2^{n-1} + 2^n \\)\n - Solution: \\( a_n = k \\cdot 3^n - 2^{n+1} \\).\n\n2. Solve \\( a_n = 3a_{n-1} + 3^n \\).\nExample\n\n1. Solve: \\( a_n = 3a_{n-1} + 2^n \\).\n - Try: \\( p(n) = c2^n \\).\n - Substitute we get: \\( c \\cdot 2^n = 3 \\cdot c \\cdot 2^{n-1} + 2^n \\)\n - Solution: \\( a_n = k \\cdot 3^n - 2^{n+1} \\).\n\n2. Solve \\( a_n = 3a_{n-1} + 3^n \\).\n - Try \\( cn3^n \\).\nExample\n\n1. Solve: \\( a_n = 3a_{n-1} + 2^n \\).\n - Try: \\( p(n) = c2^n \\).\n - Substitute we get: \\( c \\cdot 2^n = 3 \\cdot c \\cdot 2^{n-1} + 2^n \\)\n - Solution: \\( a_n = k \\cdot 3^n - 2^{n+1} \\).\n\n2. Solve \\( a_n = 3a_{n-1} + 3^n \\).\n - Try \\( cn3^n \\).\n - Substitute: \\( cn3^n = 3c(n - 1)3^{n-1} + 3^n \\).\nExample\n\n1. Solve: \\( a_n = 3a_{n-1} + 2^n \\).\n - Try: \\( p(n) = c2^n \\).\n - Substitute we get: \\( c \\cdot 2^n = 3 \\cdot c \\cdot 2^{n-1} + 2^n \\)\n - Solution: \\( a_n = k \\cdot 3^n - 2^{n+1} \\).\n\n2. Solve \\( a_n = 3a_{n-1} + 3^n \\).\n - Try \\( cn3^n \\).\n - Substitute: \\( cn3^n = 3c(n - 1)3^{n-1} + 3^n \\).\n - Solve for \\( c \\): \\( c = 1 \\)\nExample\n\n1. Solve: \\( a_n = 3a_{n-1} + 2^n \\).\n - Try: \\( p(n) = c2^n \\).\n - Substitute we get: \\( c \\cdot 2^n = 3 \\cdot c \\cdot 2^{n-1} + 2^n \\)\n - Solution: \\( a_n = k \\cdot 3^n - 2^{n+1} \\).\n\n2. Solve \\( a_n = 3a_{n-1} + 3^n \\).\n - Try \\( cn3^n \\).\n - Substitute: \\( cn3^n = 3c(n - 1)3^{n-1} + 3^n \\).\n - Solve for \\( c \\): \\( c = 1 \\)\n - General solution: \\( a_n = \\alpha 3^n + n \\cdot 3^n \\)\nExample\n\n1. Solve: \\( a_n = 3a_{n-1} + 2^n \\).\n - Try: \\( p(n) = c2^n \\).\n - Substitute we get: \\( c \\cdot 2^n = 3 \\cdot c \\cdot 2^{n-1} + 2^n \\)\n - Solution: \\( a_n = k \\cdot 3^n - 2^{n+1} \\).\n\n2. Solve \\( a_n = 3a_{n-1} + 3^n \\).\n - Try \\( cn3^n \\).\n - Substitute: \\( cn3^n = 3c(n - 1)3^{n-1} + 3^n \\).\n - Solve for \\( c \\): \\( c = 1 \\)\n - General solution: \\( a_n = \\alpha 3^n + n \\cdot 3^n \\)\n\n3. Solve: \\( a_n = 2a_{n-1} - a_{n-2} + 2n \\).\nExample\n\n1. Solve: \\( a_n = 3a_{n-1} + 2^n \\).\n - Try: \\( p(n) = c2^n \\).\n - Substitute we get: \\( c \\cdot 2^n = 3 \\cdot c \\cdot 2^{n-1} + 2^n \\).\n - Solution: \\( a_n = k \\cdot 3^n - 2^{n+1} \\).\n\n2. Solve \\( a_n = 3a_{n-1} + 3^n \\).\n - Try \\( cn3^n \\).\n - Substitute: \\( cn3^n = 3c(n - 1)3^{n-1} + 3^n \\).\n - Solve for \\( c \\): \\( c = 1 \\).\n - General solution: \\( a_n = \\alpha 3^n + n \\cdot 3^n \\).\n\n3. Solve: \\( a_n = 2a_{n-1} - a_{n-2} + 2n \\).\n - \\( 2n \\) is a solution of the homogeneous equation, so we try \\( p(n) = cn^2 \\) a polynomial of degree 2.\nExample\n\n1. Solve: \\( a_n = 3a_{n-1} + 2^n \\).\n - Try: \\( p(n) = c2^n \\).\n - Substitute we get: \\( c \\cdot 2^n = 3 \\cdot c \\cdot 2^{n-1} + 2^n \\).\n - Solution: \\( a_n = k \\cdot 3^n - 2^{n+1} \\).\n\n2. Solve \\( a_n = 3a_{n-1} + 3^n \\).\n - Try \\( cn3^n \\).\n - Substitute: \\( cn3^n = 3c(n - 1)3^{n-1} + 3^n \\).\n - Solve for \\( c \\): \\( c = 1 \\).\n - General solution: \\( a_n = \\alpha 3^n + n \\cdot 3^n \\).\n\n3. Solve: \\( a_n = 2a_{n-1} - a_{n-2} + 2n \\).\n - \\( 2n \\) is a solution of the homogeneous equation, so we try \\( p(n) = cn^2 \\) a polynomial of degree 2.\n - Substitute: \\( cn^2 = 2c(n - 1)^2 - c(n - 2)^2 + 2n \\). Does not produce a solution.\nExample\n\n1. Solve: \\( a_n = 3a_{n-1} + 2^n \\).\n - Try: \\( p(n) = c2^n \\).\n - Substitute we get: \\( c \\cdot 2^n = 3 \\cdot c \\cdot 2^{n-1} + 2^n \\).\n - Solution: \\( a_n = k \\cdot 3^n - 2^{n+1} \\).\n\n2. Solve \\( a_n = 3a_{n-1} + 3^n \\).\n - Try \\( cn3^n \\).\n - Substitute: \\( cn3^n = 3c(n - 1)3^{n-1} + 3^n \\).\n - Solve for \\( c \\): \\( c = 1 \\).\n - General solution: \\( a_n = \\alpha 3^n + n \\cdot 3^n \\).\n\n3. Solve: \\( a_n = 2a_{n-1} - a_{n-2} + 2n \\).\n - \\( 2n \\) is a solution of the homogeneous equation, so we try \\( p(n) = cn^2 \\) a polynomial of degree 2.\n - Substitute: \\( cn^2 = 2c(n - 1)^2 - c(n - 2)^2 + 2n \\). Does not produce a solution.\n - So we try a polynomial of degree 3: \\( p(n) = cn^2 + dn^3 \\).\nExample\n\n1. Solve: \\( a_n = 3a_{n-1} + 2^n \\).\n - Try: \\( p(n) = c2^n \\).\n - Substitute we get: \\( c \\cdot 2^n = 3 \\cdot c \\cdot 2^{n-1} + 2^n \\).\n - Solution: \\( a_n = k \\cdot 3^n - 2^{n+1} \\).\n\n2. Solve \\( a_n = 3a_{n-1} + 3^n \\).\n - Try \\( cn3^n \\).\n - Substitute: \\( cn3^n = 3c(n - 1)3^{n-1} + 3^n \\).\n - Solve for \\( c \\): \\( c = 1 \\).\n - General solution: \\( a_n = \\alpha 3^n + n \\cdot 3^n \\).\n\n3. Solve: \\( a_n = 2a_{n-1} - a_{n-2} + 2n \\).\n - \\( 2n \\) is a solution of the homogeneous equation, so we try \\( p(n) = cn^2 \\) a polynomial of degree 2.\n - Substitute: \\( cn^2 = 2c(n - 1)^2 - c(n - 2)^2 + 2n \\). Does not produce a solution.\n - So we try a polynomial of degree 3: \\( p(n) = cn^2 + dn^3 \\).\n - Substitute and solve for \\( c, d \\) we find that \\( \\frac{1}{3}n^3 + n^2 \\) is a particular solution.\nExample\n\n1. Solve: \\( a_n = 3a_{n-1} + 2^n \\).\n - Try: \\( p(n) = c2^n \\).\n - Substitute we get: \\( c \\cdot 2^n = 3 \\cdot c \\cdot 2^{n-1} + 2^n \\).\n - Solution: \\( a_n = k \\cdot 3^n - 2^{n+1} \\).\n\n2. Solve \\( a_n = 3a_{n-1} + 3^n \\).\n - Try \\( cn3^n \\).\n - Substitute: \\( cn3^n = 3c(n - 1)3^{n-1} + 3^n \\).\n - Solve for \\( c \\): \\( c = 1 \\).\n - General solution: \\( a_n = \\alpha 3^n + n \\cdot 3^n \\).\n\n3. Solve: \\( a_n = 2a_{n-1} - a_{n-2} + 2n \\).\n - \\( 2n \\) is a solution of the homogeneous equation, so we try \\( p(n) = cn^2 \\) a polynomial of degree 2.\n - Substitute: \\( cn^2 = 2c(n - 1)^2 - c(n - 2)^2 + 2n \\). Does not produce a solution.\n - So we try a polynomial of degree 3: \\( p(n) = cn^2 + dn^3 \\).\n - Substitute and solve for \\( c, d \\) we find that \\( \\frac{1}{3}n^3 + n^2 \\) is a particular solution.\n - So the general solution is: \\( a_n = \\alpha + \\beta n + n^2 + \\frac{1}{3}n^3 \\).\nWith every sequence $a_n$ we can associate a power series:\n\n$$f(x) = \\sum_{i=0}^{\\infty} a_n x^n$$\n\nand vice versa, every power series expansion of a function $f(x)$ gives rise to a sequence $a_n$. Are there any uses of this relationship in counting?\nWith every sequence $a_n$ we can associate a power series:\n\n$$f(x) = \\sum_{i=0}^{\\infty} a_n x^n$$\n\nand vice versa, every power series expansion of a function $f(x)$ gives rise to a sequence $a_n$. Are there any uses of this relationship in counting?\n\nIn this section we shall explore the interaction among polynomials, power series and counting.\nGenerating Functions\n\nWith every sequence $a_n$ we can associate a power series:\n\n$$f(x) = \\sum_{i=0}^{\\infty} a_n x^n$$\n\nand vice versa, every power series expansion of a function $f(x)$ gives rise to a sequence $a_n$. Are there any uses of this relationship in counting?\n\nIn this section we shall explore the interaction among polynomials, power series and counting.\n\n**Definition**\n\nThe function\n\n$$f(x) = \\sum_{i=0}^{\\infty} a_n x^n$$\n\nis the generating function of the sequence $a_n$. \n\nRecurrence Relations and Generating Functions\nLet us start with an example we visited before: how many different solutions in non-negative integers does the equation $x + y + z + t = 27$ have?\nLet us start with an example we visited before: how many different solutions in non-negative integers does the equation $x + y + z + t = 27$ have?\n\nConsider the function $f(x) = (1 + x + x^2 + \\ldots x^{27})^4$. \n\nRecurrence Relations and Generating Functions\nLet us start with an example we visited before: how many different solutions in non-negative integers does the equation $x + y + z + t = 27$ have?\n\nConsider the function $f(x) = (1 + x + x^2 + \\ldots x^{27})^4$.\n\nIt is not difficult to see that the coefficient of $x^{27}$ is the answer, but how easy is it to calculate it?\nLet us start with an example we visited before: how many different solutions in non-negative integers does the equation $x + y + z + t = 27$ have?\n\nConsider the function $f(x) = (1 + x + x^2 + \\ldots x^{27})^4$.\n\nIt is not difficult to see that the coefficient of $x^{27}$ is the answer, but how easy is it to calculate it?\n\nWell, if you have a nice math program, it will be very easy.\nLet us start with an example we visited before: how many different solutions in non-negative integers does the equation $x + y + z + t = 27$ have?\n\nConsider the function $f(x) = (1 + x + x^2 + \\ldots x^{27})^4$.\n\nIt is not difficult to see that the coefficient of $x^{27}$ is the answer, but how easy is it to calculate it?\n\nWell, if you have a nice math program, it will be very easy.\n\nBut we can do better, Consider the function $g(x) = (\\sum_{i=0}^{\\infty} x^i)^4$. \n\nRecurrence Relations and Generating Functions\nLet us start with an example we visited before: how many different solutions in non-negative integers does the equation $x + y + z + t = 27$ have?\n\nConsider the function $f(x) = (1 + x + x^2 + \\ldots x^{27})^4$.\n\nIt is not difficult to see that the coefficient of $x^{27}$ is the answer, but how easy is it to calculate it?\n\nWell, if you have a nice math program, it will be very easy.\n\nBut we can do better, Consider the function $g(x) = (\\sum_{i=0}^{\\infty} x^i)^4$.\n\nAgain, the coefficient of $x^{27}$ in the Taylor expansion of this function is the answer.\nLet us start with an example we visited before: how many different solutions in non-negative integers does the equation $x + y + z + t = 27$ have?\n\nConsider the function $f(x) = (1 + x + x^2 + \\ldots x^{27})^4$.\n\nIt is not difficult to see that the coefficient of $x^{27}$ is the answer, but how easy is it to calculate it?\n\nWell, if you have a nice math program, it will be very easy.\n\nBut we can do better, Consider the function $g(x) = (\\sum_{i=0}^{\\infty} x^i)^4$.\n\nAgain, the coefficient of $x^{27}$ in the Taylor expansion of this function is the answer.\n\nWe notice that $\\sum_{i=0}^{\\infty} x^i = \\frac{1}{1-x}$.\nLet us start with an example we visited before: how many different solutions in non-negative integers does the equation $x + y + z + t = 27$ have?\n\nConsider the function $f(x) = (1 + x + x^2 + \\ldots x^{27})^4$.\n\nIt is not difficult to see that the coefficient of $x^{27}$ is the answer, but how easy is it to calculate it?\n\nWell, if you have a nice math program, it will be very easy.\n\nBut we can do better, Consider the function $g(x) = (\\sum_{i=0}^{\\infty} x^i)^4$.\n\nAgain, the coefficient of $x^{27}$ in the Taylor expansion of this function is the answer.\n\nWe notice that $\\sum_{i=0}^{\\infty} x^i = \\frac{1}{1-x}$.\n\nSo the answer will be the coefficient of $x^{27}$ in the expansion of $(1 - x)^{-4}$.\nExamples, continued\n\nWould it be nice if we could use an extended Binomial Coefficient and write the answer:\n\n\\[\n\\binom{-4}{27} \\text{ or in general } \\binom{-4}{k}\n\\]\n\nExample\n\nA box contains 30 red, 40 blue and 50 white balls. In how many ways can you select 70 balls?\n\nThe coefficient of \\(x^{70}\\) in the product\n\n\\[\n(1 + x + \\ldots + x^{30})(1 + x + \\ldots + x^{40})(1 + x + \\ldots + x^{50})\n\\]\n\nis the answer.\n\nNote that:\n\n\\[\n(1 + x + \\ldots + x^{30})(1 + x + \\ldots + x^{40})(1 + x + \\ldots + x^{50}) = (1 - x^{31})(1 - x^{41})(1 - x^{51}) = (1 - x)^{-3}\n\\]\nExamples, continued\n\nWould it be nice if we could use an extended Binomial Coefficient and write the answer:\n\n\\[\n\\binom{-4}{27} \\quad \\text{or in general} \\quad \\binom{-4}{k}\n\\]\n\nExample\n\n1. A box contains 30 red, 40 blue and 50 white balls. In how many ways can you select 70 balls?\nExamples, continued\n\nWould it be nice if we could use an extended Binomial Coefficient and write the answer:\n\n\\[\n\\binom{-4}{27} \\quad \\text{or in general} \\quad \\binom{-4}{k}\n\\]\n\nExample\n\n1. A box contains 30 red, 40 blue and 50 white balls. In how many ways can you select 70 balls?\n\n2. The coefficient of \\(x^{70}\\) in the product\n\n\\[\n(1 + x + \\ldots + x^{30})(1 + x + \\ldots + x^{40})(1 + x + \\ldots + x^{50})\n\\]\n\nis the answer.\nExamples, continued\n\nWould it be nice if we could use an extended Binomial Coefficient and write the answer:\n\n\\[\n\\binom{-4}{27} \\quad \\text{or in general} \\quad \\binom{-4}{k}\n\\]\n\nExample\n\n1. A box contains 30 red, 40 blue and 50 white balls. In how many ways can you select 70 balls?\n\n2. The coefficient of \\(x^{70}\\) in the product\n\\[\n(1 + x + \\ldots + x^{30})(1 + x + \\ldots + x^{40})(1 + x + \\ldots + x^{50})\n\\]\n\nis the answer.\n\n3. Note that:\n\\[\n\\frac{1-x^{31}}{1-x} \\cdot \\frac{1-x^{41}}{1-x} \\cdot \\frac{1-x^{51}}{1-x} = (1 - x)^{-3}(1 - x^{31})(1 - x^{41})(1 - x^{51}).\n\\]\nExamples\n\nAll we need is to find the coefficient of $x^{70}$ in:\n\n$$\\left( \\sum_{i=0}^{\\infty} \\binom{-3}{i} x^i \\right) (1 - x^{31} - x^{41} - x^{51} + \\ldots)$$\n\nwhich turns out to be 1061 once we understand the meaning of\n\n$$\\binom{-3}{i}$$\n\nDrill\n\nUse this technique to find the number of distinct solution to:\n\n$$x_1 + x_2 + x_3 + x_4 = 50$$\n\n$$10 \\leq x_1 \\leq 25, \\ 15 \\leq x_2 \\leq 30, \\ 10 \\leq x_3, \\ 15 \\leq x_4 \\leq 25.$$\nThe Generalized Binomial Theorem\n\nTheorem (The generalized binomial theorem)\n\n\\[(1 + x)^r = \\sum_{i=0}^{\\infty} \\binom{r}{i} x^i \\quad \\binom{r}{i} = \\frac{r(r-1)\\ldots(r-i+1)}{i!}\\]\n\nFor negative integers we get:\n\n\\[\\binom{r}{i} = \\frac{r(r-1)\\ldots(r-i+1)}{i!} = (-1)^i \\binom{-r+i-1}{-r-1}\\]\nThe Generalized Binomial Theorem\n\nTheorem (The generalized binomial theorem)\n\n\\[(1 + x)^r = \\sum_{i=0}^{\\infty} \\binom{r}{i} x^i \\quad \\binom{r}{i} = \\frac{r(r-1)\\ldots(r-i+1)}{i!}\\]\n\nChứng minh.\n\nFollows directly from Taylor’s expansion of \\((1 + x)^r\\).\n\nFor negative integers we get:\n\n\\[\\binom{r}{i} = \\frac{r(r-1)\\ldots(r-i+1)}{i!} = (-1)^i \\binom{-r+i-1}{-r-1}\\]\nThe Generalized Binomial Theorem\n\nTheorem (The generalized binomial theorem)\n\n\\[(1 + x)^r = \\sum_{i=0}^{\\infty} \\binom{r}{i} x^i \\quad \\binom{r}{i} = \\frac{r(r-1)\\ldots(r-i+1)}{i!}\\]\n\nChứng minh.\n\nFollows directly from Taylor’s expansion of \\((1 + x)^r\\).\n\nFor negative integers we get:\n\n\\[\\binom{r}{i} = \\frac{r(r-1)\\ldots(r-i+1)}{i!} = (-1)^i \\binom{-r+i-1}{-r-1}\\]\n\nDrill\n\nShow that:\n\n\\[\\binom{\\frac{1}{2}}{k} = \\frac{(-1)^k}{4^k} \\binom{2k}{k}\\]\nQuestion\n\nYou need to calculate the product of $n$ matrices $A_1 \\times A_2 \\times \\ldots \\times A_n$. How do we parenthesize the expression to do it in the most economical way?\nQuestion\n\nYou need to calculate the product of $n$ matrices $A_1 \\times A_2 \\times \\ldots \\times A_n$. How do we parenthesize the expression to do it in the most economical way?\nQuestion\n\nYou need to calculate the product of $n$ matrices $A_1 \\times A_2 \\times \\ldots \\times A_n$. How do we parenthesize the expression to do it in the most economical way?\n\nIn how many ways can you parenthesize the product?\nQuestion\n\nYou need to calculate the product of $n$ matrices $A_1 \\times A_2 \\times \\ldots \\times A_n$. How do we parenthesize the expression to do it in the most economical way?\n\nIn how many ways can you parenthesize the product?\n\nWhy does it matter?\n\nDrill\n\nLet $A[m, n]$ denote an $m \\times n$ matrix ($m$ rows and $n$ columns). For each possible multiplication of the following product calculate the number of multiplications of real numbers needed to calculate the product.\n\n$$A[10, 20]A[20, 40]A[40, 50]A[50, 10]$$\nCatalan Numbers\n\nExample\n\nA \\times B \\times C can be parethesized in two different ways.\n\nA \\times B \\times C \\times D can be parethesized in 5 different ways.\n\nLet \\( m_n \\) be the number of ways to properly parenthesize the product of \\( n+1 \\) matrices.\n\n\\( m_1 = 1, m_2 = 2, m_3 = 5, m_n =? \\) (for convenience, we set \\( m_0 = 0 \\)).\n\n\\[ m_n = \\sum_{i=0}^{n} m_i \\cdot m_{n-i} \\]\nExample\n\n1. $A \\times B \\times C$ can be parenthesized in two different ways.\nExample\n\n1. $A \\times B \\times C$ can be parethesized in two different ways.\n2. $A \\times B \\times C \\times D$ can be parethesized in 5 different ways.\nExample\n\n1. $A \\times B \\times C$ can be parenthesized in two different ways.\n2. $A \\times B \\times C \\times D$ can be parenthesized in 5 different ways.\n3. Let $m_n$ be the number of ways to properly parenthesize the product of $n + 1$ matrices.\nCatalan Numbers\n\nExample\n\n1. $A \\times B \\times C$ can be parenthesized in two different ways.\n2. $A \\times B \\times C \\times D$ can be parenthesized in 5 different ways.\n3. Let $m_n$ be the number of ways to properly parenthesize the product of $n + 1$ matrices.\n4. $m_1 = 1$, $m_2 = 2$, $m_3 = 5$, $m_n = ?$ (for convenience, we set $m_0 = 0$).\nCatalan Numbers\n\nExample\n\n1. $A \\times B \\times C$ can be parenthesized in two different ways.\n2. $A \\times B \\times C \\times D$ can be parenthesized in 5 different ways.\n3. Let $m_n$ be the number of ways to properly parenthesize the product of $n + 1$ matrices.\n4. $m_1 = 1$, $m_2 = 2$, $m_3 = 5$, $m_n = ?$ (for convenience, we set $m_0 = 0$).\n5. \n\n$$m_n = \\sum_{i=0}^{n} m_i \\cdot m_{n-i}$$\n\nRecurrence Relations and Generating Functions\nThe generating function of the sequence $m_n$ is:\n\n$$A(x) = \\sum_{i=0}^{\\infty} m_i x^i$$\nThe generating function of the sequence $m_n$ is:\n\n$$A(x) = \\sum_{i=0}^{\\infty} m_i x^i$$\n\n$$A^2(x) = \\sum_{k=0}^{\\infty} b_k x^k \\quad b_k = \\sum_{j=0}^{k} m_j \\cdot m_{k-j}$$\nThe generating function of the sequence $m_n$ is:\n\n$$A(x) = \\sum_{i=0}^{\\infty} m_i x^i$$\n\n$$A^2(x) = \\sum_{k=0}^{\\infty} b_k x^k \\quad b_k = \\sum_{j=0}^{k} m_j \\cdot m_{k-j}$$\n\nFor $n = 0, 1 \\sum_{i=0}^{n} m_i \\cdot m_{n-i} = 0$. Since $m_1 = 1$ this means that:\n\n$$A^2(x) = \\sum_{i=0}^{\\infty} b_i x^i = \\sum_{i=0}^{\\infty} m_i x^i - x = A(x) - x$$\nCatalan Numbers\n\nOr: \\( A(x) = \\frac{1}{2x} (1 \\pm \\sqrt{1 - 4x}) \\).\nCatalan Numbers\n\nOr: \\( A(x) = \\frac{1}{2x} (1 \\pm \\sqrt{1 - 4x}) \\).\n\nSubstituting the initial condition \\( m_0 = A(0) = 0 \\) we get:\n\n\\[\nA(x) = \\frac{1}{2x} (1 - \\sqrt{1 - 4x})\n\\]\n\n\\[\n(1 - 4x)^{\\frac{1}{2}} = \\sum_{k=0}^{\\infty} \\binom{1/2}{k} (-4)^k x^k = \\sum_{k=0}^{\\infty} \\binom{2k}{k} x^k\n\\]\n\n\\[\n\\left( \\text{Using: } \\binom{1/2}{k} = (-1/4)^k \\binom{2k}{k} \\right).\n\\]\nCatalan Numbers\n\nOr: \\( A(x) = \\frac{1}{2x} (1 \\pm \\sqrt{1 - 4x}) \\).\n\nSubstituting the initial condition \\( m_0 = A(0) = 0 \\) we get:\n\n\\[\nA(x) = \\frac{1}{2x} (1 - \\sqrt{1 - 4x})\n\\]\n\n\\[\n(1 - 4x)^{\\frac{1}{2}} = \\sum_{k=0}^{\\infty} \\binom{1/2}{k} (-4)^k x^k = \\sum_{k=0}^{\\infty} \\binom{2k}{k} x^k\n\\]\n\n\\[\n\\left( \\text{Using : } \\binom{1/2}{k} = (-1/4)^k \\binom{2k}{k} \\right).\n\\]\n\n\\( m_n \\) is the coefficient of \\( x^n \\) in the expansion of:\n\n\\[\n(1 - \\sqrt{1 - 4x})/(1/2x)\n\\]\n\nA simple calculation yields:\n\n\\[\nm_n = \\frac{1}{n+1} \\binom{2n}{n}\n\\]\nThese are the **Catalan Numbers**. They count many other objects, for instance the number of binary trees, the number of grid paths from \\((0, 0)\\) to \\((0, 2n)\\) that stay above the \\(x\\)-axis, the number of binary sequences of length \\(2n\\) with \\(n\\) 1’s such that when scanning from left to right the number of 1’s is never less than the number of 0’s and more.\nThese are the **Catalan Numbers**. They count many other objects, for instance the number of binary trees, the number of grid paths from \\((0, 0)\\) to \\((0, 2n)\\) that stay above the \\(x\\)-axis, the number of binary sequences of length \\(2n\\) with \\(n\\) 1’s such that when scanning from left to right the number of 1’s is never less than the number of 0’s and more.\n\nIn most of these cases, we show that these sequences satisfy the same recurrence relation and initial conditions.\nThese are the **Catalan Numbers**. They count many other objects, for instance the number of binary trees, the number of grid paths from \\((0, 0)\\) to \\((0, 2n)\\) that stay above the \\(x\\)-axis, the number of binary sequences of length \\(2n\\) with \\(n\\) 1’s such that when scanning from left to right the number of 1’s is never less than the number of 0’s and more.\n\nIn most of these cases, we show that these sequences satisfy the same recurrence relation and initial conditions.\n\nRecurrence relations are a powerful tool for solving many problems. There are many types of generating function, we only scratched the surface of this beautiful theory.\n\nSome more challenging problems will be posted in our assignments folder.", "id": "./materials/360.pdf" }, { "contents": "Can a binary relation be both symmetric and anti-symmetric?\n\nRoshan Poudel\n\nInstituto Politécnico de Bragança, Bragança, Portugal\nBinary Relations\n\nCartesian Product\n\nFor any two sets $X$ and $Y$, the Cartesian product of $X$ by $Y$ is defined as:\n\n$$X \\times Y = \\{(a, b) : a \\in X \\land b \\in Y\\}$$\n\nBinary Relations\n\n- If $X$ and $Y$ are two sets, then a binary relation from $X$ to $Y$ is a subset of $X \\times Y$.\n- A subset of $X \\times X$ is called a binary relation in $X$.\n- The empty set ($\\emptyset$) and the cartesian product $X \\times Y$ are binary relations from $X$ to $Y$.\n- If $R$ is a binary relation from $X$ to $Y$ and $a \\in X$, $b \\in Y$, we write $(a, b) \\in R$ or $aRb$. \nReflexive Relations\n\n**Definition**\n\nA binary relation $R$ defined in set $X$ is reflexive if it relates every element of $X$ to itself.\n\n$$R \\text{ is reflexive iff } \\forall a \\in X \\implies aRa$$\n\n**Example**\n\nFor a set $A = \\{1, 2, 3\\}$, the relation $R$ on $A$ defined as $R = \\{(1, 1), (1, 2), (1, 3), (2, 2), (3, 3)\\}$, is reflexive because $(1, 1), (2, 2), (3, 3)$ are in the relation.\nTransitive Relations\n\n**Definition**\n\nA binary relation $R$ defined in a set $X$ such that for all $a$, $b$ and $c$ in $X$, if $aRb$ and $bRc$ then $aRc$, is said to be transitive.\n\n$$R \\text{ is transitive iff } \\forall a, b, c \\in X, aRb \\land bRc \\implies aRc$$\n\n**Example**\n\nFor a set $A = \\{1, 2, 3\\}$, $R = \\{(1, 1), (1, 2), (2, 3), (1, 3), (3, 3)\\}$ is transitive because:\n\n- For every $a$, $b$, $c$, $aRb$ and $bRc$ implies $aRc$. Actually, $(1, 2)$ and $(2, 3)$ are in $R$ and so is $(1, 3)$, $(1, 1)$ and $(1, 2)$ are in $R$ and so is $(1, 2)$, $(1, 1)$ and $(1, 3)$ are in $R$ and so is $(1, 3)$, $(2, 3)$ and $(3, 3)$ are in $R$ and so is $(2, 3)$, $(1, 3)$ and $(3, 3)$ are in $R$ and so does $(1, 3)$.\n\n**Note:** If only $aRb$ exists without $bRc$ then it is not necessary\nSymmetric Relations\n\n**Definition**\n\nA binary relation $R$ defined in a set $X$ is said to be symmetric in $X$ if and only if for any $a$ and $b$ in $X$, $aRb$ implies $bRa$.\n\n$$R \\text{ is Symmetric iff } \\forall a, b \\in X, aRb \\implies bRa$$\n\n**Example**\n\nFor a set $A = \\{1, 2, 3\\}$, relation $R = \\{(1, 2), (2, 1), (2, 3), (3, 2), (3, 3)\\}$ is symmetric because:\n\n- For every $aRb$ there exists $bRa$. Actually, $(1, 2)$ and $(2, 1)$ both exist in $R$, $(2, 3)$ and $(3, 2)$ both exist in $R$.\n- For $(3, 3)$ the symmetric is also $(3, 3) \\in R$. \nA binary relation $R$ defined in a set $X$ is said to be anti-symmetric in $X$ if and only if for any $a$ and $b$ in $X$, $aRb$, $bRa$ implies $a = b$.\n\n$$R \\text{ is anti-symmetric iff } \\forall a, b \\in X, aRb \\land bRa \\implies a = b$$\n\nIf only $aRb$ exist and $bRa$ does not, then it is not necessary for $a = b$ for the relation $R$ to be anti-symmetric.\n\nNote: anti-symmetric doesn’t mean not symmetric.\nExample\n\nFor a set $A = \\{1, 2, 3\\}$, relation $R = \\{(1, 1), (2, 1), (1, 3), (3, 3)\\}$ is anti-symmetric because:\n\n- $(1, 1)$ and $(3, 3)$ both fit in the condition if $aRb$ and $bRa$ then $a = b$.\n- Furthermore, $(2, 1)$ and $(1, 3)$, their symmetric ones doesn’t exist in $R$ so they do not need to be equal for $R$ to be symmetric.\nA binary relation that is reflexive, symmetric and transitive is called an equivalence relation.\n\nThe equivalence class of an element $a$ of $X$ is the set of the elements of $X$ that relate to $a$:\n\n$$[a]_R = \\{ x \\in A : xRa \\}$$\n\nElement $a$ is said to represent such class.\nExample\n\nLet us consider a set $A = \\{a, b, c\\}$. Is\n\n$\\{R = (a, a), (b, b), (c, c), (a, c), (c, a)\\}$ an equivalence relation in $A$?\n\n- Since $(a, a), (b, b)$ and $(c, c)$ are all in $R$, $R$ is reflexive.\n- For all the pairs in $R$, the symmetric pair is also in $R$. For example $(a, c)$ has $(c, a)$, and the same happens for the other pairs of $R$. So, $R$ is also symmetric.\n- If $aRb$ and $bRc$ there is also $aRc$. For example, there is $aRa$ and $aRc$ and there is also $aRc$. This applies for all other possible combinations of pairs so, $R$ is also transitive.\n\nAs $R$ is reflexive, symmetric and transitive, then $R$ is an equivalence Relation.\nExample - Equivalent Classes\n\nIn the relation $R$ above, what are the equivalent classes of $[a]$, $[b]$ and $[c]$?\n\n1. In $R$, $a$ is related with $a$ and $c$, so, $[a] = \\{a, c\\}$\n2. In $R$, $b$ is related with $b$ only, so $[b] = \\{b\\}$\n3. In $R$, $c$ is related with $a$ and $c$, so, $[c] = \\{a, c\\}$.\n\nTherefore, the set of all equivalence classes for the equivalence relation $R$ is $\\{\\{a, c\\}, \\{b\\}\\}$. \nPartial Order\n\nDefinition\nA binary relation that is reflexive, anti-symmetric and transitive is called a partial order.\n\nExample\nIs a relation \\( R = \\{(1, 2), (1, 1), (2, 2), (2, 3), (1, 3), (3, 3)\\} \\) in \\( X = \\{1, 2, 3\\} \\) a partial order?\n\n- \\( R \\) is reflexive as \\((1, 1), (2, 2)\\) and \\((3, 3)\\) all belong in \\( R \\).\n- The only pairs whose symmetric also exists in \\( R \\) are \\((1, 1), (2, 2), (3, 3)\\). so, here for all \\( aRb \\) and \\( bRa \\) then \\( a = b \\). so, \\( R \\) is anti-symmetric\n- If for all \\( a, b, c \\in X \\), \\( aRb \\) and \\( bRc \\), there is also \\( aRc \\). Like there is \\( aRa \\) and \\( aRc \\) then there is also \\( aRc \\), so \\( R \\) is also transitive.\n\n\\( R \\) is reflexive, anti-symmetric and transitive. so, \\( R \\) is a partial order.\nYes, a relation can be both symmetric and anti-symmetric at the same time. Or it can be neither as well.\n\n**Explanation**\n\nLet us consider a set $A = \\{1, 2, 3\\}$ and relation $R = \\{(1, 1), (2, 2), (3, 3)\\}$ in $A$. Let’s see if $R$ can be both symmetric and anti-symmetric:\n\n- For $(1, 1)$, the symmetric pair is also $(1, 1)$. The same happens for all the pairs $(x, x)$ in $R$, so the relation $R$ is symmetric.\n\n- Since the elements of $R$ are pairs of the type $(x, x)$, they satisfy the requirement ‘if $(a, b) \\in R$ and $(b, a) \\in R$ then $a = b$’ which is the condition required for anti-symmetry, so $R$ is also anti-symmetric.", "id": "./materials/361.pdf" }, { "contents": "If $|A| = 25$, $|B| = 20$, $|A \\cap B| = 10$ and $|\\mathbb{U}| = 40$. Then, Find $|A - B|$ and $|B - A|$.\n\n- $|A| = 25$, $|B| = 20$, $|\\mathbb{U}| = 40$ and $|A \\cap B| = 10$\n\n- We know, $|\\mathbb{U}| = |A| + |B| + |A \\cap B| - |A \\cup B|$\n\nSo, filling in all known values\n\n- $40 = 20 + 25 + 10 - |A \\cup B|$\n\n- so, $|A \\cup B| = 55 - 25 = 15$\n\nNow finding $|A - B|$\n\n- $|A| = |A - B| + |A \\cup B|$\n\n- $|A - B| = |A| - |A \\cup B|$\n\n- $|A - B| = 25 - 15 = 10$\n\nsimilarly,\n\n- $|B| = |B - A| + |A \\cup B|$\n\n- $|B - A| = 20 - 15 = 5$", "id": "./materials/362.pdf" }, { "contents": "What is $A \\cup \\emptyset$ and $A \\cap \\emptyset$?\n\nRoshan Poudel\n\nInstituto Politécnico de Bragança, Bragança, Portugal\nUnion\n\n1. The union of two sets is the set containing all of the elements from both of those sets. It is represented by the symbol $\\cup$.\n\n2. $A \\cup B = \\{x \\mid x \\in A \\text{ and } x \\in B\\}$\n\nIntersection\n\n1. The intersection of two sets is the set containing just the elements that are in both of those sets. It is represented by the symbol $\\cap$.\n\n2. $A \\cap B = \\{x \\mid x \\in A \\text{ or } x \\in B\\}$\nUnion and Intersection - Venn Diagram\n\nFigure 1: $A \\cup B$\n\nFigure 2: $A \\cap B$\nIf $A = \\{1, 3, 5, 7, 9, 11, 13, 15\\}$ and $B = \\{2, 3, 5, 7, 11, 13\\}$, then:\n\n- $A \\cup B = \\{1, 2, 3, 5, 7, 9, 11, 13, 15\\}$\n- $A \\cap B = \\{3, 5, 7, 11, 13\\}$\nWhat is $A \\cup \\emptyset$ and $A \\cap \\emptyset$?\n\n**Finding $A \\cup \\emptyset$.**\n\nThe empty set is the set with no elements so, the union of any set $A$ and the $\\emptyset$ is always going to be $A$.\n\n$$A \\cup \\emptyset = A$$\n\n**Finding $A \\cap \\emptyset$.**\n\nAn empty set is a set with no elements so, the intersection of any set $A$ and $\\emptyset$ is always going to be $\\emptyset$ as there is no element simultaneously belonging to both the sets.\n\n$$A \\cap \\emptyset = \\emptyset$$\nOther Set Operations\n\n**Difference**\n\n1. The difference of any two sets $A$ and $B$ written as $A - B$ which is the set containing the elements that are in $A$ but not in $B$.\n2. $A - B = \\{x \\mid x \\in A \\text{ and } x \\notin B\\}$\n3. For two disjoint sets $A$ and $B$, $A - B = A$ and $B - A = B$.\n\n**Compliment**\n\n1. For a set $A$ in a universe $U$, the compliment of $A$ or $\\overline{A}$ is set of all the elements that are in the universe but not in $A$.\n2. $\\overline{A} = \\{x \\in U \\mid x \\notin A\\}$\nFigure 3: $A - B$\n\nFigure 4: $\\overline{A}$\nConsider $A = \\{a, e, i, o, u\\}$ and $B = \\{a, b, c, d, e\\}$, and find $A - B$ and $B - A$.\n\n- $A - B = \\{i, o, u\\}$\n- $B - A = \\{b, c, d\\}$\n\nNotice that $A - B \\neq B - A$\nIf the universe $U$ is the set of letters in the English alphabet and $A$ is the set of the consonant letters of the same alphabet, what is $\\overline{A}$?\n\n- $\\overline{A} = \\{a, e, i, o, u\\}$\n\n- Also, $\\overline{U} = \\emptyset$ and $\\overline{\\emptyset} = U$\n## Properties of Set Operations\n\n### Commutative\n\n1. \\( A \\cup B = B \\cup A \\)\n2. \\( A \\cap B = B \\cap A \\)\n\n### Associative\n\n1. \\( A \\cup (B \\cup C) = (A \\cup B) \\cup C \\)\n2. \\( A \\cap (B \\cap C) = (A \\cap B) \\cap C \\)\n\n### Idempotent\n\n1. \\( A \\cup A = A \\)\n2. \\( A \\cap A = A \\)\nDe Morgan’s Laws\n\nDe Morgan’s Law\n\n1. \\((A \\cup B) = \\overline{A} \\cap \\overline{B}\\)\n\n2. \\((A \\cap B) = \\overline{A} \\cup \\overline{B}\\)\n\nClick here to check the proof here", "id": "./materials/363.pdf" }, { "contents": "Let $U = \\{1, 2, 3, 4, \\ldots, 20\\}$,\n$A = \\{x: 5 < x \\leq 10\\}$,\n$B = \\{x: 8 \\leq x \\leq 15\\}$\n$C = \\{x: 1 \\leq x \\leq 5\\}$.\nFind $(A \\cap B \\cap C) \\cup C$.\n\n- $A = \\{6, 7, 8, 9, 10\\}$\n- $B = \\{8, 9, 10, 11, 12, 13, 14, 15\\}$\n- $C = \\{1, 2, 3, 4, 5\\}$\n\nNow finding $(A \\cap B \\cap C)$\n\n- $(A \\cap B \\cap C) = \\emptyset$ as there are no common elements in $A$, $B$ and $C$\n\nSo, calculating $(A \\cap B \\cap C) \\cup C$.\n\n- $(A \\cap B \\cap C) \\cup C = \\emptyset \\cup C = C$\n\n- so, $(A \\cap B \\cap C) \\cup C = \\{1, 2, 3, 4, 5\\}$", "id": "./materials/364.pdf" } ]