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\newcommand{\otto}{\longleftrightarrow} \newcommand{\hasnor}{\vartriangleright} \newcommand{\bl}{\biggl} \newcommand{\br}{\biggr} \newcommand{\fr}{\dfrac} \DeclareMathOperator{\floor}{floor} \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\tg}{tg} \DeclareMathOperator{\cm}{cm} \undef\sin \DeclareMathOperator{\sin}{sen} \DeclareMathOperator{\ctg}{ctg} \DeclareMathOperator{\imp}{\implies} \DeclareMathOperator{\inv}{inv} \newcommand{\dx}{\ dx} \newcommand{\du}{\ du} \newcommand{\dy}{\ dy} \newcommand{\dt}{\ dt} \newcommand{\ds}{\ ds} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\arcsinh}{arcsinh} \DeclareMathOperator{\arccosh}{arccosh} \newcommand{\fdiff}{\fr{dy}{dx}} \newcommand{\sdiff}{\fr{d^2 y}{dx^2}} % lattice tools \newcommand{\edge}[1]{\ar@{-}[#1]} \newcommand{\node}{*+[o][F-]{ }} \usepackage{graphicx} \usepackage{caption} \usepackage{lipsum} % title and dudes \author{Charlie Root \& wife} \title{Notes on Calculus} \date{Jan 10, 2006} \begin{document} \fontfamily{cmr}\selectfont \maketitle %% \setlength{\parskip}{7pt} %% \setlength{\parindent}{0pt} \section{Introdução} O habilidoso uso de ferramentas é condição {\em sine qua non} pra análise matemática. Começamos por ferramentas que certamente usaremos na empreitada. \subsection{O sistema Hindu-Arábico de numerais} \newcommand{\al}{al-Khow\={a}rism\={\i}} Os numerais que usamos são um fantástico sistema de raciocínio --- provavelmente a maior contribuição matemática de todos os tempos. O sistema foi disseminado por Ab\={u}'Abd All\={a} Muhammad ibn M\={u}s\={a} al-Khow\={a}rism\={\i}'', nome que literalmente significa pai de Abdullah, Mohammed, filho de Moisés, nativo de Khow\={a}rizm. Os termos ``algarismo'' e ``algoritmo'' vêm do nome dele, \al. Por volta do ano 825, al-Khow\={a}rism\={\i} escreveu o livro {\em Kitab al jabr wa'l-muq\={a}bala}, que essencialmente significa ``regras para recuperar e equacionar''~\cite[capítulo~1, seção~1.1, página~1]{knuthv1}. O livro explica como usar os numerais. Mais tarde foi traduzido e apareceu em latim sob o título ``Liber Algorismi'', que significa ``o livro de al-Khow\={a}rism\={\i}''~\cite[página~21]{eti}. Khow\={a}rism\={\i} hoje é a cidade de Khiva no Uzbequistão. O termo {\em al} é o equivalente ao artigo ``o'' da língua portuguesa. Então o nome ``al-Khow\={a}rism\={\i}'' significa, efetivamente, ``o (livro) daquele que vem de Khiva''. A essência do sistema é usar os símbolos 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 na expressão de um número natural. Por exemplo, o número quatro mil trezentos e vinte um pode ser expresso pela soma $4000 + 300 + 20 + 1$, o que podemos (também) abreviar por $4321$. Observemos então que os naturais têm um nome completo e um nome abreviado. O nome completo de quatro mil trezentos e vinte um é % \[1\cdot10^0 + 2\cdot10^1 + 3\cdot 10^2 + 4\cdot 10^3 = 4321.\] % Quando usamos o nome completo, não precisamos nos preocupar com qualquer ambiguidade de leitura --- não precisamos informar ao leitor que o algarismo menos significativo encontra-se na direita. Já quando usamos o nome abreviado, precisamos informar o fato --- em ``4321'', o algarismo 4 representa o número quatro mil. O nome abreviado exige do leitor que conte a quantidade de algarismos usados pelo número para compreender seu algarismo mais à esquerda --- o mais significativo. Já o nome completo pode ser escrito de várias formas. Por exemplo, quatro mil trezentos e vinte um também pode ser escrito tanto por % \[4\cdot 10^3 + 3\cdot 10^2 + 2\cdot10^1 + 1\cdot10^0\] % quanto por % \[2\cdot10^1 + 1\cdot10^0 + 4\cdot 10^3 + 3\cdot 10^2,\] % dentre outras expressões. Agora fica evidente a suprema importância do algarismo zero. Ele serve pra demarcar uma potência de dez, isto é, uma ``casa decimal''. Se o número que desejamos escrever não é composto de qualquer centena, mas é composto de alguns milhares, então escrevemos ``0'' quando desejamos informar a quantidade de centenas e, assim, o algarismo que escrevermos à esquerda do zero expressa a quantidade de milhares. Com essas ideias, \al\ nos ensina a raciocinar sobre números enormes por via de considerar apenas cada uma de suas partes, ou seja, cada um de seus algarismos. Podemos raciocionar sobre trilhões como se algumas unidades fossem. A aritmética de \al, portanto, é provavelmente a mais enfática ilustração científica de modularização e análise. (``Enfática'' porque as aplicações são absolutamente pervasivas.) Por exemplo, pra somar dois números, fazemos a soma em cada algarismo correspondente, ou seja, unidade com unidade, dezena com dezena, centena com centena e assim sucessivamente, como ilustra a Figura~\ref{fg-alkw-sum} (na página~\pageref{fg-alkw-sum}). \begin{figure}[H] \centering \begin{verbatim} 1234 + 4321 ---- 5555 \end{verbatim} \caption{Como computar uma soma segundo \al. Com quatro algarismos, operamos quatro somas, algarismo a algarismo.} \label{fg-alkw-sum} \end{figure} Subtração é um caso levemente mais complicado. Subtrair significa computar a distância entre dois números. Se desejamos 49 menos 17, temos um caso fácil porque o algarismo 7 é menor que o algarismo 9 e o algarismo 1 é menor que o algarismo 4. Assim obtemos a distância 2 entre 7 e 9 bem como a distância 3 entre 1 e 4, logo $49 - 17 = 32$. O caso mais complicado é quando o subtraendo não é menor que o minuendo, que é o que ocorre em $37 - 19$, ilustrado pela Figura~\ref{fg-alkw-diff}. \begin{figure}[H] \centering \begin{verbatim} 37 20 + 17 - 19 equivalente a - 10 - 9 --- -- -- 18 10 + 8 \end{verbatim} \caption{Subtração segundo \al. Se o algarismo do subtraendo é maior que o algarismo do minuendo, use o sistema de \al\ e transfira um potência de dez para simplificar o trabalho. Basta a transferência de uma única potência de dez. A subtração desenhada à esquerda desta figura é equivalente às duas subtrações e soma desenhadas na direita. O número 37 é substituído por 20 + 17, sendo que subtraímos 9 de 17 e depois 10 de 20. Por fim, somamos as subtrações, obtendo $10 + 8 = 18$.} \label{fg-alkw-diff} \end{figure} Como é frequente no trabalho matemático, a subtração é reescrita de forma que o trabalho seja facilitado no sentido de que cada tarefa preliminar seja mais simples que o problema original, embora sob a pena de que o problema original será resolvido com um maior número de passos. Vejamos agora a multiplicação segundo \al. Para multiplicar 322 por 27, usaremos a distribuição da multiplicação sobre uma soma. Em outras palavras, como $322\times27 = 322 \times 20 + 322 \times 7$, computaremos cada problema em separado e somaremos o resultado, como ilustra a Figura~\ref{fg-alkw-prod}. Os algarismos {\tt 644} representam o número $6440 = 322 \times 20$, mas tipicamente omiti-se o zero já que $322 \times 20 = 322\times 2 \times 10$ e multiplicação por dez é trivial. \begin{figure}[H] \centering \begin{verbatim} 322 x 27 --- 2254 + 644 ------ 8694 \end{verbatim} \caption{Como computar um produto segundo \al. Dado que 322 vezes 27 é o mesmo que 322 vezes 7 mais 322 vezes 2 vezes 10, podemos modularizar a computação de um produto por via de uma série de somas.} \label{fg-alkw-prod} \end{figure} Como exercício, considere computar a multiplicação entre 123456789 e 987654321. É um trabalho perfeitamente executável com a ajuda de \al, apesar de trabalhoso. A Figura~\ref{fg-alkw-long-prod} ilustra o que acontece. Os fatores têm nove algarismos cada um, o que nos gera uma soma envolvendo nove termos, cada um deles com nove algarismos --- sugerindo um crescimento quadrático do volume de trabalho da multiplicação convencional, isto é, da multiplicação de \al. \begin{figure}[H] \centering \begin{verbatim} (*) A multiplicação de 123456789 por 987654321 123456789 x 987654321 --------- 123456789 246913578 370370367 + 493827156 617283945 740740734 864197523 1987654312 1111111101 001221433341222100 transferências ----------------- 121932631112635269 \end{verbatim} \caption{A multiplicação entre 123456789 por 987654321 segundo \al. Dedicamos esta conta a Leonardo Pisano. Obrigado, Fibonacci!} \label{fg-alkw-long-prod} \end{figure} Pra tornar tolerável o trabalho ilustrado pela Figura~\ref{fg-alkw-long-prod}, é preciso auxiliar a memória humana. Usamos uma linha de transferências onde anotamos as transferências entre casas decimais. Por exemplo, na segunda casa decimal temos a soma de 8 + 8 = 16; anotamos o algarismo 6 em seu lugar no produto final e escrevemos o algarismo 1 na linha das transferências, indicando assim que a soma da casa decimal anterior resultou-se em 16. Em outras palavras, toda a informação necessária pra se continuar o trabalho está no papel. Sendo assim, podemos trabalhar com números de quaisquer tamanhos porque podemos interromper a operação depois de um dia de trabalho e continuar no dia seguinte depois de um merecido descanso. A divisão entre inteiros segundo \al\ é reduzida a um problema de multiplicação e subtração. A Figura~\ref{fg-alkw-long-div} (na página \pageref{fg-alkw-long-div}) ilustra o algoritmo chamado divisão longa. Similar à estratégia exibida pela multiplicação longa da Figura~\ref{fg-alkw-long-prod}, queremos uma estratégia que exponha no papel toda a informação necessária para a execução da operação em múltiplas etapas, já que os números não são minúsculos. Os símbolos {\tt x} que marcamos no topo do dividendo representam os algarismos do dividendo que já operamos na divisão. Por exemplo, se estivéssemos no momento que em calculamos apenas o primeiro algarismo do quociente --- que é 1 ---, teríamos marcado um {\tt x} só nos algarismos {\tt 123456}, indicando que ainda consideraremos os algarismos {\tt 789123456789} ao longo da computação do algoritmo. Por que se chama divisão longa? Pra ser constrastada com divisão curta. Na divisão longa, escrevemos a multiplicação parcial do quociente pelo divisor logo abaixo do dividendo parcial; em seguida, efetuamos a subtração. Na divisão curta, omitimos a subtração tal que escrevemos abaixo do dividendo apenas o resultado da subtração, tornando a verificação do trabalho mais difícil. Também exibida pela Figura~\ref{fg-alkw-long-div} (na página \pageref{fg-alkw-long-div}), desenhamos uma seta indicando o produto do algarismo corrente do quociente (corrente durante a execução do algoritmo) que deu origem ao minuendo. Assim, o quociente escrito abaixo do divisor poderia ser omitido da Figura~\ref{fg-alkw-long-div}. \begin{figure}[H] \centering \begin{verbatim} (*) A divisão de 123456789123456789 por 123123 xxxxxxxxxxxxxxxxxx 123456789123456789 / 123123 -123123 --> 1 1002711021689 ------ 3337 -0 --> 0 ---- 33378 -0 --> 0 ----- 333789 -246246 --> 2 ------- 875431 -861861 --> 7 ------- 135702 -123123 --> 1 ------- 125793 -123123 --> 1 ------- 26704 -0 --> 0 ----- 267045 -246246 --> 2 ------- 207996 -123123 --> 1 ------- 848737 -738738 --> 6 ------- 1099998 -984984 --> 8 -------- 1150149 -1108107 --> 9 -------- 42042 --> resto \end{verbatim} \caption{A multiplicação entre 123456789 por 987654321 segundo \al. O quociente é 1002711021689 e o resto, 42042. Se tivéssemos interromido o algoritmo em qualquer momento, teríamos no papel tudo que precisamos pra continuá-lo. Por falta de espaço, exibimos a tabuada de 123123 na Figura~\ref{fg-alkw-tabuada-123123}, que deve ser pré-computada pra que evitemos a busca na escolha de algarismos do quociente.} \label{fg-alkw-long-div} \end{figure} \begin{figure}[!htb] \centering \begin{verbatim} (*) A tabela de multiplicação por 123123 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | |--------+--------+--------+--------+--------+--------+--------+--------+---------| | 123123 | 246246 | 369369 | 492492 | 615615 | 738738 | 861861 | 984984 | 1108107 | \end{verbatim} \caption{Uma base de dados útil à divisão por 123123. Sem ela, precisaríamos buscar cada algarismo do quociente a cada etapa, correndo o risco de fazer escolhas incorretas, acarretando em mais perda de tempo e momento.} \label{fg-alkw-tabuada-123123} \end{figure} \begin{figure}[!htb] \centering \begin{verbatim} (*) A verificação do quociente de 123456789123456789 / 123123 1002711021689 x 123123 ------------- 3008133065067 2005422043378 + 1002711021689 3008133065067 2005422043378 1002711021689 000110111222321100 transferências ------------------ 123456789123414747 + 42042 resto ------------------ 123456789123456789 \end{verbatim} \caption{A verificação da divisão de 123456789123456789 por 123123.} \end{figure} Estudemos agora a divisão entre 322 e 37. A tabuada (Figura~\ref{fg-alkw-322-37} na página \pageref{fg-alkw-322-37}) revela imediatamente que o primeiro algarismo do quociente é 8. O resto de $26 < 37$ revela que terminamos. Desejando a parte fracionada da expansão decimal da divisão, continuamos. O zero que justamos ao 26 representa o fato de que multiplicamos o resto 26 por 10. Por causa disso, precisaremos corrigir o quociente por via de uma divisão por 10 --- o que é feito imediatamente com a adição do ponto decimal. Encontramos o algarismo 7 e como resto obtemos $1 < 37$. Justapomos então um zero ao algarismo 1, obtendo o dividendo $10 < 37$, revelando o algarismo zero da parte fracionada. Adicionamos então novo zero, produzindo o algarismo 2 na parte fracionada. Computando-se o resto, obtemos 26 novamente, revelando a dízima periódica. O termo ``dízima'' significa a décima parte de alguma coisa, mas no contexto em que estamos significa uma quantidade menor que 1. O termo ``periódica'' é evidenciado pelo fato de que temos um ciclo, o que representamos pelos colchetes --- ou seja, $322 \div 37 = 8.702[702] = 8.[702]$. A justificativa para esse sistema de computação em papel é % \begin{align*} \frac{322}{37} &= 8 + \frac{26}{37}\\ &= 8 + \frac{260}{37} \cdot\frac{1}{10}\\ &= 8 + \left(7 + \frac{1}{37}\right) \cdot\frac{1}{10}\\ &= 8 + 7\cdot\frac{1}{10} + \frac{1}{37} \cdot\frac{1}{10}\\ &= 8 + 7\cdot\frac{1}{10} + \left(\frac{100}{37} \cdot\frac{1}{100}\right) \cdot\frac{1}{10}\\ &= 8 + 0.7 + (2 \cdot \frac{1}{100} + \frac{26}{37} \cdot\frac{1}{100}) \cdot\frac{1}{10}\\ &= 8 + 0.7 + 0.002 + \frac{26}{37} \cdot\frac{1}{1000}\\ &= 8 + 0.7 + 0.002 + 0.702 \cdot\frac{1}{1000}\\ &= 8 + 0.702 + 0.000702 + \cdots \end{align*} % \begin{figure}[!htb] \centering \begin{verbatim} (*) A divisão de 322 por 37 322 / 37 -296 --> 8.[702] ---- 260 --> ponto decimal -259 --> 7 ---- 10 --> 0 -0 -- 100 --> 2 -74 --- 260 --> dízima periódica (*) A tabuada de 37 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | |----+----+-----+-----+-----+-----+-----+-----+-----| | 37 | 74 | 111 | 148 | 185 | 222 | 259 | 296 | 333 | \end{verbatim} \caption{A divisão de 322 por 37.}\label{fg-alkw-322-37} \end{figure} \subsection{A aritmética dos números racionais} Uma fração é uma espécie de número inteiro com uma unidade de medida diferente da unidade universal. A ``unidade universal'' a que nos referimos é o número 1. Por exemplo, o numeral 2 representa duas unidades universais. Suponha que desejamos trabalhar com uma unidade diferente --- por exemplo, seja nossa unidade a metade da unidade universal. Então duas da nossa nova unidade faz uma unidade universal. Se tenho 7 unidades da nossa nova unidade e quero somá-la com mais 7 unidades da nossa nova unidade, então tenho 7 unidades da unidade universal. Em outras palavras, existe uma clara conversão possível entre unidades desde que todas sejam definidas em função da unidade universal. O fato % $$\frac{a}{b} \times \frac{c}{d} = \frac{ab}{cd}$$ % decorre de definição matemática~\cite[capítulo~1, seção~3, página~5]{mccoy}. O que fazemos a seguir é motivar essa definição, mostrando que ela pode ser vista como natural --- pra alguma definição de ``natural''. Definamos, por exemplo, que $1/4$ represente o inverso multiplicativo do natural $4$, ou seja, $\inv(4) = 1/4$ é o número $x$ que soluciona a equação $4x = x + x + x + x = 1$, o que é o mesmo que dizer que se agruparmos quatro partes dessa parte chamada $1/4$, obtemos a unidade universal inteira. Decorre imediatamente daí que o inverso multiplicativo de $1/4$ é $\inv(1/4) = 4$. Assim temos % $$ 4 \times \inv(4) = \inv(4) \times 4 = 1,$$ % assumindo que multiplicação é comutativa. \begin{notation*} Adotamos a convenção de não precisar escrever o símbolo da multiplicação, que é o símbolo $\times$, sempre que estiver claro que a justaposição de símbolos expressa o produto entre dois ou mais termos. \end{notation*} Assim, podemos deduzir % \begin{align*} 4 \inv(4) &= 1\\ (4 \inv(4)) (4 \inv(4)) &= 1\\ 16 (\inv(4))^2 &= 1, \end{align*} % e logo $\inv^2(4)$ é o inverso multiplicativo de 16, que é o mesmo que dizer $(1/4)^2 = (1/4)(1/4) = 1/16$, já que $1/4$ é só uma segunda notação para $\inv(4)$. Estendemos a definição pra que $5/4$, por exemplo, seja uma forma de escrever cinco vezes o inverso multiplicativo de quatro, o que é o mesmo que tomar cinco unidades universais e dividi-las em quatro partes iguais. Fica determinado assim o significado de se tomar um número natural $m$ e multiplicá-lo pelo inverso multiplicativo de um número natural $n$. Em símbolos, a definição determina o significado de % $$m \inv(n) = m \times \frac{1}{n}.$$ % Agora perguntamos --- quanto é $5/4 \times 7/3$? Por definição, $5/4$ é a soma de cinco peças de tamanho $1/4$, ou seja, $5 \times (1/4)$. Similarmente, $7/3 = 7 \times (1/3)$. Usemos uma barra de sabão como ilustração da unidade universal. Repartindo a barra em três partes iguais, qualquer uma delas é representada pelo numeral $1/3$. Se partirmos cada um desses $1/3$ em quatro partes iguais, teremos um total de $4 + 4 + 4 = 12$ partes novas, menores que as anteriores --- o tamanho exato delas é $1/12$ porque com 12 delas obtemos a barra de sabão inteira. Observamos que $1/3$, expresso por essas partes novas menores, exige quatro delas, ou seja, $1/3 = 1/12 + 1/12 + 1/12 + 1/12$. Similarmente, repartimos as partes de tamanho $1/4$ em três partes, obtendo uma novo pedaço da barra de sabão que vem a ser também de tamanho $1/12$. Logo, $(1/4) = 1/12 + 1/12 + 1/12$. Assim, computamos % \begin{align*} \frac{5}{4} \times \frac{7}{3} &= 5 \times \frac{1}{4} \times 7 \times \frac{1}{3}\\ &= 5\times7\times \frac{1}{4} \times \frac{1}{3}\\ &= 35 \times 3 \times \frac{1}{12}\times 4\times\frac{1}{12}\\ &= 35 \times 3 \times 4 \times \frac{1}{12}\times \frac{1}{12}\\ &= 420 \times \frac{1}{12}\times \frac{1}{12}\\ &= 420 \times \inv(12) \times \inv(12)\\ &= 420 \times \inv^2(12) = 420 \times \inv(12^2) = 420 \times \inv(144)\\ &= 420 \times \frac{1}{144}\\ &= \frac{420}{144}\\ &= \frac{35}{12}, \end{align*} % como desejado. Com a multiplicação definida, a divisão de frações é reduzida à multiplicação. Em símbolos, definimos que % $$\frac{a/b}{c/d} = \frac{a}{b}\frac{d}{c},$$ % o que é natural, visto que $\inv(c/d) = d/c$. Agora perguntamos --- quanto é $3/4 + 7/6$? Com unidades diferentes, não podemos somar. Mas $3/4 = 18/24$ e $7/6 = 28/24$, logo $3/4 + 7/6 = 46/24$. Subtração é similar. Isso mostra que a aritmética de frações não é diferente da aritmética de inteiros. Quando colocamos as frações em unidades equivalentes, podemos operará-las como se inteiros fossem. \subsection{A aritmética dos números irracionais} Os números irracionais são de difícil aritmética. A matemática nunca conseguiu oferecer uma linguagem expressiva o suficiente pra manipular os números irracionais respeitando a tradição matemática em que computações são executadas por via da escrita em função de uma linguagem composta por um conjunto finito de símbolos. O número $\sqrt{2}$, por exemplo, é um caso feliz no sentido de que usamos os símbolos $\sqrt{\cdot}$ e $2$ como forma de expressá-lo. Em outras palavras, apesar dos símbolos 0--9 não serem suficientes para escrever $\sqrt{2}$ de forma finita, como a tradição matemática efetivamente normatiza, conseguimos um resultado satisfatório ao introduzir ``novos verbos'' em nossa linguagem --- como ``$\sqrt{\cdot}$''. É, aliás, por isso que os matemáticos definem que $\sqrt{2}$ é um número algébrico: pode ser expresso por uma linguagem algébrica. Outros irracionais como $e$ ou $\pi$ são chamados de transcendentais porque transcendem qualquer linguagem algébrica, ou seja, cada um deles precisaria de um novo símbolo. É notório, entretanto, que os trancendentais são vastamente mais abundantes na linha real (pra alguma definição de ``abundantes'') que os algébricos. (Matemáticos costumam dizer ``quase todos os números da linha real são transcendentais'', sendo ``quase todos'' um termo definido na Teoria da Medida.) Portanto, não ter uma linguagem com a qual escrever esses números significa que a vasta maior parte dos números escapa a nosso poder de expressão, ou seja, nossos sistemas operam um conjunto ínfimo de números. Em 1936--1937, por exemplo, Alan Turing surpreende o mundo com a descoberta dos números não-computáveis, números transcendentais cujos algarismos jamais poderiam ser obtidos por um programa de computador. A relevância do fato é que programa de computador é apenas uma realização da expressibilidade dos nossos sistemas de cálculo. Logo, a limitação se aplica a nós --- porque os computadores são apenas uma realização do nossos sistemas de raciocínio. Temos números ``demais'' na linha real no sentido de que não temos (e não teremos) uma linguagem expressiva o suficiente pra alcançá-los, o que nos leva a questionar nossa concepção intelectual de ``número'' --- problema difícil. \subsection{O teorema fundamental da aritmética} Decorre do lema de Euclides. Essa é maior evidência de que Euclides já conhecia o teorema fundamental. Ele provavelmente não o escreveu como Gauss o fez em 1801 porque ele não tinha linguagem suficiente pra isso, mas devia ter uma clara noção de que a teoria dos números estaria implicada em seu lema. \begin{axiom*}[da boa ordem] Todo subconjunto não-vazio dos números naturais possui um elemento mínimo. \end{axiom*} \begin{lemma*}[de Euclides] Se $p$ é um primo que divide $ab$, então $p$ divide $a$ ou $p$ divide $b$. \end{lemma*} \begin{theorem*}[fundamental da aritmética] Qualquer número natural é expresso unicamente por um produto de números primos exceto pela ordem desses fatores. \end{theorem*} \begin{proof} Primeiro provamos a expressão por primos. Só depois nos preocupamos em mostrar que é unica. Seja $E$ o conjunto de todos os naturais que não podem ser expressos por um produto de primos. ($E$ é o conjunto dos naturais \underline{e}stranhos.) Pelo axioma da boa ordem, existe um natural mínimo $m$ em $E$. Esse $m$ não é primo: se fosse, não estaria em $E$. Mas se $m$ é composto, então $m = m_1 m_2$ e $1 < m_1 < m$ e $1 < m_2 < m$, implicando que nem $m_1$ nem $m_2$ estão em $E$. Se não estão em $E$, então eles podem ser expressos por um produto de primos, implicando que $m$ também pode. Logo, $m \in E$ é uma contradição. A parte difícil do teorema é provar que são únicos. É difícil porque a escrita dela é difícil, mas não é difícil compreender o que acontece nela. Queremos mostrar que a fatoração é única no sentido especificado pelo teorema. Assuma que existam naturais que possuam pelo menos duas fatorações primas distintas. Seja $E$ (de ``\underline{e}stranhos'') o conjunto composto por esses naturais. Então $E$ possui um menor elemento, chame-o de $m$. Pela definição de $E$, o número $m$ pode ser escrito como % $$p_1^{e_1} \cdots p_r^{e_r} = m = q_1^{d_1} \cdots q_s^{d_s},$$ % sendo $p_i, q_i$ primos tal que tenhamos acima duas formas distintas de escrever $m$. \begin{remark*} Observe que podemos escrever a equação acima sem potências --- por exemplo, se $e_1 = 3$, podemos escrever $p_1 p_1 p_1$ em vez de $p_1^3$, mas aí o segundo $p_1$ deve ser escrito com índice $2$, já que é o segundo a aparecer e, logo, $p_1^3$ é escrito como $p_1 p_2 p_3$, sendo certo neste exemplo que $p_1 = p_2 = p_3$. Assim obtemos % $$p_1 p_2 \cdots p_r = m = q_1 q_2 \cdots q_s$$ \end{remark*} \begin{remark*} Comentário. Note que $r$ pode ser igual a $s$ ou diferente --- isto é, nada se afirma sobre a quantidade de fatores de $m$ nessas duas fatorações. \end{remark*} Agora observe que $p_1$ é um primo que divide $m$. Portanto, $p_1$ tem que dividir o lado direito da equação também, implicando que $p_1$ divide tanto $m$ quanto a fatoração prima do lado direito --- ou seja, acabamos de satisfazer as condições do Lema de Euclides. \begin{remark*} Veja por quê. Escreva o Lema de Euclides na sua forma típica. Se $p$ divide $ab$ e $p$ não divide $a$, então $p$ divide $b$. Seja $m$ o produto $ab$. (Basta pegar qualquer uma das fatorações de $m$ e escrevê-la como apenas dois fatores $a$, $b$ de forma que o fator $a$ não contenha o primo $p$.) Então $p$ não divide $a$, mas vai ter que dividir $b$ e $b$ é composto de primos $q_i$. Portanto, $p$ tem que dividir algum primo $q_i$, que é a dedução que fazemos em seguida. \end{remark*} Logo, Euclides garante que $p_1$ divide algum $q_i$, implicando que $p_1 = q_i$ pra algum $i$ satisfazendo $1 \leq i \leq s$. \begin{remark*} Podemos tranquilamente assumir que $i = 1$. Basta pegar o primo $q_i$ e escrevê-lo primeiro na fatoração-$q$ de $m$. \end{remark*} Logo, $p_1$ aparece em ambas fatorações primas de $m$, o que nos permite cancelá-lo em ambas, obtendo % $$\frac{m}{p_1} = p_2 ... p_r = q_2 ... q_s$$ % implicando, assim, que $m/p_1$ tem duas fatorações primas distintas e, logo, $m/p_1$ pertence ao conjunto $E$. Mas $m/p_1$ é certamente menor que $m$, violando o fato de que $m$ é o menor elemento de $E$. Então $E$ não possui um menor elemento e, portanto, tem que estar vazio. Logo, a fatoração de $m$ é única, como desejado. \end{proof} \subsection{A versão {\em inversa} do teorema de Pitágoras} O teorema de Pitágoras afirma que se um triângulo tem um ângulo de noventa graus, então seus lados satisfazem $c^2 = a^2 + b^2$, sendo $c$ a hipotenusa. O {\em converso} do teorema de Pitágoras afirma que se os lados de um triângulo qualquer satisfizerem $c^2 = a^2 + b^2$, então o triângulo possui um ângulo de noventa graus. Infelizmente, o converso do teorema também é chamado no Brasil e em Portugal de ``o recíproco do Teorema de Pitágoras''. O problema é que existe um terceiro teorema chamado ``{\em the reciprocal Pythagorean theorem}'' que relaciona a altura $h$ de um triângulo retângulo relativo ao ângulo reto, afirmando que % $$\frac{1}{h^2} = \frac{1}{a^2} + \frac{1}{b^2}.$$ A tradução (para português) do livro ``Calculus'' de James Stewart, feita pela Cengage Learning em 2009, que é uma tradução da sexta edição norte-americana, traduz ``{\em converse of the Pythagorean}'' para ``a recíproca do teorema de Pitágoras''~\cite[problema~12, apêndice~B, página~A14]{calculo}, o que me fez inicialmente achar que se tratava do ``{\em the reciprocal Pythagorean theorem}''. Felizmente, o mundo anglofônico também chama o {\em reciprocal Pythagorean theorem} de {\em inverse Pythagorean theorem}, o que traduzo --- a partir de agora --- para o ``inverso do teorema de Pitágoras'', evitando assim uma tremenda confusão. O inverso do teorema de Pitágoras relaciona o inverso multiplicativo do quadrado de cada cateto com a altura uma altura do triângulo. Pra conhecê-lo, precisamos desenhar um triângulo para nomear seus lados e escrever o teorema. Desenhe um triângulo com lados $a, b, c$ de forma que $c$ seja a hipotenusa e $a, b$ os catetos. Trace a altura do triângulo relativa ao ângulo reto e chame-a de $d$, ou seja, trace um segmento de reta perpendicular à hipotenusa a partir do ângulo reto. Agora olhe para o triângulo de duas perspectivas diferentes: na perspectiva (1), tome o segmento $d$ como a altura do triângulo e, assim, sua área é $cd/2$, já que $c$ é a base; na perspectiva (2), tome $b$ como a altura do triângulo e, assim, sua área é $ab/2$. As duas perspectivas dizem a área de um mesmo triângulo, logo $cd/2 = ab/2$, que implica $cd = ab$. O teorema de Pitágoras afirma que $c^2 = a^2 + b^2$, o que implica $1/b^2 + 1/a^2 = c^2(a^2 b^2)$. Mas como inferimos acima que $ab = cd$, então $a^2b^2 = c^2 d^2$. Logo, $1/b^2 + 1/a^2 = c^2(c^2 d^2) = 1/d^2$, que é a afirmação do teorema. Em palavras, a soma dos inversos (multiplicativos) do quadrado dos catetos é o inverso (multiplicativo) do quadrado da altura $d$. %% Uma aplicação da recíproca é provar que um ângulo é reto? O %% Teorema de Pitágoras já não faz isso? Faz. Não sei pra que serve %% esse teorema. Eis uma versão trigonométrica da obtenção da recíproca do teorema de Pitágoras. Seja $x$ o ângulo entre $a$ e $c$. Então $\sin x = d/a$ e $a = d/\sin x$. Como uma segunda perspectiva, $\sin x = b/c$ e $b = c \sin x$. Agora $$ab = \frac{d c\sin x }{\sin x} = cd,$$ como desejado. \begin{exercise*} Um triângulo retângulo tem catetos $y$ e $4$, hipotenusa $h$ e altura $a$ perpendicular à hipotenusa. Expresse a altura com respeito a $h$. \sol. O inverso do teorema de Pitágoras nos dá $1/a^2 = 1/y^2 + 1/16$. O teorema de Pitágoras nos dá $h^2 = y^2 + 16$, logo $y = \sqrt{h^2 - 16}$. Assim, % \begin{align*} \frac{1}{a^2} &= \frac{16 + y^2}{16y^2}\\ a^2 &= \frac{16y^2}{16 + y^2}\\ a &= \frac{4y}{\sqrt{16 + y^2}}\\ &= \frac{4 \sqrt{h^2 - 16}}{\sqrt{h^2}}, \end{align*} % como desejado. \end{exercise*} \begin{exercise*} Um triângulo retângulo tem uma altura de 12 cm relativa ao ângulo reto, ou seja, a altura de 12 cm é perpendicular à hipotenusa. Expresse o tamanho da hipotensua com respeito ao perímetro do triângulo. O perímetro $P = a = b + c$, sendo $c$ a hipotenusa. Então $c = P - a - b$. Queremos eliminar $a$ e $b$. A versão inversa do teorema de Pitágoras nos dá $1/12^2 = 1/a^2 + 1/b^2$. E agora? \end{exercise*} \subsection{Fatoração {\em et cetera}} Descrevemos aqui uma série de fatos que usaremos inúmeras vezes. \subsubsection{Fatoração de polinômios quadráticos por completar o quadrado} Uma técnica pra se fatorar um polinômio quadrático é completar o quadrado. Alguns polinômios são quadrados perfeitos. Por exemplo, $x^2 + 2x + 1 = (x + 1)^2$. Logo, são fatoráveis por ``produtos notáveis'', como são tipicamente chamados. Outros não são perfeitos como $$p(t) = t^2 + \frac{7}{2}t + \frac{3}{2},$$ mas podemos adicionar uma constante a eles que os torna produtos notáveis. A ideia é a seguinte. O termo $t^2$ é claramente um quadrado de lado $t$. Se você estender um retângulo pra cima e outro pra direita do quadrado $t$, você obterá dois retângulos em que um lado tem tamanho $t$. O outro lado precisa ser um número em que a área ocupada pelos retângulos resultem em $\sfrac{7}{2}t$. Esse número é a metade de $\sfrac{7}{2}$. Assim temos $t^2 + \sfrac{7}{4}t$, que ainda não é um quadrado perfeito porque está faltando a área de um quadrado de lado $\sfrac{49}{16}$, que é o quadrado da metade de $\sfrac{7}{2}$. Daí terminamos com $$t^2 + \frac{7}{2}t + \frac{49}{16} = \left(t + \frac{7}{4}\right)^2.$$ Como o polinômio original possui o termo $\sfrac{3}{2}$, precisamos adicioná-lo ao quadrado e também subtrair o termo \sfrac{49}{16} que adicionamos. Assim obtemos a igualdade % $$t^2 + \frac{7}{2}t + \frac{3}{2} = p(t) = \left(t + \frac{7}{4}\right)^2 + \frac{3}{2} - \frac{49}{16},$$ % como desejado. \subsubsection{Fatoração de polinômios quadráticos com coeficiente líder} Precisaremos calcular números como % $$\lim \frac{t^2 - 9}{2t^2 + 7t + 3}.$$ % Será preciso fatorar numerador e denominador aí. O numerador é uma {\em diferença entre dois quadrados} em que se tem o produto dos conjugados $(t + 3)(t - 3)$. %% \footnote{Não se deve chamar isso de um produto de conjugados. É %% uma terminologia problemática nesse caso. De uma forma geral, t^2 - %% n = (t - \sqrt{n})(t + \sqrt{n}) pra todo $n$ inteiro e daí podemos %% definir uma função $f$ tal que $f(a + b\sqrt{n}) := a -b\sqrt{n}$, %% sendo $a, b$ números racionais --- digamos. Essa função é %% tipicamente chamada de conjugação. Ela pode ser vista como uma %% generalização da conjugação complexa restrita a inteiros %% gaussianos, que não números complexos da forma $a + bi$, sendo $a, %% b$ inteiros. A generalização, portanto, é relaxar os inteiros %% gaussianos e permitir racionais. Se definirmos $f$ assim, obtemos %% uma contradição quando $n$ é um quadrado perfeito. Por exemplo, %% seja $n = 9$. Então $f(1 + 2\sqrt{9}) := 1 - 2\sqrt{9} = 1 - 6 = %% -5$. Agora compute $f(1 - 2\sqrt{9}) := 1 + 2\sqrt{9} = 1 + 6 = %% 7$.} O denominador é menos claro. Quando se tem uma constante no coeficiente líder, pode-se procurar por uma fatoração da forma $(2t + a)(t + b)$. Observe que o problema é o mesmo que fatorar um polinômio quadrático mônico, que é um polinômio de coeficiente líder igual a 1. Em mais detalhes, se % $$Lt^2 + 7t + 3 = (Lt + a)(t + b) = Lt^2 + (Lb + a)t + ab,$$ % então, pra encontrarmos uma fatoração (se existir), a pista mais relevante é o termo $ab$. Se, por exemplo, $ab = 3$, deduzimos que só há duas possibilidades --- $a = 1, b = 3$ ou $a = 3, b = 1$ porque 3 é primo. É o caso acima. Queremos $ab = 3$ e $(2b + a) = 7$. Se tentarmos $b=1$, não conseguimos porque $2b + a$ vai dar 5. Se tentarmos $a=1$, conseguimos: $2b + a = 6 + 1 = 7$. Logo, $2t^2 + 7t + 3 = (2t + 1)(t + 3)$. Em outras palavras, o coeficiente líder maior que 1 não chega a modificar o método de fatoração --- é só uma pequena perturbação. O maior problema é quando o termo $ab$ oferece muitas possibilidades. Por exemplo, fosse $ab = 24$, a coisa se complica um pouco mais porque $24 = 1\times24 = 2\times12 = 4\times6 = 8\times3$, sendo que cada produto aí oferece duas possibilidades. Várias possibilidades a verificar --- oito nesse caso. \subsubsection{Fatoração de soma e diferença de cubos} Outros produtos notáveis são a {\em soma de cubos} e a {\em diferença de cubos}. Por exemplo, % \begin{align*} a^3 + b^3 &= (a + b)(a^2 - ab + b^2)\\ a^3 - b^3 &= (a - b)(a^2 + ab + b^2) \end{align*} % Se lembrarmos de um, provavelmente lembramos do outro por inverter os sinais nos lugares certos e fazendo a expansão pra confirmar o resultado. Se esquecermos de ambos, podemos recuperá-los efetuando a divisão longa de, por exemplo, $a^3 + b^3$ dividido por $a + b$: % \begin{verbatim} a^3 + b^3 dividido por a + b a^3 + a^2b a^2 + b^2 - ab ---------- b^3 - a^2 b b^3 + a b^2 Logo, a^3 + b^3 = (a + b)(a^2 - ab + b^2). ------------- -a^2 b - a b^2 -a^2 b - a b^2 -------------- 0 \end{verbatim} % Se for útil, observe que quando queremos a soma, o termo linear é uma soma. Quando queremos a diferença, o termo linear é uma subtração. Se o termo linear é uma soma, o quadrático é uma subtração. Se o termo linear é uma subtração, o quadrático é uma soma. Os sinais têm que se inverter que é pra ocorrer uma cancelamento de termos não cúbicos: de outra forma, não teríamos a identidade. \subsubsection{Fatoração envolvendo cubos e quadrados} Às vezes encontramos mais de uma identidade na mesma expressão. Por exemplo, $x^6 - y^6 = (x^2)^3 - (y^2)^3$. É uma diferença de cubos, mas também é uma diferença de quadrados, afinal % $$(x^3)^2 - (y^3)^2 = x^6 - y^6 = (x^2)^3 - (y^2)^3,$$ % isto é, exponenciação é comutativa --- embora não associativa. Percebendo um caso desses, é preciso ver qual fatoração é mais interessante. Tente ambas. \subsubsection{O teorema das raízes racionais} Dado um polinômio de coeficientes inteiros, o teorema garante que qualquer raiz racional $p/q$ reduzida implica que $p$ é fator do termo constante e que $q$ é fator do coeficiente líder. É possível, então, --- fazendo uma busca por fatores --- saber se há raízes racionais. Por exemplo, suponha que desejamos pelo máximo local de $f(x) = -x^3 + 7x^2 -11x + 5$. Se houver uma raiz racional $\pm p/q$ de $f(x)$, então $p$ divide $5$ e $q$ divide 1, logo $p \in \{\pm 1, \pm 5\}$, $q = \pm 1$, o que nos dá candidatos $r \in \{\pm 1, \pm 5\}$ e assim descobrimos que são raízes $r = 1$ e $r = 5$. Então $(x - 1)$ fatora $f(x)$. Dividindo $f(x)$ por $(x - 1)$, obtemos % \begin{align*} -x^3 + 7x^2 -11x + 5 &= (x - 1)(-x^2 + 6x - 5)\\ &= -(x - 1)(x^2 - 6x + 5)\\ &= -(x - 1)(x - 1)(x - 5)\\ &= -(x - 1)^2 (x - 5), \end{align*} % o que significa que $x = 1$ é uma raiz de multiplicidade 2, o que significa que $x = 1$ é um mínimo local, logo o máximo local está entre em (1, 5). Pra descobrir o máximo local exatamente, precisamos das raízes de $f'(x) = -3x^2 + 14x - 11$. O fato de $x = 1$ ser um mínimo local revela que $x=1$ é raiz de $f'(x)$, o que nos permite fatorar $f'(x)$. Obtemos % $$-3x^2 + 14x - 11 = (x - 1)(-3x + 11),$$ % implicando que o máximo local tem que estar em $x = 11/3$. \subsubsection{O método de Horner} \lipsum[1] \subsubsection{Manipulação algébrica por racionalização} Para calcular % $$\lim_{h \to 0} \frac{\sqrt{4 + h} - 2}{h},$$ % podemos multiplicar pelo conjugado do numerador, que é $\sqrt{4 + h} + 2$. Obtemos % $$\lim_{h \to 0} \frac{\sqrt{4 + h} - 2}{h} = \lim \frac{4 + h - 4}{h\sqrt{4+h}+2} = \lim \frac{1}{\sqrt{4+h}+2}.$$ % Compreenda a cultura matemática. Por que se chama ``racionalização''? Seria talvez por que a operação transforma a expressão numa função racional? (Qual a definição de função racional?) O termo ``racionalização'' poderia se chamar ``derradicalização'' porque o objetivo é remover um {\em radical} e não transformar a expressão numa função racional. Entretanto, às vezes essa ``derradicalização'' ocorre no numerador e às vezes, no denominador. Muitas vezes o executor da álgebra não explicita exatamente que racionalização ele deseja, assumindo que a audiência compreenderá. Trocamos um radical no numerador por um radical no denominador, o que parece lucro nenhum. Obtém-se o lucro ao perceber que, no exemplo acima, a racionalização é claramente desejada no numerador porque o objetivo ali é computar um limite, o que não é possível quando ambos numerador e denominador se aproximam de zero. Com a racionalização, o limite é trivialmente obtido --- $\sfrac{1}{4}$. \begin{remark*} Quando se racionaliza uma expressão, sempre se obtém uma diferença de quadrados. Memorizando o fato, não precisamos calculá-lo. \end{remark*} \subsection{Desigualdades} Precisamos de bastante intimidade com desigualdades. Por que não podemos multiplicar uma desigualdade por $-1$? Porque não produz verdades: por exemplo, $2 < 3$ não implica $-2 < -3$, mas certamente implica $-3 < -2$: a relação se inverte no lado negativo. Multiplicação por números positivos maiores que um pode ser feita. Já computar inversos multiplicativos inverte as desigualdades, mas só se o inverso for positivo. Por exemplo, $2 < 3$, mas multiplicar essa desigualdade por $-1/2$ produziria a falsidade $-1 > -3/2$. \subsubsection{A desigualdade triangular} Uma desigualdade famosa é a tringular~\cite[teorema~7, apêndice~A, página~A8]{calculo}. Se $x$ e $y$ são dois números positivos, então certamente $|x + y| = |x| + |y|$. Se ambos forem negativos, também teremos $|x + y| = |x| + |y|$. Agora, se os sinais deles se invertem, isso faz com com que $x + y$ seja um número menor que $|x| + |y|$. Assim, % $$|x + | \leq |x| + |y|,$$ % como desejado. \subsubsection{O uso de desigualdades em limites} Provar limites requer habilidade com desigualdades. Por exemplo, provar que $\lim 4 - \sfrac{3x}{5} = 7$ à medida que $x\to -5$ requer mostrar que $|4 - \sfrac{3x}{5} - 7| < e$. Some $4 - 7$, fatore o $-3$, coloque-o em evidência fora do valor absoluto, divida a desigualdade por $3$. Abra o valor absoluto em duas desigualdades. Multiplique ambas por 5 e obtenha $|x - (-5)| < \sfrac{5e}{3}$. Agora considere provar $\lim 1/x = 1$ com $x\to 1$. Teremos que nos restringir ao redor de $x=1$, mas --- sem tomar cuidado --- terminaremos indo ao lado negativo do domínio, o que faz a função descer sem limite, prejudicando nossas desigualdades. Por exemplo, restrinja $|x - 1| < 1$ e observe que a restrição implica $0 < x < 2$, produzindo $1/2 < 1/x < \text{{\tt indefinido}}$. Em outras palavras, a restrição $|x - 1| < 1$ é folgada demais e nos leva ao ponto $x = 0$ de discontinuidade da função. Basta nos restringirmos de forma mais apertada --- um raio de 1/2 resolveria. Mas podemos sempre ser mais conservador usando, por exemplo, um raio de 1/4. Obteríamos $-1/4 < x - 1 < 1/4$ implicando $4/5 < 1/x < 4/3$, o que nos dá a restrição $1/|x| < 4/3$. Daí $d = \min(4/3, 3e/4)$ servirá bem. Estamos fora de contexto aqui: nem definimos o que é limite e já estamos falando das dificuldades com eles. A mensagem é --- desenvolva-se com desigualdades porque elas são fundamentais na análise matemática, especialmente se você se interessa por um estudo além-Cálculo. No problema 36 da Seção 2.3~\cite[página~96]{stewart}, precisamos provar que $\lim 1/x = 1/2$ com $x \to 2$. Uma estratégia pra este é muito parecida com a do problema anterior. Restrinja $|x - 2| < 1/4$ e obtenha $1/|2x| < 2/7$. Daí $d = \min(2/7, 7e/2)$ servirá bem. Aproveite o momento e observe a dica do problema 37~\cite[página~96]{stewart}. O problema é provar que $\sqrt{x} \to \sqrt{a}$ com $x \to a$, assumindo $a > 0$. A dica é que $|\sqrt{x} - \sqrt{a}| = |x - a|/(\sqrt{x} + \sqrt{a})$. O problema não elabora porque elaborar consome parágrafos e parágrafos, mas a dica pode até ser omitida porque ela é uma simples manipulação algébrica. Estamos interessados em mostrar $|\sqrt{x} - \sqrt{a}| < e$, mas precisamos de uma conexão com $|x - a|$. Racionalize a expressão $\sqrt{x} - \sqrt{a}$, obtendo $(x - a)/(\sqrt{x} + \sqrt{a})$. O numerador pode ser negativo, mas sempre que ele for, também será a expressão $\sqrt{x} - \sqrt{a}$. Essas duas expressões também são sempre positivas sempre que uma delas for positiva. Então $|\sqrt{x} - \sqrt{a}|$ é positiva se e somente se $|x - a|$ também for, o que nos dá a equação da dica. Eis uma prova elegante do fato. Seja $d = e\sqrt{a}$. Então % $$|\sqrt{x} - \sqrt{a}| = \left|\frac{x - a}{\sqrt{x} + \sqrt{a}}\right| = \frac{|x-a|}{\sqrt{x} + \sqrt{a}} \leq \frac{|x - a|}{\sqrt{a}} < \frac{d}{\sqrt{a}} < e.$$ % Essa prova nos leva a conclusões importantes. A restrição não precisa ser simétrica: observe que a prova acima toma $d$ {\em menor} que um fator de $e$. Devemos ler a prova da direita pra esquerda. É lá na direita que está a assunção de que $|x - a| < d$. Daí manipulação algébrica nos leva ao lado esquerdo, provando que $|\sqrt{x} - \sqrt{a}| < e$, como desejado. Elaboremos a observação sobre simetria relativa a $d$. Suponha que digamos que $d = 1$. Então assumimos que $|x - a| < 1$, o que implica $-1 < x - a < 1$, mas isso não nos ajuda porque valores negativos não pertencem ao domínio de $\sqrt{x}$. A definição de limite não nos obriga a trabalhar com restrições simétricas como $-1 < x - a < 1$. Podemos tranquilamente restringir $a/2 < |x - a| (a + 1)/2$, que não é simétrica --- o raio direito da desigualdade é maior que o raio esquerdo. Essa restrição nos garante que % $$\frac{1}{\sqrt{x} - \sqrt{a}} < \frac{1}{\sqrt{4a/3} - \sqrt{a}},$$ % o que eventualmente nos daria $|\sqrt{x} - \sqrt{a}| < e$, usando-se a dica do problema. Mas observe que essa solução é muito menos elegante que a primeira. Essa prova bem arrumadinha, lida da direita pra esquerda, é sempre importante à educação matemática porque bagunça faz tudo mais difícil de entender. Se a solução elegante é difícil de ler, então devemos estudar cada detalhe dela e aos poucos vê-la como natural porque essa prática facilita a análise (ao invés de complicá-la). \begin{exercise*} Prove que $\lim \sqrt[4]{9 -x} = 0$ com $x \to 9$ pela esquerda. Agora a ausência de simetria no intervalo delta é solicitada pelo problema. Queremos assumir $9 - d < x < 9$ e mostrar que $|\sqrt[4]{9 -x}| < e$, ou seja, $|\sqrt[4]{9-x}| < e$ implica $0 < \sqrt[4]{9-x} < e$, que implica $0 < 9-x< e^4$, que implica $-e^4 < x - 9 < 0$, que implica $-e^4 +9 < x < 9$. Tomando-se $d = e^4$, assumimos $9 - e^4 < x < 9$ de forma assimétrica: estamos provando um limite pela esquerda. Manipulação algébrica nos leva a $|\sqrt[4]{9 -x} - 0| f(x_2)$ whenever $x_1 < x_2$ in I.\defend \masciidef{} Let $f:A \to Y$, $g: B \to W$ be functions. Then we define $(f+g)(x) = f(x) + g(x)$ where the domain is $A \cap B$. We define $(f - g)(x) = f(x) - g(x)$ where the domain is $A \cap B$. We define $(fg)(x) = f(x)g(x)$, where the domain is $A \cap B$. We define $(f/g)(x) = f(x)/g(x)$, where the domain is $\{x \in A \cap B : g(x) \neq 0\}$.\defend \masciidef{} Let $f: A \to B$, $g: X \to Y$. We define the composite function $f \circ g$ as $(f \circ g)(x) = f( g(x) )$. The domain of the composition is all $x \in X$ in which $g(x)$ is defined which goes to an $a \in A$ in which $f(a)$ is defined. So, the domain is the intersection of a subset of $Y$, the range of $g(x)$, and the set $A$; that is, the domain of $f \circ g$ is $\text{Range}(g) \cap A$.\defend \subsection{Limits} \masciidef{} We say that $f$ approaches $L$ as $x$ approaches $a$ if and only if for each $e > 0$ given, there is a $d > 0$ such that % $$0 < |x - a| < d \implies |f - L| < e.$$\defend \begin{remark*} Em vez de letras $d$ e $e$, o típico é escrever $\delta$ e $\epsilon$, mas não temos essas letras num teclado usual de hoje. Quando começarmos a trabalhar com o número transcendental $2.7182845904523536...$, também usaremos a letra $e$ pra representá-lo. Não haverá nenhuma confusão porque o contexto deixa bem claro a quem nos referimos. \end{remark*} \masciidef{} We say that $f$ approaches $L$ from the left as $x$ approaches $a$ from the left if and only if for each $e > 0$ given, there is a $d > 0$ such that $$a - d < x < a \implies |f - L| < e.$$\defend \masciidef{} We say that $f$ approaches $L$ from the right as $x$ approaches $a$ from the right if and only if for each $e > 0$ given, there is a $d > 0$ such that $$a < x < a + d \implies |f - L| < e.$$\defend Uma coisa se aproxima de outra se e somente se a distância entre elas disaparece ``à medida que o tempo passa''. Note que se a distância entre $x$ e $a$ se tornar pequena, então torna-se pequena também a distância entre $f$ e $L$. Se isso ocorre pra qualquer pequena distância arbitrada, então o limite de $f$ é $L$. \begin{exercise*} Provar limites pela definição é muito difícil na maioria das vezes. Por exemplo, prove que $\lim x^2 - 1 = 3$ com $x \to -2$. Precisamos de um $d$ que possa nos levar à desigualdade $$|x^2 - 1 - 3| = |x^2 - 4| = |(x + 2)(x - 2)| = |x + 2||x - 2| < e$$ pra todo $e$ positivo. Pra ultrapassar o obstáculo desse produto, observe que estamos interessados em valores de $x$ {\em muito} próximos de $-2$. Portanto, podemos tranquilamente nos restrigir ao universo de busca em que $d$ não é maior que 1, ou seja, podemos assumir que $|x + 2| < 1$, obtendo $-1 < x + 2 < 1$, o que nos dá $-5 < x - 2 < -3 < 5$. Daí deduzimos que $|x - 2|<5$, que é fator do produto que nos complica. O outro fator é $|x + 2|$ e queremos que esse produto de distâncias seja menor que $e$. Agora compreender a estratégia ficou mais fácil: se assumirmos que $|x + 2| < e/5$, o que ocorre com o produto? O produto nos dá o que desejamos: $$|x + 2||x - 2| < \frac{e}{5} \times 5 = e.$$Logo, podemos tomar $d = \min(e/5, 1)$. Eis a prova. Seja $d = \min(e/5, 1)$. Assuma que $|x - (-2)| < d = \min(e/5, 1)$. Se $e > 5$, então $d = 1$ e aí $|x - (-2)| = |x + 2| < 1$ e $|x+2||x-2| < 1 < e$, como desejado. Já se $e \leq 5$, então $|x - (-2)| < e/5$ e aí $|x + 2| < 1$. Como estamos interessamos em valores muito próximos de $-2$, podemos restringir $x$ tal que $|x - (-2)| = |x + 2| < 1$, ou seja, não nos afastamos uma unidade do valor $-2$. Nesse intervalo restrito, podemos afirmar que $-1 < x + 2 < 1$ e $-5 < x - 2 < -3 < 5$, logo $|x - 2| < 5$. Multiplicando ambas desigualdades importantes aqui, obtemos % $$|(^2 - 1) -3|= |(x + 2)(x - 2)|= |x + 2||x - 2| < e/5 \times 5 = e,$$ % como desejado. Isso prova. A leitura da prova também requer cuidados. Como ler a prova? O delta escolhido é o menor valor dentre $e/5$ e $1$. Se o adversário escolhe um $e$ muito grande --- maior ou igual a 5, por exemplo ---, escolhemos $d = 1$, partindo de $|x - (-2)| < 1$, que implica $|x - 2| < 5$ e, assim, obtemos $|x + 2||x - 2| < 5$, como desejado. Se o adversário escolhe $e$ pequeno --- menor que 5, por exemplo ---, então tomamos um quinto desse $e$, partindo de $|x - (-2)| < e/5$. Precisamos também da desigualdade $|x + 2| < 1$. Quando próximo de $-2$, temos $|x + 2| < 1$ e podemos chegar perto o quanto quisermos. Essa restrição implica $|x - 2| < 5$ --- subtraia 4 e observe que $-3 < 5$. Daí multiplicamos as duas desigualdades, obtendo $$|x + 2||x - 2| < 5\times\frac{e}{5} = e.$$ Isso é uma segunda escrita da prova, sem dúvida. Enfim, a função é simples, mas a prova envolve habilidade com desigualdades, valor absoluto e a definição epsilon-delta. \qed\end{exercise*} \masciidef{} The line $x = a$ is called a vertical asymptote of the curve $y = f(x)$ if at least one of the following statements is true: $\lim f = \infty$ as $x \to a$, $\lim f = \infty$ as $x \to a^-$, $\lim f = \infty$ as $x \to a^+$, $\lim f = - \infty$ as $x \to a$, $\lim f = - \infty$ as $x \to a^-$, $\lim f = \infty$ as $x \to a^+$.\defend \begin{remark} Em palavras, pra descobrir assíntotas, basta procurar por quais valores do domínio a função tende a $\pm \infty$, ou seja, quais valores (do domínio) fazem a função crescer ou decrescer sem limite. \end{remark} \masciith{} Let $c$ be a constant and suppose $\lim f$ and $\lim g$ as $x \to a$ exist. Then the following are true: (1) $\lim(f + g)$ = $\lim f + \lim g$; (2) $\lim(f - g)$ = $\lim f - \lim g$; (3) $\lim(c f)$ = $c \lim f$; (4) $\lim(fg)$ = $\lim f \lim g$; (5) $\lim f/g$ = $(\lim f)/(\lim g)$ as long as $\lim g \neq 0$; (6) $\lim f^n = (\lim f)^n$, where $n > 0$; (7) $\lim c = c$; (8) $\lim x = a$ (as $x \to a$); (9) $\lim x^n = a^n$, as $x \to a$ and $n > 0$; (10) $\lim \sqrt[n]{x}$ = $\sqrt[n]{a}$ as $x \to a$ and $n > 0$, and once $n > 0$, assume $a > 0$. (11) $\lim \sqrt[n]{f}$ = $\sqrt[n]{\lim f}$ as $x \to a$ and $n > 0$, and once $n$ is even, assume $f > 0$.\theend \begin{remark*} Observe que essas propriedades só se aplicam quando $\lim f$ e $\lim g$ existem. Se não existirem, o teorema não procede. Em outras palavras, antes de sair aplicando essas verdades, verifique que as condições procedem. Por exemplo, suponha que $\lim f(x)/x^2 = 5$ com $x \to 0$. Encontre $\lim f(x)$. O que se pode fazer aí? Não se pode assumir que o limite de $f(x)$ existe porque isso não foi dado; o que foi dado é que o quociente tem limite 5. Entretanto, sabemos que $\lim x^2 = 0$ existe e é zero. Assim, podemos deduzir: % \begin{align*} 0 &= 5 \times 0\\ &= \lim \frac{f(x)}{x^2} \lim x^2\\ &= \lim \frac{f(x)}{x^2}x^2 \qquad\text{porque ambos limites existem}\\ &= \lim f(x). \end{align*} % Pode-se pedir também por $\lim f(x)/x$, mas a mesma estratégia acima resolve: sabemos que $\lim x = 0$. \end{remark*} \begin{remark*} Esse é o problema 56 da Seção 2.3 de Stewart, edição 6. Na versão brasileira do livro, há um erro de digitação em que se dá um limite com $x\to 1$ e se pede por $x\to 0$. Esse erro é interessante porque é preciso se convencer de que o problema não tem solução. O que acontece num ponto do domínio não garante nada do que acontece num outro ponto {\em qualquer}. Provar o fato é fácil --- só mostrar um contra-exemplo. \end{remark*} \masciiexe{} If $p$ is a polynomial, show $\lim p(x)$ = $p(m)$ as $x \to m$. \pit The general formula for a polynomial is % $$p(x) = a_0 + a_1 x + a_2 x^2 + ... + a_n x^n.$$ % We now apply the limit laws % \begin{align*} \lim p(x) &= \lim a_0 + \lim a_1 x + \lim a_2x^2 + ... + \lim a_n x^n\\ &= a_0 + a_1 m + a_2 m^2 + ... + a_n m^n\\ &= p(m) \end{align*} % This shows that every polynomial is continuous, but we define continuity a little later. Before, we do exercises to practice our understanding of limits.\exeend \masciiexe{} If $r$ is a rational function, show that $\lim r(x) = r(m)$, as $x \to m$, for every number $m$ in the domain of $r$. \pit From the previous exercise, we have that every polynomial is continuous. A rational function is the quotient of two polynomials; so, the problem is very similar. We can represent the limit of a rational function as % \begin{align*} \lim r(x) &= \lim \bigg( \frac{a_0 + a_1 x + a_2 x^2 + ... + a_n x^n} {b_0 + b_1 x + b_2 x^2 + ... + b_n x^n} \bigg)\\ &= \frac{\lim a_0 + \lim a_1x + \lim a_2 x^2 + ... + \lim a_n x^n} {\lim b_0 + \lim b_1x + \lim b_2 x^2 + ... + \lim b_n x^n}\\ &= \frac{a_0 + a_1 m + a_2 m^2 + ... + a_n m^n} {b_0 + b_1 m + b_2 m^2 + ... + b_n m^n}\\ &= r(m) \end{align*} \exeend \masciith{} A function $f \to L$ as $x \to a$ if and only if $f \to L$ as $x \to a^-$ and $f \to L$ as $x \to a^+$.\theend \masciith{} If $f \leq g$ when $x$ is near $a$, but $x \neq a$, and the $\lim f$ and $\lim g$ both exist as $x \to a$, then $\lim_{x \to a} f \leq \lim_{x \to a} g$.\theend \begin{theorem*}[do Confronto] Se $f(x) \leq g(x) \leq h(x)$ quando $x$ está próximo de $a$, exceto possivelmente em $a$, e $\lim f(x) = L = \lim h(x)$, então $\lim g(x) = L$. \end{theorem*} \begin{remark*} Esse é o Teorema do Confronto, também chamado de Teorema do Sanduíche. Parece que alguns também o chamam de Teorema do Imprensamento. \end{remark*} Um caso bem fácil de aplicação desse teorema é quando se pede o limite de uma função trigonométrica multiplicada por um outra função --- cujo limite seja fácil calcular. Por exemplo, calcule $\lim q(x) \cos 20\pi x$ com $x\to 0$ sabendo-se que $\lim q(x) = 0$. O trabalho é fácil porque sabemos que $ -1 < \cos 20 \pi x < 1$. Logo, se multiplicarmos essa desigualdade por $q(x)$, obtemos $-q(x) < q(x)\cos 20\pi x < q(x)$ e daí o Teorema do Confronto implica $\lim q(x) \cos 20\pi x = 0$, como desejado. \begin{remark*} Às vezes um problema nos pede um limite que podemos reconhecer com experiência. Por exemplo, % $$\lim \frac{(4 + h)^2 - 16}{h}$$ % é a derivada de $x^2$ no ponto $x = 4$, ou seja, em $(4, 16)$. Então o limite é $f'(4) = 2x = 8.$ O que Stewart espera, entretanto, é que cancelemos o $h$ do denominador. \end{remark*} \begin{exercise*} Existe número real $a$ tal que % $$\lim_{x \to -2} \frac{3x^2 +ax + a + 3}{x^2 + x - 2}$$ % exista? O exercício pede que identifiquemos em que situação um limite existe ou não existe. Se o denominador tende a zero e o numerador não tende a zero (com $x \to -2$), então essa função racional sobe ou desce sem limite. A única chance pro limite existir então é quando o numerador é igual a zero, ou seja, $3x^2 +ax + a + 3 = 0$ com $x = -2$. Queremos então $12 -a + 3 = 0$ ou $a = 15$. Essa é a chance. Precisamos agora ter certeza de que o limite existe quando $a = 15$, ou seja, investiguemos % $$\lim_{x \to -2} \frac{3x^2 +15x + 18}{x^2 + x - 2}.$$ % O denominador é fatorável com $(x + 2)(x - 1)$ e o numerador é fatorável com $(3x + 9)(x + 2)$. (Divida $3x^2 +15x + 18$ por $x + 2$ por divisão longa e obtenha $3x + 9$.) Assim obtemos % $$\lim_{x \to -2} \frac{3x + 9}{x - 1} = \frac{3}{-3} = -1.$$ % Existe $a = 15$ em que o limite é $-1$. \end{exercise*} \begin{exercise*} Prove $\lim f(x) = 0$ com $x \to 0$ sendo % \[f(x) = \begin{cases} x^2 & \text {se $x$ é racional},\\ 0 & \text {se $x$ é irracional}. \end{cases}\] % O exercício parece querer uma aplicação do Teorema do Confronto. A função é claramente descontínua em todos os pontos, mas o limite na origem existe provavelmente porque a reta de pontos isolados $f(x) = 0$ se encontra com a parábola de pontos isolados $f(x) = x^2$. Os pontos se aglutinam infinitamente lá, sem deixar buracos. É o que sugere minha intuição. São fatos que $\lim 0 = 0 = \lim x^2$. Logo, % $$0 \leq f(x) \leq x^2$$ % pra todo $x$ real. Assim, % $$0 = \lim 0 \leq \lim f(x) \leq \lim x^2 = 0$$ % e então o Teorema do Confronto implica $\lim f(x) = 0$, como desejado. \end{exercise*} \begin{exercise*} Seja % \[f(x) = \begin{cases} 0 & \text {se $x$ é racional},\\ 1 & \text {se $x$ é irracional}. \end{cases}\] % Prove que $\lim f(x)$ não existe quando $x \to a$, sendo $a$ um número real. O autor quer que a gente saiba como provar que um limite não existe pela definição precisa de limite. Fazemos isso por contradição. Se o limite exitisse, a desigualdade-delta implicaria a desigualdade-epsilon pra qualquer epsilon. Como o limite não existe, algum epsilon fracassa a desigualdade. Suponha que o limite existe e é $L$. Seja $e = 1/2$. Então $|x| < d$ implica $|f(x) - L| < 1/2$, o que implica $- 1/2 < f(x) - L < 1/2$. Tome um $x$ racional em $(-d, d)$. Daí $f(x) = 0$ e obtemos $L < 1/2$. Agora tome um $x$ irracional e deduza $1/2 < L$. Nenhum número $L$ é capaz de satisfazer essa contradição, então $L$ não existe. Logo, não há limite. \end{exercise*} \begin{remark*} Quando um limite não existe, há a evidência de que --- à medida que nos aproximamos de um ponto $a$ do domínio ---, a função $f(x)$ tende a uma coisa de um jeito e tende a outra coisa de outro jeito. Na linha real, os dois jeitos típicos de se aproximar de um ponto no domínio é pela esquerda e pela direita. \end{remark*} \begin{exercise*} Mostre um exemplo em que $\lim f(x)g(x)$ existe, mas $\lim f(x)$ e $\lim g(x)$ não existem. Pegue uma função $f(x)$ que seja $-1$ pra $x$ negativo e 1 pra $x$ positivo. O limite em zero não existe. Inverta o sinal na definição de $g(x)$. Agora, se multiplicamos $f(x)g(x)$, obtemos a função constante $-1$, cujo limite em zero obviamente existe. \end{exercise*} \begin{exercise*} Um disco precisa ser fabricado com área $1000 \cm^2$. Logo seu raio é $10\sqrt{10/\pi} \cm$. Queremos permitir um erro de $5 \cm^2$ na fabricação, ou seja, pode-se produzir com área no intervalo $(995, 1005) \cm^2$. Qual o erro que se permite no raio? A área tem que satisfazer $995 < \pi r^2 < 1005$. Isolando $r$, obtemos $$\sqrt{\frac{995}{\pi}} 0$. Por hipótese, $\lim f(a + h) = f(a)$, então existem $d > 0$ tal que $0 < |h| 0$. Compute $f(0) = -1 < 0$. A função é contínua, então ela tem que cruzar o eixo $x$ em $[-2, 0]$. \end{exercise*} \begin{exercise*} Suponha $f$ contínua em $[1,5]$ e que as únicas soluções da equação $f(x) = 6$ são $x = 4$ e $x = 1$. Se $f(2) = 8$, prove que $f(3) > 6$. Prove por contradição. Suponha que $f(3) \leq 6$. Faça um desenho do gráfico. Desenho o ponto (1, 6), (2, 8) e (4, 6). A função é contínua. Se $f(3) = 6$, violamos a hipótese das únicas soluções. Se $f(3) < 6$, então o Teorema~\ref{thm:ivt} implica que tem que haver $c$ em $(2, 3)$ tal que $f(c) = 6$: a função teria que descer de $f(2)=8$ até $f(3) < 6$, o que faz com que ela necessariamente passe por um ponto $c$ satisfazendo $f(c) = 6$. \end{exercise*} \section{Derivatives} \masciidef{} The derivative of a function $f$ at a point $x = a$, written $f'(a)$ is defined to be %v \begin{equation}\label{def:derivative} f'(a) = \lim_{h \to 0} \fr{f(a + h) - f(a)}{h}. %e \end{equation} \defend This definition comes from a similar one. The derivative is the function that describes the slope of a tangent to a curve in some point. If we consider a curve $f$ at some point $a$ and draw a tangent to $f$ at the point $a$, then you can estimate the slope by $$\fr{f(x) - f(a)}{x - a},$$ for some $a$ near $x$. That's an estimate. We would get a closer estimate if we choose a number $x$ closer to $a$ than the previous. If we choose $x$ exactly equal to $a$, we get a division by zero and we go beyond what the definition of limit lets us do. Considering the limit of the expression as $x$ approaches $a$, we get $$f'(a) = \lim_{x \to a} \fr{f(x) - f(a)}{x - a},$$ which is the derivative of the function $f$ at a point $a$ with a substitution of $h$ for $x - a$. Let $h = x - a$, then $x = h + a$. Notice that as $x$ approaches $a$, $x - a$ approaches 0. So if $x$ approaches $a$, then $h$ approaches 0. So replace $x$ by $h + a$ noticing that $h$ approaches 0 as $x$ approaches $a$ and we get Equation~\ref{def:derivative}. So both definitions are equivalent. \newcommand{\Dx}{\Delta x} \newcommand{\Dy}{\Delta y} \newcommand{\D}[1]{\Delta{#1}} \newcommand{\Du}{\Delta u} \newcommand{\Dv}{\Delta v} \begin{theorem*} A derivada de $ax$ é $a$. \end{theorem*} \begin{proof} Por definição, \begin{align*} \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} &= \lim \frac{a(x + h) - ax}{h}\\ &= \lim \frac{ax + ah - ax}{h}\\ &= \lim \frac{a h}{h}\\ &= a \lim \frac{h}{h}\\ &= a, \end{align*} como desejado. \end{proof} \begin{theorem*} A derivada de $ax^2$ é $2ax$. \end{theorem*} \begin{proof} Por definição, \begin{align*} \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} &= \lim \frac{a(x + h)^2 - ax^2}{h}\\ &= \lim \frac{a(x^2 + 2xh + h^2) - ax^2}{h}\\ &= \lim \frac{\cancel{ax^2} + 2axh + ah^2 - \cancel{ax^2}}{h}\\ &= \lim \frac{\cancel{h}(2ax + ah)}{\cancel{h}}\\ &= \lim 2ax + ah\\ &= \lim 2ax\\ &= 2ax, \end{align*} como desejado. \end{proof} \begin{theorem*} A derivada de $ax^3$ é $3ax^2$. \end{theorem*} \begin{proof} Por definição, \begin{align*} \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} &= \lim \frac{a(x + h)^3 - ax^3}{h}\\ &= \lim \frac{a(x^3 + 3x^2h + 3xh^2 + h^3) - ax^3}{h}\\ &= \lim \frac{\cancel{ax^3} + 3ax^2h + 3axh^2 + ah^3 - \cancel{ax^3}}{h}\\ &= \lim \frac{\cancel{h}(3ax^2 + 3axh + ah^2)}{\cancel{h}}\\ &= \lim 3ax^2 + 3axh + ah^2\\ &= 3ax^2, \end{align*} como desejado. \end{proof} \begin{remark*} Relevante é o Teorema Binomial. \end{remark*} \begin{remark*} Os dados coletados são \begin{table}[htb] \centering \begin{tabular}{r|c} função & derivada \\\toprule $ax^1$ & $ax^0 = a$\\ $ax^2$ & $3ax^1 = 2ax$\\ $ax^3$ & $3ax^2$\\ $\vdots$ & $\vdots$\\ $ax^n$ & $nax^{n-1}$\\\bottomrule \end{tabular} \caption{A regra da potência.} \label{tb:regra-potência} \end{table} \end{remark*} \undef\d\newcommand{\d}{\frac{d}{\dx}} \begin{theorem*}[regra da soma] Se $f$ e $g$ são deriváveis, então % $$\d (f(x) + g(x)) = \d f(x) + \d g(x)$$ % \end{theorem*} \begin{proof} Seja $F(x) = f(x) + g(x)$. Por definição, % \begin{align*} \d (f(x) + g(x)) &= \lim_{h \to 0} \frac{F(x + h) - F(x)}{h}\\ &= \lim \frac{f(x + h) + g(x + h) - (f(x) + g(x))}{h}\\ &= \lim \frac{f(x + h) + g(x + h) - f(x) - g(x)}{h}\\ &= \lim \left(\frac{f(x + h) - f(x) + g(x + h) - g(x)}{h}\right)\\ &= \lim \frac{f(x + h) - f(x)}{h} + \lim \frac{g(x + h) - g(x)}{h}\\ &= \d f(x) + \d g(x), \end{align*} % como desejado. \end{proof} A regra do produto é mais envolvente. Stewart oferece uma analogia que é bem útil. Suponha que $f$ e $g$ sejam sempre positivas, o que nos permite pensar em $fg$ como uma área, como uma a área de um retângulo. É uma analogia pra dar um sentido geométrico à variação $\Delta (fg)$ e nos ajudar a escrever a prova do fato. \begin{theorem*}[regra do produto] Se $u = f(x)$ e $v = g(x)$ são diferenciáveis, então % $$\d f(x)g(x) = (uv)' = u'v + u v' = f'(x)g(x) + f(x)g'(x).$$ % \end{theorem*} \begin{proof} Se $x$ variar uma quantidade pequena, como $\Dx$, então as variações de $u$ e $v$ são $\Du = f(x + \Dx) - f(x)$ e $\Dv = g(x + \Dx) - f(x)$. Logo, a variação do produto é % \begin{align*} \D{(uv)} &= (u + \Du)(v + \Dv) - uv\\ &= \cancel{uv} + u\Dv + v\Du + \Du\Dv - \cancel{uv}\\ &= u\Dv + v\Du + \Du\Dv, \end{align*} % Agora divida essa equação por $\Dx$ e compute o limite à medida que $\Dx \to 0$. Obtemos % \begin{align*} (uv)' &= \lim_{\Dx \to 0} \frac{\D{(uv)}}{\Dx}\\ &= \lim_{\Dx \to 0} \frac{u\Dv + v\Du + \Du\Dv}{\Dx},\\ &= \lim_{\Dx \to 0} \left(\frac{u\Dv}{\Dx} + \frac{v\Du}{\Dx} + \frac{\Du\Dv}{\Dx}\right),\\ &= \lim_{\Dx \to 0} \left(u\frac{\Dv}{\Dx} + v\frac{\Du}{\Dx} + \Du\frac{\Dv}{\Dx}\right),\\ &= u\lim_{\Dx \to 0}\frac{\Dv}{\Dx} + v\lim_{\Dx \to 0}\frac{\Du}{\Dx} + \left(\lim_{\Dx \to 0}\Du\right)\frac{\Dv}{\Dx},\\ &= u\lim_{\Dx \to 0}\frac{\Dv}{\Dx} + v\lim_{\Dx \to 0}\frac{\Du}{\Dx} + \cancel{0 \cdot \frac{\Dv}{\Dx}},\\ &= uv' + vu', \end{align*} % como desejado.\end{proof} \begin{remark*} Por que $\Du \to 0$? Porque $\Du = f(x + \Dx) - f(x)$. Em palavras, $\Du$ vai pra zero quando $\Dx$ vai pra zero. \end{remark*} \begin{remark*} Observe que parte da sequência de equações foi escrita de trás pra frente. Pra expressar o limite de uma soma como uma soma de limites, precisamos saber se o limite dos termos de fato existe. Mas, se soubermos que o limite dos termos de fato existe, a prova está essencialmente terminada. \end{remark*} A regra do quociente também aparece com aplicação direta da definição de derivada como o limite da razão de variações. Suponha uma pequena variação em $x$, o que implica uma pequena variação em $u$ e $v$. Compute então a variação do quociente $u/v$ e deixe a álgebra trabalhar --- que é computar a subtração das frações já que temos uma diferença de quocientes envolvida aqui. \begin{theorem*}[regra do quociente] Se $u = f(x)$ e $v = g(x)$ são diferenciáveis, então % $$\left(\frac{u}{v}\right)' = \frac{u'v - u v'}{v^2}.$$ % \end{theorem*} \begin{proof} Comece considerando a variação do quociente, que por definição é % \begin{align*} \D{\left(\frac{u}{v}\right)} &= \frac{u + \Du}{v + \Dv} - \frac{u}{v}\\ &= \frac{v(u + \Du) - u(v + \Dv)}{v(v + \Dv)}\\ &= \frac{\cancel{uv} + v\Du - \cancel{uv} - u\Dv}{v(v + \Dv)}\\ &= \frac{v\Du - u\Dv}{v(v + \Dv)}. \end{align*} % Agora divida tudo por $\Dx \to 0$ e obtenha % \begin{align*} \lim \frac{\D{\left(\frac{u}{v}\right)}}{\Dx} &= \lim \left(\frac{1}{\Dx} \frac{v\Du - u\Dv}{v(v + \Dv)}\right)\\ &= \lim \left(\frac{v\frac{\Du}{\Dx} - u\frac{\Dv}{\Dx}}{v(v + \Dv)}\right)\\ &= \frac{\lim\left(v\frac{\Du}{\Dx} - u\frac{\Dv}{\Dx}\right)}{\lim v(v + \Dv)}\\ &= \frac{v\lim\frac{\Du}{\Dx} - u\lim\frac{\Dv}{\Dx}}{v\lim(v + \Dv)}\\ &= \frac{vu' - uv'}{v^2}, \end{align*} % como desejado. \end{proof} \begin{exercise*} Encontre o ponto onde duas retas perpendiculares entre si, tangentes à parábola $x^2$, se encontram no eixo $y$. O ponto existe, como ilustra a Figura~\ref{fg-x2-perpendicular-tangents}. % \begin{figure}[!htb] \centering \includegraphics[width=0.8\textwidth, angle=0]{images/x2-perpendicular-tangents.pdf} \caption{O ponto $(0, -1/4)$ é a interseção entre as tangentes à curva $x^2$ que são perpendiculares.} \label{fg-x2-perpendicular-tangents} \end{figure} % \sol. Sendo perpendiculares, o produto das inclinações das tangentes é $-1$, ou seja, no lado positivo a inclinação é $2a$ e no lado negativo, $- 1/(2a)$. Seja $a > 0$. A reta tangente no lado positivo é $y - a^2 = 2a(x - a) = 2ax - 2a^2$, implicando $y = 2ax - a^2$. Similarmente, no lado negativo, obtemos a reta tangente $y - (-a)^2 = (-1/(2a))(x - (- a))$, implicando $y = (-1/(2a)x) + a^2 -1/2$. Estamos interessados no ponto de encontro entre as tangentes no eixo $y$, se elas de fato se encontram lá. Iguale as equações de ambas tangentes e coloque $x = 0$, obtendo % \begin{align*} 2ax - a^2 &= -\frac{1}{2a}x + a^2 -\frac{1}{2}\\ 2a^2 &= \frac{1}{2}\\ a^2 &= \frac{1}{4}\\ a &= \pm \frac{1}{2}, \end{align*} % o que nos revela que as tangentes relativas aos pontos $x = - 1/2$ e $x = 1/2$ são perpendiculares se encontram no eixo $y$. Elas têm inclinação $-1$ e $1$, logo quando $x=0$, obtemos $y = - 1/4$, como desejado. \end{exercise*} \begin{exercise}\label{exe:hiperbole} Considere a hipérbole $xy = c$. Marque qualquer ponto na curva. Trace a reta tangente à curva ao ponto. Essa reta forma um triângulo retângulo sendo os eixos $x, y$ os catetos. Mostre que a coordenada $x$ do ponto é o ponto médio entre a origem a inserseção entre a reta tangente e eixo $x$. Em seguida, mostre que a área do triângulo é uma constante~\cite[exercício~76, seção~3.1]{calculo}. % \begin{figure}[!htb] \centering \includegraphics[width=0.8\textwidth]{images/hyperbole-triangle-constant-area.pdf} \caption{Não importa que ponto na tangente escolhemos, a área do triângulo é sempre $2c$.} \label{fg-hyperbole-triangle} \end{figure} \sol. Mostraremos primeiro que a coordenada da interseção entre a tangente e o eixo $x$ é $2p$, sendo $p$ a coordenada $x$ do ponto escolhido. Seja o que ocorrer no lado positivo da hiperbóle, ocorrerá no lado negativo porque a hipérbole é uma função ímpar. Seja $p > 0$. A derivada é $y' = - c/x^2$, logo $y'(p) = -c/p^2$. A reta tangente tem equação % \begin{align*} y - \frac{c}{p} &= - \frac{c}{p^2}(x - p)\\ y &= -\frac{cx}{p^2} + \frac{2c}{p} \end{align*} % e intercepta o eixo $x$ quando $y = 0$. Logo, % \begin{align*} 0 &= -\frac{cx}{p^2} + \frac{2c}{p}\\ \frac{cx}{p^2} &= \frac{2c}{p}\\ x &= 2p, \end{align*} % como desejado. Pra mostrar que a área do triângulo é constante, precisamos da altura dele. A altura é obtida quando $x = 0$. Obtemos % \begin{equation}\label{eq:hiperbole-altura} y = 2c/p. \end{equation} % Logo, a área do triângulo é % $$\frac{1}{2} 2p \frac{2c}{p} = 2c,$$ % uma constante. \end{exercise} \begin{exercise}\label{exe:hiperbole2} Let $t(x)$ be the equation of the tangent line to the curve $1/x$ at the point $(x, 1/x)$. Let $u(x)$ be the function that maps $x$ to the intersection that $t(x)$ makes with the $y$-axis. Compute the derivative of $u(x)$ at the point $x=1$. % \begin{figure}[!htb] \centering \includegraphics[width=0.8\textwidth, angle=0]{images/tangent-intercepts-vertical-axis.pdf} \caption{À medida que o ponto $A$ se move pra direita, o ponto $B$ desce em direção à origem. O problema está perguntando --- quão rápido $B$ está descendo quando $A$ passa pelo ponto $x = 1$?} \label{fg:how-fast-B-goes-down} \end{figure} % The problem is asking me to find out how fast the intersection between the $y$-axis and the tangent $t(x)$ is going down as it passes by $x = 1$ as we move $x$ to the right, as depicted by Figure~\ref{fg:how-fast-B-goes-down}. \sol. The derivative of $1/x$ is $- 1/x^{2}$. It looks like I need to use a general point $x = a$, otherwise I can't get an algebraic formula for $u(x)$. The slope of the tangent at $(a, 1/a)$ will be $-1/a^2$. Now the tangent line is % \begin{align*} t(x) - \frac{1}{a} &= -\frac{1}{a^2}(x - a)\\ t(x) &= -\frac{x}{a^2} + \frac{1}{a} + \frac{1}{a} = \frac{-x + 2a}{a^2}. \end{align*} % It turns out the function $u(x)$ being the function that measures how high $t(x)$ intercepts the vertical axis, we'll end up having $u(a)$ and not $u(x)$ because we'll fix $x$ and let $a$ vary, since the interception happens at $x = 0$. Therefore, % $$u(a) = t(0) = \frac{2a}{a^2} = \frac{2}{a}.$$ % I want the derivative of $u(a)$ at $a = 1$, % \begin{align*} u'(a) = \frac{-2}{a^2}. \end{align*} % At $a=1$, we get $u'(1) = 1$. The slope of $u(a)$ looks reasonable because it is always negative and because it goes to zero as $a$ grows without a bound. \end{exercise} \begin{remark*} Usando o exercício~\ref{exe:hiperbole}, sabemos que a tangente à curva $1/x$ toca o eixo $y$ na altura $y = 2/p$ já que $c=1$ na hipérbole $1/x$. Assim, o exercício~\ref{exe:hiperbole2} é solucionado meramente computando a derivada de $2/p$, que é $-2/p^2$ e quando $p=1$, a derivada é $-2$, como desejado. \end{remark*} \subsection{The chain rule} Suppose I write % $$f(x) = x^2.$$ % This means that $f$ is a function of $x$. Now, if I say: differentiate $f$ with respect to $t$, someone would write % $$\fr{df}{dt} = 2x \fr{dt}{dx}$$ % because % $$\fr{df}{dt} = \fr{df}{dx} \fr{dx}{dt}.$$ % I'd like to justify this. Looking at Stewart, he says that it's reasonable to consider derivatives as rates of change; that's alright with me. Also, he says that the derivative of a composition of functions $h(g(x))$ is the product of the derivatives $h$ and $g$. So % $$[h(g(x)]' = h'(g(x)) g'(x) x'$$ % but since we're differentiating with respect to $x$, we get % $$[h(g(x)]' = h'(g(x)) g'(x)$$ % because $x' = 1$. In the function $f$, however, where's $t$? I think that we have to suppose that $x$ is a function of $t$: $x(t)$. So we rewrite $f$ as % $$f(x) = x(t)^2.$$ % Now, differentiating it, we need the chain rule which would say % $$\fr{df}{dt} = 2x(t) \fr{dx}{dt}.$$ % But, the truth is that here $x$ has no relationship with $t$ --- we didn't specify any, so it doesn't make any sense to draw such a relationship here. However, these relationships exist in related rates problems, so there to express the chain rule using Leibniz notation is useful and that's why we do it. Now, I'd like to investigate why the chain rule is as it is, without resorting to a formal proof of it. I want an intuitive idea of why the rate of change of a composition is the product of the rate of change of each of the functions involved in the composition. It's funny that the for a function $a(x) = b(x)c(x),$ the derivative of $a(x)$ is not the product of the derivatives of $b(x)$ and $c(x)$, but if $a(x) = b(c(x))$, then it is --- the derivative of $a(x)$ is the product of the derivatives of $b(x)$ and $c(x)$. Consider $f(x(t))$; that is, the composition of $f$ and $x$ and $t$. I want the rate of change of $f$ with respect to $t$. But $f$ is related to $t$ through $x(t)$. So I reason: if $x$ changes twice as fast compared to the changes in $t$, and $f$ changes three times as fast compared to the changes in $x(t)$, then we would expect $f$ to change six times as fast compared to $t$. If $t$ causes a double change in $x$, and $x$ causes a triple change in $f$, then $t$ causes a ``sixple'' change in $f$. This is a principle of mathematics. I think that this is called the Counting Principle. The chain rule expresses a chain of dependences between functions. Such dependence is resolved by the operation of multiplication because the dependence is an instance of the counting principle. The same does not happen to the product of two functions. The derivative of $f(x)g(x)$ cannot be obtained by an application of the chain rule. If so, then in order for us to be able to differentiate any function, we need the product rule, and the chain rule. \subsection{Related rates} Related rates involve direct applications of the chain rule. Suppose that oxygen is being pumped into a balloon --- let it be spherical, so that we can use the volume of a sphere which is widely known. Suppose that the rate which oxygen goes in is 10 cubic meters per second --- whatever. Now, as the balloon grows in volume, the volume grows --- evidently --- and the radius of the balloon also grows. We may ask: how fast is the radius growing when the diameter of the balloon is 1 meter? The rate at which oxygen is pumped is a derivative % $$\fr{dV}{dt} = 10 \fr{m^3}{s}.$$ % We know that volume of a sphere is described by % $$V = \fr{4}{3} \pi r^3$$ % which connects volume to the radius. We want the rate of growth of the radius, so we're looking for a derivative of the volume with respect to the radius; we were given the derivative of the volume with respect to time. We then express the change in volume with respect to the change in the radius by using the chain rule. That is, % \begin{align*} \fr{dV}{dt} &= \fr{dV}{dr} \fr{dr}{dt}\\ &= 4 \pi r^2 \fr{dr}{dt} \end{align*} % Connection established. We now solve for $dr/dt$. % $$\fr{dV}{dt} \fr{1}{4 \pi r^2} = \fr{dr}{dt}.$$ % But since $\fr{dV}{dt} = 10 \fr{m^3}{s}$, we have % $$\fr{{10 m^3/s}}{4 \pi r^2} = \fr{dr}{dt}.$$ % When the diameter is 1 meter, the radius is 0.5 meters, so % \begin{align*} \fr{{10 \frac{m^3}{s}}}{4 \pi (0.5)^2} &= \fr{dr}{dt}\\ \fr{10}{\pi} &= \fr{dr}{dt}, \end{align*} % so the radius of the balloon is growing at a speed of $\fr{10}{\pi}$ meters per second. \subsection{Linearizations}\label{sec:linearization} Suppose that we want to find the tangents to $$\fr{x}{x + 1}$$ that touch the point (1,2). We would first compute the derivative of the curve which is % $$\fr{1}{(x + 1)^2}.$$ % Now, these tangents will look like % $$y - f(a) = f'(a) (x - a).$$ % This is the point-slope formula of a line. We know that the slope of a line is found by using two points of the line and computing % $$\fr{y_2 - y_1}{x_2 - x_1} = s,$$ % so $y_2 - y_1 = s (x_2 - x_1).$ Here, however, we only have one point, but we also have a form for the slope. At a point $x = a$, we have % $$y_2 - f(a) = \fr{1}{(a + 1)^2} (x_2 - a),$$ % so for these tangents to touch (1,2) it must be true that % $$2 - \fr{a}{(a + 1)} = \fr{1}{(a + 1)^2} (1 - a).$$ % Now, which numbers $a$ are solutions to this equation? They're $-2 - \sqrt{3}$, and $-2 + \sqrt{3}$. This gives us two lines; they are % \begin{align*} y - \fr{(-2 - \sqrt{3})}{(-2 - \sqrt{3} + 1)} &= \fr{(x - (-2 - \sqrt{3}))}{(-2 - \sqrt{3} + 1)}^2\\ y - \fr{(-2 + \sqrt{3})}{(-2 + \sqrt{3} + 1)} &= \fr{(x - (-2 + \sqrt{3}))}{(-2 + \sqrt{3} + 1)}^2. \end{align*} % Now, because these two lines are tangents to the curve, they share nearby points when $x$ is close enough to 1. So, under these circumstances, we may say that % $$f(x) = L(x) = f(a) + f'(a)(x - a),$$ % for $x$ nearby $x = a.$ \subsection{Sobre a palavra ``tangente''} A linearização de uma curva, o que analisamos na Seção~\ref{sec:linearization}, é a obtenção de uma reta tangente a uma curva $f(x)$. O nome ``tangente'' é usado primeiramente trigonometria. Por que a palavra ``tangente'' reaparece na linearização local de uma curva? A palavra vem do Latim {\em tangens} que significa tocar. Usando o círculo unitário, desenhe um ângulo agudo no primeiro quadrante e chame-o de $\alpha$. A linha reta que determina $\alpha$ tem uma inclinação, que é % $$\frac{\sin\alpha}{\cos\alpha} = \tg \alpha.$$ % Quando calculamos a inclinação de uma reta tangente a uma curva $f(x)$, obtemos exatamente a tangente de um ângulo $\alpha$ formado pela reta tangente à curva $f(x)$ e o eixo $x$. Quando computamos uma aproximação da inclinação da reta tangente, fazemos referência a retas ``secantes''. O termo ``secante'' vem do Latim ``secans'', que significa cortar. No círculo unitário, a reta secante corta o círculo. Ao estimarmos a inclinação de uma reta tangente a uma curva, uma secante tende toca a curva em dois pontos, ou seja, corta um pedaço da curva. Veja também a Seção~\ref{sec:sine-cosine} pra compreender o ``co'' em ``cosseno''. \subsection{Differentials} If $y = f(x)$, where $f$ is a differentiable function, then the differential $dx$ is an independent variable; that is, $dx$ can be given the value of any real number. The differential $dy$ is then defined in terms of $dx$ by the equation % $$dy = f'(x) dx$$ % So $dy$ is a dependent variable; it depends on the values of $x$ and $dx.$ If $dx$ is given a specific value and $x$ is taken to be some specific number in the domain of $f$, then the numerical value of $dy$ is determined. Suppose that $P$ is a point on a curve $f(x)$ with coordinates ($x$, $f(x)$) and that $Q$ is some point a little further on the curve with coordinates $(x + \Delta x, f(x + \Delta x)).$ Let $dx = \Delta x$. Then, the change in $y$ is % $$\Delta y = f(x + \Delta x) - f(x).$$ % Certainly, because $f(x + \Delta x)$ is the height of the point $Q$; then, if we subtract the height of the point $P$, we get the change in $y$ as we moved from $P$ to $Q$. Now, $dy$ expresses the change in the linearization of $f(x)$, which for a concave down curve with positive slope, would be larger than $\Delta y$. In a concave down graph with positive slope, for example, let $L(x)$ be a tangent to the curve at $P$. So $\Delta y < dy$. The value of $\Delta y$ is simply $f(x + \Delta x) - f(x)$, so the value of $dy$ should be something like $L(x + \Delta x) - L(x)$. However, the value of $dy$ is $f'(x) dx$. So it must follow that %v $$L(x + \Delta x) - L(x) = f'(x) \dx.$$ %e %% \subsection{Newton's method} %% Newton's method is a clever way to extract good approximations of roots %% of equations if you have a good clue of where the root is. Without the a %% good clue, there may not be any good result. %% The point slope form of a line is %cm %% y - f(x_1) = f'(x_1)(x - x_1), %% where x_1 is a given number. If we set y = 0 in this equation, then we %% find the root of it; the number x in which y = 0. If the slope is not %% zero, we may write %align by = %% 0 - f(x_1) = f'(x_1)(x - x_1) %% - f(x_1) = f'(x_1)(x - x_1) %% - f(x_1)/f'(x_1) = x - x_1 %% - f(x_1)/f'(x_1) + x_1 = x %% \fr{- f(x_1)}{f'(x_1)} = x - x_1 %% \fr{- f(x_1)}{f'(x_1)} + x_1 = x - x %% 0 - f(x_01 %% Exercise. Start with x_1 = -1, and find the real root of f(x) = x^3+x+1=0. %% Solution. A formula for the slope is 3x^2 + 1, so a tangent line is %cm %% y - f(-1) = (3x^2 + 1)(x - x_1), %% which in turn gives us easily a root x = -3/4 if we set y = 0. So -3/4 %% becomes our next guess. We write %cm %% y - f(-3/4) = (3(-3/4)^2 + 1)(x - (-3/4)) %% 0 - 0.671875 = 2.6875x + 2.015625 %% 2.6875 = 2.6875x %% 1 = x \subsection{More on differentiation} \masciith{extreme value theorem} If $f$ is continuous on a closed interval $[a,b]$, then $f$ attains an absolute maximum value $f(c)$ and an absolute minimum value $f(d)$ at some numbers $c$ and $d$ in $[a,b]$.\theend \masciith{fermat's theorem} If $f$ has a local maximum or minimum at $c$, and if $f'(c)$ exists, then $f'(c) = 0$. \theend \masciiproof We need to assume a local maximum and show the conclusion, then a local minimum and show the conclusion, but both phases are similar, so we'll write only one. Let $f(c)$ be a local minimum. So $f(c) \leq f(x).$ Later.\profend \masciidef{} A critical number of a function $f$ is a number $c$ in the domain of $f$ such that $f'(c)$ or $f'(c)$ does not exist. \defend \masciith{fermat's theorem} If has a local maximum of minimum at $c$, then $c$ is a critical number of $f$. \theend From this, and supposing that we have implemented an algorithm to compute derivatives of functions, then we can write a program that can answer the critical numbers of a function. \masciith{} Let $f(x)$ be continuous on a closed interval $[a,b]$. (1) Compute all critical points in $(a,b)$. (2) Compute the values of $f$ at % $$x = a \text{ and } x = b.$$ % Now, the largest computed value in (2) and (3) is the absolute maximum; the smallest is the absolute minimum. \theend %e \masciiexe{} Show that 5 is a critical number of %v $$g(x) = 2 + (x - 5)^3,$$ %e and show that $g(x)$ has no local extremum at $5.$ {\em Solution.} A critical number $c$ is one in which $g'(c) = 0$ or $g'(c)$ does not exist. We have $g'(5) = 0,$ so $5$ is a critical number. Assume that $5$ is a local maximum. Then $g(5) \geq g(x)$ for all $x$ in some open interval $(5 - d, 5 + d),$ where $d > 0.$ Let $x < 5.$ Then %align by < %v \begin{align*} x - 5 &< 0\\ (x - 5)^3 &< 0\\ 2 + (x - 5)^3 &< 2. \end{align*} %e Let $x > 5.$ Then %v \begin{align*} x - 5 &> 0\\ (x - 5)^3 &> 0\\ 2 + (x - 5)^3 &> 2. \end{align*} %e So there's no neighborhood of $5$ such that $g(5) \geq 2.$ \exeend \masciiexe{} Prove that $x^3 + x - 1 = 0$ has exactly one real root. {\em Solution.} Let $f(x) = x^3 + x - 1.$ Notice that $f(0) = -1$ and that $f(1) =$ $1.$ Since polynomials are continuous, then there is a $c$ in $(0,1)$ such that $f(c) = 0,$ so $c$ is a root. Now, suppose that there are two roots, a and $b.$ Since they're roots, it's true that $f(a) = 0 = f(b).$ Notice that $f$ is continuous on $[a,b]$ and differentiable at $(a,b)$ so there's a number $d$ in $[a,b]$ such that $f'(d) =$ $0.$ But $f'(x) = 3x^2 + 1 \geq 0$ for all $x,$ so no real $d$ is possible, and therefore, it's false that there are two roots, so there's only one. \exeend \masciiexe{} If $f(x)$ has a minimum value at $c,$ show that $-f(x)$ has a maximum value at $c.$ {\em Solution.} If $f$ has a minimum at $c,$ then $f(c) \leq f(x)$ for all $x$ in the domain of $f.$ Multiply this inequality by $-1$ to get $-f(c) \geq -f(x)$ for all $x$ in the domain of $f.$ So $c$ is a maximum for $-f(x).$ \exeend \masciiexe{} A cubic function is a polynomial of degree $3;$ that is, it has the form $f(x) = mx^3 + nx^2 + ox + p$, where $m \neq 0.$ Show that a cubic function can have no more than two critical numbers. How many local extrema can a cubic function have? {\em Solution.} Polynomials are differentiable, so there is no real number $c$ to which $f'(c)$ does not exist. Notice that $f'(x) = 3mx^2 + 2nx + o$ which is is a quadratic equation which has only two roots maximum, so no more than two critical numbers is possible. Every local extremum is a critical number, so no more than two would be possible. \exeend \subsection{The mean value theorem} Rolle's theorem says that a differentiable function that goes up and down has some $c$ such that $f'(c) = 0.$ The up and down criteria is that $f(a) = f(b)$ for a domain $[a,b].$ So let's write the theorem. \masciith{} Let $f$ be a function that satisfies: $f$ is continuous on the closed interval $[a,b], f$ is differentiable on the open interval $(a,b),$ $f(a) = f(b).$ Then there is a number $c$ in $(a,b)$ such that $f'(c) = 0.$ \masciiproof{} We prove in three cases. Case one: let $f(x)$ be a constant $k.$ Then $f'(x) = 0$ and so any number $c$ in $(a,b)$ can be taken. Case two: let $f(x) > f(a)$ for some $x$ in $(a,b).$ I'm bored here. Let's move on. \profend The Mean Value Theorem should be called perhaps the Same Slope Theorem. I don't really know why the word ``mean'' was chosen; perhaps there's more to it than what I know right now. The theorem says that if $f$ is very nice on $[a,b],$ then there is a number in the interval to which a tangent to the curve has the same slope as the secant $AB$ --- that is, a secant line from the $a$ to $b.$ The tangent would be parallel to this secant line, then. \masciith{mean value theorem} Let $f$ be continuous on $[a,b]$ and differentiable on $(a,b).$ There is a $c$ in $(a,b)$ such that $$f'(c) = \fr{f(b)-f(a)}{b-a}.$$ \theend \masciiexe{} Suppose that $f(0) = -3$ and $f'(x) \leq 5$ for all $x.$ How large can $f(2)$ possibly be? {\em Solution.} We're given one point and a restriction for the slope of the function. Then we're asked how large can $f(2)$ be? Geometrically, plot the number $(0,-3)$ and take your ruler and position it as if you were going to draw a line with slope $5$ --- the maximum slope you can. Then, see how large $f(2)$ could be. Now, precisely. In the interval $[0,2],$ there is a number $c$ such that %align by = %v \begin{align*} f'(c) &= \fr{f(2)-f(0)}{2-0}\\ &= \fr{f(2)+3}{2}. \end{align*} %e But we're looking for the highest value of $f(2),$ so we will use our highest slope possible: $5.$ That is, %v \begin{align*} 5 &= \fr{f(2)+3}{2}\\ 10 &= f(2) + 3\\ 7 &= f(2). \end{align*} %e Très mignon, non? \exeend \masciith{} Let $f'(x) = 0$ for all $x$ in $(a,b).$ Then $f$ is constant. \theend \masciiproof{} Let $x_1$ and $x_2$ be any two distinct numbers in $(a,b),$ so $x_1 < x_2.$ Since $f$ is differentiable on $(a,b)$ it is differentiable on $(x_1, x_2)$ and therefore continuous on $[x_1, x_2].$ By the mean value theorem, there's $c$ in $[x_1, x_2]$ such that $$f'(c) = \fr{f(x_2)-f(x_1)}{x_2-x_1}.$$ Since $f'(x) = 0$ for all $x,$ then $f'(c) = 0.$ So %v \begin{align*} 0 &= \fr{f(x_2)-f(x_1)}{x_2-x_1}\\ &= f(x_2) - f(x_1)\\ f(x_1) &= f(x_2). \end{align*} %e So $f$ has the same value at any two distinct numbers $x_1, x_2$ in $(a,b),$ which by definition means that $f$ is constant. \profend \masciicor{} Let $f'(x) = g'(x)$ for all $x$ in $(a,b).$ Then $f - g$ is constant on $(a,b).$ That is, $f(x) = g(x) + c$ where $c$ is a constant. \corend \masciiproof{} Let $F(x) = f(x) - g(x).$ Then %v $$F'(x) = f'(x) - g'(x) = 0$$ %e for all $x$ in $(a,b).$ So $f(x) - g(x)$ is constant.\profend \masciiexe{} Let $f(x) = 1 - x^{2/3}.$ Show that $f(-1) = f(1)$ but there is no number $c$ in $(-1, 1)$ such that $f'(c) = 0.$ \exeend \masciiexe{} Let $f(x) = \sqrt[3]{x}$ be defined on $[0,1].$ Find all values that satisfy the mean value theorem. {\em Solution.} The function is differentiable on $(0,1),$ so by the mean value theorem there is a $c$ in $(0,1)$ such that %v \begin{align*} f'(c) &= \fr{f(0)-f(1)}{-1}\\ f'(c) &= 1. \end{align*} %e Which number $c$ would be this one? It's true that $f'(x) = \fr{x^{-2/3}}{3}.$ So %v \begin{align*} \fr{c^{-2/3}}{3} &= 1\\ c^{-2/3} &= 3\\ \fr{1}{\sqrt[3]{c^2}} &= 3\\ \fr{1}{c^2} &= 27\\ \fr{1}{27} &= c^2\\ \pm \sqrt{\fr{1}{27}} &= c. \end{align*} %e But only $\sqrt{\fr{1}{27}}$ is applicable because $0 < c < 1.$ \exeend \masciiexe{} Show that $x^5 + 10x + 3 = 0$ has exactly one real root. {\em Solution.} Let $p(x) = x^5 + 10x + 3.$ We first show that there is a real root. Notice that $p(-1) = -8$ and $p(1) = 14.$ By the intermediate value theorem, there is a $c$ in $(-1,14)$ such that $p(c) = 0,$ so $c$ is a root. Now suppose that there are two roots: $a$ and $b.$ Since they're roots, it's true that $f(a) = 0 = f(b).$ So, by Rolle's theorem, there is a $d$ in $(a,b)$ such that $f'(d) = 0.$ Let's find $d.$ Since $p'(x) = 5x^4 + 10 \geq 0$ for all $x,$ it turns out that no such $d$ is possible. Therefore, there is no two real roots $a$ and $b.$ So there's only one. \exeend \masciiexe{} Show that $x^5 -6x + c = 0$ has only one root in $[-1,1].$ {\em Solution.} Let $f(x) = x^5 -6x + c.$ Suppose $f(x) = 0$ has two roots a and $b$ in $[-1,1].$ Then $f(a) = 0 = f(b)$ so by Rolle's theorem there is a $d$ in $(a,b)$ with $f'(d) = 0.$ So $0 = f'(d) = 5d^4 - 6$ which implies $$d = \pm \sqrt[4]{\fr{6}{5}}$$ which are both outside $(a,b)$ because $(a,b)$ subset $[-1,1]$ and $d$ is not in $[-1,1].$ So $f(x)$ can have at most one root in $[-1,1].$ This problem is different from the previous one because in the previous one, we're supposed to show that there is one root, and here there could be none in that particular interval. For example, if $c = 5,$ we have $f(x) = x^5 - 6x + 5,$ so $x = 1$ is a root, but if $c > 5,$ then there are no more roots in that interval. Similarly, if $c = -5,$ we have $f(x) = x^5 -6x - 5$ and so $x = -1$ is a root, but if $c < -5,$ then there are no more roots in that interval. A good exercise here is to ponder how someone would come up with a question like that. That is, how to construct equations from roots in a particular interval such that the equation has a particular degree? Take a function such as $f(x) = x^2 - 1.$ We know that the roots of this function are $-1$ and $1,$ so % $$F(x) = \int f(x)\ dx = \fr{x^3}{3} -x + c$$ % will have local extrema at $-1$ and $1.$ Now, if $F(-1) > 0$ and $F(1) < 0,$ then by the intermediate value theorem, we know that there is a number $-1 < d < 1$ such that $F(d) = 0.$ Because we constructed $F(x)$ from $f(x),$ we know that $F(x)$ has only the local extrema we want, so no more roots can exist in that interval. Now, how can we make sure that $F(-1) > 0$ and $F(1) < 0?$ Well, at $-1,$ we have % $$\fr{-1^3}{3} + 1 + c > 0,$$ % so % $$c > \fr{1}{3} - 1 = \fr{-2}{3},$$ % and at $1,$ we have % $$\fr{1^3}{3} - 1 + c < 0,$$ % so % $$c < - \fr{1}{3} + 1 = \fr{2}{3}.$$ % So we need a number $c$ such that $$- \fr{2}{3} < c < \fr{2}{3}.$$ Now, suppose that you want your final equation to have a particular degree $n.$ Then, look for a derivative of degree $n - 1.$ For example, suppose that $n = 101.$ Then we need a function of degree $100.$ For example, $x^{100} - 1$ will have roots at $-1$ and $1.$ What if you want to work with an arbitrary interval for the root to fall in? Suppose you want an interval like $(-m,m).$ Then let $x = m$ and %align by = %v \begin{align*} x &= m\\ x - m &= 0\\ (x - m)^2 &= 0. \end{align*} %e So evidently, we have $\pm m$ as a root. Then we integrate the function to get %v $$F(x) = \fr{x^3}{3} - m x^2 + x m^2 + c.$$ %e Now by the intermediate value theorem, this is a function which has a root in $(-m, m)$ for some particular values of $c.$ To find out which values of $c$ we need to choose, we compute %v \begin{align*} \fr{-m^3}{3} -m (-m^2) + (-m) m^2 + c &< 0\\ \fr{-m^3}{3} -2 m^3 + c &< 0\\ c &< \fr{m^3}{3} + 2m^3. \end{align*} %e For the other part, %v \begin{align*} \fr{m^3}{3} -m (m^2) + (m) m^2 + c &> 0\\ \fr{m^3}{3} - m^3 + m^3 + c &> 0\\ c &> \fr{-m^3}{3}. \end{align*} %e So, % $$\fr{-m^3}{3} < c < \fr{m^3}{3} + 2m^3.$$ I wouldn't believe this if I were you. This seems rather fallacious. \exeend \masciiexe{} Show that $x^4 + 4x + c = 0$ has at most two real roots. {\em Solution.} A similar reasoning. Assume it has three: $a, b, c.$ Then it's true that $0 = f(a) = f(b) = f(c).$ So by Rolle's theorem, there is a $d$ in $(a,b)$ such that $f'(d) = 0,$ and there is an $e$ in $(b,c)$ such that $f'(e) = 0.$ The derivative of the function is $g(x) = x^3 + 4,$ and we must find two reals roots of this function; they would be $d$ and e. However, one root of $g(x)$ is $$x = \sqrt[3]{-4},$$ because %v \begin{align*} x^3 + 4 &= 0\\ x^3 &= -4\\ x &= \sqrt[3]{-4}. \end{align*} %e Now, let $x < \sqrt[3]{-4}.$ Then %v \begin{align*} x^3 &< -4\\ x^3 + 4 &< 0, \end{align*} %e and let $x > \sqrt[3]{-4},$ so %v \begin{align*} x^3 &> -4\\ x^3 + 4 &> 0. \end{align*} %e So to the left of $\sqrt[3]{-4}, g(x)$ is negative, and to the right $g(x)$ is positive. So, $g(x)$ will touch the $x$-axis only once; there can't be two real roots, then. Therefore, $f(x)$ has at most two real roots. \exeend \masciiexe{} Show that a polynomial of degree $3$ has at most $3$ real roots. {\em Solution.} A polynomial of degree $3$ is of the form %v $$ax^3 + bx^2 + cx + d.$$ %e Assume that it has four real roots. Then it's true that %v $$0 = f(a) = f(b) = f(c) = f(d).$$ %e So by Rolle's theorem, there are numbers $p$ in $(a,b), q$ in $(b,c),$ $r$ in $(c,d)$ such that $0 = f'(p) = f'(q) = f'(r).$ But $f'(x)$ is a second degree polynomial with three real roots which is not possible. \exeend \masciiexe{} Show that a polynomial of degree $n$ has at most $n$ real roots. {\em Solution.} We know that a polynomial of degree $1$ has $1$ real root. So, if we try to prove this by induction, at least we know that the proposition is true for $n = 1.$ Now, suppose that the proposition is true for some $n = k.$ Let $p(x)$ be of degree $k + 1.$ Suppose that $p(x)$ has more than $k + 1$ real roots; they are $r_1, r_2,$ ..., $r_{k+1}, r_{k+2}.$ Then %v $$0 = p(r_1) = ... = p(r_{k+2})$$ %e and by Rolle's theorem it's true that %v $$0 = p'(c_1) = ... = p'(c_{k+1}).$$ %e But if $p$ is of degree $k + 1,$ then $p'$ is of degree $k,$ and cannot have $k + 1$ roots. So it's not true that $p(x)$ has more than $k + 1$ real roots. Therefore, it has less or equal to $k + 1$ roots. Proofs by induction may seem fallacious sometimes. One could object here that we cannot base our argument on the fact that $p'$ of degree $k$ cannot have $k + 1$ roots because it is the same thing that we're trying to prove. That's an unfair objection, however, because we tested for $n = 1$ and it was true; then we assumed to be true for all polynomials of $k$th degree that they don't have $k + 1$ roots, so it's true under that scope of the proof. From that assumption, we showed that the same could not happen to $p$ of degree $k + 1$ and with everything altogether, we concluded that a polynomial of degree $n$ has at most $n$ roots. In fact, I seldom use the letter $n$ twice in a proof by induction; instead, I use $k$ for the assumption when I'm trying to show that something is true for $k + 1.$ This way I avoid getting confused with the $n$ of the proposition and raise the objection that I'm assuming what I'm trying to prove. \exeend \masciiexe{} Suppose that $f$ is differentiable on $\reals$ and has two roots. Show that $f'$ has at least one root. {\em Solution.} If this proposition is true, then it means that a differentiable function of degree $2$ or more changes direction at at least in one point. Call the roots of $f$ by $a$ and $b.$ So %v $$f(a) = 0 = f(b).$$ %e By Rolle's theorem, we have that $f'(c) = 0$ for some $c$ in $(a,b).$ \exeend \masciiexe{} Suppose that $f$ is twice differentiable on $\reals$ and $f$ has three roots. Show that $f''$ has at least one real root. {\em Solution.} Let $a, b,$ and $c$ be in $\complex$. So %v $$0 = f(a) = f(b) = f(c).$$ %e By Rolle's theorem, we have $f'(d) = 0$ for some $d$ in $(a,b)$ and $f'(e) = 0$ for some $e$ in $(b,c).$ So these are two real roots of $f'.$ Since $f$ is twice differentiable, we can differentiate $f'$ to get $f''.$ Since $f'(d) = 0 = f'(e),$ then by Rolle's theorem, there is an $i$ in $(d,e)$ such that $f''(i) = 0,$ so $f''$ has at least one real root. We may try to generalize this. \exeend \masciith{} Suppose that $f$ has $n$ roots, and is differentiable on $\reals n - 1$ times. Then $f^{(k)}$ has $n - k$ real roots where $0 < k \leq n - 1.$ \theend Let's test our generalization. Let $f(x) = x^7 - 1.$ This polynomial has $7$ roots and is $7 - 1$ times differentiable on $\reals$. Then, by our generalization, it must be true that $f'(x) = x^6$ has $7 - 1$ real roots, $f''(x) = x^5$ must have $7 - 2$ real roots, $f^{(3)}(x) = x^4$ must have $7 - 3$ real roots and so on until $f^{(6)} = x$ which must have one real root. The generalization seems to work. Can we prove this now? Hmm, I don't know. \theend %v-- \subsection{Even more on differentiation} It's interesting to notice that the mean value theorem is used in many applications. There is a variety of problems in which the mean value theorem can be used; one challenge is to recognize when you can use it; we'll describe some interesting problems in which the mean value theorem can be used as a path to the solution. Let's start by looking at the proof of the following theorem. \masciith{} If $f'(x) > 0$ on some interval $B,$ then $f$ is increasing on $B.$ If $f'(x) < 0$ on $B,$ then $f$ is decreasing on $B.$ \theend \masciiproof{} Let $x_1, x_2$ be in $B$ such that $x_1 \neq x_2,$ so we may assume that $x_1 < x_2.$ We must show that $f(x_1) < f(x_2).$ Assume that $f'(x) > 0,$ and notice that $f$ is differentiable on $[x_1,x_2].$ So by the mean value theorem, there is a $c$ in $[x_1,x_2]$ such that %v $$f'(c)(x_2 - x_1) = f(x_2) - f(x_1).$$ %e Notice that $f'(c) > 0$ and notice that $x_2 - x_1 > 0$ too. So the left side of the equation above is positive and therefore the right side must be as well; that is, $f(x_2) > f(x_1),$ and we're done. For a decreasing function, we need only repeat the proof flipping the notation around. \profend Let's see more problems in which I think that the mean value theorem would provide a way to a solution. The next three are of a pretty interesting type of problem, and we already solved one of the same kind, I think. \masciiexe{} Suppose $f(1) = 10$ and $f'(x) \geq 2$ for $1 \leq x \leq 4,$ how small can $f(4)$ possibly be? \exeend \masciiexe{} Suppose $f$ is continuous on $[2,5]$ and $1 \leq f'(x) \leq 4$ for all $x$ in $(2,5).$ Show that $3 \leq f(5) - f(2) \leq 12.$ \exeend \masciiexe{} Does there exist a function $f$ such that $f(0) = -1, f(2) = 4,$ and $f'(x) \leq 2$ for all $x?$ \exeend %v++ \masciiexe{} Show that $\tan{x} > x$ for $0 < x < \pi/2.$ {\em Solution.} We're going to show that $\tan(x) - x > 0.$ We know that % $$\tan(x) = \fr{\sin(x)}{\cos(x)}.$$ % As $x \to 0, \tan(x) \to 0.$ As $x \to \pi/2, \tan(x) - x \to \infty$ so if I can show that $\tan(x) - x$ is increasing on $(0, \pi/2),$ then we know that % $$f(x) = \tan(x) - x > 0$$ % is true for all $x$ in $(0, \pi/2)$ which makes $\tan(x) > x$ true for all $x$ in $(0, \pi/2).$ We look into $f'(x)$ then. We have % $$f'(x) = \sec^2(x) - 1 = 1/\cos^2(x) - 1$$ % which is only zero when $\cos^2(x) = 1$ which happens at $x = 0,$ but 0 is not in $(0, \pi/2).$ So, for $x$ in $(0, \pi/2),$ $f'(x)$ is always positive. Therefore, $\tan(x) - x > 0$ for all $x$ in $(0, \pi/2)$ and so $\tan(x) > x$ for all $x$ in $(0, \pi/2).$ \exeend \masciiexe{} Prove that for all $x > 1,$ % $$2 \sqrt{x} > 3 - \fr{1}{x}.$$ %v-- {\em Solution.} We will show that $2 \sqrt{x} - 3 + 1/x > 0$ for $x > 1.$ When $x = 1$ we have $f(x) = 0.$ As $x \to \infty, f(x) \to \infty.$ It's true that %v $$f'(x) = \fr{1}{\sqrt{x}} - \fr{1}{x^2}.$$ %e Now, when $f'(x) = 0?$ Never. We have that %v $$\fr{1}{\sqrt{x}} > \fr{1}{x^2},$$ %e for all $x > 1.$ So $f'(x) > 0$ for all $x > 1.$ So $f(x)$ is always increasing on $(1,\infty),$ so $2 \sqrt{x} > 3 - 1/x$ for all $x > 1.$ \exeend \masciiexe{} Suppose that $f$ and $g$ are continuous on $[a,b]$ and differentiable on $(a,b).$ Suppose also that $f(a) = g(a)$ and $f'(x) < g'(x)$ for $a < x < b.$ Prove that $f(b) < g(b).$ {\em Solution.} We're going to consider $h(x) = f(x) - g(x).$ What's $h(a)?$ It's %v $$h(a) = f(a) - g(a) = 0,$$ %e because $f(a) = g(a).$ It's true that $h(x)$ is continuous and differentiable on $(a,b),$ so we can apply the mean value theorem to say that there exists a number $c$ in $(a, b)$ such that %align by = %v \begin{align*} h'(c)&= f'(c) - g'(c)\\ &= \fr{h(b) - h(a)} {b - a}\\ &= \fr{f(b) - g(b) - f(a) + g(a)} {b - a}. \end{align*} %e We know that $b - a > 0.$ We also know that $f'(c) - g'(c) < 0$ because $f'(x) < g'(x)$ for $a < x < b.$ So it must be true that % $$f(b) - g(b) - f(a) + g(a) < 0,$$ % otherwise $f'(c) - g'(c)$ would be positive. So, we write %align by < %v \begin{align*} f(b) - g(b) - f(a) + g(a) &< 0\\ f(b) - g(b) - (f(a) - g(a)) &< 0\\ f(b) - g(b) - 0 &< 0\\ f(b) &< g(b), \end{align*} %e as desired. \exeend \masciiexe{} Show that $\sqrt{ 1 + x} < 1 + x/2$ if $x > 0.$ {\em Solution.} We'll look into $f(x) = \sqrt{1 + x} - 1 - x/2 < 0.$ At $0,$ we have $\sqrt{1} - 1 = 0.$ As $x \to \infty,$ we have $f(x) \to -\infty.$ So we must show that $f'(x)$ is always decreasing. We have %v $$f'(x)= \fr{1}{2\sqrt{1+x}}-\fr{1}{2}.$$ %e Is it ever zero? It is when $\sqrt{1 + x} = 1$ which only happens when $x = 0,$ but we don't care about $x = 0.$ We must show then %v $$2 \sqrt{1 + x} > 2.$$ %e That is, we can show that $\sqrt{1 + x} > 1$ for all $x > 0.$ It indeed is because $\sqrt{1 + x} > 1$ implies $1 + x > 1$ which is true whenever $x >$ $0.$ So $f'(x) < 0$ for all $x > 0.$ So we're done.\exeend \masciiexe{} Suppose $f$ is an odd function and is differentiable everywhere. Prove that for every positive number $b,$ there exists a number $c$ in $(-b,b)$ such that %v $$f'(c) = \fr{f(b)}{b}.$$ %e {\em Solution.} By the mean value theorem, there's a $c$ in $(-b, b)$ such that %v $$f'(c) = \fr{f(b)-f(-b)}{b+b}.$$ %e But since $f$ is odd, then we may rewrite that as %align by = %v \begin{align*} f'(c) &= \fr{f(b)+f(b)}{b+b}\\ &= \fr{2f(b)}{2b}\\ &= \fr{f(b)}{b}, \end{align*} %e as desired. \exeend \masciiexe{} Prove that $|\sin{a} - \sin{b}| \leq |a - b|$ for all $a$ and $b$ in $\reals$. {\em Solution.} By the mean value theorem, there's a $c$ in $(a, b)$ such that %v $$\sin'(c) = \fr{\sin(b)-\sin(a)}{b-a}.$$ %e We know that $\sin'(c) = \cos(c)$ and that $|\cos(c)| \leq 1.$ Now, take the absolute value of the equation above to get %align by = %v \begin{align*} |\cos(c)| &= \bl|\fr{\sin(b)-\sin(a)}{b-a}\br|\\ &= \fr{|{\sin(b) - \sin(a)}|}{|{ b - a}|}\\ |b - a||\cos(c)| &= |\sin(b) - \sin(a)|. \end{align*} %e Since $|\cos(c)| \leq 1,$ then it's true that %v $$|b - a| \geq |\sin(b) - \sin(a)|.$$ %e \exeend \masciiexe{} If $f'(x) = m$ for all $x,$ where $m$ is a constant, show that $f(x) = mx + n$ for some constant $n.$ \exeend \masciiexe{} A number $m$ is called a fixed point of $f$ if $f(m) = m.$ Prove that if $f'(x) \neq 1$ for all real numbers $x,$ then $f$ has at most one fixed point. \exeend \masciith{} Suppose $f''$ is continuous near $c.$ Then if $f'(c) = 0$ and $f''(c) > 0,$ then $f$ has a local minimum at $c,$ and if $f'(c) = 0$ and $f''(c) < 0,$ then $f$ has a local maximum at $c.$ \theend Notice that the theorem does not mention any case when $f''(c) =$ $0.$ That's because we cannot say much when this is the case. The same situation happens when $f''(c)$ does not exist. But the first derivative handles all these difficulties. \masciiexe{} Assume that all functions in this proposition are twice differentiable and that their second derivatives are never zero. If $f$ and $g$ are concave upward on $B,$ show that $f + g$ is concave upward on $B.$ \exeend \masciiexe{} Assume that all functions in this proposition are twice differentiable and that their second derivatives are never zero. If $f$ is positive and concave upward on $B,$ show that $[f(x)^2]$ is concave upward on $B.$ \exeend \masciiexe{} Assume that all functions in this proposition are twice differentiable and that their second derivatives are never zero. If $f$ and $g$ are positive, increasing, concave upward functions on $B,$ show that fg is concave upward on $B.$ \exeend \masciiexe{} Assume that all functions in this proposition are twice differentiable and that their second derivatives are never zero. Suppose that $f$ and $g$ are both concave upward on $(-\infty, \infty).$ Under what condition on $f$ will the composite function $h(x) = f(g(x))$ be concave upward? \exeend \masciiexe{} Show that a cubic function always has exactly one point of inflection. If it has three roots, show that the $x$ coordinate of the inflection point is $(x_1 + x_2 + x_3)$ / $3.$ \exeend \masciiexe{} Prove that if $(m,f(m))$ is a point of inflection of $f,$ and $f''$ exists in an open interval containing $m,$ then $f''(m) = 0.$ Stewart suggests: apply the first derivative test and fermat's theorem to the function $g = f'.$ \exeend \masciiexe{} Show that if $f(x) = x^4,$ then $f''(0) = 0,$ but $(0,0)$ is not an inflection point of the graph of $f.$ \exeend \masciiexe{} Show that $g(x) = x |x|$ has an inflection point at $(0,0)$ but $g''(0)$ does not exist. \exeend \masciiexe{} Suppose that $f''$ is continuous and $f'(m) = f''(m) = 0,$ but $f''(m) > 0.$ Does $f$ have a local maximum or minimum at $m?$ Does $f$ have a point of inflection at $m?$ \exeend \masciiexe{} Find a function $f$ whose $f'$ and $f''$ are always negative. \exeend % % Solution. Let f'' be -1, for example. It's always negative. Now, f' % could be something like -x, which is always negative too. Now, f can be % something like (-x^2)/2. %end \masciiexe{} Find a cubic function $f(x) = mx^3 + nx^2 + px + q$ that has a local maximum of $3$ at $-2$ and a local mininum of $0$ at $1.$ {\em Solution.} The requirements for $f'(x)$ are $f'(-2) = 0, f'(1) = 0$ and $f'(x) = 3mx^2 + 2nx + p.$ Also, $f(-2) = 3$ and $f(1) = 0.$ So we need %align by = %v \begin{align*} -8m + 4n - 2p + q &= 3\\ m + n + p + q &= 0\\ 12m - 4n + p + 0 &= 0\\ 3m + 2n + p + 0 &= 0. \end{align*} %e This system gives us $m = 2/9, n = 1/3, p = -4/3, q = 7/9.$ So the function that we want is $f(x) = (2/9) x^3 + (1/3) x^2 - (4/3) x + (7/9).$ \exeend \subsection{Limits at infinity} \masciidef{} Let $f$ be a function defined on some interval $(m,\infty).$ Then %v $$\lim_{x \to \infty} f(x) = L$$ %e means that for every $e > 0,$ there is a number $n$ such that %v $$\text{if } x > n \text{ then } |f(x) - L| < e. $$ %e \defend \masciidef{} Let $f$ be a function defined on some interval $(-\infty,m).$ Then %v $$\lim_{x \to -\infty} f(x) = L$$ %e means that for every $e > 0,$ there is a number $n$ such that %v $$\text{if } x < n \text{ then } |f(x) - L| < e. $$ %e \defend \masciidef{} Let $f$ be a function defined on some interval $(m,\infty).$ Then %v $$\lim_{x \to \infty} f(x) = \infty$$ %e means that for every positive $M,$ there is a number $n$ such that %v $$\text{if } x > n \text{ then } f(x) > M. $$ %e \defend \masciith{} If $r > 0$ is a rational number, then %v $$\lim_{x \to \infty} \fr{1}{x^r} = 0.$$ %e If $r > 0$ is a rational number such that $x^r$ is defined for all $x,$ then %v $$\lim_{x \to \infty} \fr{1}{x^r} = 0.$$ %e \theend \subsection{Drawing functions} We have enough tools to draw functions now. Let's look into the things which we need to draw functions. We need the critical numbers. With the critical numbers, we break the real line up into intervals delimited by the critical numbers. We then look at the first derivative of the function at each of these intervals; we then determine whether $f$ is increasing on each of these intervals. So we now have a clue of the way that the function travels in the plane of Descartes, but we need to know when it goes up and when it comes down and when it simply slows down and keeps going: we the extreme values; we need the local maxima, local minima and the absolute ones if they exist. But to find the local extrema we need only the first derivative. We observe the critical points in which the function changes sign; that is, when it changes from increasing to decreasing or from decreasing to increasing. \subsubsection{An algorithm} Let $f(x)$ be a function. Determine the domain of $f(x).$ Then look for $y$-intercepts, if possible, and find out where the curve touches the $y$-axis. Mark these points. Look for symmetry. Is it odd? Is it even? Find out, if you can, and keep that in mind making wise decisions: if it's even or odd, we draw only for $x \geq 0$ and we use a mirror later. Look for periodicity. Is there a period? Which one? Look for asymptotes. Are there vertical ones? Are there horizontal ones? Find them. Are there slant asymptotes? If so, find them. Compute $f'(x).$ Find all critical numbers. Solve $f'(x) = 0.$ Break the real line into as many intervals as possible from the solutions. Determine where $f'(x)$ is positive and where $f'(x)$ is negative. Find all local maxima and all local minima. Compute $f''(x).$ Find all inflection points by computing whether $f''(x) > 0$ or $f''(x) < 0.$ Look at $f'(x)$ when $f''(x) = 0.$ Convince yourself that you need not a new division of the real line to look into $f''(x).$ Draw $f(x);$ sketch the asymptotes; plot the intercepts, maximum and minimum points, and inflection points. Make the curve pass through these points, rising and falling according to the information given by $f'(x),$ and according to the information given by $f''(x).$ If additional accuracy is desired near any point, you may compute the value of $f'(x)$ there, since they should be close enough; notice that the tangent indicates the direction in which $f(x)$ goes. \subsection{Slant asymptotes} \masciidef{} A slant asymptote is a line $y = mx + b$ of a function $f(x)$ if and only if %v $$\lim_{x \to \infty} (f(x) - (mx + b)) = 0.$$ %e This definition says that if the two curves --- $f(x)$ and $y$ --- get closer and closer as $x \to \infty,$ then $y$ is a slant asymptote of $f(x).$ One way to compute where two curves get closer and closer is to find out whether the distance from the points in one curve to the other --- that's what the definition does, in fact. \defend Perhaps the challenge is when to recognize that we have a slant asymptote. For example, we'll have slant asymptotes when the degree of the numerator of a rational function is one more than the degree of the denominator. \masciiexe{} Show that %v $$f(x) = \fr{x^3}{x^2 + 1}$$ %e has a slant asymptote. {\em Solution.} Perform long division to find that %v $$f(x) = \fr{x^3}{x^2 + 1} = x - \fr{x}{x^2 + 1}.$$ %e Notice that %v $$- \fr{x}{x^2 + 1} \to 0 \text{ as } x \to \infty.$$ %e So $f(x) - x \to 0$ as $x \to \infty,$ and $y = x$ is a slant asymptote of $f(x).$ \exeend \masciiexe{} Draw $f(x) = x^{2/3} (6 - x)^{1/3}.$ {\em Solution.} We'll follow exactly what we defined above. We need the critical numbers, so we will find out where $f'(x)$ fails to exist, and where $f'(x) = 0.$ Let $u = x^{2/3}, v = (6 - x)^{1/3}.$ So %align by = %v \begin{align*} f'(x)&= u'v + uv'\\ &= (2/3) x^{-1/3} (6 - x)^{1/3} + x^{2/3} (-1/3) (6 - x)^{-2/3}\\ &= \fr{4-x}{\sqrt[3]{x(6-x)^2}}. \end{align*} %e So we now know that the we have critical at $x = 0, x = 4,$ and $x = 6.$ Our intervals are %table %v++ % % interval | 4 - x | cbrt{x} | cbrt{(6 - x)^2} | f'(x) | f(x) % x < 0 | + | - | + | - | % 0 < x < 4 | + | + | + | + | % 4 < x < 6 | - | + | + | - | % 6 < x | - | + | + | - | % %v-- Now we know where it goes up and down, and so we have the local minima and the local maxima. We have a local minimum at $x = 0$ because it goes down in the negative side of $x$ and goes up after touching $x = 0.$ We have a local maximum at $x = 4$ because note that as it goes over $x = 4,$ it changes from going up to going down. Now, what does $f(x)$ look like? So far, I can say that it comes down from the left side and it will have a cusp at $0$ because the $f$ is continuous at $x = 0,$ but not differentiable there --- we have a vertical tangent there ---, so that's a fair guess. From that point, it goes up and to %v $$f(4) = (2 \cdot 2)^{2/3} 2^{1/3} = 2^{5/3} \approx 3.1748,$$ %e and it is differentiable at that $f(4),$ so I can say that it has a ``smooth'' high point at $y \approx 3.1748,$ and then it comes down forever with a nondifferentiable point at $x = 6$ which gives us a vertical tangent there as well. Let's see if we can get more information. Can $f''(x)$ gives us any more clues? Yes, $f''(x)$ can tell us about the concavities of $f(x).$ Computing $f''(x),$ we get %v $$f''(x) = \fr{-8}{x^{4/3}(6-x)^{5/3}}.$$ %e Now we check where $f''(x)$ is positive and where it is negative. It is never zero, but it may be undefined at $x = 0$ and $x = 6.$ Building a table, we see the following behavior. %table %v++ % % interval | -8 | x^{4/3} | (6 - x)^{5/3} } | f''(x) % x < 0 | - | + | + | - % 0 < x < 6 | - | + | + | - % 6 < x | - | + | - | + % %v-- So it's positive when $x > 6,$ and negative to the left side. So $f$ is concave down when $x < 0,$ concave down when $0 < x < 6$ and concave up when $x > 6.$ Now we're able to draw a accurate enough graph.\exeend \masciiexe{} Suppose you're given a formula for a function $f.$ How can you tell where $f$ is increasing or decreasing? How do you determine where the graph of $f$ is concave upward or concave downward? How do you locate the inflection points? {\em Solution.} First, look at $f'.$ Break the real line in as many intervals as necessary given by the zeroes of $f'$ and where $f'$ is not defined --- the critical numbers of $f.$ If $f'$ has $n$ critical numbers, you get $n + 1$ intervals. Pick a number in each of these intervals and notice where $f'$ is positive and negative. If positive, then $f$ is increasing, if negative, then $f$ is decreasing. Second, look at $f''.$ Find the critical numbers of $f'.$ Break the real line into as many intervals as required by the number of critical numbers of $f'.$ Notice where it is positive, and where it is negative. Where it is positive, $f$ is concave upward; otherwise, downward. Third, using the second part, notice where it changes sign; we have an inflection point only at the points where it changes sign. \exeend \subsection{Optimizing things} If you have a function that describes something, then you can look for the absolute maximum and mininum and you'll have a clue of where in the input of the function you have a maximization or a minimization of the output. Finding minima and maxima values is actually not too hard, but finding the functions is sometimes tricky, as we should expect, because to describe phenomena in the universe requires considerable skills. \masciiexe{} Find two integers whose difference is $100$ and whose product is a minimum. {\em Solution.} Let $x, y$ be integers with $x < y.$ Then $y - x = 100.$ We want to minimize $f(x) = xy.$ We know that $y = 100 + x.$ So $f(x) = x(100 + x)$ and $f'(x) = 2x + 100.$ Now $f'(x)$ is zero when $x = -50,$ and there $y = 150.$ So $150 - 50 = 100$ and the product is $150 \cdot 50 = 7500.$ We're done. We could object to this solution saying that we're not sure whether $x = -50$ is a minimum, but possibly a local extremum. That's futile, though, because $f(x)$ is a parabola concave upward, so I know that where $f'(x) = 0,$ we do have a local mininum. \exeend \masciiexe{} Find the point on the line $y = 4x + 7$ that is closest to the origin. {\em Solution.} Here we're required to develop a distance function, and for the plane of Descartes, we have the famous %v $$d = \sqrt{(a - 0)^2 + (b - 0)^2}.$$ %e We want the distance from a point on y curve to the origin. To find the distance, we need the point $(a,b)$ which is common for both functions; that is %v $$d = \sqrt{x^2 + (4x + 7)^2}.$$ %e Now, differentiating $d$ is harder than differentiating $d^2 = x^2 + (4x + 7)^2$, so we will work with $d^2$ because they both share the same minimum --- but notice that this is not true in general; for example, $f(x) = x^2 - 1$ and $f(x)^2 = (x^2 - 1)^2$ don't share minima. We have $(d^2)' = 2x + 8(4x + 7)$ which has a zero when $x = -28/17,$ so the point is $(-28/17, 7/17).$ \exeend \masciiexe{} Find the points on the hyperbola $y^2 - x^2 = 4$ that are closest to the point $(2,0).$ {\em Solution.} I'd like to see the hyperbola first. So I'm going to graph it. This hyperbola is symmetric about the $x$-axis and about the $y$-axis because $y^2 - x^2 = 4 = (-y)^2 - (-x)^2.$ Let's consider just the $y > 0$ of the plane of Descartes. We know that the curve touches $y = 2$ because if $x = 0,$ then $y = \sqrt{4}.$ As $x \to \infty,$ $y \to \infty,$ so there's no horizontal asymptote --- nor any vertical one. There may be slant asymptotes. We compute %v $$y' = \fr{x}{\sqrt{4 + x^2}},$$ %e which is zero when $x = 0,$ so $x = 0$ is the only critical number. We compute %v $$y''= \fr{\sqrt{4 + x^2} - x^2 (4 + x^2)^{-1/2}}{4 + x^2},$$ %e which is never negative. From $y'$ and $y'',$ we conclude the following table. %table % interval | y' | y'' % x < 0 | - | + % 0 < x | + | + So, there's a local minimum at $0,$ and no inflection points. What about a slant asymptote? If there's one, and it's $g(x),$ then it must be true that %v $$\lim_{x \to \infty} \sqrt{4 + x^2} - g(x) = 0.$$ %e Suspecting that $g(x) = x$ will do, we let $q(x) = \sqrt{4 + x^2} + x.$ Then %align by = %v \begin{align*} \sqrt{4 + x^2} - x &= (\sqrt{4 + x^2} - x) \cdot \fr{q(x)}{q(x)}\\ &= \fr{4 + x^2 - x^2 } {\sqrt{4 + x^2} + x}\\ &= \fr{4 }{\sqrt{4 + x^2} + x}\\ &= 0, \end{align*} %e as $x \to \infty$ because $(4 + x^2) \to \infty,$ and $\sqrt{4 + x^2}$ as well, and certainly does $x.$ So $g(x) = x$ is a slant asymptote. Now we have a pretty good idea of what it looks like. Let's begin then. The distance from $(2,0)$ to the hyperbola is given by %v $$d = \sqrt{ (x - 2)^2 + y^2 }.$$ %e To work with only one variable, we must notice the points we are interested in considering are simply the points of the hyperbola, so we may replace $y^2$ in $d$ by $4 + x^2.$ So %v $$d = \sqrt{ (x - 2)^2 + 4 + x^2 }.$$ %e But it's easier to work with %v $$d^2 = (x - 2)^2 + 4 + x^2,$$ %e and we can since they share the same minimum. We now minimize $d^2$ by computing $(d^2)' = 4x - 4,$ which is zero when $x = 1.$ So the closest points to $(2,0)$ are $(1,\sqrt{5})$ and $(1,-\sqrt{5}).$ \exeend \masciiexe{} Find the dimensions of the rectangle of largest area that can be inscribed in a circle of radius $r.$ {\em Solution.} The formula of a circle is $x^2 + y^2 = r^2.$ We can position this circle at the origin of the plane of Descartes. Then we can work with $y = \sqrt{r^2 - x^2}$ which is half of the circle. Notice that the height of the rectangle will touch the circle --- it's inscribed, and it only makes sense this way, since we want to maximize the area. So, if a rectangle has area $A(x,y) = xy,$ then we will be working with %v $$A(x) = 2x \sqrt{r^2 - x^2}.$$ %e Now we have a function of one variable which we wish to maximize. To maximize it, we take its derivative which is %v $$A'(x) = 2 \sqrt{r^2 - x^2} - \fr{2 x^2}{\sqrt{r^2 - x^2}}.$$ %e This function is zero when $x = \fr{\sqrt{2} r}{2}.$ So the largest area is %v $$A(x) = \sqrt{2} r \sqrt{r^2 - \fr{r^2}{2}}.$$ %e \exeend \masciiexe{} Find the area of the largest isosceles triangle that can be inscribed in the ellipse %v $$\fr{x^2}{a^2} + \fr{y^2}{b^2} = 1.$$ %e {\em Solution.} Without computing too much, let's develop the algorithm required. The area of a triangle is %v $$A(x,h) = \fr{xh}{2}.$$ %e In order to work with a single variable, we need h with respect to $x.$ We know that one side of the triangle will have length $\sqrt{h^2 + x^2}.$ We know that the height of the triangle will always touch the ellipse, so %v $$h = y.$$ %e We can replace $y$ by computing %align by = %v \begin{align*} x^2 + \fr{a^2 y^2} {b^2} &= a^2\\ b^2 x^2 + a^2 y^2 &= b^2 a^2\\ a^2 y^2 &= b^2 a^2 - b^2 x^2\\ y^2 &= b^2 - \fr{b^2 x^2} {a^2}\\ y &= \sqrt{b^2 - \fr{b^2 x^2}{a^2}}. \end{align*} %e Therefore, %v $$A(x) = \fr{ x \sqrt{b^2 - \fr{b^2 x^2 }{ a^2}} }{2}.$$ %e Now maximize it to get all $x$ such that $A'(x) = 0.$ Hopefully, there's only one that makes sense. The answer will be with respect to the constants $a$ and $b.$ Plug the answer in $A(x)$ to get the largest area. Hmm, nope. This solution is fallacious. The solution assumes that the triangle will only take the space possible in half of the ellipse. We may get a larger triangle by using the other half too; I don't know if that wouldn't happen, though. I don't know. \exeend \subsection{Applications to economics} Suppose that the cost of feeding $x$ people in Brazil is described by %v $$C(x) = 0.3 x^3 - 2x^2 + 20.$$ %e This function has a local minimum at $x = 4 + 4/9.$ We could, therefore, think that this is the exact point that we should produce --- so that we can take advantage of low cost level, but as we shall see, there is another interesting point of this function which seems to be the an optimized way to produce with respect to cost per unit. Let's consider the function %v $$c(x) = C(x)/x,$$ %e which is called the average cost function. By dividing $C(x)$ by $x,$ we are looking at the cost per unit. That is, if a supplier gives us a discount for buying a lot, we simply pretend that the price per unit is lower. Therefore, looking at $c(x),$ we would want to produce $x$ units such that $x$ is a local minimum of $c(x).$ Correct? We see that the minimum of $c(x)$ is not the same minimum of $C(x).$ Why would it be? They are different functions. Also, let's take both minima and compute how much it would cost to produce $x$ units in both functions. The average cost function is %v $$c(x) = C(x)/x = 0.3 x^2 - 2x + 20/x$$ %e and it intersects the marginal cost function at $x \approx 4.78.$ Now, let's compare. We have $C(4.78) \approx 7.06,$ while $C(4 + 4/9) \approx 6.83.$ But we have $c(4.78) \approx 1.47,$ while $c(4 + 4/9) \approx 1.53.$ So, it is worth producing a little more in order to get a lower unit cost --- even though the cost of production is increasing a bit. Perhaps because this function is a little unusual, it shows this behavior, or perhaps it doesn't make much sense otherwise. Notice that picking a local minimum at the cost function doesn't make too much sense. Suppose, for instance, that the minimum shows up close to $0.$ Will us produce almost nothing just to get the best price? Probably not. Also, notice that $C(x)$ is a very unusual cost function; most cost functions don't have local minima, I'd say --- it's very unusual to buy, say, $10$ and pay $10,$ and then getting a cost of $5$ for buying $20$ from the same supplier. \subsection{Sigma notation} We're about to study about integrals. It will be useful, therefore, to look into sigma notation used in Riemann sums. If we want to write $2^3 +$ $3^3 + ... + n^3,$ we pick an index $i$ and we replace the sum by %v $$\sum_{i = 2}^n i^3.$$ %e The letter $n$ used in the sum must not be confused with the letter $n$ used in limit expressions which is seen in Riemann sums. Suppose that we're computing an approximation of the area under some curve $f.$ We write %v $$A\approx f(x_1) \Delta x + f(x_2) \Delta x + f(x_3) \Delta x + f(x_4) \Delta x.$$ %e Using $n$ rectangles, we get %v $$A \approx f(x_1) \Delta x + ... + f(x_n) \Delta x.$$ %e To write this as a sum, we let $i$ be an iterator, and so $i$ goes from $1$ to $n.$ That is %v $$A \approx \sum_{i = 1}^n f(x_i) \Delta x.$$ %e Now, if we want to increase $n$ forever, we say so by writing %v $$A = \lim_{n \to \infty} \sum_{i = 1}^n f(x_i) \Delta x.$$ %e Notice now that we have an exact area. These sums are called Riemann sums because Bernhard Riemann was calculating ``areas'' this way, I think. It is indeed a very remarkable technique. What we call a Riemann sum, though, is always an approximation to an area; that's what we call it today. Perhaps Riemann was already taking limits to find exact areas. By the way, the word ``exact'' used in conjunction with area is a mathematical exactness. We defined the area under a curve to be the integral of the curve over some interval --- so, the function must be continuous on that interval. So it is exact because it is by definition anyway. The definition of an integral, as we shall see, is what we see above: the limit, as $n \to \infty$, of a Riemann sum with $n$ rectangles. \section{Integrals} We start with the definition of a definite integral. Let's first observe what the definition is; let's look at its words and symbols, and after that, let's talk about ways of thinking of it. \masciidef{} Let $f$ be a continuous function defined on $[a, b].$ Break this interval into $n$ equal pieces so that each piece will have a length of $\Delta x = (b - a)/n.$ Let $x_0, ..., x_n$ be the endpoints of these subintervals and we choose arbitrary points $x_1^*, ..., x_n^*$ in these subintervals an define the integral of $f$ from $a$ to $b$ as %v $$\int_a^b f(x)\ dx = \lim_{n \to \infty} \sum^n_{i=1} f(x_1^*) \Delta x.$$ \defend %e When $f(x) \geq 0$ for all $x$ in $[a,b],$ then a definite integral is an area; that is, a definite integral is a number; it signifies a quantity; an amount of squares of some unit such as meters. If that's not the case, then we may get a difference of areas, or perhaps a completely negative area. The latter case if when $f(x) \leq 0$ for all $x$ in $[a,b].$ None of these cases is any problem; it's a matter of interpretation. Mathematics is not really concerned with these interpretation; mathematics is concerned with definitions and theorems, but I don't mean mathematicians. I believe that mathematicians are very much concerned with these interpretations, but ``negative area'' is not problem for a mathematician. If a negative balance in your account if money that you owe, then a negative area is an area that you don't have. We talk about these arbitrary points in the definition because they are really arbitrary; that is, we can pick any points in these subintervals whatsoever. In cases when our function is a table of values, then we will of course pick the points in which we have values; a table of values is a very discontinuous function. What I mean is that we have a finite number of points in which the function is defined, so the function is not continuous anywhere. So we can't integrate it; but we can get an approximation, anyway; as Riemann used to do. In situations such as this one described, we are indeed picking arbitrary points. In an abstract case; in cases where we're dealing with mathematical functions, we will indeed choose equally spaced points if we're going to integrate directly through Riemann's technique. The truth is that for mathematical functions, we integrate using our integration methods, but numerical integration is always useful for when we don't have a continuous function to work with. Some discontinuities can be resolved. For example, suppose that a function is continuous on an interval $[a,c)$ and continuous on an interval $(c,b].$ That is, the function is continuous on $[a,b]$ except at $c.$ Suppose now that the limit of the function as we approach $c$ exists and has value $L.$ Then we may, simply... hah, I don't know if we can do what I'm thinking. For a non-abstract case, indeed; but for an abstract function, I don't know. What I had in mind was to patch the function, but if we patch it, we have now a continuous function and then of course we can integrate. When we use Riemann's technique to integrate, we often pick the midpoints of the $n$ subintervals we choose. That's because for some common functions, we actually get a pretty good approximation of the area by picking the midpoints. \subsection{Properties of integrals} In our definition of the definite integral, we assumed that the lower limit is a point on the $x$-axis which is more to the left of the upper limit, but a Riemann sum makes sense even if the lower limit is greater than the upper limit. Suppose we want to integrate over an interval $[a,b],$ but we want to start from the right and add to the left; then $\Delta x$ changes from $(b - a)/n$ to $(a - b)/n.$ So %v $$\int_b^a f(x)\ dx = - \int_a^b f(x)\ dx.$$ %e Also, if a $= b,$ then $\Delta x = 0$ so %v $$\int_a^a f(x)\ dx = 0.$$ %e Also, integrating a constant, is the same as computing the area of a rectangle because a constant function is a straight line. So, if the interval is $[a,b],$ then the integral will be equal to $c (b - a),$ where $c$ is any constant. Also, the sum of two integrals is equal to the integral of the sum of two functions over the same interval. You can always concatenate adjacent intervals as says the property %v $$\int_a^c f(x)\ dx + \int_c^b f(x)\ dx = \int_a^b f(x)\ dx.$$ %e If $f(x) \geq 0$ for $[a,b],$ then %v $$\int_a^b f(x)\ dx \geq 0.$$ %e This makes total sense, no? Similarly, if $f(x) \geq g(x)$ for $[a,b],$ then %v $$\int_a^b f(x)\ dx \geq \int_a^b g(x)\ dx.$$ %e Suppose now that $m \leq f(x) \leq M$ for $[a,b],$ where $m$ and $M$ are constants. Then, by using the property defined above, we can write %v $$\int_a^b m\ dx \leq \int_a^b f(x)\ dx \leq \int_a^b M\ dx,$$ %e which in turn implies that %cm %v $$m (b - a) \leq \int_a^b f(x)\ dx \leq M (b - a).$$ %e which has a meaningful geometric interpretation. This property is very useful to estimate integrals. It also allows us to start a chain of reasoning if we're dealing with areas or integrals, since all integrals have this property. Suppose for instance, that we want to estimate the area %v $$\int_1^4 \sqrt{x}\ dx.$$ %e Since $\sqrt{x}$ is defined on $[1,4],$ then we can use the ``closed interval theorem'' to get the absolute minimum and absolute maximum of the function in this interval. So it's true that %align by \leq %v \begin{align*} 1 (4 - 1) &\leq \int_1^4 \sqrt{x}\ dx \leq 2 (4 - 1)\\ 3 &\leq \int_1^4 \sqrt{x}\ dx \leq 6. \end{align*} %e So the area we're looking for indeed lies between the $3$ and $6$ units square, or is $3$ or $6$ units square. Let's estimate some integrals now, using Riemann sums. Suppose that $f(x)$ $= 2 - x^2$ for $0 \leq x \leq 2.$ We will take the right endpoints as heights for the rectangles. We have each $\Delta x = 2/4 = 1/2.$ So our heights will be computed from $1/2, 2/2, 3/2, 4/2.$ The $i$th height will be $i/2$ from $i=1$ to $i=4.$ So %v $$\sum_{i=1}^4\fr{1}{2} \bl[ 2 - \bl( \fr{i}{2} \br)^2 \br] = \fr{1}{4}$$ %e Now, if we would use $n$ rectangles, then we would get a $\Delta x =$ $2/n.$ Each height will be computed from $2/n, 4/n, 6/n, ..., 2n/n.$ So, we get a formula for a Riemann sum, which when taking the limit, we find %v $$\lim_{n\to \infty} \sum_{i=1}^n \fr{2}{n} \bl[2 - \bl(\fr{2i}{n}\br)^2 \br] = \fr{4}{3} = 1 + \fr{3}{9}.$$ %e Suppose that $f(x) = 3x - 7$ for $0 \leq x \leq 3.$ Let's compute the integral by using Riemann sums. We will break the interval down in $n$ pieces, so each piece will be $3/n.$ Using the right endpoints, we will compute the heights from $3/n, 6/n,$ 9n, ..., 3n/n. In general, we will compute the heights from 3i/n from $1$ to $n.$ In symbols, we have %v $$\lim_{n\to \infty} \fr{3}{n} \sum_{i=1}^n \fr{9i}{n} = \fr{27}{2}$$ %e Suppose that $f(x) = x - 2 \sin(2x)$ for $0 \leq x \leq 3.$ In general, $\Delta x$ $= 3/n.$ So the heights will be computed from $3i/n$ from $1$ to $n.$ In symbols, we have the integral equal to %v $$\lim_{n\to \infty} \fr{3}{n} \sum_{i=1}^n \fr{3i}{n} - 2 \sum_{i=1}^n \sin\bl(\fr{6i}{n} \br)$$ %e I don't know how to compute the second sum. The sine function seems to complicate matters to me. I would, of course, start by looking for a pattern there, and rewrite the sine function as a new sum involving the pattern of values that I found by computing many sine function calls from the values that interests us; perhaps this way we would be able to get rid of the sine expression. Let us now turn ourselves to the midpoint rule. Suppose that we want to compute the area under $2 - x^2$ for $0 \leq x \leq 2$ with $4$ rectangles. Dividing a distance of $2$ in $4$ pieces will yield a $1/2$ distance for each piece. We are interested in the midpoints of each of these pieces, so we will start with the midpoint between $0$ and $1/2$ which is $1/4;$ the next will be $(1/2 + 1)/2 = 3/4;$ the next will be 5/4. In general, we have $(2i + 1)/4$ where $i$ ranges from 0 to 3. Therefore, we have %v $$\sum_{i=0}^3 \fr{1}{2} \bl[2 - \bl( \fr{2i + 1}{4} \br)^2 \br] = 1.375$$ %e which is a pretty good approximation, considering the amount of rectangles we're using. \masciiexe{} Estimate $\int_0^{10} \sin( \sqrt{x} )\ dx.$ {\em Solution.} We will compute a Riemann sum with $5$ rectangles by using midpoints to compute heights. So we write %v $$\int_0^{10}\sin( \sqrt{x} )\ dx \approx \sum_{i=0}^4 2 ( \sin( \sqrt{2i+1} )) \approx 6.64.$$ %e Because the interval goes from $0$ to $10,$ we chose to divide it in $5$ pieces so that we can get simple looking expressions; we have, for example, $\Delta x = 2,$ and the heights, computed from midpoints, are expressed 2i $+ 1$ where $i$ goes from $0$ to $4.$ The integral itself is approximately $6.28,$ so we have a pretty good approximation. \exeend \masciiexe{} Estimate $\int_1^2 \sqrt{1 + x^2}\ dx.$ {\em Solution.} We will compute a Riemann sum using $10$ rectangles. We will have a $\Delta x = 1/10.$ We will compute the heights by using the midpoints of each subinterval. The first left endpoint is $1,$ and the second is $1 + 1/10 = 11/10.$ The second left endpoint is $11/10 + 1/10 = 12/10,$ and it goes on like that. The first midpoint is $(1 + 11/10)/2 = 21/20.$ The second midpoint is $(12/10 + 13/10)/2 = 25/20.$ It is not clear to me what the $i$th midpoint will be. We will then develop some algebra here to see if we can have a general form for midpoints, and then come back to solve this problem. \exeend \subsection{Interval patterns in Riemann sums} The length of each rectangle is usually computed by dividing the interval $[r,s]$ in $n$ pieces. So $\Delta x = (s - r)/n.$ Now, what is the first point of the interval? It is certainly $r.$ The next point will be $r + (s - r)/n,$ and the next will be $r + 2(s - r)/n,$ and the next will be $r + 3(s - r)/n.$ Therefore, the $i$th will be $r + i(s - r)/n.$ We are interested as well in the midpoints. So we need to half of the sum of two adjacent points. That is, %v $$\fr{\bl[r+ \fr{(i - 1)(s - r)}{n}\br] + \bl[r + \fr{i (s - r)}{n}\bl]}{2}.$$ %e But notice that this is a formula for midpoints, not a pattern. We want a pattern as well, so that we can write a Riemann sum right away. In order to find a pattern, we will look at some values of $i.$ When $i = 1,$ we have %v $$\fr{1}{2}\bl( \bl[r + 0 \cdot \fr{s - r}{n}\br] + \bl[r + \fr{s - r}{n}\br] \br) = \fr{1}{2} \bl(2r + \fr{s - r}{n}\br) = r + \fr{s - r}{2n}$$ %e and the sequence goes as %v $$r+ \fr{(s - r)}{2n}, r + \fr{3(s - r)}{2n}, r + \fr{5(s - r)}{2n}, ..., r + \fr{(2i + 1)(s - r)}{2n},$$ %e where $i$ starts at zero. Now, we must discover what is the last value of $i.$ Would it be simply $n?$ Or perhaps $n - 1?$ I say that we will have $n$ midpoints; I also say that the first is obtainable by $i = 0,$ so the last must be obtainable by $i = n - 1.$ No? I think so. We're now with enough tools to work on that previous problem. %% For that, we will look at an example to see what kind of last value we %% end up with. Suppose, for instance, that we are working with an interval %% [1,2]. Then, the last right endpoint will be 1 + n(2 - 1)/n = 2. So, if %% we're computing right endpoints, i goes from 1 to n. If we are computing %% left endpoints, then the last endpoint is 1 + (n - 1)(2 - 1)/n. %% Therefore, the last midpoint will be the midpoint between the last %% left endpoint and the last right endpoint. That is, %align by = %% r + n(s - r)/n + r + (n - 1)(s - r)/n %% = 2r + (s - r) + (n - 1)(s - r)/n %% = 2r + (n(s - r) + (n - 1)(s - r))/n %% = 2r + ( (s - r)(n + n - 1) )/n %% = 2r + ( (s - r)(2n - 1) )/n \masciiexe{} Estimate $\int_1^2 \sqrt{1 + x^2}\ dx.$ {\em Solution.} We will compute a Riemann sum using $10$ rectangles. We will have a $\Delta x = 1/10.$ By the work we've done above, we will have the heights as %v $$r + \fr{(2i + 1)(s - r)}{2n}$$ %e where $r = 1, s = 2,$ and $n = 10.$ Then, we write the Riemann sum as %v $$\fr{1}{10} \sum_{i=0}^{9} \sqrt{1 + \bl(1 + \fr{2i + 1}{20}\br)^2} \approx 1.81$$ %e \exeend \masciiexe{} Evaluate $\int_{-1}^5 1 + 3x\ dx.$ {\em Solution.} We now want the integral really; but since we don't have techniques yet, we will compute it as Riemann used to do. By the work we developed above, and knowing that we're working with $n$ subinterval of length $6/n,$ our heights will be %v $$r + \fr{(2i + 1)(s - r)}{2n},$$ %e where $r = -1, s = 5,$ so %v $$-1 + \fr{3(2i + 1)}{n},$$ %e and then we can compute %v $$\lim_{n \to \infty} \fr{6}{n} \sum_{i=1}^n \bl[1 + 3\bl(-1 + \fr{3(2i + 1)}{n}\br)\br] = 42,$$ %e Now, if we're suspicious of the formula above for the midpoints, let's pick the right endpoints which will be the following sequence $-1 + 6/n, -1 + 2 \cdot 6/n, -1 + 3 \cdot 6/n,$ ..., $-1 + 6i/n.$ So we write %v $$\lim_{n \to \infty} \fr{6}{n} \sum_{i=1}^n \bl[1 + 3\bl(-1 + \fr{6i}{n}\br)\br] = 42,$$ %e and as a final check, let's carefully compute as %align by = %v \begin{align*} \int_{-1}^5 1 + 3x\ dx &= \lim_{n\to \infty} \fr{6}{n} \sum_{i=1}^n \bl[1 + 3\bl(-1 + \fr{6i}{n}\br)\br]\\ &= \lim_{n\to \infty} \fr{6}{n} \sum_{i=1}^n \bl[1 - 3 + \fr{18i}{n}\br] \\ &= \lim_{n\to \infty} \fr{6}{n} \sum_{i=1}^n -2 + \fr{6}{n} \sum_{i=1}^n \fr{18i}{n}\\ &=\lim_{n\to \infty} \bl(-12 + \fr{60 + 48}{n^2} \sum_{i=1}^n i\br) \\ &=\lim_{n\to \infty} \bl[-12 + \fr{108}{n^2} \bl(\fr{n(n+1)}{2}\br)\br] \\ &=\lim_{n\to \infty} \bl[-12 + 108 \bl(\fr{1}{2} + \fr{1}{n}\br)\br] \\ &=-12 + 54 = 42. \end{align*} %e So perhaps we may now trust the pattern developed above for midpoints. \exeend \masciiexe{} Prove that %v $$\int_0^{\pi/4} \sin^3{x}\ dx \leq \int_0^{\pi/4} \sin^2{x}\ dx.$$ %e {\em Solution.} We know that $\sin(x) \geq 0$ for all $x$ in $[0,\pi/4].$ So $\sin^3(x) \geq 0.$ Since $\sin(x) \leq 1$ for all $x$ real, then $\sin^3(x) \leq 1$ for all $x$ real. So $\sin^3(x) \leq \sin^2(x)$ for all $x$ in $[0,\pi/4].$ So %v $$\int_0^{\pi/4} \sin^3{x}\ dx \leq \int_0^{\pi/4} \sin^2{x}\ dx.$$ %e \exeend I have to prove that $ab \leq a$ if and only if $0 \leq b \leq 1.$ I don't know how to prove that right now. \masciiexe{} Prove that %v $$\int_1^2 \sqrt{5 - x}\ dx \geq \int_1^2 \sqrt{x + 1}\ dx.$$ %e {\em Solution.} We must show that $f(x) = \sqrt{5 - x} \geq g(x) = \sqrt{x + 1}$ for all $x$ in $[1,2].$ We have $f(1) = 2 \geq g(1) = \sqrt{2}.$ At $x = 2,$ we have $f(2) = \sqrt{3} \geq \sqrt{3} = g(2).$ Now notice that %v $$f'(x)= \fr{-1}{2 \sqrt{5 - x}} \text{ and } g(x) = \fr{1}{2\sqrt{x+1}}.$$ %e So $f(x)$ is an always decreasing function, while $g(x)$ is an always increasing function. This means that if they cross each other, they only do so once, and we will know that to the left of the crossing point, we have $f(x) \geq g(x),$ and to the right, we have $f(x) \leq g(x).$ This interesting point is $x = 3.$ So, $f(x) \leq g(x)$ for all $x$ in $[1,2].$ \exeend \masciiexe{} Prove that %v $$2 \leq \int_{-1}^1 \sqrt{1 + x^2}\ dx \leq 2 \sqrt{2}.$$ %e {\em Solution.} Let $f(x) = \sqrt{1 + x^2}.$ So $f(-1) = \sqrt{2}.$ Notice that our domain is $-1 \leq x \leq 0$ union $0 \leq x \leq 1.$ Start with $0 \leq x \leq 1.$ Square it to get $0 \leq x^2 \leq 1$ which implies %v $$1 \leq x^2 + 1 \leq 2.$$ %e Since every term of the inequality above is positive, we may take the square root of each of them to get %v $$1 \leq \sqrt{x^2 + 1} \leq \sqrt{2}.$$ %e Now integrate that to get %v $$\int_{-1}^21\ dx \leq \int_{-1}^1 \sqrt{x^2 + 1}\ dx \leq \int_{-1}^1 \sqrt{2}\ dx,$$ %e which implies %v $$2\leq \int_{-1}^1 \sqrt{x^2 + 1}\ dx \leq 2 \sqrt{2}.$$ %e as desired. \exeend \masciiexe{} Prove that %v $$\fr{\pi}{6} \leq \int_{\pi/6}^{\pi/2} \sin(x)\ dx \leq \fr{\pi}{3}.$$ %e {\em Solution.} We'll try constructing the result. The distance that we will integrate is $\pi/2 - \pi/6 = \pi/3.$ We will work backwards to find which constants we should integrate. For instance, we need a constant $m$ such that $m(\pi/2 - \pi/6) = \pi/6;$ so $m = 1/2.$ Also, we need a constant $M$ such that $M(\pi/2 - \pi/6) = \pi/3;$ so $M = 1.$ Now, consider the inequality %v $$m = 1/2 \leq x \leq 1 = M.$$ %e We know that $\sin(\pi/6) = 1/2$ and that $\sin(\pi/2) = 1.$ So it's true that $1/2 \leq \sin(x) \leq 1$ for all $x$ in $[\pi/6,\pi/2].$ So we now integrate this inequality above to write %align by \leq %v \begin{align*} \int_{\pi/6}^{\pi/2}\fr{1}{2}\ dx &\leq \int_{\pi/6}^{\pi/2} \sin(x)\ dx \leq \int_{\pi/6}^{\pi/2} 1\ dx\\ \fr{\pi}{6}&\leq \int_{\pi/6}^{\pi/2} \sin(x)\ dx \leq \fr{\pi}{3}. \end{align*} %e So we have a procedure here. We discovered the constants that would work by computing $m(u - l) = d,$ where $u$ is the upper limit and l is the lower limit, and $d$ is the desired constant. We solve for $m,$ and we're ready to set up the inequality of integrals, and integrate each term. \exeend \masciiexe{} Estimate %v $$\int_1^2 \fr{1}{x}\ dx.$$ %e {\em Solution.} The method here is to look for the absolute minimum and maximum. We may use the ``extreme value theorem'' of continuous functions. Mininum and maximum, respectively, are, $1/2$ and $1.$ So %v $$\fr{1}{2} (2 - 1) \leq \int_1^2 \fr{1}{x}\ dx \leq 1 (2 - 1).$$ %e \exeend \masciiexe{} Estimate %v $$\int_0^2 \sqrt{x^3 + 1}\ dx.$$ %e {\em Solution.} Minimum and maximum, respectively, are $1$ and $3.$ So %v $$2 \leq \int_0^2 \sqrt{x^3 + 1}\ dx \leq 6.$$ %e \exeend \masciiexe{} Estimate %v $$\int_{\pi/4}^{\pi/3} \cos(x)\ dx.$$ %e {\em Solution.} We have $\cos(\pi/4) = \sqrt{2}/2,$ and $\cos(\pi/3) = 1/2.$ So %v $$\fr{1}{2}\bl(\fr{\pi}{3}- \fr{\pi}{4}\br) \leq \int_{\pi/4}^{\pi/3} \cos(x)\ dx \leq \fr{1}{2}\bl(\fr{\pi}{3} - \fr{\pi}{4}\br)$$ %e which implies %v $$\fr{\pi}{48} \leq \int_{\pi/4}^{\pi/3} \cos(x)\ dx \leq \fr{\pi}{24}.$$ %e \exeend \masciiexe{} Estimate %v $$\int_{\pi/4}^{3\pi/4} \sin^2(x)\ dx.$$ %e {\em Solution.} The minimum is $1/2$ at $\pi/4$ and $3\pi/4,$ and the maximum is $1$ at $\pi/2.$ I'm afraid I need the property %v $$\int_{\pi/4}^{3\pi/4}\sin^2{x}\ dx = 2 \int_{\pi/4}^{\pi/2} \sin^2{x}\ dx.$$ %e I'm pretty confident that this is true, but I have no formal argument. I say that the derivative of $\sin^2(x)$ is $2 \sin(x) \cos(x).$ Now I can prove, using the derivative, that from $\pi/4$ to $\pi/2,$ the function is increasing, and decreasing from $\pi/2$ to $3\pi/4.$ I can also prove that in these two intervals, the slope is only different by a sign. That is, in the same velocity that it goes up, it also goes down on the other interval. So, the area it covers is the same in both intervals. Therefore, the property above follows. Now, instead of estimating the left side of the equation, we estimate the right side. So we write %v $$\fr{1}{4}\bl(\fr{\pi}{2} - \fr{\pi}{4}\br) \leq \int_{\pi/4}^{\pi/2} \sin^2{x}\ dx \leq 1 \bl(\fr{\pi}{2} - \fr{\pi}{4} \br)$$ %e We multiply the inequality by $2$ so that %align by \leq %v \begin{align*} \fr{\pi}{8}&\leq \int_{\pi/4}^{\pi/2} \sin^2{x}\ dx + \int_{\pi/4}^{\pi/2} \sin^2{x}\ dx \leq \fr{\pi}{2} \\ \fr{\pi}{8}&\leq \int_{\pi/4}^{3\pi/4} \sin^2{x}\ dx \leq \fr{\pi}{2}. \end{align*} %e \exeend The idea of looking for absolute minimum and absolute maximum helps quite a bit, but it will become troublesome as we work with not so well behaved functions; we may, nevertheless, break the badly behaved function into as many pieces as necessary and then look for minimum and maximum in these smaller closed intervals in which the function behaves somewhat well. So, as we can see, we cannot work with too complicated functions; only with somewhat bladly behaved ones. \masciiexe{} Prove that %v $$\int_a^b x\ dx = \fr{b^2 - a^2} {2}.$$ %e {\em Solution.} We start with the definition of the integral and $try$ to find the result. The definition says that %align by = %v \begin{align*} \int_a^bx\ dx &= \lim_{n \to \infty} \sum_{i=1}^n \fr{b-a}{n} \bl[a + \fr{i(b-a)}{n}\br]\\ &=\lim_{n \to \infty} \fr{b-a}{n} \bl[\sum_{i=1}^n a + \sum_{i=1}^n \fr{i(b-a)}{n}\br]\\ &=\lim_{n \to \infty} \fr{b-a}{n} \bl[an + \fr{b-a}{n} \fr{n (n+1)}{2} \br]\\ &=\lim_{n \to \infty} a (b-a) + \lim_{n \to \infty} \bl(\fr{b-a}{n}\br)^2 \fr{n (n+1)}{2}\\ &=a (b-a) + \lim_{n \to \infty} \fr{(b-a)^2}{2} \bl(1 + \fr{1}{n}\br)\\ &=a (b-a) + \fr{(b-a)^2}{2}\\ &=\fr{2a (b-a) + (b-a)^2} {2}\\ &=\fr{2ab - 2a^2 + b^2 - 2ab + a^2}{2}\\ &=\fr{- 2a^2 + b^2 + a^2}{2}\\ &=\fr{b^2 - a^2}{2}. \end{align*} %e \exeend \masciiexe{} Prove that %v $$\int_1^3 \sqrt{x^4 + 1}\ dx \geq \fr{26}{3}.$$ %e {\em Solution.} I need the fact that %v $$\int_a^bx^2\ dx = \fr{b^3 - a^3}{3},$$ %e to prove the above. Assume so. Now notice that $f(x) = x^2$ has minimum and maximum, respectively, as $1$ and $9$ in $[1,3].$ Also, we have $\sqrt{2}$ and $\sqrt{82}$ as a minimum and maximum, respectively, of $\sqrt{x^4 + 1}.$ So, we know now that %v $$\int_1^3\sqrt{x^4 + 1}\ dx \geq \int_1^3 x^2\ dx = \fr{80}{3} \geq \fr{26}{3}.$$ %e \exeend %% Exercise. Prove that %cm %% \int_0^{\pi/2} x sin x\ dx <= \fr{\pi^2}{8}. %% Solution. We know that %cm %% \int_0^{\pi/2} x^2\ dx = \fr{\pi^3}{24}. %% My computer tells me that %cm %% \fr{\pi^3}{24} =~ 1.29 \text{ and } \fr{\pi^2}{8} =~ 1.23. \masciiexe{} Prove that %v $$\int_a^b c f(x)\ dx = c \int_a^b f(x)\ dx,$$ %e where $c$ is a constant. {\em Solution.} Straightforwadly, we write %align by = %v \begin{align*} \int_a^bc f(x)\ dx &= \lim_{n \to \infty} \sum_{i=1}^n c f(x_i) \Delta x\\ &=\lim_{n \to \infty} c \sum_{i=1}^n f(x_i) \Delta x\\ &=c \lim_{n \to \infty} \sum_{i=1}^n f(x_i) \Delta x\\ &=c \int_a^b f(x)\ dx, \end{align*} %e where $\Delta x = (b -a)/n.$ \exeend \masciith{} If $f(x)$ is continuous, then %v $$\bl|\int_a^b f(x)\ dx \br| \leq \int_a^b | f(x) |\ dx $$ %e \theend \masciiproof{} The function $f(x)$ takes values on the interval $[a,b].$ At each of these values, the function gives us a real number. So we may use the property of real numbers to say that $-|f(x)| \leq f(x) \leq |f(x)|$, so we may write %align by \leq %v \begin{align*} \int_a^b- |f(x)|\ dx &\leq \int_a^b f(x)\ dx \leq \int_a^b |f(x)|\ dx. \end{align*} %e Notice now that $f(x)$ can yield negative and positive values, and $|f(x)|$ only positive values. So we may write %v $$\bl|\int_a^bf(x)\ dx \br| \leq \int_a^b |f(x)|\ dx.$$ %e In other words, if $f(x) \geq$ for all $x$ in its domain, then we stricly have an equation; but if $f(x)$ takes a negative value anywhere in its domain, then we will stricly have an inequality. \profend \masciiexe{} Prove that %v $$\bl|\int_0^{2\pi} f(x) \sin(2x)\ dx \br| \leq \int_0^{2\pi} |f(x)|\ dx.$$ %e {\em Solution.} We now know that %v $$\bl|\int_0^{2\pi} f(x) \sin(2x)\ dx \br| \leq \int_0^{2\pi} |f(x) \sin(2x)|\ dx.$$ %e Since the integral is definite, it is a real number; so %align by \leq %v \begin{align*} \bl|\int_0^{2\pi} f(x) \sin(2x)\ dx \br| &\leq \int_0^{2\pi} |f(x)| |\sin(2x)|\ dx. \end{align*} %e Now notice that $|\sin(2x)| \leq 1,$ so because of this, it follows that %v $$\int_0^{2\pi} |f(x)| |\sin(2x)|\ dx \leq \int_0^{2\pi} |f(x)|\ dx.$$ %e We actually haven't proved that $|a| |b| \leq |a|$ when $|b| \leq 1.$ \exeend It is interesting to identify Riemann sums as integrals so that we can easily compute them with the fundamental theorem of calculus. The next exercise illustrates this. \masciiexe{} Compute %v $$\lim_{n\to \infty} \sum_{i=1}^n \fr{i^4}{n^5}.$$ %e {\em Solution.} In this case here, we may write %v $$\lim_{n \to \infty} 1/n \sum_{i=1}^n \fr{i^4}{n^4},$$ %e and claim that the interval is from $0$ to $1,$ and the integrand is $x^4.$ In symbols, %v $$\lim_{n \to \infty} 1/n \sum_{i=1}^n \fr{i^4}{n^4} = \int_0^1 x^4\ dx. $$ %e \exeend As we have mentioned in previous pages, the sample points picked in a Riemann sum need not be right endpoints, left endpoints, or midpoints. We may pick whatever points in each subinterval of the Riemann sum to compute the height of the rectangle --- as long as the $\Delta x \to 0.$ The next exercise illustrates an unusual choice. \masciiexe{} Compute %v $$\int_1^2x^{-2}\ dx. $$ %e Hint: choose $x_i^*$ to be the geometric mean of $x_{i-1}$ and $x_i.$ The geometric mean is the function $\sqrt{x_{i-1} x_i}.$ Also, use the identity %v $$\fr{1}{m (m+1)} = \fr{1}{m} - \fr{1}{m+1}.$$ %e {\em Solution.} Let $x_i = 1 + i/n.$ This is our first endpoint. It is a right enpoint. By the hint, we will choose the geometric mean as the sample points. So let $x_i^* = \sqrt{x_{i-1} x_i}$. So %align by = %v \begin{align*} \int_1^2x^{-2}\ dx &= \lim_{n \to \infty} \fr{1}{n} \sum_{i=1}^n \fr{1}{\sqrt{\bl[\bl(1 + \fr{i-1}{n}\br)\bl(1 + \fr{i}{n}\br)\br]^2}}\\ &=\lim_{n \to \infty} \fr{1}{n} \sum_{i=1}^n \fr{1} { \bl(1 + \fr{i-1}{n}\br) \bl(1 + \fr{i}{n}\bl) }\\ &=\lim_{n \to \infty} \fr{1}{n} \sum_{i=1}^n \fr{1} {\bl(\fr{n + i-1}{n}\br) \bl(\fr{n+ i}{n}\br) }\\ &=\lim_{n \to \infty} \fr{1}{n} \sum_{i=1}^n \fr{n^2} { (n + i-1) (n+ i) }\\ &=\lim_{n \to \infty} n \sum_{i=1}^n \fr{1} { (n + i-1) (n+ i) }. \end{align*} %e Now let $m = n+i-1$ so that $m + 1 = n + i.$ By the hint, then, we have %align by = %v \begin{align*} \lim_{n\to \infty} n \sum_{i=1}^n \bl( \fr{1} { n + i-1} - \fr{1} {n+ i} \br) &=\lim_{n \to \infty} n \sum_{i=1}^n \fr{1} { n + i-1} - n \sum_{i=1}^n \fr{1} {n+ i} \\ &=\lim_{n \to \infty} n \sum_{i=0}^{n-1} \fr{1} { n + i} - n \sum_{i=1}^n \fr{1} {n+ i} \\ &=\lim_{n \to \infty} n \bl( \fr{1}{n} - \fr{1}{2n} \br)\\ &=\lim_{n \to \infty} 1 - \lim_{n \to \infty} \fr{1}{2} = \fr{1}{2}. \end{align*} %e \exeend \subsection{The fundamental theorem} The fundamental theorem of calculus is remarkable. The fundamental theorem makes two claims, and so we will break it down in two theorems. \masciith{} Let $f$ be continuous on $[a,b].$ Define %v $$g(x) = \int_a^x f(t)\ dt,$$ %e for $a \leq x \leq b.$ Then $g(x)$ is continuous on $[a,b]$ and differentianble on $(a,b)$ such that $g'(x) = f(x).$ \theend \masciiproof{} Let $x$ be in $(a,b).$ Let $x + h$ be in $(a,b)$ with $h \neq 0.$ Then %align by = %v \begin{align*} g(x+h)- g(x) &= \int_a^{x+h} f(t)\ dt - \int_a^x f(t)\ dt\\ &=\bl( \int_a^{x} f(t)\ dt + \int_x^{x+h} f(t)\ dt \br) - \int_a^x f(t)\ dt\\ &=\int_a^{x} f(t)\ dt + - \int_a^x f(t)\ dt + \int_x^{x+h} f(t)\ dt\\ &=\int_x^{x+h} f(t)\ dt. \end{align*} %e Since h $\neq 0,$ we may multiply this equation by $1/h.$ So %v $$\fr{g(x+h)- g(x)}{h} = \fr{1}{h} \int_x^{x+h} f(t)\ dt.$$ %e We now consider two cases; one in which $h > 0$ and one in which $h < 0.$ First, let $h > 0.$ Consider now the interval $[x, x + h].$ Since $f$ is continuous, then by the extreme value theorem, there is a number $u$ such that $f(u) = m$ is the absolute minimum of $f$ in $[x, x + h].$ Similarly, there is a number $v$ such that $f(v) = M$ is the absolute maximum of $f$ in $[x, x + h].$ Therefore, %align by \leq %v \begin{align*} \int_x^{x+h} m\ dt &\leq \int_x^{x+h} f(t)\ dt \leq \int_x^{x+h} M\ dt\\ mh &\leq \int_x^{x+h} f(t)\ dt \leq Mh\\ f(u) h &\leq \int_x^{x+h} f(t)\ dt \leq f(v)h. \end{align*} %e Since $h > 0,$ we may write %v $$f(u) \leq \fr{1}{h} \int_x^{x+h} f(t)\ dt \leq f(v),$$ %e which is the same as %v $$f(u) \leq \fr{ g(x + h) - g(x) }{h} \leq f(v).$$ %e Now, if we let $h \to 0^+,$ then we will be squeezing the interval $[x,x+h],$ and so we would expect that eventually $f(u) = f(x),$ and similarly for $f(v).$ In symbols, %v $$\lim_{h \to 0^+} f(u) = \lim_{u \to x^+} f(u) = f(x).$$ %e Similarly for $v,$ we have %v $$\lim_{h \to 0^+} f(v) = \lim_{v \to x^+} f(v) = f(x).$$ %e So we may write %align by \leq %v \begin{align*} \lim_{h\to 0^+} f(u) &\leq \lim_{h\to0^+} \fr{g(x+h) - g(x)}{h} \leq \lim_{h\to0^+} f(v)\\ f(x)&\leq \lim_{h\to0^+} \fr{g(x+h) - g(x)}{h} \leq f(x). \end{align*} %e Therefore, by the squeeze theorem, we have %v $$\lim_{h\to0^+} \fr{g(x+h) - g(x)}{h} = f(x).$$ %e Now let $h < 0,$ and consider the interval $[x + h, x].$ Since $f$ is continuous, then by the extreme value theorem, there is a number $u$ in $[x+h,x]$ such that $f(u) = m$ is the absolute minimum of $f$ in $[x+h,x].$ Similarly, there is a number $v$ in $[x+h,x]$ such that $f(v)=M$ is the the absolute maximum of f in $[x+h,x].$ So, %align by \leq %v \begin{align*} \int_{x+h}^{x} m\ dt &\leq \int_{x+h}^{x} f(t)\ dt \leq \int_{x+h}^{x} M\ dt\\ mh &\leq \int_{x+h}^{x} f(t)\ dt \leq Mh\\ f(u) h &\leq \int_{x+h}^{x} f(t)\ dt \leq f(v)h. \end{align*} Since $h < 0,$ we may write %v $$f(u) \geq \fr{1}{h} \int_{x+h}^{x} f(t)\ dt \geq f(v),$$ %e which is the same as %v $$f(v) \leq \fr{ g(x + h) - g(x) }{h} \leq f(u).$$ %e Now, if we let $h \to 0^-,$ then we will be squeezing the interval $[x+h,x],$ and so we would expect that eventually $f(v) = f(x),$ and similarly for $f(u).$ In symbols, %v $$\lim_{h\to0^-} f(v) = \lim_{v\to x^-} f(v) = f(x).$$ %e Similarly for $u,$ we have %v $$\lim_{h\to0^-} f(u) = \lim_{u\to x^-} f(u) = f(x).$$ %e So we may write %align by \leq %v \begin{align*} \lim_{h\to0^-}f(v) &\leq \lim_{h\to0^-} \fr{ g(x+h)-g(x)} {h} \leq \lim_{h\to0^-}f(u)\\ f(x)&\leq \lim_{h\to0^-} \fr{ g(x+h)-g(x)} {h} \leq f(x). \end{align*} %e Therefore, by the squeeze theorem, we have %v $$\lim_{h\to0^-} \fr{ g(x+h)-g(x)} {h} = f(x).$$ %e Together with our previous result, we have %v $$\lim_{h\to0} \fr{g(x+h) - g(x)} {h} = f(x),$$ %e which shows that $f(x)$ is the derivative of $g(x),$ and therefore $g(x)$ is differentiable on $(a,b)$ which implies that $g(x)$ is continuous on $[a,b].$ \profend \masciith{} If $f$ is continuous on $[a,b],$ then %v $$\int_a^b f(x)\ dx = F(b) - F(a),$$ %e where $F$ is any antiderivative of $f;$ that is, $F$ is any function such that $F'(x) = f(x).$ \theend \masciiproof{} Let %v $$g(x) = \int_a^x f(t)\ dt.$$ %e We know that $g'(x) = f(x).$ That is, $g(x)$ is an antiderivative of $f(x).$ Suppose that $F(x)$ is any antiderivative of $f(x).$ We have a theorem which says that any two antiderivatives of a function $f(x)$ must differ only by a constant. So it is true that %align by = %v \begin{align*} F(x) - g(x) &= c\\ F(x) &= g(x) + c, \end{align*} %e for $a \leq x \leq b.$ Both $F(x)$ and $g(x)$ are continuous on $[a,b].$ Since both are continuous, we may write %align by = %v \begin{align*} \lim_{x\to a^+} F(x) &= \lim_{x\to a^+} \bl( g(x) + c\br)\\ F(a) &= g(a) + c. \end{align*} %e Similarly, as $x \to b^-,$ we have %align by = %v \begin{align*} \lim_{x\to b^-} F(x) &= \lim_{x\to b^-} \bl( g(x) + c \br)\\ F(b) &= g(b) + c. \end{align*} %e But notice that %v $$g(a)= \int_a^a f(t)\ dt = 0 \text{ and } g(b) = \int_a^b f(t)\ dt,$$ %e and $g(b)$ describes the area under $f$ from $a$ to $b$ which is what we're interested in. We then write %align by = %v \begin{align*} F(b) - F(a) &= (g(b) + c) - (g(a) + c)\\ &= g(b) + c - g(a) - c\\ &= g(b) + c - c\\ &= g(b). \end{align*} %e Therefore, %v $$F(b) - F(a) = g(b) = \int_a^b f(t)\ dt,$$ %e as desired. \profend So if we know the antiderivative of a function $f,$ then we can integrate it easily using the second theorem of the fundamental theorem of calculus. Outside mathematics, this theorem makes a lot of sense. If $v(t)$ is the velocity of an object and $s(t)$ is its position at a time $t,$ then $s'(t) =$ $v(t),$ so $s(t)$ is an antiderivative of $v(t).$ If we integrate $v(t),$ then what we're doing is computing the area under $v(t)$ which is the distance traveled. If the object moved from a to $b,$ then $s(b) - s(a)$ is the distance traveled; that is, the integral of $v(t)$ from $a$ to $b.$ In symbols, %v $$\int_a^b v(t)\ dt = s(b) - s(a).$$ %e \masciiexe{} Differentiate %v $$h(x)= \int_0^{x^2} \sqrt{1 + r^3}\ dr.$$ %e {\em Solution.} Let $u = x^2.$ Since du/dx $= 2x$ an$d$ since %v $$\fr{dh}{dx}= \fr{dh}{du} \fr{du}{dx},$$ %e we have %align by = %v \begin{align*} h'(x)&= \fr{d}{dx} \int_0^u \sqrt{1 + r^3}\ dr\\ &= \sqrt{1 + u^3} \fr{du}{dx}\\ &= 2x \sqrt{1 + x^6}. \end{align*} %e \exeend \masciiexe{} Differentiate %v $$h(x) = \int_2^{1/x} \sin^4(t)\ dt.$$ %e {\em Solution.} Let $u = 1/x.$ Then %align by = %v \begin{align*} h'(x) &= \fr{d}{dx} \int_2^u \sin^4(t)\ dt\\ &= \sin^4(u) \fr{du}{dx}\\ &= - \fr{\sin^4(1/x)}{x^2}. \end{align*} %e \exeend \masciiexe{} Differentiate %v $$y = \int_3^{\sqrt{x}} \fr{\cos t}{t}\ dt.$$ %e {\em Solution.} Let $u = \sqrt{x}.$ Then $du/dx = 1/2\sqrt{x}.$ So %align by = %v \begin{align*} y'&=\fr{d}{dx} \int_3^u \fr{\cos t}{t}\ dt\\ &= \fr{\cos u}{u} \fr{du}{dx}\\ &= \fr{\cos \sqrt{x}}{2 \sqrt{x} \sqrt{x}}\\ &= \fr{\cos\sqrt{x}}{2x}. \end{align*} %e \exeend \masciiexe{} Convince yourself that %v $$\int_{2x}^{3x}f(u)\ du = \int_{2x}^0 f(u)\ du + \int_0^{3x} f(u)\ du.$$ %e {\em Solution.} Suppose $x = 1.$ Then we have %align by = %v \begin{align*} \int_{2}^{3}f(u)\ du &= \int_{2}^0 f(u)\ du + \int_0^{3} f(u)\ du\\ &= - \int_{0}^2 f(u)\ du + \int_0^{3} f(u)\ du, \end{align*} %e which does seem to be true. \exeend \masciiexe{} Find a function $f$ and a number $a$ such that %v $$6+ \int_a^x \fr{f(t)}{t^2}\ dt = 2 \sqrt{x},$$ %e for all $x > 0.$ {\em Solution.} The fundamental theorem gives us an easy differentiation of the integral; so, we differentiate both sides of the equation to get %align by = %v \begin{align*} \fr{f(x)}{x^2} &= \fr{1}{x^{1/2}}\\ f(x)&= x^2 \cdot x^{-1/2}\\ f(x)&= x^{3/2}. \end{align*} %e So this a function that seems to be the correct one. The problem here is that we also must find $a$ because it is not true in general. How would we find $a$? We assume we have found it already. Call it $a.$ So there is such number; it is $a.$ Since it exists, we may compute %v $$6+ \int_a^a \fr{f(t)}{t^2}\ dt = 2 \sqrt{a},$$ %e and it's easy to do so because we know the integral is zero. So %align by = %v \begin{align*} 6+ 0 &= 2 \sqrt{a}\\ 3&= \sqrt{a}\\ 9&= a. \end{align*} %e It turns out that 9 is a unique solution to this problem. \exeend \masciiexe{} Suppose $h$ is a function such that $h(1) = -2, h'(1) = 2, h''(1) = 3, h(2) = 6, h'(2) = 5, h''(2) = 13$ and $h''$ is continuous everywhere. Evaluate %v $$\int_1^2 h''(u) du.$$ %e {\em Solution.} We know that %v $$\int_1^2 h''(u) du = h'(u) \br|_1^2 = 5 - 3 = 2,$$ %e don't we? So, we're done. This problem presents some superfluous information; possibly to confuse us. \exeend \subsection{The substitution theorem} \masciith{} If $u = g(x)$ is a differentiable function whose range is an interval $I$ and $f$ is continuous on $I$, then %v $$\int f(g(x)) g'(x) \dx = \int f(u) \du.$$ %e \theend \masciiproof{} Consider a function $F(g(x)),$ where $F'= f.$ By the chain rule, we know that the derivative of $F(g(x))$ is $F'(g(x)) g'(x) = f(g(x)) g'(x),$ so if we let $u = g(x),$ then %align by = %v \begin{align*} \int F'(g(x)) g'(x) \dx &= F(g(x)) + c \\ &=F(u) + c \\ &=\int F'(u) \du\\ &=\int f(u) \du, \end{align*} %e as desired.\profend %% It is interesting to consider the substitution rule graphically. Let us %% pick an example. Consider $y = \sqrt{2x+1}$ from $0$ to $1.$ If we want %% the area in terms of some function of $u,$ we must consider changing %% three things: $(1)$ the function $y,$ indeed; $(2)$ the limits; $(3)$ %% the $\dx$ of the integral. One puzzling thing is the change in the %% $\dx;$ we tend to think of the $\dx$ as a mere piece of notation. %% In the definition of the integral, we see that the dx represents the %% $\Delta x$ of the Riemann sum. The Riemann sum evaluates to a different %% number if we change the $\Delta x.$ Here, if we change the limits, we %% change the interval in which we're integrating; so, if you think of a %% Riemann sum, it's easier to realize that the $\Delta x$ may also change; %% we might be increasing the width of the $\Delta x$ or decreasing it. %% Let us compute the area under $f$ from $0$ to $1.$ Let $f(x) = %% \sqrt{2x+1}.$ Let $u(x) = 2x + 1.$ So $f(x) = f(u(x)).$ We know that %% %v %% $$\int f(u(x)) u'(x) \dx = \int f(u) \du$$ %% %e %% by the substitution theorem. So this means that if we want to integrate %% $f(x)$ with respect to $x,$ we may integrate $f(u(x)) u'(x)$ with %% respect to $x$ instead, which is the same as integrating $f(u)$ with %% respect to $u.$ In symbols, %align by = %% %v %% \begin{align*} %% \int_0^1\sqrt(2x+1) \dx &= \int \sqrt{u(x)} u'(x) \dx\\ %% &= \int_{u(0)}^{u(1)} \sqrt{u} \fr{du}{2}\\ %% \end{align*} %% %e %% This is somewhat puzzling. What's the difference between a variable %% named $x$ and a variable named $u?$ In general, none. However, here we %% have a particular case; the context here makes $x$ and $u$ quite %% different. For instance, as $x$ goes from $1$ to $2, u$ goes from $3$ to %% $5.$ Clearly, $u$ moves faster than $x.$ They are different axis. They %% form different planes. This is an example of the ability that %% mathematics gives us. We mirror one problem on another and we create %% analogous systems; of course, the analogy must be ideal. %% The analogy here is ideal because what we have done is changed the plane %% where the function $f(x)$ is defined. We moved $f$ from the yx-plane to %% the yu-plane. When we do that, we have to adjust $x$ to $u.$ In order to %% keep $f$ as it really is, we have to adjust $x$ to $u$ in the right way; %% the right way is described the the relationship between $x$ and $u;$ the %% relationship is that $u = 2x + 1,$ as we defined it. So, the yu-plane is %% distorted when compared to the yx-plane. The yu-plane is the %% $y(2x+1)$-plane. \subsection{Volumes} We define volume as the limit of the sums of the areas of slices of the solid where each slice has a width of $\Delta x$ and $\Delta x \to 0.$ That is, %v $$V= \lim_{n \to \infty} \sum_{i=1}^n A(x_i^*) \Delta x = \int_a^b A(x) \dx.$$ %e \subsection{Um limite legal} Compute % $$\lim_{t \to \infty} \frac{\sqrt{t}}{\exp(t)} \exp\left(\int_0^\infty \floor(t \exp(-x)) \dx\right) $$ % sabendo que % $$t! \approx \sqrt{2\pi t} \left(\frac{t}{e}\right)^t$$ % sempre que $t \to \infty$. A brincadeira aqui é que esse é um problema que é fácil de compor, mas difícil de resolver porque não se consegue facilmente desfazer a convolução que foi feita passo a passo pelo autor do problema. Mesmo que você conhecesse a aproximação acima, você não pensaria em usá-la porque ela implícita no problema não de forma aparente. \section{Studying functions} We're going to look into a series of topics now. We are mostly using Calculus to study functions. Let's look at inverse functions first. Then we look into exponential functions and slowly and scatteredly we will be looking into some interesting functions. \subsection{On inverse functions} If $f(b) = a,$ then $f^{-1}(a) = b.$ What would $f'(b)$ be? We want to express $f'(b)$ with respect to $(f^{-1})'.$ For that, we write a theorem. \masciith{} Let $f$ be one-to-one and differentiable with $g = f^{-1}$ and $f'(g(a)) \neq 0.$ Then $g$ is differentiable at $a$ and %v $$g'(x)= \fr{1}{f'(g(x))},$$ %e where $g = f^{-1}.$ \theend \masciiproof{} Let $y = g(x) = f^{-1}(x).$ Then $f(y) = x.$ Now, %align by = %v \begin{align*} f'(y) y' &= 1\\ y' &= \fr{1} {f'(y)}\\ g'(x) &= \fr{1}{f'(g(x))}, \end{align*} %e as desired. \profend Now, let us define exponential functions. Exponential functions are $a^x$ where $a$ is some positive number, usually $a > 1.$ If $a$ is between $0$ and $1,$ then the function exponentially decays. Exponential functions are defined for all reals, but then one wonders what $2^\pi$ means. It means, by definition, %v $$a^x = \lim_{r \to x} a^r,$$ %e for some rational $r.$ So to answer the question, $2^\pi$ is the limit of the sequence of numbers $2^3, 2^{3.1}, 2^{3.14},$ and so on. In fact, this explains why the base of an exponential function cannot be negative. If it were, then irrational numbers could not be powers of these bases. If they were, complex numbers would arise. For instance, consider $(-2)^\pi$; the limit doesn't converge because when $r$ is $3.1,$ the answer is not real; when $r$ is $3.14,$ the answer is real; when $r$ is $3.141,$ the answer is not real. It depends on the evenness or oddness of the last digit of $r.$ So it does not converge to any real value; so the value is undefined. \subsection{The exponential number } Here's one way to define $e.$ Consider $2^x$ and $3^x.$ Take the derivative of them. We get, for $f(x) = 2^x,$ %align by = %v \begin{align*} f'(x)&= \lim_{h \to 0} \fr{2^{x+h} - 2^x}{h}\\ &=\lim_{h \to 0} \fr{2^x 2^h - 2^x}{h}\\ &=\lim_{h \to 0} \fr{2^x(2^h - 1)}{h}\\ &=2^x \lim_{h \to 0} \fr{2^h - 1}{h}. \end{align*} %e For $g(x) = 3^x,$ we have %v $$g'(x) = 3^x \lim_{h \to 0} \fr{3^h - 1}{h}.$$ %e So the difficulty here is to find the limit that we left above. But what's the derivative of $2^x$ and $3^x$ at $x = 0?$ It's %align by = %v \begin{align*} f'(0)&= 2^0 \lim_{h \to 0} \fr{2^h - 1}{h}\\ &= \lim_{h \to 0} \fr{2^h - 1}{h}. \end{align*} %e And for $3^x?$ It's %align by = %v \begin{align*} g'(0)&= 3^0 \lim_{h \to 0} \fr{3^h - 1}{h} \\ &=\lim_{h \to 0} \fr{3^h - 1}{h}. \end{align*} %e That is, $f'(x) = f'(0) 2^x,$ and $g'(x) = g'(0) 3^x.$ Observe, then, that the derivative of these exponential functions are themselves times a constant. What would this constant be? By letting $h \to 0,$ we see that $f'(0) \approx 0.693$ and $g'(0) \approx 1.09.$ So we wonder if there is a number $p$ between $2$ and $3$ such that $h(x) = p^x$ has $h'(0) = 1.$ There is such number; we call it $e.$ That is, $e$ is the number such that %v $$\lim_{h \to 0} \fr{e^h -1}{h} = 1.$$ %e Now, consider the exponential function $f(x) = a^x$ where $a = e.$ It has the remarkable property that $f'(x) = f'(0) a^x. = 1 \cdot a^x = a^x.$ And, because of this interesting derivative, it also follows that %v $$\int e^x\ dx = e^x + c.$$ %e \masciiexe{} Find a root of $f(x) = e^x + x = 0.$ {\em Solution.} We'll use Newton's method. We know that $f(0) = 1,$ and $f(-1) =$ $1/e - 1 < 0.$ So, there's a root between $-1$ and $0.$ We'll initiate Newton's method with $a = 1/2.$ We may write %align by = %v \begin{align*} f'(a)&= \fr{y - f(a)} {x - a}\\ (x- a) f'(a) &= y - f(a)\\ xf'(a) - a f'(a) &= y - f(a)\\ xf'(a) &= y - f(a) + a f'(a). \end{align*} %e But, we're looking for what $x$ gives $y = 0.$ So we set $y = 0,$ and solve for $x.$ So %v $$x = \fr{a f'(a) - f(a)}{ f'(a) }$$ %e After we compute $x$ for $a = 1/2,$ we would feed this value back into this equation, and again and again. If the value seems to converge, we jump to the conclusion that we're pretty close to the root. We find that the value seems to converge to $-0.56714329,$ so we take that as close enough to the root. \exeend \subsection{Logarithmic functions} How is $y = \log_a x$ defined? What we want it to mean is $y$ being the number such that $a^y = x.$ So $y = \log_a x$ is defined as $y = \log_a x$ if and only if $a^y = x.$ What would its domain be? They would be positive real numbers, so zero is not there. But why not? Let's see: $a^y = 0$ is not satisfied by any base $a.$ We could try $a = 0$ and $y = 0,$ but $0^0$ is actually undefined because there seems to be no grounds for the convergence of this expression. But why not negative numbers? See the section on exponential and their inverses; the exponential functions do not have negative range, and so the logarithmic ones don't have negative domain --- after all, one is the inverse of the other. What would its range be? We know that for $\log_a x = y,$ we have a is positive and $x$ is positive. So what values for $y$ are possible? Consider $x$ large. Then $y$ must be large too. Probably, then, $y \to \infty$ as $x \to$ $\infty.$ Now consider $x$ small; for example, $a^y = 0.1.$ Then $y$ will have to be negative so that, for $p = -y,$ %v $$\fr{1}{a^p} = 0.1 = \fr{1}{10},$$ %e so $a^p = 10.$ Therefore, between $0$ and $1,$ $\log_a x < 0.$ So the range is $(-\infty, \infty).$ The domain of $a^x$ is all real numbers, so naturally the range of $\log_a x$ is all real numbers. If $a = e,$ then $\log_e x$ is called the natural logarithm function, and we write $\ln x.$ This function is so used in mathematics that some authors prefer to write $\log x$ to mean the natural logarithm. Others prefer to let $\log x$ mean $\log_{10} x$ or $\log_2 x.$ I would prefer $\log x$ to mean the natural logarithm because to me it is the most interesting one, and I use it a lot more often then the others. I will stick to $\ln x$ because many authors do so, and I actually type one letter less than I would with $\log x$, so I will let that be my choice. Now, notice that although we write $\ln x$ instead of $\ln(x)$, it still is a function, and its inverse is $e^x$; so $\ln e^x = x$. We will later define these functions again, in some other way, which seems to be more interesting, and some facts about these functions will be clarified. \masciith{} For a $\neq 1,$ a $> 0,$ it is true that %v $$\log_a x = \fr{\ln x} {\ln a}. $$ %e \theend \masciiproof{} Writing $\log_a x$ means to say that there is a number $y$ such that $a^y = x.$ Now take natural logarithm of both sides of this equation to get %align by = %v \begin{align*} y\ln a &= \ln x\\ y&= \fr{ \ln x}{ \ln a}, \end{align*} %e as we wanted. \profend This proof shows that we can rewrite any logarithmic function in terms of $\ln x.$ But also, notice that we can rewrite any logarithmic function in terms of the logarithm of any valid base. A machine, therefore, needs only to implement an algorithm for one base, and rewrite all the others in terms of the chosen base. A quick note is that we have not shown why it's true that $\ln a^y = y \ln a.$ Let us do that now; let's prove the theorems of logarithms. We're going to need the derivative of $\ln x,$ so let's show it first. \masciith{} The derivative of $\ln x$ is $1/x.$ \theend \masciiproof{} Let $y = \ln x.$ So %align by = %v \begin{align*} y&= \ln x\\ e^y&= x\\ e^yy'&= 1\\ y'&= \fr{1}{e^y}\\ &=\fr{1}{e^{\ln x}}\\ &=\fr{1}{x}. \end{align*} \profend \subsection{The theorems of logarithms} There are three important theorems; the product theorem, the division theorem and the power theorem. The product theorem lays a foundation for the other two, so we prove it first. The proof of the first comes out by considering a function $f(x) = \ln ax.$ We differentiate it and we obtain a function equal to the $g(x) = \ln x.$ If two functions have the same derivative, they must differ by a constant. This fact is an important theorem of calculus which is a corollary of Fermat's theorem. Fermat's theorem is proved using another important theorem: the mean value theorem. The mean value theorem can be generalized and it was; by Cauchy. Cauchy's generalization is used to prove another important theorem; l'Hospital's theorem. \masciith{} For $x > 0, y > 0,$ it's true that $\ln xy = \ln x + \ln y.$ \theend \masciiproof{} Let $f(x) = \ln yx.$ Notice, however, that $y$ is a real number and $x$ is a variable. Now differentiate $f$ to get %align by = %v \begin{align*} f'(x)&= \fr{y}{yx}\\ &= \fr{1}{x} \end{align*} %e So $f$ and $\ln x$ have the same derivative; so $\ln x - \ln yx = c.$ Now, if the theorem is true, then $c = - \ln y.$ We cannot assume any value for $y$ because $y$ is an arbitrary constant, but $x$ is variable and takes all values of the domain of $\ln x.$ Since $x = 1$ is in the domain, we may consider the equation when $x = 1.$ We get %v $$0- \ln y = c,$$ %e as expected. Therefore $\ln x - \ln y = \ln yx.$ \profend Now we use this result to prove the others. \masciith{} For $x > 0$, it's true that $\ln x^y = y \ln x.$ \theend \masciiproof{} We have that $x^y = x_1 x_2$ ... $x_y.$ Apply the product theorem as much as necessary to write $\ln x^y = \ln x_1 \ln x_2 ... \ln x_y = y \ln x.$ This isn't a formal argument, however. Perhaps we can simply prove it by induction. Consider $\ln x^n,$ where $n$ is a natural number. Show that it works for $n= 1,$ as it clearly does. Assume it works for some natural $k:$ $\ln x^k = k \ln x.$ Then notice that %align by = %v \begin{align*} \ln x^{k+1} &= \ln x^k x\\ &= \ln x^k + \ln x \\ &= k \ln x + \ln x\\ &= (k + 1) \ln x. \end{align*} %e Therefore, it's true for all naturals. I guess we can similarly show for all negative integers, and take the case where $n = 0$ for granted. What about the rest of the reals, though? \profend By the way, do not mistake $\ln (x^y)$ for $(\ln x)^y;$ only the former has the property above. When I write without parentesis, I mean the first. \masciith{} For $x > 0, y \neq 0,$ it's true that $\ln x/y = \ln x - \ln y.$ \theend \masciiproof{} We know that $\ln x/y = \ln xy^{-1}.$ So %align by = %v \begin{align*} \ln xy^{-1} &= \ln x + \ln y^{-1}\\ &= \ln x - \ln y, \end{align*} %e as desired. \profend \subsection{The prime number theorem} Let $\pi(n)$ be the number of primes less or equal to $n.$ So $\pi(15) = 6$ because $2$, $3$, $5$, $7$, $11$, $13$ are the $6$ primes before $15.$ Also, we find that $\pi(25) = 9,$ and $\pi(100) = 25.$ There is a theorem, called the Prime Number Theorem, which asserts that %v $$\lim_{n \to \infty} \fr{ \pi(n) } { n \ln n } = 1.$$ %e \masciiexe{} Find the inverse of %v $$y= \fr{ e^x + 1 } { -e^x + 1 }.$$ %e {\em Solution.} Try to group the exponential terms together, and factor them. %align by = %v \begin{align*} y&= \fr{ e^x + 1 } { -e^x + 1 }\\ y- y e^x &= e^x + 1\\ -ye^x - e^x &= 1 - y\\ e^x(-y - 1) &= 1 - y\\ e^x&= \fr{ 1 - y} {- y - 1}\\ x&= \ln \fr{ 1 - y} {-y - 1}\\ x&= \ln - \fr{ 1 - y} {y + 1}\\ x&= \ln \fr{ 1 + y} {1 - y}. \end{align*} %e It works. \exeend Now, let us look into another technique. Any function $f(x) = g(x)^{h(x)}$ where $g(x) > 0$ can be analyzed as a power of $e$ because we can write $g(x)$ as $e^{\ln g(x)}$ so that $f(x) = h(x) e^{\ln g(x)}.$ This allows us to compute, for instance, %v $$\lim_{x \to \infty } x^{ \ln x }.$$ %e To compute, we write %align by = %v \begin{align*} \lim_{x\to \infty } x^{ \ln x } &= \lim_{x \to \infty } e^{\ln x^{ \ln x} }\\ &= \lim_{x \to \infty } e^{\ln x \ln x}\\ &= \lim_{x \to \infty } e^{(\ln x)^2}\\ &= \lim_{x \to \infty } e^{2 \ln x} = \infty. \end{align*} %e We know that $\ln x \to \infty$ as $x \to \infty$ and $e^x \to \infty$ as $x \to \infty.$ So the limit above goes to $\infty$ too. An interesting observation is that we did not fallaciously used the property of logarithms to write the last step. We use a property of exponents; notice that $(a^x)^y = a^{xy}.$ \masciiexe{} Find $f^{(n)}$ if $f(x) = \ln(x - 1).$ {\em Solution.} The first derivative is %v $$f'(x)= \fr{1} {x - 1}.$$ %e The second is %v $$f''(x)= - \fr{1}{(x -1)^2}.$$ %e The third is %v $$f^{(3)}(x)= \fr{2}{(x -1)^3}.$$ %e The forth is %v $$f^{(4)}(x)= - \fr{6}{(x -1)^4}.$$ %e Let us let $n$ stand for the power of the denominator because it matches it straightforwardly. For the sign of the expression, let $(-1)^{n+1}$ determine the sign, because it is negative when $n$ is even, and positive otherwise. So far, then, we have %v $$f^{(n)} = \fr{ (-1)^{n+1} k } {(x - 1)^n}.$$ %e We have to figure out the term that will replace $k$ more precisely. For $k,$ we have that it goes as $1, 1, 2, 6, 24,$ ... so we may guess that they can be expressed as $n!$ with $n = 0, 1, 2,$ ... So we need match this list with respect to the $n$ as we have already established in the formula above. For the first derivative, $n = 1,$ and $k = 0!;$ when $n = 2, k = 1!;$ when $n = 3, k = 2!.$ So we jump to the conclusion that %v $$f^{(n)} = \fr{ (-1)^{n+1} (n - 1)! } {(x - 1)^n}.$$ %e And we're done. \exeend \subsection{The general antiderivative of the reciprocal function} We know that $\ln x + c$ has $1/x$ as its derivative, but when we integrate $1/x,$ we may have negative numbers as limits of integration and $\ln x + c$ is only defined for positive numbers. But it's valid to ask about the area ``under $1/x$'' on the negative side since $1/x$ is defined everywhere but at zero. So for limits on the negative side, the antiderivative $\ln x + c$ will not be helpful. Consider $\ln |x| + c.$ To differentiate this function, we consider $x > 0,$ and $x < 0.$ When $x > 0$, $\ln |x| + c = \ln x + c,$ so the derivative is $1/x.$ When $x < 0,$ then $\ln |x| + c = \ln -x + c,$ so the derivative is $(-1)/(-x) = 1/x.$ Therefore, % \[ \fr{d}{dx} \ln |x| = \begin{cases} 1/x & \text {if $x > 0$},\\ 1/x & \text {if $x < 0$}. \end{cases}\] % In other words, the general antiderivative of $1/x$ is $\ln |x| + c.$ \subsection{A nice definition of the natural log} Some people prefer to define $\ln x$ not as the inverse of exponential function $e^x,$ but rather as the area under $1/x.$ This definition is really nice because we can nicely show that it is the inverse of the exponential function, and it becomes quite apparent how to compute $\ln x$ for any $x > 0.$ Let us define it then. \masciidef{} Define %v $$\ln x = \int_1^x \fr{1}{t} \dt,$$ %e for all $x > 0.$ \defend The derivative of $\ln x$ is trivially obtained by the fundamental theorem. Let us now consider the inverse function of $\ln x.$ We will call it $\exp x,$ but it is not to be assumed yet that $\exp x = e^x.$ We will show so. \masciidef{} Define %v $$\exp x = y \text{\quad if and only if\quad} \ln y = x. $$ %e \defend Now, because they are inverse functions, we can write $\exp \ln x = x$ and $\ln \exp x = x.$ Also, we have identities such as $\exp 0 = 1$ because $\ln 1 = 0.$ We have $\exp 1 = e$ because $\ln e = 1.$ Now if $r$ is any rational number, then consider %v $$\ln e^r = r \ln e = r.$$ %e Therefore, $\exp r = e^r$ because $\ln e^r = r.$ This is how the exponential function comes to life as the inverse function of $\ln x.$ We have not considered irrational numbers, however. Let $r$ be irrational. Consider $a^r,$ where $a > 0.$ We define %v $$a^r = \lim_{x \to r} a^x.$$ %e Now consider $\ln e^r.$ Since $r$ is irrational, we have %align by = %v \begin{align*} \ln e^r &= \ln \lim_{x \to r} e^x\\ &= \lim_{x \to r} \ln e^x\\ &= \lim_{x \to r} x \ln e\\ &= \lim_{x \to r} x\\ &= r. \end{align*} %e But notice that we had trouble proving that the power theorem of logarithms works for all real numbers. Here we're assuming so. We assumed that $\ln x$ is continuous when we pulled the limit out from the $\ln x$ function. \subsection{All other exponential functions} Let us now define all the other exponential functions; we will define them in terms of the exponential function $e^x.$ \masciidef{} Let $m > 0.$ Define %v $$m^x = e^{ \ln a^x } = e^{ x \ln a },$$ %e for all $x.$ \defend This definition makes some proofs easy to be written. The trouble here is to trust that this definition makes sense. For instance, how do we know that $2^2 = e^{2 \ln 2}$? \masciith{} Let $m$ be a real number. Then $m^{x+y} = m^x m^y.$ \theend \masciiproof{} Apply the definition above in %align by = %v \begin{align*} m^{x+y}&= e^{ \ln m^{x+y} }\\ &= e^{ (x+y) \ln m }\\ &= e^{ x \ln m + y \ln m}\\ &= e^{ x \ln m } e^{y \ln m}\\ &= m^x m^y, \end{align*} %e as desired. \profend \subsection{The inverse trigonometric functions} Now we will use the knowledge acquired so far to look at new functions, their properties and possibly some applications of them. For instance, we will begin to finding the derivative of the inverse trigonometric functions. Let $y = \arcsin x,$ so $\sin y = x.$ The domain of $\arcsin x$ is $[-1, 1],$ and the range of $\arcsin x$ is $-\pi/2 \leq y \leq \pi/2.$ Implicitly, we differentiate it to get %align by = %v \begin{align*} \cos y y'&= 1\\ y'&=\fr{1}{\cos y}. \end{align*} %e With the range above, we have that $\cos y \geq 0.$ We know that $\cos^2 y + \sin^2 y = 1$, so $\cos^2 y = 1 - \sin^2 y,$ so $\cos y =$ $\sqrt{1-\sin^2 y}.$ But notice that $\sin y = x,$ so $\sin^2 y =$ $x^2.$ Therefore, $\cos y = \sqrt{1 - x^2}$. So %v $$y'= \fr{1}{\sqrt{1 - x^2}}.$$ %e We perform the same strategy for $\arccos x.$ The function $\cos x$ is one-to-one from $0$ to $\pi.$ So the domain of $\arccos x$ is $[-1,1];$ the range is $[0, \pi].$ Let y = $\arccos x,$ so $\cos y = x.$ Then %align by = %v \begin{align*} -\sin y y'&= 1\\ y'&= - \fr{1}{\sin y}. \end{align*} %e Since $\sin y = \sqrt{1 - \cos^2 y}$ and $\cos^2 y = x^2,$ we have %v $$y' = - \fr{1}{\sqrt{1-x^2}}.$$ %e Let us now look into the inverse tangent function. We have to restrict $\arctan x$ on $(-\pi/2, \pi/2)$ because we need an one-to-one function. The domain for $\arctan x$ is $(-\infty, \infty)$ and range is $(-\pi/2, \pi/2).$ To find its derivative, we write %align by = %v \begin{align*} \arctan x &= y\\ x &= \tan y. \end{align*} %e Now differentiate with respect to $x.$ We get %align by = %v \begin{align*} 1&= \sec^2 y y'\\ y'&= \fr{1}{\sec^2 y}. \end{align*} %e We wish now to express $\sec^2 y$ in terms of $x.$ Notice that $\sec^2 y = \tan^2 y + 1.$ Since $x = \tan y,$ then $\sec^2 y = x^2 + 1.$ Therefore %v $$y'= \fr{1}{x^2 + 1},$$ %e and we're done. Now we know a few antiderivatives. For instance, %v $$\int \fr{1}{\sqrt{1-x^2}} \dx = \arcsin x + c,$$ %e and %v $$\int - \fr{1}{\sqrt{1-x^2}} \dx = \arccos x + c,$$ %e and %v $$\int \fr{1}{x^2 + 1} \dx = \arctan x + c.$$ %e Notice that sometimes we have integrands similar, though not exactly one of the above; but it's a matter of algebraically working it out for that to be so. For example, %align by = %v \begin{align*} \int\fr{1}{x^2 + a^2} \dx &= \int \fr{\dx}{a^2 \bl(\fr{x^2}{a^2} + 1\br)}\\ &=\fr{1}{a^2} \int \fr{\dx}{ \bl(\fr{x}{a}\br)^2 + 1}. \end{align*} %e Now let $u = x/a,$ so $\dx = a \du.$ Then %align by = %v \begin{align*} \fr{1}{a}\int \fr{\du}{u^2 + 1} &= \fr{1}{a} \arctan u + c\\ &=\fr{1}{a} \arctan \fr{x}{a} + c. \end{align*} %e A similar work can be done for all other inverse trigonometric functions. \masciiexe{} Evaluate %v $$\int \fr{x \dx}{x^4 + 9}.$$ %e {\em Solution.} First, factor the number $9$ from the denominator. We get %v $$\fr{1}{9} \int \fr{x \dx}{ \bl( \fr{x^4}{9} + 1\br)}.$$ %e Now let $u = (x^2)/3$, so that $\dx = (3/2x) du.$ Then we write %align by = %v \begin{align*} \fr{1}{9}\fr{3}{2} \int \fr{\du}{ u^2 + 1} &= \fr{1}{6} \arctan u + c\\ &=\fr{1}{6} \arctan \fr{x^2}{3} + c, \end{align*} %e as desired. \exeend \masciiexe{} Simplify $\cos (\arctan x).$ {\em Solution.} Let $y = \arctan x$ so that $\tan y = x.$ If $\tan y = x,$ then thinking of a right triangle, we have an angle $y$ and $y$'s opposite side has length $x$, while $y$'s adjacent side has length $1.$ Therefore, the hypothenuse has length $\sqrt{1 + x^2}$, and $\cos y = 1/\sqrt{1 + x^2}.$ So %v $$\cos(\arctan x) = \fr{1}{\sqrt{1 + x^2}}$$ %e is simplified as desired. \exeend \masciiexe{} Prove that $\arctan x + \arccot x = \pi/2.$ {\em Solution.} Let $f(x) = \arctan x + \arccot x.$ Notice that %v $$f'(x) = \fr{1}{x^2 + 1} - \fr{1}{x^2 + 1} = 0.$$ %e This means that $\arctan x$ has the same derivative as $\arccot x.$ Therefore, they must differ by a constant. That is, $f(x) = c,$ for all $x.$ Notice that %v $$\arctan 1 + \arccot 1 = \fr{\pi}{4} + \fr{\pi}{4} = \fr{\pi}{2}.$$ %e So $f(x) = \arctan x + \arccot x = \pi/2.$ \exeend \masciiexe{} Find $\arcsin \sqrt{3}/2.$ {\em Solution.} We know that $\sin \pi/3 = \sqrt{3}/2,$ so $\arcsin \sqrt{3}/2 = \pi/3.$ \exeend \masciiexe{} Draw the following diagram: mark a point $A$ and draw a straight line to the right up to a point $P;$ continue the straight line up to a point $B.$ From $B$ draw a perpendicular line going up to a point $Q.$ From $Q$ draw a line that meets the point $P.$ This procedure creates a right triangle where the angle $PBQ$ is a right angle. Now, from $A$, draw a perpendicular line going up to a point $R.$ From $R$ draw a line that meets the point $P;$ so now that we have two right triangles where $P$ is a point that they have in common. Now, let $AB = 3$ units. Let $RA = 5$ units, and let $QB = 2$ units. Let the angle $RPA$ be called $\gamma$ and let the angle $QPB$ be called $\alpha.$ There's an angle $\beta$ such that $\beta = \pi/2 - (\gamma + \alpha).$ Assuming we can move the point $P$ between $A$ and $B$ adjusting the size of the hypothenuses of the triangles, how far from $A$ should we position $P$ as to maximize the angle $\beta?$ {\em Solution.} Let $PB = x$ so that $AP = 3 - x.$ So RP = $\sqrt{25 + (3-x)^2},$ and PQ = $\sqrt{4 + x^2}.$ Now, we need to express $\beta$ with respect to $x.$ So, naturally we will use trigonometric functions; but the ones that express angles are the inverse trigonometric ones, so we must find out which ones will give us a simple enough expression. Also, notice that $\beta$ is maximized, when the sum $\gamma + \alpha$ is minimized, so we may attack the problem either way; I choose the latter; I'm going to minimize the sum $\gamma + \alpha.$ Notice that %align by = %v \begin{align*} \tan\gamma &= \fr{5}{3 - x}\\ \tan\alpha &= \fr{2}{x}. \end{align*} %e So, it's certainly true that %align by = %v \begin{align*} \gamma&= \arctan \fr{5}{3 - x}\\ \alpha&= \arctan \fr{2}{x}. \end{align*} %e Therefore %v $$\gamma+ \alpha = \arctan \fr{5}{3 - x} + \arctan \fr{2}{x}.$$ %e Now the work is to differentiate this function, and find its local minimums, and identify which one is the one we are looking for. Notice that because $AB = 3$ units, $P$ can only range from $A$ to $B;$ that is, the value of $x$ ranges from 0 to 3. In fact, I can see that the value where $\beta$ is maximized lies between 2 and 3. It might even be $x = 2.5$ --- so $P$ is closer to $B$ than to $A.$ That's what I have to say; I don't want to compute the value. We could have used any of the inverse trigonometric functions, but I believe that $\arctan x$ would give us the simplest case; I initially tried $\arcsin x$ and I found very complicated derivatives for us to manually look for roots, so I tried $\arctan x$ instead and it became simple enough --- though I didn't want to find its roots anyway. \exeend \subsection{The hyperbolic functions} The hyperbolic functions have nothing to do with trigonometric functions, but they have very similar properties. Let begin with their definitions. Define %v $$\sinh x = \fr{e^x - e^{-x}}{2}$$ %e as the hyperbolic sine function. Define %v $$\cosh x = \fr{e^x + e^{-x}}{2}$$ %e as the hyperbolic cosine. Now, like the trigonometric functions, we define $\csch x = 1/\sinh x,$ $\sech x = 1/\cosh x,$ $\tanh x = \sinh x / \cosh x,$ and finally $\coth x = \cosh x / \sinh x.$ Notice that $\cosh x$ is the average of the functions $e^x$ and $e^{-x},$ and $\sinh x$ is half of the difference between them. These functions are pretty popular in scientic contexts; they are used to describe some important phenomena. For instance, the effect of gravity on a wire hanging on posts --- such as eletricity posts --- on the streets is described by %v $$y= c + d \cosh \fr{x}{d},$$ %e where $c$ and $d$ are constants. By the way, when a wire hangs between two posts, the the system satisfies the differential equation %v $$\sdiff= \fr{pq}{T} \sqrt{1 + \bl(\fdiff\br)^2},$$ %e where $p$ is the linear density of the wire, $g$ is the acceleration due to gravity and $T$ is the tension in the cable, at its lowest point. A solution to this equation is the function %v $$y = \fr{T}{pg} \cosh \fr{pgx}{T}.$$ %e I think that linear density is a term used to describe the density of objects that have a shape that we may consider one dimensional. No objects that we know in the world are one dimensional, of course, but a wire can be considered one dimensional, so the density of a wire is a measure of mass per unit length, and we call it linear density. These hyperbolic functions satisfy some identies which look very much like the trigonometric functions --- though not always they are too similar. For instance, while $\cos^2 x + \sin^2 x = 1,$ we have $\cosh^2 x - \sinh^2 x = 1.$ Let us now define the inverse of the $\sinh x$ function, and then find its derivative. To define it, let $y = \arcsinh x$ so that $\sinh y = x.$ Then %align by = %v \begin{align*} x&= \sinh y \\ &= \fr{e^y - e^{-y}}{2}\\ 2x&= e^y - e^{-y}\\ e^y- e^{-y} - 2x &= 0. \end{align*} %e Multiply this equation by $e^y$ to get %align by = %v \begin{align*} (e^y)^2- e^y e^{-y} - 2xe^y &= 0\\ (e^y)^2- 2xe^y - 1&= 0. \end{align*} %e Now let $p = e^y,$ and notice that we have a quadratic equation, whose solution is $p = x \pm \sqrt{x^2 + 1}.$ But since $e^y > 0$ for all $y,$ and since $x < \sqrt{x^2 + 1}$ for all $x,$ we need not consider the solution $p = x - \sqrt{x^2 + 1}.$ Therefore %align by = %v \begin{align*} e^y&= x + \sqrt{x^2 + 1}\\ y&= \ln (x + \sqrt{x^2 + 1})\\ &= \arcsinh x. \end{align*} %e The inverse hyperbolic cosine can be defined in a similar manner. Their derivatives are easily obtained by using the chain rule. \subsection{The theorem of l'Hospital} L'Hospital's theorem is a pretty interesting one. Before we state it, let us consider something. Suppose that $f'$ and $g'$ are continuous everywhere. Then let us consider the expression %v $$\lim_{x \to c} \fr{ f'(x) } {g'(x) }.$$ %e Since they are both continuous, we use the theorem of limits of continuous function to write that %v $$\lim_{x \to c} \fr{ f'(x) } {g'(x) } = \fr{f'(c)}{g'(c)}.$$ %e Now, let us write the equation in terms of the definition of derivatives, and algebraically manipulate it. We write %align by = %v \begin{align*} \lim_{x \to c} \fr{ f'(x) } {g'(x) } &= \fr{f'(c)}{g'(c)}\\ &=\fr{\lim_{x \to c} \fr{f(x)-f(c)}{x-c}} {\lim_{x \to c} \fr{g(x)-g(c)}{x-c}}\\ &=\lim_{x \to c} \fr{\fr{f(x)-f(c)}{x-c}} {\fr{g(x)-g(c)}{x-c}}\\ &=\lim_{x \to c} \fr{f(x)-f(c)} {g(x)-g(c)}. \end{align*} %e Now, suppose that $f(c) = g(c) = 0.$ Then %align by = %v \begin{align*} \lim_{x \to c} \fr{f(x)-f(c)} {g(x)-g(c)} &= \lim_{x \to c} \fr{f(x)} {g(x)}. \end{align*} %e Let us look back now. We supposed $f'$ and $g'$ are continuous, and that $f(c) = g(c) = 0;$ this last supposition means that $f' \to 0$ as $x \to c;$ similarly for $g';$ that is, $g' \to 0$ as $x \to c.$ And, we concluded that %v $$\lim_{x \to c} \fr{f(x)}{g(x)} = \lim_{x \to c} \fr{f'(x)}{g'(x)}.$$ %e In other words, the limit of the ratio of two functions is equal to the limit of the ratio of their derivatives, provided that the derivatives are continuous over the interval where $c$ is, and provided that the derivatives approach zero as $x$ approaches $c.$ This is sort of what l'Hospital's theorem is about. We have proved a particular case of l'Hospital's theorem. The theorem is more general than this: it applies to the case when the derivatives approach infinity too. It is harder to prove l'Hospital's theorem more generally, so I'm going to leave it at this for now. Using algebra, the theorem of l'Hospital can be used to evaluate limits of other indeterminated forms such as $0 \cdot \infty, \infty - \infty, 0^0,$ $\infty^0,$ and $1^\infty.$ The products, subtractions and divisions can be treated by plain algebra, and the powers can be treated by logarithms. \subsection{A função exponencial com a hipérbole no expoente} Qual é o gráfico de $e^{1/x}$? No lado positivo, quando $x \in (0, 1]$, a função desce do infinito até chegar a $f(1) = e$ porque um $x$ muito pequeno produz um exponente grande pra função exponencial. Então perto de zero, a função tem altíssimos valores. Afastando-se de zero e chegando a $x=1$, a função chega a $y = e$ exatamente. Dali pra direita, o expoente decai a zero, o que faz com que $y$ se aproxime de $1$. Então $y=1$ é uma assíntota horizontal. No lado negativo, quando $x \in [-1, 0)$, o expoente vai de $-1$ a números altíssimos de magnitude, mas negativos. Qual é o efeito de um expoente negativo de grande magnitude? O efeito é $1/e^M$, sendo $M$ grande e positivo. Logo o efeito é levar a função a zero. Já quando $x \to -\infty$, o expoente tende a zero, fazendo com que a função tenda a 1. Então no lado negativo, a função parte de $y=0$ e tende a $y=1$ no sentido negativo. \section{Sequências e séries} Sequências e séries são importantes em Cálculo porque elas servem como os fundamentos da teoria. Quão bom são esses fundamentos é algo questionável e investigável, mas não faremos isso aqui. Estamos apenas interessados na análise por enquanto. A primeira com que se começar é a série geométrica. A série geométrica é também chamada de progressão geométrica. Por exemplo, se começamos uma sequência com 1 e tomamos metades em sequência, obtemos $1, 1/2, 1/4, 1/8$, ..., $1/2^n$, sendo que o índice inicial é $n = 0$. Seja $r$ a razão da sequência. Então a série gerada por essa sequência é % $$S = 1 + r + r^2 + \cdots + r^i + \cdots.$$ % Qual é a soma dessa série? Compute % $$rS = r + r^2 + r^3 + \cdots + r^i + \cdots$$ % e então compute $S - rS = 1$. Logo, obtemos a soma por computar % $$S(1 - r) = 1 \implies S = \frac{1}{1 - r},$$ % que converge se e somente $|r| < 1$. Nesse exemplo, o primeiro termo é $1$, mas a série inteira poderia ter um fator $a$ em cada termo, quando obteríamos % $$\sum_{i=0}^\infty ar^i = a + ar + ar^2 + \cdots + ar^i + \cdots$$ % Usando o mesmo raciocínio acima, obtemos a soma % $$S = \frac{a}{1 - r}.$$ % Agora note que essas são séries infinitas. Se truncarmos essa série tal que obtenhamos, por exemplo, % $$\sum_{i=0}^n ar^i = a + ar + ar^2 + \cdots + ar^n,$$ % então temos uma situação nova em que só aparecem $n + 1$ termos aí. Qual é a soma dessa série finita? O mesmo raciocínio procede, mas o resultado é diferente. Obtemos % $$S = \frac{a(1 - r^{n+1})}{1 - r} = \frac{a}{1-r} - \frac{ar^{n+1}}{1 - r}.$$ % É claro que se agora permitirmos que $n \to \infty$, obtemos o resultado anterior. Então fica a observação de que o caso finito é um e o infinito, outro. A série geométrica serve como primeiro exemplo de uma série convergente. Vejamos agora uma primeira série divergente, que é a série harmônica 1 + 1/2 + 1/3 + 1/4 + ... Por que essa linda série diverge? Porque é possível compará-la a uma outra série que claramente diverge---uma série com infinitas metades. Veja---é óbvio que $1 + 1/2 + 1/2 + 1/2 + \cdots$ diverge. Agora, note na harmônica que $1/3 + 1/4 > 1/4 + 1/4 = 1/2$. Note que $1/5 + 1/6 + 1/7 + 1/8 > 4 \cdot 1/8 = 1/2$. Reconheça que é possível continuar a agrupar os termos em metades eternamente. Portanto, a harmônica diverge. Agora que temos uma convergente e uma divergente, podemos usá-las como base de comparação para outras. Uma série muito famosa de Euler é % $$\sum_{i=1}^\infty \frac{1}{i^2} = \frac{\pi^2}{6}.$$ % Loucura, não? A intuição de Euler aí foi brilhante, mas a gente retorna eventualmente pra contar o que ele fez. Pro momento, só queremos deduzir que a série converge---não importa pra onde. Por que ela converge? Usaremos uma série geométrica. Ignore o $1 + 1/2^2$ por agora. Observe que $1/3^2 + 1/4^2 > 2 \cdot 1/4^2$. Okay. Observe também que $1/5^2 + 1/6^2 + 1/7^2 + 1/8^2 > 4 \cdot 1/8^2$. Vê a estratégia? Conseguimos comparar a de Euler com a série geométrica $1 + 1/4 + 1/8 + 1/16 + \cdots + 1/2^i + \cdots$, que já sabemos convergir pra $2 = 1/(1 - 1/2)$. Portanto, a de Euler converge e converge mais rápido que essa geométrica. O mesmo argumento pode ser aplicado de uma forma levemente diferente, ilustrando que podemos tomar mais de um ponto de vista. Em vez da estratégia acima, considere a seguinte. Note que $1/2^2 + 1/3^2 < 2 \cdot 1/2^2 = 1/2$. Note também que $1/4^2 + 1/5^2 + 1/6^2 + 1/7^2 < 4 1/4^2 = 1/4$. Obtemos a mesma série geométrica de razão $1/2$, mas de um jeito levamente diferente---e também mais claro. \begin{exercise} O que podemos concluir sobre % $$\sum_{i=2}^\infty \frac{1}{i^2 - 1}?$$ % Primeiro note que ela é muito parecida com a de Euler. Se a de Euler converge, seria dramático se essa não convergisse: no infinito, não $i^2$ e $i^2 - 1$ são praticamente a mesma coisa. Mas não é assim que se argumenta. Primeiro observe que a série começa em $i=2$. Compare-a com a de Euler a partir de $i=2$ com a série % $$\sum_{i=2}^\infty \frac{1}{(i - 1)^2} = 1 + 1/4 + 1/9 + \cdots.$$ % Já sabemos que a de Euler converge com início em $i=1$ e portanto converge com início em $i=2$. Isso nos dá % $$\sum_{i=2}^\infty \frac{1}{i^2} < \sum_{i=2}^\infty \frac{1}{(i - 1)^2}$$ % A pergunta é---a da direita converge? Claro: ela é idêntica a de Euler exceto que ela tem um termo a mais, que é o termo 1. Portanto, ela converge. Agora note que a nossa série de interesse pode ser comparada a essa da direita, garantindo que ela converge também. É óbvio que $i^2 - 1 < i^2$. Agora precisamos perceber que $(i - 1)^2 < i^2 - 1$. Note que $i=1$ nem nos interessa. Quando $i=2$, obtemos $1 < 3$. Quando $i = 3$, obtemos $4 < 8$. É óbvio que a desigualdade é verdadeira. Agora, é possível manipular a série de interesse e obter o limite dela. Note que $i^2 - 1 = (i + 1)(i - 1)$, que é o denominador da série de interesse. Agora experimente calcular % $$\frac{1}{i - 1} -\frac{1}{i + 1} = \frac{i + 1 - (i - 1)}{i^2 - 1} = \frac{2}{i^2 - 1}.$$ % Aparece um $2$ no numerador. Multiplique a série por $1/2$ e obtenha % $$\sum_{i=2}^n \frac{1}{i^2 - 1} = \frac{1}{2} \sum_{i=2}^n \left( \frac{1}{i - 1} - \frac{1}{i + 1}\right).$$ % Se você expandi-la, você obtém % $$1/2 - \cancel{1/6} + 1/4 - \cancel{1/8} + \cancel{1/6} - 1/10 + \cancel{1/8} - 1/12 + \cdots$$ % Na aproximação acima, estamos com o erro $-(1/10 + 1/12)$, mas esse erro tende a zero: quando chegarmos a $i=6$, obteremos um termo $1/10$ que cancelerá o $- 1/10$. Quando chegarmos a $i=7$, obteremos o termo $1/12$ que cancelerá o $-1/12$. Certo---novos erros aparecerão, que serão eventualmente cancelados por $i$ maiores. Os termos que sobram são $1/2 + 1/4 = 3/4$, que é a soma da série. Então uma estratégia interessante é procurar substituir uma série por uma expressão equivalente que seja telescópica. \end{exercise} \subsection{Método de Newton e raízes quadradas} Use o método de Newton. Por exemplo, se você quer uma sequência que convirja pra $\sqrt{2}$, então construa o polinômio $x^2 - 2 = 0$ e agora use o método de Newton pra solucionar o polinômio. Obtemos % \begin{align*} a_{n + 1} = a_n - \frac{f(a_n)}{f'(a_n)} = a_n - \frac{a_n^2 - 2}{2a_n} = \frac{2a_n^2 - a_n^2 + 2}{2 a_n} = \frac{a_n^2}{2a_n} + \frac{2}{2a_n} = \frac{1}{2}\left(a_n + \frac{2}{a_n}\right). \end{align*} % Experimente. Seja $a_0 = 3$, por exemplo. Então $a_1 \approx 1.8333$, $a_2 \approx 1.4621$, $a_3 \approx 1.4149$, $a_4 \approx 1.4142$. E quanto a $e$? Você precisa encontrar uma equação que defina o número implicitamente. Por exemplo, $\ln x = 1$. Então o método de Newton nos dá a sequência % $$a_{n + 1} = a_n - \frac{f(a_n)}{f'(a_n)} = a_n(2 - \ln a_n)$$ % e daí obtemos $a_0 = 3$, $a_1 \approx 2.7041$, $a_2 \approx 2.7182$, obtendo uma precisão de pelo menos quatro casas decimais. %\subsection{} \newpage \tableofcontents \newpage \printbibliography \end{document}