Debate 2 - Notebook - Método da Bissecção.ipynb

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    "# Capítulo 2 - Soluções de equações em uma variável #\n",
    "\n",
    "Os métodos numéricos discutidos neste capítulo visam aproximar soluções para equações de uma variável quando as soluções exatas destas equações não podem ser obtidas através de métodos algébricos (solução explícita), como é o caso da equação $90000 e^{\\lambda} + \\frac{10500}{\\lambda} (e^{\\lambda} -1) - 125000 = 0$.\n",
    "\n",
    "Estes métodos utilizam teoremas matemáticos para criar algoritmos iterativos que convergem para a solução da equação quando os parâmetros iniciais são adequadamente escolhidos. Cada método apresentado neste capítulo possui suas especificidades e critérios de convergência.\n",
    "\n",
    "Os métodos serão implementados para resolver uma das questões mais básicias da aproximação numérica: a determinação de raízes (ou zeros) de uma função do tipo $f(x) = 0$."
   ]
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    "## Método da Bissecção #\n",
    "\n",
    "Em computação, os processos que usam da técnica de divisão contínua de conjuntos ao meio para obter um resultado são chamados de procedimentos de busca binária. O Método da Bissecção é um desses procedimentos, e baseia-se no **Teorema do Valor Intermediário** para buscar a raiz de uma função em um determinado intervalo $[a,b]$ através de sucessivas subdivisões deste. \n",
    "\n",
    "---\n",
    "<p style=\"text-align: center;\">**Teorema do Valor Intermediário**</p>\n",
    "\n",
    "Supondo que $f$ é uma função contínua definida no intervalo $[a, b]$, e que $f(a)$ e $f(b)$ possuem sinais opostos, ou seja, $f(a) . f(b) < 0$, então existirá pelo menos um número $p$ em $(a, b)$ com $f(p) = 0$. \n",
    "\n",
    "---\n",
    "\n",
    "O Método da Bissecção inicia com a definição de um intervalo $[a, b]$ para o qual uma determinada função $f$ atenda aos critérios do teorema do valor intermediário, garantindo assim a existência de uma raiz da equação $f(x)=0$.\n",
    "\n",
    "Definido tal intervalo calcula-se o valor da função $f(x)$ em seu ponto médio $(p)$, sendo:\n",
    "\n",
    "$${p}={a}_{n}+\\frac{{b}_{n}-{a}_{n}}{2}=\\frac{{a}_{n}+{b}_{n}}{2}$$.\n",
    "\n",
    "a cada subdivisão, o ponto médio do intervalo, $p$, é testado, podendo uma das 3 seguintes situações ocorrer:\n",
    "\n",
    "<p style=\"text-align: center;\"> 1. $f(p)=0$; </p>\n",
    "<p style=\"text-align: center;\"> 2. $f(p)>0$; </p>\n",
    "<p style=\"text-align: center;\"> 3. $f(p)<0$. </p>\n",
    "\n",
    "A ocorrência da situação 1 (ou se atende aos critérios de tolerância estabelecidos) significa que $p$ é a raiz da equação $f(x)=0$ e o método é encerrado. Já as situações 2 e 3 indicam a necessidade de uma nova iteração que avalie os pontos $a$, $b$, e o ponto médio entre eles, $p$: se $f(a) . f(p) < 0$ significa que a raiz da função está no intervalo $[a_n,b_n]=[a, p]$. Neste caso, $a_n$ mantém o valor de $a$ e $b_n$ assume o valor de $p$; caso o contrário, se $f(p) . f(b) < 0$, $a_n$ assume o valor de $p$ e $b_n$ mantém o valor de $b$.\n",
    "\n",
    "Como o intervalo $[{a}_{n}, {b}_{n}]$ atende aos critérios do Teorema do Valor Intermediário, a cada iteração $n$ o intervalo estudado vai fechando em direção à raiz, até que o ponto $p$ resulte em $f(p) = 0$ ou que os critérios de tolerância sejam atendidos para uma aproximação considerada suficientemente boa ($f(p) \\approx 0$ ou outro critério de parada do método). \n",
    "\n",
    "O método da bisseccção pode adotar diversos critérios de parada. Alguns deles são:\n",
    "\n",
    "* $\\frac{{b}_{n}-{a}_{n}}{2}<\\varepsilon$ (tamanho limite da bissecção do intervalo $[{a}_{n}, {b}_{n}]$ após $n$ iterações)\n",
    "\n",
    "* $\\left|{p}_{n}-{p}_{n-1}\\right|<\\varepsilon$\n",
    "\n",
    "* $\\frac{\\left|{p}_{n}-{p}_{n-1}\\right|}{{p}_{n}}<\\varepsilon$, com ${p}_{n}\\neq0$ (seu uso é o mais indicado, por se aproximar de um erro relativo)\n",
    "\n",
    "* $\\left|f({p}_{n})\\right|<\\varepsilon$\n",
    "\n",
    "* $n>N_0$ ($n$ atingiu o número máximo de iterações $N_0$. Comumente usado para evitar loops infinitos em métodos iterativos que não convergem)\n",
    "\n",
    "---\n",
    "\n",
    "<p style=\"text-align: center;\"> **Teorema - limitante para o erro de aproximação no método da Bissecção** </p>\n",
    "\n",
    "Sejam $f \\in C[a, b]$ e $f(a).f(b)<0$, o método da Bissecção gera uma sequência $\\left \\{ {p}_{n} \\right \\}_{n=1}^{\\infty}$ que se aproxima de um zero $p$ de $f$ com $$\\left|{p}_{n}-{p}\\right| \\leqslant  \\frac{b-a}{2^{n}}\\textrm{, quando n}\\geqslant 1$$\n",
    "**Demonstração:**\n",
    "\n",
    "![Teorema_2_1](https://lh3.googleusercontent.com/_FFpatuJfuNL_cEpb7gg_bwrAq3MJ5km93j_FArpcJ42LhzeFP3m2zmREekOTec-5wu7ACc0JmY03oK9HaqcT9TEVWLn3otczck9dmxTlbuSNi-GecizvgRqbA2DCWoR6lTMONXHE1_x8RzaaazmWewsAGazGK0ns4PXf8PwGIBlt_lG2detMNWfQWHKXHokZrj_mv1ixECIrNw_HEjpNwUk2ijqMGnp1w5B6D8JTW3pdKpmRFGza7JcQVBicPnbyhPSYk0DDyQYXDumo_VhOJE7M4i0HxNkBq1ajTfvzBJBv6e5G4Y77F-4Qi59WanvW_x_fTKF0ZNEvaXE-4AOhfQurer0Sy2X-a4c-fb4uQvkdKLCOeqPT82lyzMljnEfoSlp87bxTyqusW2aC3bCXtWAnP6N5DaRRJ6TiyIdrD79__WRLp9Vsqk7BkTNZPqvWMWZqn5IkVJMM1JbwtrNdbMk8UsxLBdROapWqLqh74ddqqsRGPXol6uogU1FvnWnf-nGzJoSnthcZF6uGi8-Pstb881bBA4Rr9EeoRwHooyO96QpiUpyigjneDEOd2mLrfUFq1NxNhYb9fL1ArCQ7-6C48S8mk__AkQAop9rRpu9g5-jm3f13mZxDTNGXoizIfaQau9A7L2cUUh35fyIf820vcRLVZho=w273-h143-no)\n",
    "\n",
    "Ou seja, a raiz $p$ sempre estará no intervalo $[{a}_{n}, {p}_{n}]$ ou $[{p}_{n}, {b}_{n}]$ (ou no máximo em um dos extremos: ${a}_{n}=p$; ou ${b}_{n}=p$; ou $p={p}_{n}$). Pode-se escrever: $$\\left|{p}_{n}-{p}\\right| \\leqslant  \\frac{{b}_{n}-{a}_{n}}{2}\\textrm{, para todo n}\\geqslant 1,$$como: $$\\frac{{b}_{n}-{a}_{n}}{2} = \\frac{b-a}{2^{n}},$$a equação fica: $$\\left|{p}_{n}-{p}\\right| \\leqslant (b-a)\\frac{1}{2^{n}}$$.\n",
    "\n",
    "Esta equação pode ser escrita como uma equação de taxa de convergência $\\left|{\\alpha}_{n}-{\\alpha}\\right| \\leqslant K\\left| {\\beta}_{n}\\right| \\therefore {\\alpha}_{n} = \\alpha + O({\\beta}_{n})$. Assim a sequência $\\left \\{ {p}_{n} \\right \\}_{n=1}^{\\infty}$ converge para a raiz $p$ com uma taxa de convergência de $O(\\frac{1}{2^n})$, ou seja: $${p}_{n} = p + O(\\frac{1}{2^{n}})$$\n",
    "\n",
    "como exemplo: $$\\left|{p}_{9}-{p}\\right| \\leqslant  \\frac{2-1}{2^{9}} \\approx 2 x 10^{-3},$$é o limitante (erro máximo) para o método da bissecção com 9 iterações e $a=1$ e $b=2$.\n",
    "\n",
    "---\n",
    "\n",
    "### Vantagem do Método da Bissecção #\n",
    "\n",
    "A maior vantagem da utilização deste método para buscar a solução de uma equação do tipo $f(x)=0$ é que, atendidas as condições estabelecidas para sua aplicação, a convergência para uma solução é garantida. Por esse motivo importante, este método é bastante utilizado como um primeiro passo na busca por uma solução.\n",
    "\n",
    "### Desvantagens do Método da Bisseção #\n",
    "\n",
    "O Método da Bissecção possui algumas desvantagens notáveis quando comparado com outros métodos, sendo as mais importantes:\n",
    "\n",
    "* Convergência lenta\n",
    "\n",
    "Quando comparado a outros métodos, este método costuma precisar de uma quantidade maior de iterações para alcançar um resultado satisfatório e com uma boa precisão. \n",
    "\n",
    "Observe que, como destacado anteriormente, a sequência {$p_n$}$_{n=1}^\\infty$ converge para o valor da raiz $p$, com uma taxa de convergência de O$(\\frac{1}{2^n})$. Quando comparamos com a sequência ${(\\beta_n)}_{n=1}^{\\infty}$ definida na seção \"Convergência\" do capítulo anterior, percebemos que o valor aqui assumido pela variável n é baixo (2), o que indica uma baixa taxa de convergência para o resultado.\n",
    "\n",
    "$$Colocar \\, um \\,  exemplo \\, de \\, convergência \\, lenta$$\n",
    "\n",
    "* Resultados interessantes para o problema podem ser descartados\n",
    "\n",
    "Devido à maneira como o método funciona, e às suas condições iniciais e critérios de parada, é possível que soluções com erros menores sejam descartadas. Uma outra possibilidade é quando há mais de uma solução no intervalo, sendo uma delas ignorada durante as iterações.\n",
    "\n",
    "$$Colocar \\, um \\,  exemplo \\, de \\, função \\, trigonométrica$$\n",
    "\n",
    "* Limitação da aplicação do método \n",
    "\n",
    "O método da Bissecção falha para uma equação que não troca de sinal em sua imagem, ou seja, que o seu gráfico não cruza o eixo x, mas somente o tangencia nos pontos da suas raízes. Exemplos destas equações são $f(x)=3x^{4}$ e $g(x)=-2x^{2}$.\n",
    "\n",
    "$$Colocar \\, um \\, exemplo$$\n",
    "\n",
    "* Sensibilidade quanto ao intervalo escolhido\n",
    "\n",
    "Como o método trabalha com sucessivas subdivisões de um intervalo, a escolha do intervalo inicial possui um impacto considerável no número de iterações que serão necessárias para se chegar a um resultado preciso. Além disso, escolhendo-se um intervalo que contenha mais de uma raiz, o método ignorará pelo menos uma delas.\n",
    "\n",
    "$$Colocar \\, um \\,  exemplo$$\n",
    "\n",
    "### Observações quanto à aplicação do Método da Bissecção #\n",
    "\n",
    "Critérios de parada e tamanho do intervalo escolhido\n",
    "\n",
    "### Implementação do método da Bissecção em linguagem Python #\n",
    "\n",
    "O programa abaixo implementa o método da Bissecção para a obtenção de uma raiz de uma função, gerando uma tabela com os parâmetros do método nas sucessivas subdivisões do intervalo $[a, b]$ que contém a raiz, e a ilustração gráfica do método, representando a função (linha verde) e as subdivisões do intervalo $[a, b]$ em linhas azuis e vermelhas."
   ]
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  {
   "cell_type": "markdown",
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   "source": [
    "Adicionar vantagens e desvantagens do método da bissecção. Um dos problemas do método da bissecção é que ele pode ignorar uma raiz, por exemplo. Conselhos para otimização."
   ]
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  {
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   "execution_count": 20,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Digite a função f(x): cos(x)+sqrt(x)\n",
      "Tolerância para o tamanho do intervalo (b-a)/2 (digite 0 para desativar): 0\n",
      "Tolerância para o erro relativo de p (ponto médio do intervalo) (digite 0 para desativar): 0\n",
      "Tolerância para o valor de f(p) ~= 0 (digite 0 para desativar): 0.001\n",
      "Número máximo de iterações: 10\n",
      "Defina a: 0\n",
      "Defina b: 1\n",
      "f(a) e f(b) devem ter sinais opostos!\n",
      "Defina a: -2\n",
      "Defina b: 0\n"
     ]
    },
    {
     "ename": "TypeError",
     "evalue": "can't convert complex to float",
     "output_type": "error",
     "traceback": [
      "\u001b[1;31m---------------------------------------------------------------------------\u001b[0m",
      "\u001b[1;31mTypeError\u001b[0m                                 Traceback (most recent call last)",
      "\u001b[1;32m<ipython-input-20-f47b75009186>\u001b[0m in \u001b[0;36m<module>\u001b[1;34m()\u001b[0m\n\u001b[0;32m     39\u001b[0m     \u001b[0ma\u001b[0m \u001b[1;33m=\u001b[0m \u001b[0mfloat\u001b[0m\u001b[1;33m(\u001b[0m\u001b[0minput\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;34m\"Defina a: \"\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0;32m     40\u001b[0m     \u001b[0mb\u001b[0m \u001b[1;33m=\u001b[0m \u001b[0mfloat\u001b[0m\u001b[1;33m(\u001b[0m\u001b[0minput\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;34m\"Defina b: \"\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[1;32m---> 41\u001b[1;33m     \u001b[1;32mif\u001b[0m \u001b[1;33m(\u001b[0m\u001b[0mf\u001b[0m\u001b[1;33m(\u001b[0m\u001b[0ma\u001b[0m\u001b[1;33m,\u001b[0m\u001b[0mx\u001b[0m\u001b[1;33m,\u001b[0m\u001b[0mfx\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m*\u001b[0m\u001b[0mf\u001b[0m\u001b[1;33m(\u001b[0m\u001b[0mb\u001b[0m\u001b[1;33m,\u001b[0m\u001b[0mx\u001b[0m\u001b[1;33m,\u001b[0m\u001b[0mfx\u001b[0m\u001b[1;33m)\u001b[0m \u001b[1;33m>\u001b[0m \u001b[1;36m0.0\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m:\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0m\u001b[0;32m     42\u001b[0m         \u001b[0mprint\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;34m\"f(a) e f(b) devem ter sinais opostos!\"\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0;32m     43\u001b[0m         \u001b[1;32mcontinue\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n",
      "\u001b[1;32m<ipython-input-20-f47b75009186>\u001b[0m in \u001b[0;36mf\u001b[1;34m(a, x, fx)\u001b[0m\n\u001b[0;32m     16\u001b[0m \u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0;32m     17\u001b[0m \u001b[1;32mdef\u001b[0m \u001b[0mf\u001b[0m\u001b[1;33m(\u001b[0m\u001b[0ma\u001b[0m\u001b[1;33m,\u001b[0m \u001b[0mx\u001b[0m\u001b[1;33m,\u001b[0m \u001b[0mfx\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m:\u001b[0m \u001b[1;31m#Retorna o valor numérico da função digitada pelo usuário\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[1;32m---> 18\u001b[1;33m     \u001b[1;32mreturn\u001b[0m \u001b[0mfloat\u001b[0m\u001b[1;33m(\u001b[0m\u001b[0mfx\u001b[0m\u001b[1;33m.\u001b[0m\u001b[0msubs\u001b[0m\u001b[1;33m(\u001b[0m\u001b[0mx\u001b[0m\u001b[1;33m,\u001b[0m \u001b[0ma\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0m\u001b[0;32m     19\u001b[0m \u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0;32m     20\u001b[0m \u001b[0mx\u001b[0m \u001b[1;33m=\u001b[0m \u001b[0msp\u001b[0m\u001b[1;33m.\u001b[0m\u001b[0msymbols\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;34m'x'\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n",
      "\u001b[1;32mC:\\Anaconda3\\lib\\site-packages\\sympy\\core\\expr.py\u001b[0m in \u001b[0;36m__float__\u001b[1;34m(self)\u001b[0m\n\u001b[0;32m    236\u001b[0m             \u001b[1;32mreturn\u001b[0m \u001b[0mfloat\u001b[0m\u001b[1;33m(\u001b[0m\u001b[0mresult\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0;32m    237\u001b[0m         \u001b[1;32mif\u001b[0m \u001b[0mresult\u001b[0m\u001b[1;33m.\u001b[0m\u001b[0mis_number\u001b[0m \u001b[1;32mand\u001b[0m \u001b[0mresult\u001b[0m\u001b[1;33m.\u001b[0m\u001b[0mas_real_imag\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m[\u001b[0m\u001b[1;36m1\u001b[0m\u001b[1;33m]\u001b[0m\u001b[1;33m:\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[1;32m--> 238\u001b[1;33m             \u001b[1;32mraise\u001b[0m \u001b[0mTypeError\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;34m\"can't convert complex to float\"\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0m\u001b[0;32m    239\u001b[0m         \u001b[1;32mraise\u001b[0m \u001b[0mTypeError\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;34m\"can't convert expression to float\"\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0;32m    240\u001b[0m \u001b[1;33m\u001b[0m\u001b[0m\n",
      "\u001b[1;31mTypeError\u001b[0m: can't convert complex to float"
     ]
    }
   ],
   "source": [
    "#Método da Bissecção para determinar uma raiz de uma função em uma variável do tipo f(x) = 0\n",
    "\n",
    "#Utiliza a biblioteca sympy, disponível em: http://docs.sympy.org/latest/index.html\n",
    "#Utiliza a biblioteca matplotlib, disponível em: https://matplotlib.org/\n",
    "\n",
    "import sympy as sp\n",
    "import matplotlib.pyplot as plt\n",
    "\n",
    "def sgn(s): # função que testa o sinal de f(x)\n",
    "    if (s < 0.0):\n",
    "        return -1\n",
    "    elif (s == 0.0):\n",
    "        return 0\n",
    "    else:\n",
    "        return 1\n",
    "\n",
    "def f(a, x, fx): #Retorna o valor numérico da função digitada pelo usuário\n",
    "    return float(fx.subs(x, a))\n",
    "\n",
    "x = sp.symbols('x')\n",
    "fx = sp.sympify(input(\"Digite a função f(x): \"))\n",
    "\n",
    "TOL = abs(float(input(\"Tolerância para o tamanho do intervalo (b-a)/2 (digite 0 para desativar): \"))) # Tolerância para o tamanho do intervalo (b-a)/2\n",
    "e = abs(float(input(\"Tolerância para o erro relativo de p (ponto médio do intervalo) (digite 0 para desativar): \"))) # Tolerância para o erro relativo de p\n",
    "efp = abs(float(input(\"Tolerância para o valor de f(p) ~= 0 (digite 0 para desativar): \"))) # Tolerância para f(p) ~= 0\n",
    "N0 = abs(int(input(\"Número máximo de iterações: \"))) # Número máximo de iterações\n",
    "\n",
    "# Variáveis extras para o cálculo\n",
    "TOL_val=0.0\n",
    "e_val=0.0\n",
    "FP=0.0\n",
    "pn=0.0\n",
    "n=0\n",
    "i=0.0\n",
    "passo=0.0\n",
    "\n",
    "# Obtém a e b e testa se f(a) e f(b) possuem sinais opostos\n",
    "while (True):\n",
    "    a = float(input(\"Defina a: \"))\n",
    "    b = float(input(\"Defina b: \"))\n",
    "    if (f(a,x,fx)*f(b,x,fx) > 0.0):\n",
    "        print(\"f(a) e f(b) devem ter sinais opostos!\")\n",
    "        continue\n",
    "    elif (abs(f(a,x,fx)) <= efp):\n",
    "        print(\"a já é uma raiz da função para a tolerância desejada!\")\n",
    "        N0 = -1 # Para finalizar o programa\n",
    "    elif (abs(f(b,x,fx)) <= efp):\n",
    "        print(\"b já é uma raiz da função para a tolerância desejada!\")\n",
    "        N0 = -1 # Para finalizar o programa\n",
    "    break\n",
    "\n",
    "# Plota o gráfico da função no intervalo [a, b]\n",
    "plt.rcParams['figure.figsize'] = 10, 10 # Tamanho da figura\n",
    "graf1 = sp.plotting.plot(fx,show=sp.false) # Plota o gráfico da função\n",
    "if (a<b): # Determina os limites do eixo x\n",
    "    graf1.xlim = [a, b]\n",
    "else:\n",
    "    graf1.xlim = [b, a]\n",
    "# Determina os limites do eixo y do gráfico e calcula o passo com que a sequência de subdivisões de [a, b] serão plotadas no eixo y\n",
    "if (f(a,x,fx) < 0):\n",
    "    graf1.ylim = [f(a,x,fx), f(b,x,fx)]\n",
    "    passo = abs(f(a,x,fx))/N0\n",
    "else:\n",
    "    graf1.ylim = [f(b,x,fx), f(a,x,fx)]\n",
    "    passo = abs(f(b,x,fx))/N0    \n",
    "graf1[0].line_color = 'green' # Cor da linha do gráfico da função\n",
    "graf1.legend = sp.false # Não mostrar legenda\n",
    "\n",
    "# Realiza os cálculos do método da biseccção\n",
    "print ('\\nn\\t a\\t         b\\t         p\\t         f(p)\\t         (b-a)/2\\t e. relat. p\\tMundança')\n",
    "for n in range (1, N0+1):\n",
    "    p = a + (b-a)/2\n",
    "    i -= passo # i vai descendo no eixo y e plotando as subdivisões do intervalo [a, b] uma abaixo da outra\n",
    "    graf1.append(sp.plotting.plot(i, (x, a, p), show=sp.false, line_color='blue')[0]) # Plota a próxima subdivisão do intervalo [a, p]\n",
    "    graf1.append(sp.plotting.plot(i, (x, p, b), show=sp.false, line_color='red')[0]) # Plota a próxima subdivisão do intervalo [p, b]\n",
    "    FP = f(p,x,fx)\n",
    "    TOL_val = abs((b-a)/2)\n",
    "    if(p != 0.0): \n",
    "        e_val = abs((pn-p)/p)\n",
    "    print (n, '\\t', \"{:.10f}\".format(a), '\\t', \"{:.10f}\".format(b), '\\t', \"{:.10f}\".format(p), '\\t', \"{:.10f}\".format(FP), '\\t', \"{:.10f}\".format(TOL_val), '\\t', \"{:.10f}\".format(e_val), end=\"\")\n",
    "    if (abs(FP) <= efp or TOL_val < TOL or  e_val < e): #Se F(p) ~= 0 ou (b-2)/2 < erro ou erro relativo de p < erro, encerra o programa\n",
    "        print (\"\\n\\nA raiz da função, dentro dos parâmetros de tolerância, é\", p)\n",
    "        break\n",
    "    if (sgn(f(a,x,fx)) * sgn(FP) > 0.0):\n",
    "        a = p\n",
    "        print ('   f(a)*f(p)>0.0, a=p')\n",
    "    else:\n",
    "        b = p\n",
    "        print ('   f(a)*f(p)<0.0, b=p')\n",
    "    pn = p\n",
    "\n",
    "print (\"\\nO método da Bissecção foi encerrado após\", n, \"iterações.\")\n",
    "graf1.show() # Mostra o gráfico ao final"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Conjunto de exercícios 2.1 #\n",
    "\n",
    "**(1) Use o método da Bissecção para encontrar $p_3$ para $f(x)=\\sqrt{x} - \\cos x=0$ em [0,1].**\n",
    "\n",
    "Este problema pode ser resolvido atraves do uso do algoritmo anteriormente desenvolvido para demonstrar o método da bissecção, reproduzido abaixo."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 21,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Digite a função f(x): sqrt(x)-cos(x)\n",
      "Tolerância para o tamanho do intervalo (b-a)/2 (digite 0 para desativar): 0\n",
      "Tolerância para o erro relativo de p (ponto médio do intervalo) (digite 0 para desativar): 0.00001\n",
      "Tolerância para o valor de f(p) ~= 0 (digite 0 para desativar): 0.000001\n",
      "Número máximo de iterações: 5\n",
      "Defina a: 0\n",
      "Defina b: 1\n",
      "\n",
      "n\t a\t         b\t         p\t         f(p)\t         (b-a)/2\t e. relat. p\tMundança\n",
      "1 \t 0.0000000000 \t 1.0000000000 \t 0.5000000000 \t -0.1704757807 \t 0.5000000000 \t 1.0000000000   f(a)*f(p)>0.0, a=p\n",
      "2 \t 0.5000000000 \t 1.0000000000 \t 0.7500000000 \t 0.1343365349 \t 0.2500000000 \t 0.3333333333   f(a)*f(p)<0.0, b=p\n",
      "3 \t 0.5000000000 \t 0.7500000000 \t 0.6250000000 \t -0.0203937045 \t 0.1250000000 \t 0.2000000000   f(a)*f(p)>0.0, a=p\n",
      "4 \t 0.6250000000 \t 0.7500000000 \t 0.6875000000 \t 0.0563212514 \t 0.0625000000 \t 0.0909090909   f(a)*f(p)<0.0, b=p\n",
      "5 \t 0.6250000000 \t 0.6875000000 \t 0.6562500000 \t 0.0178067276 \t 0.0312500000 \t 0.0476190476   f(a)*f(p)<0.0, b=p\n",
      "\n",
      "O método da Bissecção foi encerrado após 5 iterações.\n"
     ]
    }
   ],
   "source": [
    "#Método da Bissecção para determinar uma raiz de uma função em uma variável do tipo f(x) = 0\n",
    "\n",
    "#Utiliza a biblioteca sympy, disponível em: http://docs.sympy.org/latest/index.html\n",
    "#Utiliza a biblioteca matplotlib, disponível em: https://matplotlib.org/\n",
    "\n",
    "import sympy as sp\n",
    "\n",
    "def sgn(s): # função que testa o sinal de f(x)\n",
    "    if (s < 0.0):\n",
    "        return -1\n",
    "    elif (s == 0.0):\n",
    "        return 0\n",
    "    else:\n",
    "        return 1\n",
    "\n",
    "def f(a, x, fx): #Retorna o valor numérico da função digitada pelo usuário\n",
    "    return float(fx.subs(x, a))\n",
    "\n",
    "x = sp.symbols('x')\n",
    "fx = sp.sympify(input(\"Digite a função f(x): \"))\n",
    "\n",
    "TOL = abs(float(input(\"Tolerância para o tamanho do intervalo (b-a)/2 (digite 0 para desativar): \"))) # Tolerância para o tamanho do intervalo (b-a)/2\n",
    "e = abs(float(input(\"Tolerância para o erro relativo de p (ponto médio do intervalo) (digite 0 para desativar): \"))) # Tolerância para o erro relativo de p\n",
    "efp = abs(float(input(\"Tolerância para o valor de f(p) ~= 0 (digite 0 para desativar): \"))) # Tolerância para f(p) ~= 0\n",
    "N0 = abs(int(input(\"Número máximo de iterações: \"))) # Número máximo de iterações\n",
    "\n",
    "# Variáveis extras para o cálculo\n",
    "TOL_val=0.0\n",
    "e_val=0.0\n",
    "FP=0.0\n",
    "pn=0.0\n",
    "n=0\n",
    "i=0.0\n",
    "passo=0.0\n",
    "\n",
    "# Obtém a e b e testa se f(a) e f(b) possuem sinais opostos\n",
    "while (True):\n",
    "    a = float(input(\"Defina a: \"))\n",
    "    b = float(input(\"Defina b: \"))\n",
    "    if (f(a,x,fx)*f(b,x,fx) > 0.0):\n",
    "        print(\"f(a) e f(b) devem ter sinais opostos!\")\n",
    "        continue\n",
    "    elif (abs(f(a,x,fx)) <= efp):\n",
    "        print(\"a já é uma raiz da função para a tolerância desejada!\")\n",
    "        N0 = -1 # Para finalizar o programa\n",
    "    elif (abs(f(b,x,fx)) <= efp):\n",
    "        print(\"b já é uma raiz da função para a tolerância desejada!\")\n",
    "        N0 = -1 # Para finalizar o programa\n",
    "    break\n",
    "\n",
    "# Realiza os cálculos do método da biseccção\n",
    "print ('\\nn\\t a\\t         b\\t         p\\t         f(p)\\t         (b-a)/2\\t e. relat. p\\tMundança')\n",
    "for n in range (1, N0+1):\n",
    "    p = a + (b-a)/2\n",
    "    i -= passo # i vai descendo no eixo y e plotando as subdivisões do intervalo [a, b] uma abaixo da outra\n",
    "    graf1.append(sp.plotting.plot(i, (x, a, p), show=sp.false, line_color='blue')[0]) # Plota a próxima subdivisão do intervalo [a, p]\n",
    "    graf1.append(sp.plotting.plot(i, (x, p, b), show=sp.false, line_color='red')[0]) # Plota a próxima subdivisão do intervalo [p, b]\n",
    "    FP = f(p,x,fx)\n",
    "    TOL_val = abs((b-a)/2)\n",
    "    if(p != 0.0): \n",
    "        e_val = abs((pn-p)/p)\n",
    "    print (n, '\\t', \"{:.10f}\".format(a), '\\t', \"{:.10f}\".format(b), '\\t', \"{:.10f}\".format(p), '\\t', \"{:.10f}\".format(FP), '\\t', \"{:.10f}\".format(TOL_val), '\\t', \"{:.10f}\".format(e_val), end=\"\")\n",
    "    if (abs(FP) <= efp or TOL_val < TOL or  e_val < e): #Se F(p) ~= 0 ou (b-2)/2 < erro ou erro relativo de p < erro, encerra o programa\n",
    "        print (\"\\n\\nA raiz da função, dentro dos parâmetros de tolerância, é\", p)\n",
    "        break\n",
    "    if (sgn(f(a,x,fx)) * sgn(FP) > 0.0):\n",
    "        a = p\n",
    "        print ('   f(a)*f(p)>0.0, a=p')\n",
    "    else:\n",
    "        b = p\n",
    "        print ('   f(a)*f(p)<0.0, b=p')\n",
    "    pn = p\n",
    "\n",
    "print (\"\\nO método da Bissecção foi encerrado após\", n, \"iterações.\")"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Pelo método acima, o valor obtido para $p_3$ é de 0.625. Observe que os critérios de parada foram estabelecidos aleatoriamente como muito baixos para evitar que houvessem erros indesejáveis, ao passo que garantisse iterações suficientes para que o resultado desejado ($p_3$) fosse obtido.\n",
    "\n",
    "E de fato o valor obtido para $p_3$ pode ser confirmado através de cálculos manuais, como mostrado abaixo:\n",
    "\n",
    "Definindo-se a=0, b=1 e $p_3$ para $f(x)=\\sqrt{x} - \\cos x=0$, temos:\n",
    "\n",
    "$f(a)=-1.0000$ ;  $f(b)\\approx 0.4597$\n",
    "$p_1$ = 0.50 e $f(p_1)\\approx -0.1704$\n",
    "\n",
    "Como $f(p_1)<0$, o novo intervalo é $[0.5,1]$\n",
    "\n",
    "$p_2$=0.7500 e $f(p_2) \\approx 0.1343$.\n",
    "\n",
    "Como $f(p_2)>0$, o novo intervalo é $[0.5,0.75]$\n",
    "\n",
    "Calcula-se então que $p_3 = 0.625$, o que confirma o resultado anterior.    "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 28,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Digite a função f(x): sqrt(x)-cos(x)\n",
      "ponto:  0.75\n",
      "0.134336534910618\n"
     ]
    }
   ],
   "source": [
    "import sympy as sp\n",
    "\n",
    "def f(a, x, fx): #Retorna o valor numérico da função digitada pelo usuário\n",
    "    return (fx.subs(x, a))\n",
    "\n",
    "fx = sp.sympify(input(\"Digite a função f(x): \"))\n",
    "s=float(input(\"ponto:  \"))\n",
    "print(f(s,x,fx))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**(3) Use o método da bissecção para encontrar soluções com precisão melhor que $10^{-2}$ para $x^3-7x^2+14x-6=0$ em cada um dos intervalos:**\n",
    "\n",
    "$a) [0,1] \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, b) [0,2] \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, c)[3.2,4]$\n",
    "\n",
    "Para este exercício, podemos utilizar como auxílio uma versão modificada do algoritmo do método da bissecção anteriormente desenvolvido."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 31,
   "metadata": {
    "scrolled": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Digite a função f(x): x**3-7*x**2+14*x-6\n",
      "Tolerância para o tamanho do intervalo (b-a)/2 (digite 0 para desativar): 0\n",
      "Tolerância para o erro relativo de p (ponto médio do intervalo) (digite 0 para desativar): 0\n",
      "Tolerância para o valor de f(p) ~= 0 (digite 0 para desativar): 0.01\n",
      "Número máximo de iterações: 100\n",
      "Defina a: 3.2\n",
      "Defina b: 4\n",
      "\n",
      "n\t a\t         b\t         p\t         f(p)\t         (b-a)/2\t e. relat. p\tMundança\n",
      "1 \t 3.2000000000 \t 4.0000000000 \t 3.6000000000 \t 0.3360000000 \t 0.4000000000 \t 1.0000000000   f(a)*f(p)<0.0, b=p\n",
      "2 \t 3.2000000000 \t 3.6000000000 \t 3.4000000000 \t -0.0160000000 \t 0.2000000000 \t 0.0588235294   f(a)*f(p)>0.0, a=p\n",
      "3 \t 3.4000000000 \t 3.6000000000 \t 3.5000000000 \t 0.1250000000 \t 0.1000000000 \t 0.0285714286   f(a)*f(p)<0.0, b=p\n",
      "4 \t 3.4000000000 \t 3.5000000000 \t 3.4500000000 \t 0.0461250000 \t 0.0500000000 \t 0.0144927536   f(a)*f(p)<0.0, b=p\n",
      "5 \t 3.4000000000 \t 3.4500000000 \t 3.4250000000 \t 0.0130156250 \t 0.0250000000 \t 0.0072992701   f(a)*f(p)<0.0, b=p\n",
      "6 \t 3.4000000000 \t 3.4250000000 \t 3.4125000000 \t -0.0019980469 \t 0.0125000000 \t 0.0036630037\n",
      "\n",
      "A raiz da função, dentro dos parâmetros de tolerância, é 3.4125000000000005\n",
      "\n",
      "O método da Bissecção foi encerrado após 6 iterações.\n"
     ]
    }
   ],
   "source": [
    "#Método da Bissecção para determinar uma raiz de uma função em uma variável do tipo f(x) = 0\n",
    "\n",
    "#Utiliza a biblioteca sympy, disponível em: http://docs.sympy.org/latest/index.html\n",
    "#Utiliza a biblioteca matplotlib, disponível em: https://matplotlib.org/\n",
    "\n",
    "import sympy as sp\n",
    "\n",
    "def sgn(s): # função que testa o sinal de f(x)\n",
    "    if (s < 0.0):\n",
    "        return -1\n",
    "    elif (s == 0.0):\n",
    "        return 0\n",
    "    else:\n",
    "        return 1\n",
    "\n",
    "def f(a, x, fx): #Retorna o valor numérico da função digitada pelo usuário\n",
    "    return float(fx.subs(x, a))\n",
    "\n",
    "x = sp.symbols('x')\n",
    "fx = sp.sympify(input(\"Digite a função f(x): \"))\n",
    "\n",
    "TOL = abs(float(input(\"Tolerância para o tamanho do intervalo (b-a)/2 (digite 0 para desativar): \"))) # Tolerância para o tamanho do intervalo (b-a)/2\n",
    "e = abs(float(input(\"Tolerância para o erro relativo de p (ponto médio do intervalo) (digite 0 para desativar): \"))) # Tolerância para o erro relativo de p\n",
    "efp = abs(float(input(\"Tolerância para o valor de f(p) ~= 0 (digite 0 para desativar): \"))) # Tolerância para f(p) ~= 0\n",
    "N0 = abs(int(input(\"Número máximo de iterações: \"))) # Número máximo de iterações\n",
    "\n",
    "# Variáveis extras para o cálculo\n",
    "TOL_val=0.0\n",
    "e_val=0.0\n",
    "FP=0.0\n",
    "pn=0.0\n",
    "n=0\n",
    "i=0.0\n",
    "passo=0.0\n",
    "\n",
    "# Obtém a e b e testa se f(a) e f(b) possuem sinais opostos\n",
    "while (True):\n",
    "    a = float(input(\"Defina a: \"))\n",
    "    b = float(input(\"Defina b: \"))\n",
    "    if (f(a,x,fx)*f(b,x,fx) > 0.0):\n",
    "        print(\"f(a) e f(b) devem ter sinais opostos!\")\n",
    "        continue\n",
    "    elif (abs(f(a,x,fx)) <= efp):\n",
    "        print(\"a já é uma raiz da função para a tolerância desejada!\")\n",
    "        N0 = -1 # Para finalizar o programa\n",
    "    elif (abs(f(b,x,fx)) <= efp):\n",
    "        print(\"b já é uma raiz da função para a tolerância desejada!\")\n",
    "        N0 = -1 # Para finalizar o programa\n",
    "    break\n",
    "\n",
    "# Realiza os cálculos do método da biseccção\n",
    "print ('\\nn\\t a\\t         b\\t         p\\t         f(p)\\t         (b-a)/2\\t e. relat. p\\tMundança')\n",
    "for n in range (1, N0+1):\n",
    "    p = a + (b-a)/2\n",
    "    i -= passo # i vai descendo no eixo y e plotando as subdivisões do intervalo [a, b] uma abaixo da outra\n",
    "    graf1.append(sp.plotting.plot(i, (x, a, p), show=sp.false, line_color='blue')[0]) # Plota a próxima subdivisão do intervalo [a, p]\n",
    "    graf1.append(sp.plotting.plot(i, (x, p, b), show=sp.false, line_color='red')[0]) # Plota a próxima subdivisão do intervalo [p, b]\n",
    "    FP = f(p,x,fx)\n",
    "    TOL_val = abs((b-a)/2)\n",
    "    if(p != 0.0): \n",
    "        e_val = abs((pn-p)/p)\n",
    "    print (n, '\\t', \"{:.10f}\".format(a), '\\t', \"{:.10f}\".format(b), '\\t', \"{:.10f}\".format(p), '\\t', \"{:.10f}\".format(FP), '\\t', \"{:.10f}\".format(TOL_val), '\\t', \"{:.10f}\".format(e_val), end=\"\")\n",
    "    if (abs(FP) <= efp or TOL_val < TOL or  e_val < e): #Se F(p) ~= 0 ou (b-2)/2 < erro ou erro relativo de p < erro, encerra o programa\n",
    "        print (\"\\n\\nA raiz da função, dentro dos parâmetros de tolerância, é\", p)\n",
    "        break\n",
    "    if (sgn(f(a,x,fx)) * sgn(FP) > 0.0):\n",
    "        a = p\n",
    "        print ('   f(a)*f(p)>0.0, a=p')\n",
    "    else:\n",
    "        b = p\n",
    "        print ('   f(a)*f(p)<0.0, b=p')\n",
    "    pn = p\n",
    "\n",
    "print (\"\\nO método da Bissecção foi encerrado após\", n, \"iterações.\")"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Utilizando o método acima, para a equação fornecida no enunciado, e tolerância 0.01 para o valor da raiz, observa-se que os valores estimados das raízes para cada um dos intervalos solicitados é:\n",
    "\n",
    "$a)p=0.5859375 \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, b) p=0.5859375 \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, c)p = 3.4125000000000005$\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**(7) Esboce os gráficos de $y=x$ e $y=senx$.**\n",
    "\n",
    "Abaixo encontram-se esboçados os gráficos das duas funções:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": 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\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0x11546c41e48>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "import numpy as np\n",
    "import matplotlib.pyplot as plt\n",
    "import math\n",
    "\n",
    "# EXEMPLO: f(x)=x^2+x-3\n",
    "x=np.linspace(-10,10,100)   # Estou a criar dados de input em X.\n",
    "y=np.zeros((100,1))         # Estou a criar um vector de saida em Y.\n",
    "\n",
    "i=0\n",
    "while i < 100:\n",
    "    y[i]= 2 * math.sin(x[i])     # Estou a atribuir a cada célula de Y\n",
    "    i=i+1                   # o correspondente valor de X após passar por f(x)\n",
    "\n",
    "# marker é o simbolo do marcador e color a cor do marcador.\n",
    "plt.plot(x,y,'-', 'seno')   # estou a fazer o gráfico dos dois vectores.\n",
    "plt.plot(x,x,'-', 'y=x')   # estou a fazer o gráfico dos dois vectores.\n",
    "plt.plot([-10,10],[0,0],'-')  # Aqui estou a tentar criar linhas\n",
    "plt.plot([0,0],[-20,100],'-') # para os eixos X e Y.\n",
    "\n",
    "plt.ylim(0,3)  # Aqui estou a limitar a janela dos gráficos em Y.\n",
    "plt.xlim(0,3)  # ... e em X.\n",
    "\n",
    "plt.show()    # Esta linha de código serve para mostrar o gráfico total."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Analisando o gráfico acima, podemos perceber que a raiz das funções está em algum ponto onde o x esteja entre 1.5 e 2.0. \n",
    "Logo adotaremos esse intervalo para encontrar a raiz.\n",
    "Os parâmetros de entrada do nosso método de bisseccao abaixo, são:\n",
    "a - Valor inicial do intervalo - 1.5\n",
    "b - Valor final do intervalo - 2.0\n",
    "TOL - Tolerância, foi dada no enunciado como $10^{-5}$\n",
    "N - Número máximo de iterações que devem ser executadas - Defini como 30"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 16,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "1.8954895019531248\n"
     ]
    }
   ],
   "source": [
    "import numpy\n",
    "import math\n",
    "\n",
    "def bisseccao(a, b, TOL, N):  \n",
    "    f = lambda x: x -2 * math.sin(x)\n",
    "    i = 1  \n",
    "    fa = f(a)  \n",
    "    while (i <= N):  \n",
    "        #iteracao da bissecao  \n",
    "        p = a + (b-a)/2  \n",
    "        fp = f(p)  \n",
    "        #condicao de parada  \n",
    "        \n",
    "        if ((fp == 0) or ((b-a)/2 < TOL)):  \n",
    "            return p  \n",
    "        #bissecta o intervalo  \n",
    "        i = i+1  \n",
    "        if (numpy.sign(fa) * numpy.sign(fp) > 0):  \n",
    "            a = p  \n",
    "            fa = fp  \n",
    "        else:  \n",
    "            b = p  \n",
    " \n",
    "    raise NameError('Num. max. de iter. excedido!');\n",
    "\n",
    "saida = bisseccao(1.8,2.0,pow(10,-5),40);\n",
    "print(saida);"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Exercicio 9 - 2.1\n",
    "\n",
    "<p>a) Esboce os gráficos d $y=e^{x}-2$ e $y=cos(e^{x}-2)$.</p>\n",
    "\n",
    "<p>b) Utilize o método da Bissecção para determinar uma aproximação com precisão $10^{-5}$ (0.00001), de um valor em $[0.5,1.5]$ com $e^{x}-2=cos(e^{x}-2)$.</p>"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "import matplotlib.pyplot as plt\n",
    "import math\n",
    "\n",
    "\n",
    "#######################\n",
    "## RESOLUÇÃO LETRA A ##\n",
    "#######################\n",
    "\n",
    "#variáveis utilizadas na plotagem do gráfico\n",
    "def fatorial(num):\n",
    "    if (num==0):\n",
    "        return 1\n",
    "    return (num * fatorial(num-1))\n",
    "\n",
    "def cosseno(x):\n",
    "    cosseno=0\n",
    "    for i in range(0,20):\n",
    "        cosseno += (((-1)**i/fatorial(2*i))*(x**(2*i)))\n",
    "    return cosseno\n",
    "\n",
    "def exponencial(x):\n",
    "    exponencial=0\n",
    "    for i in range(0,41):\n",
    "        exponencial += ((x**i)/(fatorial(i)))\n",
    "    return exponencial\n",
    "\n",
    "def funcao1(x):\n",
    "    return (exponencial(x)-2)\n",
    "\n",
    "def funcao2(x):\n",
    "    return (cosseno((exponencial(x)-2)))\n",
    "\n",
    "#declaração das variáveis\n",
    "j=0\n",
    "lim1 = -1 #float(input(\"Informe o limite inferior(LIM1):\")) #limite inferior\n",
    "lim2 = 2 #float(input(\"Informe o limite superior(LIM2):\")) #limite superior\n",
    "\n",
    "#variáveis utilizadas na plotagem do gráfico\n",
    "i=lim1 \n",
    "valoresFuncao1 = []\n",
    "valoresFuncao2 = [] \n",
    "valoresX = []\n",
    "\n",
    "#Construção do gráfico da função para os intervalos designados\n",
    "while i<lim2:\n",
    "    valoresX.append(i)\n",
    "    valoresFuncao1.append(funcao1(i))\n",
    "    valoresFuncao2.append(funcao2(i))\n",
    "    i = i + 0.01 #aumentar o número de pontos, para definir melhor a tragetória do gráfico\n",
    "\n",
    "#excução da plotagem\n",
    "#configuração dos eixos retirado de:http://www.scipy-lectures.org/intro/matplotlib/matplotlib.html\n",
    "plt.figure(figsize=(9,9), dpi=80)\n",
    "ax = plt.gca()\n",
    "\n",
    "ax.spines['right'].set_color('none')\n",
    "ax.spines['top'].set_color('none')\n",
    "ax.xaxis.set_ticks_position('bottom')\n",
    "ax.spines['bottom'].set_position(('data',0))\n",
    "ax.yaxis.set_ticks_position('left')\n",
    "ax.spines['left'].set_position(('data',0))\n",
    "\n",
    "\n",
    "plt.grid(True, which='both')\n",
    "plt.plot(valoresX, valoresFuncao1, color=\"blue\", linestyle=\"-\", label=\"(e^x)−2 \")\n",
    "plt.plot(valoresX, valoresFuncao2, color=\"red\", linestyle=\"-\", label=\"cos((e^x)−2)0.\")\n",
    "plt.legend(loc='upper left')\n",
    "\n",
    "plt.show()\n",
    "\n",
    "\n",
    "#######################\n",
    "## RESOLUÇÃO LETRA B ##\n",
    "#######################\n",
    "\n",
    "#declaração das variáveis\n",
    "j=0\n",
    "pos1 = float(input(\"Informe o limite inferior(POS1):\")) #limite inferior\n",
    "pos2 = float(input(\"Informe o limite superior(POS2):\")) #limite superior\n",
    "tol  = float(input(\"Informe a tolerancia:\"))#tolerancia\n",
    "\n",
    "print(\"\\n\\n\")\n",
    "\n",
    "def funcao(x):\n",
    "    return (cosseno((exponencial(x)-2))-(exponencial(x)-2))\n",
    "\n",
    "#Implementação do metodo da Bissecção\n",
    "def metodoBisseccao(pos1,pos2, tol):\n",
    "    print(\"novo intervalo [\",pos1,\",\",pos2,\"]\")\n",
    "    print(\"f(pos1)=\",funcao(pos1))\n",
    "    print(\"f(pos2)=\",funcao(pos2))\n",
    "    \n",
    "    if(funcao(pos1)*funcao(pos2)<0):\n",
    "        \n",
    "        p = (pos1+pos2)/2\n",
    "        print(\"valor p=\",p)\n",
    "        print(\"f(p)=\",funcao(p))\n",
    "        print(\"\\n\")\n",
    "        \n",
    "        if(funcao(p)==0 or math.fabs((pos2-pos1)/2)<tol):\n",
    "            print(\"Sucesso!\")\n",
    "            print(\"A raiz procurada é: \",p)\n",
    "            return p\n",
    "        else:\n",
    "            if(funcao(pos1)*funcao(p)>0):\n",
    "                pos1 = p\n",
    "            else:\n",
    "                pos2 = p\n",
    "                \n",
    "        metodoBisseccao(pos1,pos2,tol)\n",
    "    else:\n",
    "        print(\"não possui zero para o intervalo!\")\n",
    "\n",
    "metodoBisseccao(pos1,pos2,tol)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**(13) Encontre uma aproximação para $\\sqrt[3]{25}$ com erro menor que $10^{-4}$ usando o Algoritmo da Bissecção. [Dica: Considere $f(x)=x^3-25$]**\n",
    "\n",
    "Utilizando o algoritmo do método da bisseccção mostrado abaixo, é possível calcular que o valor aproximado para $\\sqrt[3]{25}$ é 2.924017906188965"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 32,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Digite a função f(x): x**3-25\n",
      "Tolerância para o tamanho do intervalo (b-a)/2 (digite 0 para desativar): 0\n",
      "Tolerância para o erro relativo de p (ponto médio do intervalo) (digite 0 para desativar): 0\n",
      "Tolerância para o valor de f(p) ~= 0 (digite 0 para desativar): 0.0001\n",
      "Número máximo de iterações: 100\n",
      "Defina a: 0\n",
      "Defina b: 5\n",
      "\n",
      "n\t a\t         b\t         p\t         f(p)\t         (b-a)/2\t e. relat. p\tMundança\n",
      "1 \t 0.0000000000 \t 5.0000000000 \t 2.5000000000 \t -9.3750000000 \t 2.5000000000 \t 1.0000000000   f(a)*f(p)>0.0, a=p\n",
      "2 \t 2.5000000000 \t 5.0000000000 \t 3.7500000000 \t 27.7343750000 \t 1.2500000000 \t 0.3333333333   f(a)*f(p)<0.0, b=p\n",
      "3 \t 2.5000000000 \t 3.7500000000 \t 3.1250000000 \t 5.5175781250 \t 0.6250000000 \t 0.2000000000   f(a)*f(p)<0.0, b=p\n",
      "4 \t 2.5000000000 \t 3.1250000000 \t 2.8125000000 \t -2.7526855469 \t 0.3125000000 \t 0.1111111111   f(a)*f(p)>0.0, a=p\n",
      "5 \t 2.8125000000 \t 3.1250000000 \t 2.9687500000 \t 1.1650085449 \t 0.1562500000 \t 0.0526315789   f(a)*f(p)<0.0, b=p\n",
      "6 \t 2.8125000000 \t 2.9687500000 \t 2.8906250000 \t -0.8467674255 \t 0.0781250000 \t 0.0270270270   f(a)*f(p)>0.0, a=p\n",
      "7 \t 2.8906250000 \t 2.9687500000 \t 2.9296875000 \t 0.1457095146 \t 0.0390625000 \t 0.0133333333   f(a)*f(p)<0.0, b=p\n",
      "8 \t 2.8906250000 \t 2.9296875000 \t 2.9101562500 \t -0.3538593650 \t 0.0195312500 \t 0.0067114094   f(a)*f(p)>0.0, a=p\n",
      "9 \t 2.9101562500 \t 2.9296875000 \t 2.9199218750 \t -0.1049103215 \t 0.0097656250 \t 0.0033444816   f(a)*f(p)>0.0, a=p\n",
      "10 \t 2.9199218750 \t 2.9296875000 \t 2.9248046875 \t 0.0201903982 \t 0.0048828125 \t 0.0016694491   f(a)*f(p)<0.0, b=p\n",
      "11 \t 2.9199218750 \t 2.9248046875 \t 2.9223632812 \t -0.0424122176 \t 0.0024414062 \t 0.0008354219   f(a)*f(p)>0.0, a=p\n",
      "12 \t 2.9223632812 \t 2.9248046875 \t 2.9235839844 \t -0.0111239791 \t 0.0012207031 \t 0.0004175365   f(a)*f(p)>0.0, a=p\n",
      "13 \t 2.9235839844 \t 2.9248046875 \t 2.9241943359 \t 0.0045299415 \t 0.0006103516 \t 0.0002087247   f(a)*f(p)<0.0, b=p\n",
      "14 \t 2.9235839844 \t 2.9241943359 \t 2.9238891602 \t -0.0032978357 \t 0.0003051758 \t 0.0001043732   f(a)*f(p)>0.0, a=p\n",
      "15 \t 2.9238891602 \t 2.9241943359 \t 2.9240417480 \t 0.0006158486 \t 0.0001525879 \t 0.0000521839   f(a)*f(p)<0.0, b=p\n",
      "16 \t 2.9238891602 \t 2.9240417480 \t 2.9239654541 \t -0.0013410446 \t 0.0000762939 \t 0.0000260926   f(a)*f(p)>0.0, a=p\n",
      "17 \t 2.9239654541 \t 2.9240417480 \t 2.9240036011 \t -0.0003626108 \t 0.0000381470 \t 0.0000130461   f(a)*f(p)>0.0, a=p\n",
      "18 \t 2.9240036011 \t 2.9240417480 \t 2.9240226746 \t 0.0001266158 \t 0.0000190735 \t 0.0000065230   f(a)*f(p)<0.0, b=p\n",
      "19 \t 2.9240036011 \t 2.9240226746 \t 2.9240131378 \t -0.0001179983 \t 0.0000095367 \t 0.0000032615   f(a)*f(p)>0.0, a=p\n",
      "20 \t 2.9240131378 \t 2.9240226746 \t 2.9240179062 \t 0.0000043085 \t 0.0000047684 \t 0.0000016308\n",
      "\n",
      "A raiz da função, dentro dos parâmetros de tolerância, é 2.924017906188965\n",
      "\n",
      "O método da Bissecção foi encerrado após 20 iterações.\n"
     ]
    }
   ],
   "source": [
    "#Método da Bissecção para determinar uma raiz de uma função em uma variável do tipo f(x) = 0\n",
    "\n",
    "#Utiliza a biblioteca sympy, disponível em: http://docs.sympy.org/latest/index.html\n",
    "#Utiliza a biblioteca matplotlib, disponível em: https://matplotlib.org/\n",
    "\n",
    "import sympy as sp\n",
    "\n",
    "def sgn(s): # função que testa o sinal de f(x)\n",
    "    if (s < 0.0):\n",
    "        return -1\n",
    "    elif (s == 0.0):\n",
    "        return 0\n",
    "    else:\n",
    "        return 1\n",
    "\n",
    "def f(a, x, fx): #Retorna o valor numérico da função digitada pelo usuário\n",
    "    return float(fx.subs(x, a))\n",
    "\n",
    "x = sp.symbols('x')\n",
    "fx = sp.sympify(input(\"Digite a função f(x): \"))\n",
    "\n",
    "TOL = abs(float(input(\"Tolerância para o tamanho do intervalo (b-a)/2 (digite 0 para desativar): \"))) # Tolerância para o tamanho do intervalo (b-a)/2\n",
    "e = abs(float(input(\"Tolerância para o erro relativo de p (ponto médio do intervalo) (digite 0 para desativar): \"))) # Tolerância para o erro relativo de p\n",
    "efp = abs(float(input(\"Tolerância para o valor de f(p) ~= 0 (digite 0 para desativar): \"))) # Tolerância para f(p) ~= 0\n",
    "N0 = abs(int(input(\"Número máximo de iterações: \"))) # Número máximo de iterações\n",
    "\n",
    "# Variáveis extras para o cálculo\n",
    "TOL_val=0.0\n",
    "e_val=0.0\n",
    "FP=0.0\n",
    "pn=0.0\n",
    "n=0\n",
    "i=0.0\n",
    "passo=0.0\n",
    "\n",
    "# Obtém a e b e testa se f(a) e f(b) possuem sinais opostos\n",
    "while (True):\n",
    "    a = float(input(\"Defina a: \"))\n",
    "    b = float(input(\"Defina b: \"))\n",
    "    if (f(a,x,fx)*f(b,x,fx) > 0.0):\n",
    "        print(\"f(a) e f(b) devem ter sinais opostos!\")\n",
    "        continue\n",
    "    elif (abs(f(a,x,fx)) <= efp):\n",
    "        print(\"a já é uma raiz da função para a tolerância desejada!\")\n",
    "        N0 = -1 # Para finalizar o programa\n",
    "    elif (abs(f(b,x,fx)) <= efp):\n",
    "        print(\"b já é uma raiz da função para a tolerância desejada!\")\n",
    "        N0 = -1 # Para finalizar o programa\n",
    "    break\n",
    "\n",
    "# Realiza os cálculos do método da biseccção\n",
    "print ('\\nn\\t a\\t         b\\t         p\\t         f(p)\\t         (b-a)/2\\t e. relat. p\\tMundança')\n",
    "for n in range (1, N0+1):\n",
    "    p = a + (b-a)/2\n",
    "    i -= passo # i vai descendo no eixo y e plotando as subdivisões do intervalo [a, b] uma abaixo da outra\n",
    "    graf1.append(sp.plotting.plot(i, (x, a, p), show=sp.false, line_color='blue')[0]) # Plota a próxima subdivisão do intervalo [a, p]\n",
    "    graf1.append(sp.plotting.plot(i, (x, p, b), show=sp.false, line_color='red')[0]) # Plota a próxima subdivisão do intervalo [p, b]\n",
    "    FP = f(p,x,fx)\n",
    "    TOL_val = abs((b-a)/2)\n",
    "    if(p != 0.0): \n",
    "        e_val = abs((pn-p)/p)\n",
    "    print (n, '\\t', \"{:.10f}\".format(a), '\\t', \"{:.10f}\".format(b), '\\t', \"{:.10f}\".format(p), '\\t', \"{:.10f}\".format(FP), '\\t', \"{:.10f}\".format(TOL_val), '\\t', \"{:.10f}\".format(e_val), end=\"\")\n",
    "    if (abs(FP) <= efp or TOL_val < TOL or  e_val < e): #Se F(p) ~= 0 ou (b-2)/2 < erro ou erro relativo de p < erro, encerra o programa\n",
    "        print (\"\\n\\nA raiz da função, dentro dos parâmetros de tolerância, é\", p)\n",
    "        break\n",
    "    if (sgn(f(a,x,fx)) * sgn(FP) > 0.0):\n",
    "        a = p\n",
    "        print ('   f(a)*f(p)>0.0, a=p')\n",
    "    else:\n",
    "        b = p\n",
    "        print ('   f(a)*f(p)<0.0, b=p')\n",
    "    pn = p\n",
    "\n",
    "print (\"\\nO método da Bissecção foi encerrado após\", n, \"iterações.\")"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**(19) Considere que {$p_n$} é a sequência definida por $p_n=\\sum_{k=1}^{n} \\frac{1}{k}$. Mostre que {$p_n$} diverge apesar do $\\lim_{n\\to \\infty}(p_n-p_{n-1})=0$.**\n",
    "\n",
    "A sequência definida por $p_n=\\frac{1}{k}_{k=1}^{n} $ gera o seguinte conjunto de resultados:\n",
    "\n",
    "$$1, \\frac{1}{2}, \\frac{1}{3}, \\frac{1}{4}, \\frac{1}{5}, \\frac{1}{6}, \\frac{1}{7}, \\frac{1}{8} ...$$\n",
    "\n",
    "Observa-se que, apesar do valor dos termos seguir diminuindo à medida que o valor de n cresce, o valor da série $p_n=\\sum_{k=1}^{n} \\frac{1}{k}$ segue aumentando, pois o valor de $\\frac{1}{k}$ nunca é, de fato, 0. Para facilitar a visualização desse conceito, vamos considerar uma série semelhante, cujos termos são:\n",
    "\n",
    "$$ 1, \\frac{1}{2}+ \\frac{1}{4}+ \\frac{1}{4}+ \\frac{1}{8}+ \\frac{1}{8}+ \\frac{1}{8}+ \\frac{1}{8} + ...$$\n",
    "\n",
    "Note que todos os valores desta segunda série são iguais ou menores que os valores da série original, de maneira que possibilita garantir que o valor do somatório de seus termos será sempre menor ou igual ao dos termos da primeira série. Observe ainda que, podemos agrupar os termos iguais, de forma a obter sempre o resultado 1/2, como mostrado a seguir.\n",
    "\n",
    "$$1, \\frac{1}{2}+ 2*1\\frac{1}{4} + 4*\\frac{1}{8}+...^=1+\\frac{1}{2}+\\frac{1}{2}+\\frac{1}{2}+...$$\n",
    "\n",
    "Desta maneira, infere-se que, para os $2^3$ primeiros termos da série, o valor da soma será equivalente a $1+n*\\frac{1}{2}$, cujo $\\lim_{n\\to \\infty} \\to \\infty$, ou seja, essa série não converge.\n",
    "\n",
    "Como já foi garantido anteriormente que o valor dos termos da série $\\lim_{n\\to \\infty}(p_n-p_{n-1})=0$ será sempre maior ou igual ao valor dos termos da segunda série, pode-se concluir que $\\lim_{n\\to \\infty}${$(p_n)$}$\\to \\infty$ e a série diverge.\n",
    "\n",
    "No entanto, quando trata-se a diferença $(p_n-p_{n-1})$, apenas o termo $n$ da série definida por $\\sum_{k=1}^{n} \\frac{1}{k}$ é considerado, e esse termo possui a forma $\\frac{1}{k}$, cujo limite quando $k\\to \\infty$ é 0."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Seção 2.1 DISCUSSION QUESTIONS - 3"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Ele é sensível a valores iniciais. Apenas para comparar o metódo de newton é muito mais sensível a esse tipo de valores.\n",
    "\n",
    "MIRANDA, Mario J.; FACKLER, Paul R. Lecture Notes In Computational Economic Dynamics. Unpublished, The Ohio State University, 1996.\n",
    "\n",
    "'Erros iniciais são os cometidos no arredondamento dos dados iniciais. Os erros intermediários são\n",
    "decorrentes dos erros cometidos durante a aplicação do método numérico e os erros finais decorrentes da\n",
    "apresentação final do resultado.'\n",
    "\n",
    "JOSÉ, Ionildo José Sanches Prof Ionildo; FURLAN, Diógenes Cogo. MÉTODOS NUMÉRICOS. 2006. Tese de Doutorado. Universidade Federal do Paraná."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Iteração de Ponto Fixo ##\n",
    "\n",
    "Um número $p$ é um **ponto fixo** de uma função $g$ se $g (p) = p$. \n",
    "\n",
    "### Teorema 2.3 ##\n",
    "**a.** Se $g ∈ C [a,b], e g(x) ∈ [a,b]$, pata todo $x ∈ [a,b], g$ terá um ponto fixo em $[a,b]$.\n",
    "\n",
    "**b.** Além disso, se $g'(x)$ existir em $(a,b)$ e existir uma constante positiva $k < 1$ com\n",
    "<p style=\"text-align: center;\"> $|g'(x)|≤ k$, para todo $x ∈ (a,b)$ <p>\n",
    "    \n",
    "então o ponto fixo em $[a,b]$ será único.\n",
    "\n",
    "#### Demostração####\n",
    "\n",
    "**a.** Se $g(a)=a$ ou $g(b)=b$, $g$ tem um ponto fixo em una extremidade. Caso contrário, $g(a) > a$ e $g(b) < b$. A função $h(x) = g(x) - x$ é contínua em $[a,b]$, com \n",
    "\n",
    "<p style=\"text-align: center;\"> $h(a) = g(a) - a > 0$ e $h(b) = g(b) - b < 0$.<p>\n",
    "\n",
    "O Teorema do Valor Intermediário implica a existência de $p ∈ (a,b)$ para o qual $h(p) = 0$. Esse número $p$ é um ponto fixo de $g$, uma vez que\n",
    "<p style=\"text-align: center;\"> $ 0 =h(p) = g(p) - p$ implica que $g(p) = p$ <p>\n",
    "\n",
    "**b.** Suponha também que $g'(x) ≤ k < 1$ e que $p$ e $q$ sejam pontos fixos em $[a,b]$. Se $p ≠ q$, o Teorema do Valor Médio impliqca a existência de um número ε entre $p$ e $q$ e, portanto, em $[a,b]$, com \n",
    "<p style=\"text-align: center;\"> $(g(p)-g(q))/(p-q) = g'(ε) $. \n",
    "\n",
    "PAra determinas a aproximação do ponto fixo de uma função $g$, escolhemos uma aproximação inicial $p_0$ e geramos a seqüência $[p_n]_{n=0}^{\\infty}$, definido $p_n = g(p_n-1)$, para cada $n ≥ 1.$ Se a seqüência converfir para $p$ e $q$ for cntínua, então\n",
    "<p style=\"text-align: center;\"> $p = lim_{n\\to\\infty} g(p_n-1) = g(lim_{n\\to\\infty} p_n-1) = g(p)$, <p>\n",
    "    \n",
    "e será obtida uma solução para $x=g(x).$ Essa técnica é denominada **iteração de ponto fixo** ou **iteração funcional**. \n",
    "\n",
    "### Teorema 2.4 (Teorema do Ponto Fixo)###\n",
    "\n",
    "Seja $g ∈ C[a,b]$ tal que $g(x) ∈ [a,b]$, para todo $x$ em $[a,b]$. Além disso, supoha que $g'$ exista em $(a,b)$ e que exista uma constante $0 < k < 1$ com \n",
    "<p style=\"text-align: center;\">  $|g'(x)|≤ k$, para todo $x ∈ (a,b)$. <p>\n",
    "    \n",
    "Então, para qualquer número $p_0$ em $[a,b]$, a seqüência definida por\n",
    "<p style=\"text-align: center;\"> $p_n = g(p_n-1), n≥1$ <p>\n",
    "\n",
    "converge para o único ponto fixo em $[a,b]$.\n",
    "\n",
    "\n",
    "\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Conjunto de exercícios 2.2 #"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**(7) Utilize um método de iteração de ponto fixo para determinar uma solução com precisão $10^{-2}$ para $x^{4}-3x^{2}-3=0$ em $[1,2]$. Utilize $p_{0}=1$.**"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {
    "scrolled": true
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Informe o limite inferior(POS1):1\n",
      "Informe o limite superior(POS2):2\n",
      "Informe o chute para X0:1\n",
      "Informe a tolerancia(formato 0.xxxx1):0.01\n",
      "Informe o número maximo de iterações:10\n",
      "F( 1.0 )= -5.0\n",
      "1 ª Previsão:  1.5650845800732873\n",
      "0.5650845800732873\n",
      "2 ª Previsão:  1.79357287863573\n",
      "0.22848829856244257\n",
      "3 ª Previsão:  1.8859437430173158\n",
      "0.09237086438158593\n",
      "4 ª Previsão:  1.9228478439840826\n",
      "0.03690410096676677\n",
      "5 ª Previsão:  1.9375075399574273\n",
      "0.01465969597334471\n",
      "6 ª Previsão:  1.9433169298986768\n",
      "0.005809389941249465\n"
     ]
    },
    {
     "data": {
      "text/plain": [
       "1.9433169298986768"
      ]
     },
     "execution_count": 10,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "import math \n",
    "\n",
    "#declaração da váriavel\n",
    "pos1 = float(input(\"Informe o limite inferior(POS1):\")) #limite inferior\n",
    "pos2 = float(input(\"Informe o limite superior(POS2):\")) #limite inferior\n",
    "x0 = float(input(\"Informe o chute para X0:\")) #chute para x0\n",
    "tol  = float(input(\"Informe a tolerancia(formato 0.xxxx1):\"))#tolerancia\n",
    "nMax = float(input(\"Informe o número maximo de iterações:\"))#iteração\n",
    "\n",
    "def funcao1(x):\n",
    "    return ((x**4)-3*(x**2)-3)\n",
    "\n",
    "def funcao2(x):\n",
    "    return ((3*(x**2)+3)**(1/4))\n",
    "\n",
    "def metodoDoPontoFixo(pos1, pos2, x0, tol, nMax):    \n",
    "    if(funcao1(pos1)*funcao1(pos2)<0):\n",
    "        i=0\n",
    "        while i<=nMax:\n",
    "            p = funcao2(x0)\n",
    "            \n",
    "            print(i+1,\"ª Previsão: \",funcao2(x0))\n",
    "            print(p-x0)\n",
    "            if(math.fabs(p-x0)<tol):\n",
    "                return p\n",
    "            i+=1\n",
    "            x0 = p\n",
    "        print(\"O procedimento não foi bem sucedido!\")\n",
    "    else:\n",
    "        print(\"Não possui raiz\")\n",
    "    \n",
    "print(\"F(\",x0,\")=\",funcao1(x0))\n",
    "metodoDoPontoFixo(pos1, pos2, x0, tol, nMax)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**(9) Use o teorema 2.3 para mostrar que $g(x)=\\pi+0.5* sen(\\frac{x}{2})$ tem um  único ponto fixo no intervalo [0,2$\\pi$]. Use a iteração de ponto fixo para encontrar um valor aproximado do ponto fixo com erro menor que $10^{-2}$. Use o Corolário 2.5 para estimar o número de iterações necessárias para se alcançar a precisão de $10^{-2}$ e compare essa estimativa teórica com o número realmente necessário de iterações.**\n",
    "\n",
    "O Teorema 2.3 assume duas condições para afimrar que uma determinada função possui exatamene um ponto fixo em um determinado intervalo, sendo elas:\n",
    "\n",
    "I - A função g(x) deve ser contínua em [a,b], e sua imagem para valores entre a e b devem estar contida no próprio intervalo [a,b]. \n",
    "\n",
    "Observa-se que a função $g(x)=\\pi+0.5* sen(\\frac{x}{2})$ de fato possui todos os valores de sua imagem para valores entre [0,2$\\pi$], já que $\\pi$ é uma constante, e os valores que podem ser assumidos por $0.5* sen(\\frac{x}{2})$ estão entre o intervalo $(-0.5,0.5)$, o que significa que a imagem da função $g(x)$ no intervalo [0,2$\\pi$] é um subconjunto desse intervalo, como mostra o gráfico 1 abaixo.\n",
    "\n",
    "II - A função $g'(x)$ deve existir no intervalo [a,b], além de existir uma constante positiva k<1 tal que:\n",
    "$$|g'(x)|\\leqslant k, \\,\\,\\, para \\, todo \\, x\\in (a,b)$$\n",
    "\n",
    "observa-se que, como $g'(x)=0.25 * cos(\\frac{x}{2})$, depende somente do valor do cosseno (que possui módulo limitado entre os valores 1 e -1), todos os valores de $|g'(x)|$ estarão no intervalo [0,0.25], o que significa que a constante $k=0.25$, por exemplo, que é menor que 1, existe e  atende ao  requisito 2.\n",
    "\n",
    "Já que a função $g(x)=\\pi+0.5* sen(\\frac{x}{2})$ atende a ambos os requisitos. O teorema 2.3 permite afirmar que $g(x)$ possui um único ponto fixo no intervalo [0,2$\\pi$].\n",
    "\n",
    "Utilizando o Corolário 2.5, pode-se estimar o número de iterações necessárias para se atingir uma precisão de $10^{-2}$ pela inequação:\n",
    "\n",
    "$$|p-p_n|\\leqslant \\frac{k^n}{1-k}*|p_1-p_0|$$\n",
    "\n",
    "Escolhendo-se $p_0 = 0$, temos que $p_1=g(0)=\\pi$. Como já demonstrado anteriormente, k=0.25 é uma escolha válida para k. Logo:\n",
    "\n",
    "$$10^{-2}\\leqslant \\frac{0.25^n}{1-0.25}*|\\pi|$$\n",
    "\n",
    "$$\\frac{0.75}{\\pi} *10^{-2}\\leqslant 0.25^n$$\n",
    "\n",
    "$$\\log(0.04188790)\\leqslant \\log(0.25^n)$$\n",
    "\n",
    "$$-1.37791139 \\leqslant -0.60205999*n$$\n",
    "\n",
    "$$n\\geqslant 2.28886613$$\n",
    "\n",
    "Ou seja, segundo o corolário 2.5, serão necessárias pelo menos 3 iterações para se alcançar a precisão de $10^{-2}$.\n",
    "\n",
    "Aplicando o método da iteração de ponto fixo ao problema, chegamos à conclusão de que na terceira iteração, a precisão buscada já é alcançada, o que concorda com o calculado segundo o corolário 2.5."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 82,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": 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\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0x115472a7dd8>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "import matplotlib.pyplot as plt #importa o módulo que permite fazer gráficos\n",
    "import numpy as np #importa o módulo necessário para o cálculo da função sen(x/2) e o número pi\n",
    "\n",
    "t=np.linspace(0, 2,endpoint=True) #cria um conjunto de números igualmente espaçados entre 0 e 2, incluindo o limite do intervalo\n",
    "\n",
    "x=np.pi*t #Transforma a lista que vai de 0 a 2, em uma lista que vai de 0 a 2pi\n",
    "plt.plot(x,np.pi+0.5*np.sin(x))\n",
    "plt.title('GRÁFICO 1')\n",
    "plt.show()\n",
    "\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 93,
   "metadata": {
    "scrolled": false
   },
   "outputs": [
    {
     "name": "stderr",
     "output_type": "stream",
     "text": [
      "C:\\Anaconda3\\lib\\site-packages\\matplotlib\\axes\\_base.py:2961: UserWarning: Attempting to set identical left==right results\n",
      "in singular transformations; automatically expanding.\n",
      "left=0, right=0\n",
      "  'left=%s, right=%s') % (left, right))\n",
      "C:\\Anaconda3\\lib\\site-packages\\matplotlib\\axes\\_base.py:3285: UserWarning: Attempting to set identical bottom==top results\n",
      "in singular transformations; automatically expanding.\n",
      "bottom=0, top=0\n",
      "  'bottom=%s, top=%s') % (bottom, top))\n"
     ]
    },
    {
     "data": {
      "image/png": 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k8fE0HUkTEZFSITIykqSkJJKTkzl9+jQxMTG57iQ8Y+/eve6vFy1axNVXX11gnzt27KBevXrcfffdjBo1ih9++KHI9Vx99dW5wsurr77K1Vdfzbx58xg5ciQZGRkkJibSrFkzypcvD9ghaMyYMSxatIh69erl6u+3336jRYsWRV5/QZKSktxfL1q0iGbNmhX5s3369GHfvn1s376d7du3U7Vq1fMGtEOHDpGeng7AgQMHWLdunfvRP/v37wcgPT2dF198kXvuuQewj7zdcsstvP/+++5r1gD+8pe/0LBhQ/e1fStXrnT3dfb3deHChe6xSk5Odt81umPHDrZs2ULjxo05ceIEx44dA+DgwRN89tnnBAW14Kqr4NQpu67s7GyeffZZd13eRkfSRESkVPDz82Pq1Kn06NGDrKwsRo4cSWhoKADR0dFEREQQFRXF66+/zqJFi/Dz8yMgIKDQWTfWrFnDyy+/TIUKFahevTpz584tcj19+vTh7bff5q677uK3337jnXfeIT4+nho1atCxY0eeffZZatSokevRGhMmTOD48eMMGDAAgCuvvJJFixYB9qm/Pn36XMTI5DZp0iS2bNlCuXLlaNSoEW+99RYACQkJvPXWW+5TsR06dGDz5s0cP36cwMBAZs2aRY8ePQrsd/369dx8880cOnSIzz77jL///e8kJiby66+/MmbMGMqVK0d2djaTJk1yB6uXX36ZxYsXk52dzb333kvnzp0BeOaZZ0hNTWXs2LGA/b1NSEgA4I033mDIkCGcPn2a4OBgZs+eDcCjjz7Khg0bsCyLxo0b8/bbbwPw1Vdf8cILL1ChQgXKlSvHm2++yWWXXcbvv//OzTffjDFw4kQmPXvezogRPalWDWbNmse0adMAuOWWWxgxYsQlj3tJsPI7x+9hjhcgIr4nIiLC/Utfisevv/5a6FGpsuj//b//x+LFiwu8A7Fbt27MnTu3SBfhd+zYkdjYWPfF8HLp0tJgyxZ70vSmTSHn/oMiK2Cf99hUBDqSJiIicpFeeeUV/vjjjwJD2hdffFGkflJSUnjooYcU0IrRqVPw2292QLvqKsh58kmpopAmIiJykc5c4H6p6tatS79+/YqlL4GTJ+2AZlkQEgL53I9RKiikiYiIiM84ccKei/NMQKtc2emKLp7u7hQRERGfcOKEfQStXLnSH9BAR9JERETEBxw/bge0ChXsa9DOegRdqaWQJiIiIqXasWP2Kc6KFe2A5qUTCFwwhTQREREptY4eha1bfS+gga5JExGRUmL9+vWEhYWRlpbGiRMnCA0NZePGjU6XJQ46csQ+glapkn0Nmi8FNNCRNBERKSUiIyOJioriiSee4NSpU9xxxx3FMo2SlE6HD8O2bfbjNa66Cvx8MNH44CaJiIgndOpUvP2tWXP+NtHR0URGRlK5cmVef/314i1ASo1Dh+D33+0H1DZt6psBDXS6U0RESpGDBw9y/Phxjh07RlpamtPliANSU+0jaNWq+e4RtDN8eNNERKQkFeXIV3EbPXo0//jHP0hOTmbixIlMnTrV80WIYw4cgO3boXp1+wha+fJOV1SyFNJERKRUmDt3Ln5+ftx+++1kZWVx7bXXsmrVKjp37ux0aeIBKSmwYwfUqAEul+8HNFBIExGRUmLYsGEMGzYMgPLly/Pdd985XJF4yv798McfULOmHdDKlZGLtRTSRERExGv9+Sfs3Am1akFwcNkJaKCQJiIiIl5q717YvRtq14agoLIV0EAhTURERLzQnj32n4AAO6BZltMVeZ5CmoiIFJkxBqss/mspHmOMHc727oU6daBxY2cCmjHG8ys9Rxk7cCgiIhercuXKpKamesU/XuKbjLFPb+7dC5dd5mxAS01NpXLlyp5f+Vl0JE1ERIokMDCQXbt2kZKS4nQp4qMOHoRjx+zHbJw8CZs3O1dL5cqVCQwMdK4AFNJERKSIKlSoQFBQkNNliA/KzoZx4+Ctt+DBB2HKlLJ5Ddq5dLpTREREHJOVBXffbQe0iRMV0M6mkCYiIiKOyMyEO++Ed9+F6GiYPFkB7Ww63SkiIiIel5EBQ4fCRx/Bs8/C4487XZH3UUgTERERjzp9GgYPhoUL4eWX4ZFHnK7IOymkiYiIiMekpcGAAbB4Mbz2Gtx/v9MVeS+FNBEREfGIU6fg5pth+XKYPh3uucfpirybQpqIiIiUuBMnICoKVq+GWbNg5EinK/J+CmkiIiJSoo4dgz59YN06mDsX7rjD6YpKB4U0ERERKTFHjkCvXhAfDx9+CIMGOV1R6aGQJiIiIiXi4EHo0QN++gn+8x+45RanKypdFNJERESk2B04AN26waZN8MkncOONTldU+iikiYiISLH680/o2hW2boVFi+yjaXLhFNJERESk2OzZA126wB9/wJIl0Lmz0xWVXgppIiIiUix27rRD2b59sGwZdOjgdEWlm0KaiIiIXLLt2+2AlpoKn38O7ds7XVHpp5AmIiIil2TbNrjhBvt5aCtWQGSk0xX5BoU0ERERuWhbtthH0NLT7dkEwsOdrsh3KKSJiIjIRUlMtG8SMAbWrIEWLZyuyLeUc7oAERERKX1++gk6dYJy5eDLLxXQSoJCmoiIiFyQ77+3r0GrXNkOaM2aOV2Rb1JIExERkSL77jv7FGfNmrB2LTRt6nRFvkshTURERIrkq6/sqZ4uu8wOaEFBTlfk2xTSRERE5LzWrIGePaF+ffsU55VXOl2R71NIExERkUJ98QX07g2NGtkB7YornK6obFBIE5ECLVu2jJCQEFwuFy+88EKe99PT0xk0aBAul4t27dqxfft293uTJ0/G5XIREhLC8uXLz9tnhw4dCA8PJzw8nAYNGtCvXz8A1qxZg7+/v/u9Z555puQ2WETyiIuDm26yrz1bswb+8henKyo79Jw0EclXVlYW48aN44svviAwMJDIyEiioqJo3ry5u82sWbOoXbs2W7duJSYmhokTJ/LRRx+xadMmYmJiSExMZM+ePXTt2pXffvsNoMA+//vf/7r7vfXWW+nbt6/7dYcOHVi8eLHnNl5EAIiNhQEDoGVLe6qnOnWcrqhs0ZE0EclXfHw8LpeL4OBgKlasyODBg4mNjc3VJjY2luHDhwPQv39/Vq5ciTGG2NhYBg8eTKVKlQgKCsLlchEfH1+kPo8dO8aqVavcR9JExBnz50P//tC6NaxcqYDmBIU0EcnX7t27adiwoft1YGAgu3fvLrCNn58f/v7+pKamFvjZovS5cOFCunTpQs2aNd3LvvnmG1q1akWvXr1ITEws1u0Ukbw+/BAGD4Z27ewjaLVqOV1R2aTTnSKSL2NMnmWWZRWpTUHLs7Ozz9vnvHnzuOuuu9yvW7duzY4dO6hevTpxcXH069ePpKSkfGueMWMGM2bMACAlJSXfNiJSuDlzYMQIuP56+OwzqF7d6YrKLh1JE5F8BQYGsnPnTvfrXbt20aBBgwLbZGZmcuTIEQICAgr87Pn6TE1NJT4+nj59+riX1axZk+o5/0r07t2bjIwMDhw4kG/No0ePJiEhgYSEBOrWrXsJWy9SNs2caQe0rl1hyRIFNKcppIlIviIjI0lKSiI5OZnTp08TExNDVFRUrjZRUVHMmTMHgAULFtC5c2csyyIqKoqYmBjS09NJTk4mKSmJtm3bnrfP+fPnc+ONN1K5cmX3sn379rmPzMXHx5OdnU0dXRwjUuymTYPRo6FXL1i0CKpWdboi0elOEcmXn58fU6dOpUePHmRlZTFy5EhCQ0OJjo4mIiKCqKgoRo0axdChQ3G5XAQEBBATEwNAaGgoAwcOpHnz5vj5+TFt2jTKly8PkG+fZ8TExDBp0qRcdSxYsIDp06fj5+dHlSpViImJyXOKVEQuzZQp8PDD0LcvfPQRVKrkdEUCYOV37YiHOV6AiPieiIgIEhISnC5DxOtNngyPPWbfyfnhh1ChgtMVeT2P/S9RpztFRETKIGPg6aftgHb77TBvngKat9HpThERkTLGGHjiCXj+ebjzTnjnHci5IkG8iEKaiIhIGWIMTJgAr7xi3ygwfTqU03k1r6SQJiIiUkYYAw88AG+8AX/9K7z+Oug+HO+l7CwiIlIGZGfDPffYAe2hhxTQSgOFNBERER+XlQWjRsGMGfC3v8E//6mAVhrodKeIiIgPy8y0bw744AN46imIjlZAKy0U0kRERHxURgYMGQLz59t3cv7tb05XJBdCIU1ERMQHpafD4MHw6af26c2HH3a6IrlQCmkiIiI+Ji3NnkFgyRL7BoH77nO6IrkYCmkiIiI+5ORJuPlm+PxzePtt+1loUjoppImIiPiIEyfgpptgzRp4910YMcLpiuRSKKSJiIj4gKNHoU8f+PpreP99+4YBKd0U0kREREq5w4ehZ09ISLAnSh840OmKpDgopImIiJRiBw9C9+7w88+wYAH06+d0RVJcFNJERERKqZQU6NYNNm+GhQvt053iOxTSRERESqF9+6BrV9i2DRYtso+miW9RSBMRESlldu+GLl1g506Ii4MbbnC6IikJCmkiIiKlyB9/QOfO8OefsGwZdOjgdEVSUhTSRERESonkZDugHToEX3wB11zjdEVSkhTSRERESoGtW+2Advw4rFwJbdo4XZGUNIU0ERERL7d5sx3QMjJg9Wpo1crpisQTyjldgIiIiBRs40a4/nrIzlZAK2sU0kRERLzUhg3QqROUL2/Px9mihdMViScppImIiHihhAT7FGfVqrB2LTRr5nRF4mkKaSIiIl7m22/t56D5+9sBzeVyuiJxgkKaiIiIF/nvf+2pnurWtQNa48ZOVyROUUgTERHxEqtWQc+ecMUVdkBr2NDpisRJCmkiIiJe4PPP7QnSg4Lgyy+hQQOnKxKnKaSJiIg4bMkSuOkmCAmxH7Nx+eVOVyTeQCFNRETEQQsXws03Q8uW9unOunWdrki8hUKaiIiIQ/7zHxgwwJ7iacUKCAhwuiLxJgppIiIiDvj3v+G226B9e/t6tFq1nK5IvI1CmoiIiIfNng3DhtnTPS1bBjVqOF2ReCOFNBEREQ96+20YOdJ+FtrixVCtmtMVibdSSBMREfGQN96Ae+6xH7URG2tP+SRSEIU0ERERD/jnP+H++6FfP/jkE6hc2emKxNsppImIiJSw55+HCRNg4ED7js6KFZ2uSEoDhTQREZESYgw89RQ8/jjccQd88AFUqOB0VVJa+DldgIiIiC8yBh57DF54AUaMgJkzoXx5p6uS0kQhTUREpJgZAw8/DK++CmPGwJtvQjmdu5ILpF1GRESkGGVnw3332QHtvvtg+nQFNLk42m1ERESKSXa2/YiNadPgkUfgtdfAspyuSkorhTQREZFikJVlP6R25kz7RoGXXlJAk0uja9JEREQuUWYmDB8OH34ITz8N0dFOVyS+QCFNRETkEmRkwO23w4IFMHkyTJrkdEXiKxTSRERELlJ6OgwaZE/xNGUKjB/vdEXiSxTSRERELkJaGtx6K8TFwdSpMG6c0xWJr1FIExERuUAnT0LfvrByJcyYAXff7XRF4osU0kRERC7A8eNw003w5Zcwe7Z9w4BISVBIExERKaKjR6F3b/j2W/j3v+0bBkRKikKaiIhIERw+DD16wA8/QEwM9O/vdEXi6xTSREREziM1Fbp3h19+sR+10bev0xVJWaCQJiIiUoj9+6FbN9iyBT791D7dKeIJCmkiIiIF2LsXunaF5GT47DM7rIl4ikKaiIhIPnbvhs6d7b/j4qBTJ6crkrJGIU1EROQcf/xhB7T9+2H5crjuOqcrkrJIIU1EROQsv/9uB7TDh+GLL6BdO6crkrJKIU1ERCRHUpId0E6ehFWroHVrpyuSsqyc0wWIiHdbtmwZISEhuFwuXnjhhTzvp6enM2jQIFwuF+3atWP79u3u9yZPnozL5SIkJITly5eft88777yToKAgwsPDCQ8PZ8OGDQAYY7j//vtxuVyEhYXxww8/lNwGS5n166/QsaM9J6cCmngDhTQRKVBWVhbjxo1j6dKlbNq0iXnz5rFp06ZcbWbNmkXt2rXZunUr48ePZ+LEiQBs2rSJmJgYEhMTWbZsGWPHjiUrK+u8fb788sts2LCBDRs2EB4eDsDSpUv76NIGAAAgAElEQVRJSkoiKSmJGTNmcO+993puEKRM+OUXuP56MAbWrIFWrZyuSEQhTUQKER8fj8vlIjg4mIoVKzJ48GBiY2NztYmNjWV4zuSF/fv3Z+XKlRhjiI2NZfDgwVSqVImgoCBcLhfx8fFF6vNcsbGxDBs2DMuyuOaaazh8+DB79+4tse2WsuXHH+GGG6BCBXs+ztBQpysSsSmkiUiBdu/eTcOGDd2vAwMD2b17d4Ft/Pz88Pf3JzU1tcDPnq/Pxx9/nLCwMMaPH096enqR6xC5GOvX29egVasGa9dCSIjTFYn8j0KaiBTIGJNnmWVZRWpzocvBvoZt8+bNrF+/noMHD/Liiy8WuQ6AGTNmEBERQUREBCkpKQVslYjt66/tB9XWrm0fQWvSxOmKRHJTSBORAgUGBrJz50736127dtGgQYMC22RmZnLkyBECAgIK/GxhfdavXx/LsqhUqRIjRowgPj6+yHUAjB49moSEBBISEqhbt24xjID4qrVr7cnSL7/cDmiNGztdkUheCmkiUqDIyEiSkpJITk7m9OnTxMTEEBUVlatNVFQUc+bMAWDBggV07twZy7KIiooiJiaG9PR0kpOTSUpKom3btoX2eeY6M2MMn376KS1atHCvY+7cuRhj+Pbbb/H396d+/foeHAnxJStXQq9eEBhoB7SzzqSLeBU9J01ECuTn58fUqVPp0aMHWVlZjBw5ktDQUKKjo4mIiCAqKopRo0YxdOhQXC4XAQEBxMTEABAaGsrAgQNp3rw5fn5+TJs2jfLlywPk2yfAkCFDSElJwRhDeHg4b731FgC9e/cmLi4Ol8tF1apVmT17tjMDIqXe8uXQrx+4XLBihX0kTcRbWfld6+FhjhcgIr4nIiKChIQEp8sQL/LZZ9C/PzRvbs8kcNllTlckpVTeC2JLiE53ioiIz/vkE7jlFggLs093KqBJaaCQJiIiPu2jj2DgQIiMtE9xBgQ4XZFI0SikiYiIz3r/fbj9drjuOvt6NH9/pysSKTqFNBER8UnvvgvDh0OnThAXBzVqOF2RyIVRSBMREZ8zfTqMGgXdu8PixfaMAiKljUKaiIj4lNdeg7Fj4cYb4dNPoUoVpysSuTgKaSIi4jNefhkefNC+k/Pjj6FyZacrErl4CmkiIuITnn0WHn0UBg+GmBioWNHpikQujUKaiIiUasZAdDQ8+SQMHWrf0VmhgtNViVw6TQslIiKlljEwaRK89BKMHAkzZkDO7GMipZ6OpImISKlkDDz0kB3Q7r0XZs5UQBPfopAmIiKlTnY2/PWv8K9/wQMPwLRpUE7/oomP0S4tIiKlSnY2jBkDb75p3yjw6qtgeWzKaxHPUUgTEZFSIysLRoyAd96BJ56AF15QQBPfpRsHRESkVMjIgGHD7MdrPPOMfTeniC9TSBMREa93+rQ9UfrHH8OLL9qnOUV8nUKaiIh4tfR0GDAAPvvMvv7swQedrkjEMxTSRETEa506ZU/xtGyZfQfn2LFOVyTiOQppIiLilU6cgL59YdUq+xlod93ldEUinqWQJiIiXufYMbjxRvjqK3jvPfuGAZGyRiFNRES8ypEj0Ls3fPcdfPCBPWG6SFmkkCYiIl7j0CHo0QN+/BE++ghuvdXpikSco5AmIiJe4cAB6N4dEhPhk0/gppucrkjEWQppIiLiuP37oWtX+O03iI2Fnj2drkjEeQppIiLiqL17oUsX2L4dliyxvxYRhTQREXHQrl3QuTPs2QNLl8L11ztdkYj3UEgTERFH7NhhB7QDB+Dzz+Haa52uSMS7KKSJiIjHbdtmB7SjR+GLL6BtW6crEvE+CmkiIuJRW7bY152dOgUrV0Lr1k5XJOKdFNJERMRjNm2yA1pWFqxZAy1bOl2RiPcq53QBIiJSNvz8M3TqZH+tgCZyfgppIiJS4n74AW64ASpWhC+/hObNna5IxPsppImISImKj7dPcVavbge0q65yuiKR0kEhTURESsy6dfZMAgEBsHYtNGnidEUipYdCmoiIlIgvv7QnS69f3/66USOnKxIpXRTSRESk2K1YAb162cFszRoIDHS6IpHSRyFNRESK1dKlcOON4HLB6tX2kTQRuXAKaSIiUmwWLYJ+/ey7N1evhnr1nK5IpPRSSBMRkWLx8cdw660QHm7PJFCnjtMViZRuCmkiInLJ5s2DQYPsOTi/+AJq13a6IpHSTyFNREQuyZw5cMcdcN11sHw51KzpdEUivkEhTURELto778CIEfZsAnFx9gNrRaR4KKSJiMhFefNNuPtu+1lon30G1ao5XZGIb1FIExGRC/avf8G4cRAVBZ9+ClWqOF2RiO9RSBMRkQvy4oswfrx9J+f8+VCpktMVifgmhTQRESmyf/wDJk2C226DmBioWNHpikR8l0KaiIiclzHwxBMQHQ3DhsH774Ofn9NVifg2/YiJiEihjIGJE+Hll+Guu+Dtt6Gc/osvUuL0YyYiIgUyxr7+7OWXYexYBTQRT9KPmoiI5Cs72w5mr71mB7WpUxXQRDxJP24iIpJHVpb9DLS33rJPdb7yCliW01WJlC0KaSIikktmJtx5J7z7rn2jwOTJCmgiTtCNAyIi4paRAUOHwkcfwbPPwuOPO12RSNmlkCYiIgCcPg2DB8PChfaNAo884nRFImWbQpqIiJCWBgMGwOLF9o0C99/vdEUiopAmIlLGnToF/frB55/D9Olwzz1OVyQioJAmIlKmnThhT5K+ejXMmgUjRzpdkYicoZAmIlJGHTsGffrAunUwdy7ccYfTFYnI2RTSRETKoCNHoFcviI+HDz+EQYOcrkhEzqXnpIlIgZYtW0ZISAgul4sXXnghz/vp6ekMGjQIl8tFu3bt2L59u/u9yZMn43K5CAkJYfny5eftc8iQIYSEhNCiRQtGjhxJRkYGAGvWrMHf35/w8HDCw8N55plnSm6Dy4iDB6FrV0hIgPnzFdBEvJVCmojkKysri3HjxrF06VI2bdrEvHnz2LRpU642s2bNonbt2mzdupXx48czceJEADZt2kRMTAyJiYksW7aMsWPHkpWVVWifQ4YMYfPmzfzyyy+cOnWKd955x72eDh06sGHDBjZs2EB0dLTnBsEHHTgAXbrAzz/DJ5/AzTc7XZGIFEQhTUTyFR8fj8vlIjg4mIoVKzJ48GBiY2NztYmNjWX48OEA9O/fn5UrV2KMITY2lsGDB1OpUiWCgoJwuVzEx8cX2mfv3r2xLAvLsmjbti27du3y+Db7uj//hBtugM2bYdEiuPFGpysSkcIopIlIvnbv3k3Dhg3drwMDA9m9e3eBbfz8/PD39yc1NbXAzxalz4yMDN5//3169uzpXvbNN9/QqlUrevXqRWJiYrFuZ1mxZw906gS//w5LlkCPHk5XJCLnoxsHRCRfxpg8y6xzJnAsqE1By7Ozs8/b59ixY+nYsSMdOnQAoHXr1uzYsYPq1asTFxdHv379SEpKyrfmGTNmMGPGDABSUlIK2LKyZ+dO6NwZ9u2DZcsgZ2hFxMvpSJqI5CswMJCdO3e6X+/atYsGDRoU2CYzM5MjR44QEBBQ4GfP1+fTTz9NSkoKU6ZMcS+rWbMm1atXB+xTohkZGRw4cCDfmkePHk1CQgIJCQnUrVv3Erbed2zfDtdfD/v32w+rVUATKT0U0kQkX5GRkSQlJZGcnMzp06eJiYkhKioqV5uoqCjmzJkDwIIFC+jcuTOWZREVFUVMTAzp6ekkJyeTlJRE27ZtC+3znXfeYfny5cybN49y5f73q2nfvn3uI3Px8fFkZ2dTp04dD41C6bZtG3TsCIcOwYoV0L690xWJyIXQ6U4RyZefnx9Tp06lR48eZGVlMXLkSEJDQ4mOjiYiIoKoqChGjRrF0KFDcblcBAQEEBMTA0BoaCgDBw6kefPm+Pn5MW3aNMqXLw+Qb58A99xzD40aNaJ9TpK45ZZbiI6OZsGCBUyfPh0/Pz+qVKlCTExMnlOkkteWLfYpzvR0ezaB8HCnKxKRC2Xld+2IhzlegIj4noiICBISEpwuwxGJifZjNoyBlSuhRQunKxLxKR77X6JOd4qI+JCffrLv4ixXDr78UgFNpDRTSBMR8RHff28/B61yZTugNWvmdEUicikU0kREfMB339mnOGvWhLVroWlTpysSkUulkCYiUsp99RV06waXXWYHtKAgpysSkeKgkCYiUoqtWQM9e0L9+vYpziuvdLoiESkuCmkiIqXUF19A797QqJEd0K64wumKRKQ4KaSJiJRCcXFw0032tWdr1sBf/uJ0RSJS3BTSRERKmdhY6NcPQkNh1SrQDFgivkkhTUSkFJk/H/r3h9at7QfVaoYsEd+lkCYiUkp8+CEMHgzt2tmTpdeq5XRFIlKSFNJEREqB996DO+6wJ0xftsx+HpqI+DaFNBERLzdjBowYAV27wpIlUL260xWJiCcopImIeLFp02DMGPtRG4sWQdWqTlckIp6ikCYi4qWmTIG//hX69oVPPrHn5BSRskMhTUTEC02eDA8/DAMG2Hd0VqrkdEUi4mkKaSIiXsQYePppeOwxuP12+47OChWcrkpEnODndAEiImIzBp54Ap5/Hu68E955B8qXd7oqEXGKQpqIiBcwBiZMgFdegdGjYfp0KKdzHSJlmkKaiIjDjIEHHoA33rBvFHj9dbAsp6sSEafp/2kiIg7KzoZ77rED2kMPKaCJyP8opImIOCQrC0aNsh9W+7e/wT//qYAmIv+j050iIg7IzLRvDvjgA3jqKYiOVkATkdwU0kREPCwjA4YMsZ9/9vzz9lE0EZFzKaSJiHhQejoMHgyffmqf3nz4YacrEhFvpZAmIuIhaWlw660QF2ffIHDffU5XJCLeTCFNRMQDTp6Efv3giy/g7bftZ6GJiBRGIU1EpISdOAE33QRr1sC778KIEU5XJCKlgUKaiEgJOnoU+vSBr7+G99+3bxgQESkKhTQRkRJy+DD07AkJCTBvHgwc6HRFIlKaKKSJiJSAgwehe3f4+WdYsMC+Hk1E5EIopImIFLOUFOjWDTZvhoUL7dOdIiIXSiFNRKQY7dsHXbvCtm2waJF9NE1E5GIopImIFJPdu6FLF9i5034W2g03OF2RiJRmCmkiIsXgjz+gc2f4809Ytgw6dHC6IhEp7RTSREQuUXKyHdAOHbIfVnvNNU5XJCK+QCFNROQSbN1qB7Tjx2HlSmjTxumKRMRXKKSJiFykzZvtgJaRAatXQ6tWTlckIr5EIU1E5CJs3GjfJGBZ9nRPoaFOVyQivqac0wWIiJQ2GzZAp05QvrwCmoiUHIU0EZELkJBgn+KsWhXWroVmzZyuSER8lUKaiEgRffutfYrT398OaC6X0xWJiC9TSBMRKYL//tee6qluXTugNW7sdEUi4usU0kREzmPVKujZE664wg5oDRs6XZGIlAUKaSIihfj8c3uC9KAg+PJLaNDA6YpEpKxQSBMRKcCSJXDTTRASYj8H7fLLna5IRMoShTQRkXwsXAg33wxhYfbpzrp1na5IRMoahTQRkXP85z8wYIA9xdOKFRAQ4HRFIlIWKaSJiJzl3/+G226D9u3t69H8/Z2uSETKKoU0EZEcs2fDsGFw/fWwbBnUqOF0RSJSlimkiYgAb78NI0faz0JbvBiqVXO6IhEp6xTSRKTMe+MNuOce+1EbsbH2lE8iIk5TSBORMu2f/4T774d+/eCTT6ByZacrEhGxKaSJSJn1/PMwYQIMHGjf0VmxotMViYj8j0KaiJQ5xsBTT8Hjj8Mdd8AHH0CFCk5XJSKSm5/TBYiIeJIx8Nhj8MILMGIEzJwJ5cs7XZWISF4KaSJSZhgDDz8Mr74KY8bAm29COZ1PEBEvpV9PIlImZGfDfffZAe2++2D6dAU0EfFu+hUlIj4vO9t+xMa0afDII/Daa2BZTlclIlI4hTQR8WlZWfZDamfOtG8UeOklBTQRKR0U0kSkUMuWLSMkJASXy8ULL7yQ5/309HQGDRqEy+WiXbt2bN++3f3e5MmTcblchISEsHz58vP2mZycTLt27WjatCmDBg3i9OnT511HYTIz7Wme5syBp5+GZ59VQBOR0kMhTUQKlJWVxbhx41i6dCmbNm1i3rx5bNq0KVebWbNmUbt2bbZu3cr48eOZOHEiAJs2bSImJobExESWLVvG2LFjycrKKrTPiRMnMn78eJKSkqhduzazZs0qdB2FMcaeKP3DD2HyZIiOLubBEREpYQppIlKg+Ph4XC4XwcHBVKxYkcGDBxMbG5urTWxsLMOHDwegf//+rFy5EmMMsbGxDB48mEqVKhEUFITL5SI+Pr7APo0xrFq1iv79+wMwfPhwPv3000LXUZD0dNi2DRYsgClTYNKkkhgdEZGSpZAmIgXavXs3DRs2dL8ODAxk9+7dBbbx8/PD39+f1NTUAj9b0PLU1FRq1aqFn59fnnUVtI78pKXBzTfDkSMwdSqMH18MAyEi4gCrsP+NekJoaKipUqWKozV4o5SUFOrWret0GV5FY5K/khyXQ4cOcfToURo1agRAamoqJ06c4Morr3S3SUxMpGnTplTMmVPpl19+4eqrr2bPnj1Ur16dgIAAALZv346/vz9Avn02aNCAzZs306JFCwBOnz5NUlISoaGhBa7jTKA7Y//+A+zeXYXs7GqUK/cL//d/LUtkXEoz/RzlpTHJn8Ylf99//32iMaaFR1ZmjHH0T5s2bYzkpXHJS2OSv5Icl6+//tp0797d/fr55583zz//fK423bt3N19//bUxxpiMjAxTp04dk52dnaftmXYF9ZmdnW3q1KljMjIy8qy7oHWc7dgxYzp1MsayjHnvPWOqVq1ajCPhO/RzlJfGJH8al/wBCcZDGUmnO0WkQJGRkSQlJZGcnMzp06eJiYkhKioqV5uoqCjmzJkDwIIFC+jcuTOWZREVFUVMTAzp6ekkJyeTlJRE27ZtC+zTsixuuOEGFixYAMCcOXPo27dvoes44+hR6NkT/vtf+Pe/IefyNRGRUk3TQolIgfz8/Jg6dSo9evQgKyuLkSNHEhoaSnR0NBEREURFRTFq1CiGDh2Ky+UiICCAmJgYAEJDQxk4cCDNmzfHz8+PadOmUT5nksz8+gR48cUXGTx4ME888QT/93//x6hRowAKXAfA4cPQowf88APExEDOfQciIqWfpw7ZFfTn7bffLq4jkD5F45KXxiR/ZXlcDhwwpnVrYypUMObTT3O/d+WVVzpTlJcry/tLQTQm+dO45A8YbTyUkRy/cQBwvAARKX3274du3WDLFli4EHr1yv1+REQECQkJzhQnIr7MY4/E1ulOESl19u6Frl0hORkWL7a/FhHxNZd044BlWQGWZX1hWVZSzt+1C2g3PKdNkmVZw89a3qZly5a4XC7uv/9+98Mp58+fT2hoKOXKlcvzP+ELnWbGCQcPHqRbt240bdqUbt26cejQoXzbzZkzh6ZNm9K0aVP3RdEA33//PfmNS0H9vvzyy4SHhxMeHk6LFi0oX748Bw8eBKBx48a0bNmS8PBwIiIiSnjLC+fpcVmzZg3+/v7usXnmmWfcfXnL/uLpMfnggw8ICwsjLCyMa6+9lp9++sndlzfsK0WZguqmm+6hUaPf2bz5JO++u88d0JyegqokeXJqriFDhhASEkKLFi0YOXIkGRkZQOE/T07x5LjceeedBAUFubd/w4YNgH3J0P3334/L5SIsLIwffvih5Da4CDw5Jh06dHCPR4MGDejXrx9QdvYVy7LetSxrv2VZG8/uq6BsZNletyxrq2VZP1uW1fq8hV/KuVLgJWBSzteTgBfzaRMA/J7zd+2cr2vnvBf/9ddfm+zsbNOzZ08TFxdnjDFm06ZNZvPmzeb6668369evd58HTkxMNGFhYSYtLc38/vvvJjg42GRmZprMzEwTHBxstm3bZtLT001YWJhJTEwshjPPF2fChAlm8uTJxhhjJk+ebB599NE8bVJTU01QUJBJTU01Bw8eNEFBQebgwYPGGGMiIyNNfuNSlH4XLVpkbrjhBvfrRo0amZSUlGLfxovh6XFZvXq16dOnT551eNP+4ukxWbdunfuzcXFxpm3btu71OL2vFOX78swzc0zNmvtNjRrGPPXUF2bgwIHGmPx/N7Rp06bQPgcMGGDmzZtnjDFmzJgx5s033zTGGDNt2jQzZswYY4wx8+bNc6/DKUUZl4JqvpjfmUuWLDHZ2dkmOzvbDB482D0uBf08OcXT4zJ8+HAzf/78PHUsWbLE9OzZ02RnZ5tvvvkm18+Up3l6TM52yy23mDlz5hhjfH9fAcobO8N0BFoDG00RshHQG1iKfbr0GuA7c76cdb4GhX4YtgD1c76uD2zJp81twNtnvX47Z1l9YPOZAfrwww/N6NGjcw3auSHtQp+75JSrrrrK7NmzxxhjzJ49e8xVV12Vp8252zt69Gjz4Ycfmj179piQkJB82xWl39tuu83MmDHD/drpf3jP5ulxKegXhTftL07uKwcPHjQNGjRwv3Z6Xznf92XbNmMqV95rqlfPMN9+e/5nsjVr1qzEnsnmSZ58Vt25pkyZYh577DFjjPf9w+vpcSkopJ35eTzj7J89T3NqXzl69KipVauWOXLkiDHG9/cVoL35X6ZpnE9Iyzcbnck/+bUr6M+lPiftcmPM3pwjcnuBevm0uQLYedbrXTnLrsj5Gsh/uplzXeg0M075888/qV+/PgD169dn//79edoUti2BgYF5lhel35MnT7Js2TJuvfVW9zLLsujevTtt2rRhxowZxbeRF8GJcfnmm29o1aoVvXr1IjExsdB1OMGpfQXsSct7nXW1vdP7SmHfl6QkuP56yMiozH/+k0q7duefgur06dMlPgWVJ3hyaq6zZWRk8P7779OzZ0/3svx+npzixLg8/vjjhIWFMX78eNLT04tch6c4ta8sXLiQLl26ULNmTfcyX95XsPNLYQrKRgXloQKd98YBy7JWAH/J563Hz/fZM13ks8zkt/zsh1Pmx+RzJ6plWWRnZ+e7vCR17dqVffv25Vn+3HPPFenzBW1LQcuL4rPPPuO6665zT8MDsG7dOho0aMD+/fvp1q0bzZo1o2PHjkXq72J407i0bt2aHTt2UL16deLi4ujXrx9JSUmXNMYXw5vG5IzVq1cza9YsvvrqK/cyT+8r5ypoe379FTp3hsxMaNRoJC1bvp6nTX6fLazPwsbO0/vH+RSlngvdzqL8zhw7diwdO3akQ4cOQME/T07x9LhMnjyZv/zlL5w+fZrRo0fz4osvEh0d7VX7i1P7yrx587jrrrvcr319X+Hin0pRUB4q0HmPpBljuhpjWuTzJxb407Ks+gA5f+f977qdFBue9ToQ2JOz3H0YYNeuXTRo0KDQWgIDA9m5838h9MxnClpeklasWMHGjRvz/Onbty+XX345e/fuBWDv3r3Uq5f3AGNh27Jr1648y4Hz9hsTE8Ntt92Wa9mZz9arV4+bb76Z+Pj4Ytj6gnnTuNSsWZPq1asD0Lt3bzIyMjhw4IDH9xdvGhOAn3/+mbvuuovY2Fjq1KnjXu7pfeVc+W1nuXKtuP56MAbWrAGX64S7TWZmJkeOHCEgICDfz1asWLHAsbvssss4fPgwmZmZuZafW8fZ63BKUfbXgmq+2N+ZTz/9NCkpKUyZMsW9rKCfJ6d4elzq16+PZVlUqlSJESNGuH8+nPj3pyBO7CupqanEx8fTp08f9zJf31ewM0xhCspGBeWhghV2LvR8f4CXyX1x3Ev5tAkAkrFvGqid83VAznvrv/nmG/dFz0uWLMl1jvjca9I2btyY6wK+oKAgk5mZaTIyMkxQUJD5/fff3RcFbty40TjlkUceyXXR9oQJE/K0SU1NNY0bNzYHDx40Bw8eNI0bNzapqanGGGMiIiJMfuNSWL+HDx82tWvXNsePH3cvO378uDl69Kj76/bt25ulS5eWzEYXgafHZe/eve5rib777jvTsGFDk52d7VX7i6fHZMeOHaZJkyZm3bp1udbhDfvKud+Xpk0Hmlq1MkyDBsZszrl6derUqbku7h0wYIAxJv/fDW3atCn0e92/f/9cNw5Mmzat0HU4pSj764WMy/l+Z86cOdO0b9/enDx5Mtc6Cvp5coqnx+XMdWbZ2dnmgQceMBMnTjTGGLN48eJcNw5ERkZ6agjy8PSYGGPM9OnTzbBhw3Ktw9f3FXJuHDAFX5OWbzYC+pD7xoF4c76cdb4GhX4Y6gArgaScv8+ErwjgnbPajQS25vwZcdbyiNDQUBMcHGzGjRvn/iZ+8skn5oorrjAVK1Y09erVy3XB37PPPmuCg4PNVVdd5b6TzRj7DpumTZua4OBg8+yzz17SN/RSHThwwHTu3Nm4XC7TuXNn9z+o69evN6NGjXK3mzVrlmnSpIlp0qSJeffdd93L169fb/Ibl4L6NcaY2bNnm0GDBuWqY9u2bSYsLMyEhYWZ5s2bl7lxeeONN0zz5s1NWFiYadeuXa5g4i37i6fHZNSoUaZWrVqmVatWplWrVu4JlL1lXznzfWnQoK+pXPmkufJKY8aNm2JiY2ONMcacOnXK9O/f3zRp0sRERkaabdu2uT977u+GM9tW0Pd627ZtJjIy0jRp0sT079/fpKWlnXcdTslvG5588smLGpfC+jTGmPLly5vg4GD3PvL0008bYwr/eXKKJ8flhhtuMC1atDChoaFmyJAh5tixY8YYO7SNHTvWBAcHmxYtWuQ6sOAET46JMfbBlHP/Q1cG9pUzGWYesBfIyDlKNsoUno0sYBqwDfgFiDDnyVmacUBEvMrXX9uzB9SpA6tWQePGF9ePZhwQkRLisYsOL/XuThGRYrN2LXTvDpdfDl9+efEBTUTEFyikiYhXWLkSevaEhg3tgNaw4fk/IyLiyxTSRMRxy5fDjTdCkyb2XZw5j3gTESnTFNJExFGffQZRUdCsGaxebVo6sO0AACAASURBVJ/qFBERhTQRcdAnn8Att0BYmH2687LLnK5IRMR7KKSJiCNiYmDgQIiMhBUrwMHnxYqIeCWFNBHxuPffhyFD4Lrr7OvR/P2drkhExPsopImIR737LgwfDp06QVwc1KjhdEUiIoVbv349YWFhpKWlYVlWNcuyEi3LalHS61VIExGPmT4dRo2yn4W2eDFUq+Z0RSIi5xcZGUlUVBRPPPEEwEvAv435/+3deXRV5aH+8e8BRWsdClZsMNQpKd4CylKwYqsC4RAQGdTSRmqJFIoiilO1tCq/4kSstS6norloDU60KAIKhIAQuNJiDBZt4dZG60QMk0kFERFh//7YmIuGIUCSfYbvZ62u5mz28Jz8AY977/d9g3809nUPaOwLSBLAvffC1VeHU21MmQIHHxx1Ikmqv7Fjx9KlSxcIl74c3RTX9E6apEZ3111hQbvgAnj2WQuapORTXV3Nxx9/DHAY0CR/i1nSJDWq226DG26AvLxwRGeLFlEnkqS9N2LECG699VaAJ4E7m+KaljRJjSIIYOxYuPlm+OlP4Ykn4MADo04lSXtv0qRJHHDAAQwePBigAOgSi8V6NPZ1Y0EQNPY19iTyAJIaVhDAmDHw29+GAwUefhiaN2/aDJ07d6a8vLxpLyopHcSa6kIOHJDUoIIArrkmHCgwciQ88AA08569JO01/+qU1GC2bYNRo8KCdtVV8OCDFjRJ2lf+9SmpQWzbBpdeGs6FdsMNcM89EGuyhwKSlHosaZL229atMHQoTJwIN90EBQUWNEnaX76TJmm/bNkCQ4aE02vccks4mlOStP8saZL22WefweDB4QS1d94ZPuaUJDUMS5qkfbJ5MwwaBM8/H75/dvXVUSeSpNRiSZO01zZtCpd4Ki6GP/whnGpDktSwLGmS9srGjTBgAMyfHw4UGDYs6kSSlJosaZLqbcMGOO88eOkleOyxcMCAJKlxWNIk1ctHH0GfPlBWBk8+GS6YLklqPJY0SXtUUwO5ufC3v8Gf/gQXXhh1IklKfZY0Sbu1bh306gXLl8PUqdCvX9SJJCk9WNIk7dKaNdCzJ/zrXzB9OvTuHXUiSUofljRJO1VVBTk58M47MHNm+LMkqelY0iTVsXIl9OgBH3wAs2fDOedEnUiS0o8lTdKXvPtuWNDWrYOSEjjzzKgTSVJ6sqRJqvXWW2FBW78e5s6F00+POpEkpS9LmiQA3ngjfO9s0yZ48UU49dSoE0lSerOkSWLFirCgbd0KpaXQsWPUiSRJzaIOIClar78O3bqFP1vQJClxWNKkNPbqq9C9O7RoAQsXwne/G3UiSdIXLGlSmiorCx9xHnpoWNC+852oE0mSdmRJk9LQ4sXhSgKtWsGiRXDiiVEnkiR9lSVNSjMLF4aLpWdkhD8fe2zUiSRJO2NJk9LIvHnQp09YzEpLITMz6kSSpF2xpElpYvZsOO88yMqCBQvCO2mSpMRlSZPSwIwZMHBgOHpzwQJo3TrqRJKkPbGkSSnu2WfhwguhU6dwJYEjj4w6kSSpPixpUgp7+mn48Y/DNTjnzoWWLaNOJEmqL0ualKKKiuDii+EHP4A5c+Dww6NOJEnaG5Y0KQVNnAhDh0KPHjBrVjhhrSQpuVjSpBTz4IPw85+Hc6HNmAGHHBJ1IknSvrCkSSnknnvgiiugf3+YNg2+9rWoE0mS9pUlTUoRd94J114bjuScMgUOOmj/zlddXU08Hic7O5t4PE5NTc1O9ysqKiI7O5vs7GyKiopqty9dupSOHTuSlZXF6NGjCYJgt+d98sknOfnkkzn55JM588wzee2112rPddxxx9GxY0c6depE586d9++LSVKSsKRJKeDWW2HMGLjoIpg8GVq02P9zFhQUkJOTQ0VFBTk5ORQUFNTZp7q6mnHjxvHyyy9TVlbGuHHjakvXyJEjKSwspKKigoqKCoqLi3d73uOPP56FCxfy+uuvc/PNNzNixIgvXWvBggUsW7aM8vLy/f9ykpQELGlSEgsCuOkmGDsWhgyBxx+HAw5omHNPnz6d/Px8APLz85k2bVqdfebMmUM8HqdVq1a0bNmSeDxOcXExVVVVrF+/nq5duxKLxRgyZEjt8bs675lnnknL7XOEnHHGGaxcubJhvogkJSlLmpSkggBuuAFuvx2GD4c//hGaN2+4869evZqM7WtHZWRksGbNmjr7VFZW0rZt29rPmZmZVFZWUllZSeYOC4N+sb2+533kkUfo06dP7edYLEavXr047bTTKCwsbJgvKEkJroH+m1tSUwoCuPpquO8+uPxyuP9+aLYP/8nVs2dPVq1aVWf77bffXs8cQZ1tsVhsl9vrY8GCBTzyyCO89NJLtdsWL15MmzZtWLNmDfF4nJNOOomzzz67zrGFhYW1JW7t2rX1up4kJSpLmpRktm2DUaPgoYfgmmvg7ruhnv2njnnz5u3yz44++miqqqrIyMigqqqK1jtZ8DMzM5PS0tLazytXrqRbt25kZmZ+6XHlypUradOmzR7P+/rrrzN8+HBmz57NkTusX/XFsa1bt+b888+nrKxspyVtxIgRte+yOcBAUrLzcaeURLZuDedAe+gh+OUv96+g7Un//v1rR2sWFRUxYMCAOvvk5uZSUlJCTU0NNTU1lJSUkJubS0ZGBocddhhLliwhCAImTZpUe/yuzvvee+9xwQUX8Pjjj/Od73yn9hobN25kw4YNtT+XlJTQoUOHxvnSkpRAYjt7LNHEIg8gJYPPPw9XEXjiiXCgwG9+03gFDeDDDz/kRz/6Ee+99x7f/va3mTJlCq1ataK8vJyHHnqIiRMnAvDoo49yxx13AHDjjTcydOhQAMrLy7nkkkvYtGkTffr04f777ycWi+3yvMOHD+fZZ5/l2GOPBeCAAw6gvLycf//735x//vnbfwefM3jwYG688cY95u/cubMjQSU1hkb8m/crF7KkSYlvyxb46U/hT3+C226DenSUtGdJk9RImqyk+U6alOA++wzy8uC55+Cuu+AXv4g6kSSpKVjSpAT26acwaBC88ALcey+MHh11IklSU7GkSQlq0yYYOBBKSmDCBLjssqgTSZKakiVNSkAbN0K/flBaCo88Aj/7WdSJJElNzZImJZgNG6BvX1i8GCZNgosvjjqRJCkKljQpgXz0EfTpA2Vl8NRT8OMfR51IkhQVS5qUIKqrITcXXnsNpkyB7VODSZLSlCVNSgDr1kE8DitWwNSpcN55USeSJEXNkiZFbPVq6NkT3nwTZswI76ZJkmRJkyL0wQeQkwPvvQczZ0KPHlEnkiQlCkuaFJH33w9L2apVUFwMZ50VdSJJUiKxpEkReOedsKB9+GE4WW3XrlEnkiQlGkua1MTefDMsaBs2wLx50KVL1IkkSYnIkiY1oTfeCAva5s2wYAF06hR1IklSorKkSU1k+fJwkEAQhMs9degQdSJJUiJrFnUAKR289hp06wbNmsHChRY0SdKeWdKkRrZ0KXTvDgcfHBa0k06KOpEkKRlY0qRGtGRJ+Ijz8MNh0SLIzo46kSQpWVjSpEby0kvQqxd885thQTv++KgTSZKSiSVNagSlpdC7N7RpEz7i/Pa3o04kSUo2ljSpgc2dC+eeC8ceG5a1Y46JOpEkKRlZ0qQGNGsW9OsXvntWWgrf+lbUiSRJycqSJjWQ6dNh4EBo3x7mz4ejjoo6kSQpmVnSpAYwZQr88Idw6qnw4otw5JFRJ5IkJTtLmrSfnnoK8vLge98LF0v/xjeiTiRJSgWWNGk/PPYYXHwxnH02FBeH86FJktQQLGnSPioshKFDoWdPmDkTDj006kSSpFRiSZP2wQMPwKWXhlNtzJgBhxwSdSJJUqqxpEl76fe/hyuvhAEDYOrUcE1OSZIamiVN2gvjx8N118GgQeGIzoMOijqRJClVWdKkeggCGDcOfv1rGDw4HNF54IFRp5IkpbIDog4gJboggJtugjvugEsugYkToXnzqFNJklKdJU3ajSCA66+Hu++GESNgwgRo5v1nSVITsKRJuxAEcNVVcP/9cMUVcN99EItFnUqSlC68JyDtxLZtcNllYUG79loLmiSp6VnSpK/YuhWGDQsnq/3Vr+B3v7OgSZKano87pR18/jnk54ejN3/zGxg71oImSYqGJU3abssW+MlPwvnP7rgjvIsmSVJULGkSsHkz5OXBtGnh483rros6kSQp3VnSlPY+/RQuvBBmzQoHCFx5ZdSJJEmypCnNffIJDBwIc+fCww+Hc6FJkpQILGlKWx9/DP36wcKF8OijMHRo1IkkSfo/ljSlpfXroW9f+Mtf4PHHwwEDkiQlEkua0s5//gO9e0N5OUyeDIMGRZ1IkqS6LGlKK9XV0KsXvP46PPNM+D6aJEmJyJKmtLF2LcTj8M9/wnPPhY87JUlKVJY0pYVVq6BnT3jrLZgxI7ybJklSIrOkKeVVVkJODrz/fjgXWvfuUSeSJGnPLGlKae+9Bz16wOrVUFwMZ50VdSJJkurHkqaU9fbbYUGrqQknqz3jjKgTSZJUf5Y0paQ33wwfa27cCC++CKedFnUiSZL2jiVNKeef/wzvoG3ZAgsWwCmnRJ1IkqS9Z0lTSvnHP8JBArEYlJZC+/ZRJ5Ikad80izqA1FCWLYNu3aB5cwuaJCn5WdKUEsrLw0echxwCixbBSSdFnUiSpP1jSVPSW7IkfMR5xBFhQcvKijqRJEn7z5KmpPY//xMu9XTUUWFBO+64qBNJktQwLGlKWvPnQ+/ecMwxYUFr2zbqRJIkNRxLmpLSnDnhAunHHw8LF0KbNlEnkiSpYVnSlHReeAH694d27cJ50I4+OupEqam6upp4PE52djbxeJyampqd7ldUVER2djbZ2dkUFRXVbl+6dCkdO3YkKyuL0aNHEwTBbs9bWlrKEUccQadOnejUqRO33HJL7bmKi4tp164dWVlZFBQUNOK3lqTEYUlTUnnuObjgAjj55PBx51FHRZ0odRUUFJCTk0NFRQU5OTk7LUfV1dWMGzeOl19+mbKyMsaNG1dbukaOHElhYSEVFRVUVFRQXFy8x/OeddZZLFu2jGXLljF27FgAtm7dyqhRo5g9ezYrVqzg6aefZsWKFU3wG5CkaFnSlDT+/GcYNChc4mnePGjVKupEqW369Onk5+cDkJ+fz7Rp0+rsM2fOHOLxOK1ataJly5bE43GKi4upqqpi/fr1dO3alVgsxpAhQ2qPr895d1RWVkZWVhYnnHACLVq0IC8vj+nTpzfwt5WkxGNJU1J44gm46CLo2hVKSsLpNtS4Vq9eTUZGBgAZGRmsWbOmzj6VlZW03WHERmZmJpWVlVRWVpKZmVln+57O+9e//pVTTjmFPn36sHz58t1eQ5JSnctCKeH98Y8wbFi4msDzz8PXvx51otTRs2dPVq1aVWf77bffXq/jv3jPbEexWGyX23fn1FNP5d133+XQQw9l1qxZDBw4kIqKir06V2FhIYWFhQCsXbu2Pl9BkhKWJU0J7eGH4bLLoFev8H20Qw6JOlFqmTdv3i7/7Oijj6aqqoqMjAyqqqpo3bp1nX0yMzMpLS2t/bxy5Uq6detGZmYmK1eu/NL2NtuH4O7qvIcffnjt/ueeey6XX34569atIzMzk/fff3+n5/qqESNGMGLECAA6d+5cj9+AJCUuH3cqYd1/f1jQ+vaF6dMtaE2tf//+taM1i4qKGDBgQJ19cnNzKSkpoaamhpqaGkpKSsjNzSUjI4PDDjuMJUuWEAQBkyZNqj1+V+ddtWpV7V2zsrIytm3bxpFHHkmXLl2oqKjg7bff5rPPPmPy5Mn079+/KX4FkhStIAii/p9Ux113BQEEwfnnB8HmzVGnSU/r1q0LevToEWRlZQU9evQIPvzwwyAIguCVV14Jhg0bVrvfI488Epx44onBiSeeGDz66KO121955ZWgffv2wQknnBCMGjUq2LZt227Pe//99wff/e53g5NPPjn43ve+FyxevLj2XDNnzgyys7ODE044Ibjtttvqlf+0007b79+BJO1Ek3WkWLCT9z2auidGHUCJ5fbb4aab4Ec/CgcMHHhg1ImUjDp37kx5eXnUMSSlnt2/YNuAfNyphBEE8P/+X1jQLr4YnnzSgiZJSl8OHFBCCAL49a+hoACGDoX//m9o3jzqVJIkRceSpsgFAVx3HdxzD1x6KfzhD9DMe7ySpDTnP4WK1LZtcOWVYUG78kqYMMGCJkkSWNIUoW3bwjtnDz4Iv/gF3Hsv7GG+U0mS0oYlTZHYuhV+9jOYOBFuvBF++1sLmiRJO/KdNDW5zz+H/Hx46im45Ra4+eaoE0mSlHgsaWpSW7bA4MHwzDMwfjyMGRN1IkmSEpMlTU1m8+ZwgtoZM+D3v4drrok6kSRJicuSpibx6adwwQUwezY88ACMGhV1IkmSEpslTY3uk09gwAB48UUoLISf/zzqRJIkJT5LmhrVxx9Dv36wcCH88Y/hgAFJkrRnljQ1mvXr4dxzYcmScKH0wYOjTiRJUvKwpKlR1NRA797w6qsweTL88IdRJ5IkKblY0tTgPvwQevWCv/89nGpjwICoE0mSlHwsaWpQa9ZAPA5vvAHTp0OfPlEnkiQpOVnS1GCqqqBnT3j7bXjhhfBnSZK0byxpahCVldCjR/j/s2ZBt25RJ5IkKblZ0rTf3n03LGhr18KcOfD970edSJKk5GdJ037597/Dgvaf/8DcufC970WdSJKk1GBJ0z6rqAgL2iefwPz5cOqpUSeSJCl1WNK0T/73f8OC9vnnYUE75ZSoE0mSlFqaRR1Ayefvf4dzzoEggNJSC5okSY3Bkqa98re/QffucOCB4Xqc7dtHnUiSpNRkSVO9vfJK+Ijz61+HRYugXbuoE0mSlLosaaqXv/wlnJy2ZcvwDtqJJ0adSJKk1GZJ0x4tWhSuxXn00WFBO+64qBNJkpT6LGnarRdfhN69oW3bsKC1bRt1IkmS0oMlTbtUXAznnRc+2iwthYyMqBNJkpQ+LGnaqeefhwED4KSTYMGC8FGnJElqOpY01TF1KlxwQTj/2fz58M1vRp1IkqT0Y0nTl0yeDD/6EXTpEq7F2bJl1IkkSUpPljTVevxx+MlP4Pvfhzlz4Igjok4kSVL6sqQJgEcfhfx86NYNZs2Cww6LOpEkSenNkiYmTIBhw8K50F54IVxRQJIkRcuSlubuvRcuvzycamPaNPja16JOJEmSwJKW1n77W7j66nAk57PPwsEHR51IkiR9wZKWpm67DX75S8jLC0d0tmgRdSJJkrQjS1qaCQIYOxZuvhl++lN44gk48MCoU0mSpK86IOoAajpBAGPGhI85hw2Dhx+G5s2jTiVJknbGkpYmggCuuSYcKDByJDzwADTzPqokSQnLf6bTwLZtMGpUWNCuugoefNCCJklSovOf6hS3bRtcemk4F9oNN8A990AsFnUqSZK0J5a0FLZ1KwwdChMnwk03QUGBBU2SpGThO2kpassWGDIknF7jllvC0ZySJCl5WNJS0GefwUUXwdSpcOed4WNOSZKUXCxpKWbzZhg0CJ5/Pnz/7Oqro04kSZL2hSUthWzaFC7xVFwMf/hDONWGJElKTpa0FLFxIwwYAPPnhwMFhg2LOpEkSdoflrQUsGEDnHcevPQSPPZYOGBAkiQlN0takvvoI+jTB8rK4MknwwXTJUlS8rOkJbGaGsjNhb/9Df70J7jwwqgTSZKkhmJJS1Lr1kGvXrB8eTjVRr9+USeSJEkNyZKWhNasgZ494V//gunToXfvqBNJkqSGZklLMlVVkJMD77wDM2eGP0uSpNRjSUsiK1dCjx7wwQcwezacc07UiSRJUmOxpCWJd98NC9q6dVBSAmeeGXUiSZLUmCxpSeCtt8KCtn49zJ0Lp58edSJJktTYLGkJ7o03wvfONm2CF1+EU0+NOpEkSWoKzaIOoF1bsSJ87+yzz6C01IKmplVdXU08Hic7O5t4PE5NTc1O9ysqKiI7O5vs7GyKiopqty9dupSOHTuSlZXF6NGjCYJgt+e966676NSpE506daJDhw40b96c6upqAI477jg6duxIp06d6Ny5cyN/c0lKDJa0BPX669CtG8RiYUHr2DHqREo3BQUF5OTkUFFRQU5ODgUFBXX2qa6uZty4cbz88suUlZUxbty42tI1cuRICgsLqaiooKKiguLi4t2e9/rrr2fZsmUsW7aM8ePHc84559CqVavaay1YsIBly5ZRXl7eBN9ekqJnSUtAr74K3btDixawcCF897tRJ1I6mj59Ovn5+QDk5+czbdq0OvvMmTOHeDxOq1ataNmyJfF4nOLiYqqqqli/fj1du3YlFosxZMiQ2uPrc96nn36aiy66qBG/nSQlPktagikrC99BO/RQWLQIvvOdqBMpXa1evZqMjAwAMjIyWLNmTZ19Kisradu2be3nzMxMKisrqaysJDMzs872+pz3k08+obi4mAt3WOcsFovRq1cvTjvtNAoLCxvuS0pSAnPgQAJZvDhcLP2oo2D+fDj22KgTKdX17NmTVatW1dl+++231+v4L94z21EsFtvl9vp4/vnn+f73v/+lR52LFy+mTZs2rFmzhng8zkknncTZZ59d59jCwsLaErd27dp6XU+SEpUlLUEsXAh9+8Ixx4SjOHe4CSE1mnnz5u3yz44++miqqqrIyMigqqqK1q1b19knMzOT0tLS2s8rV66kW7duZGZmsnLlyi9tb9OmTb3OO3ny5DqPOr84tnXr1px//vmUlZXttKSNGDGCESNGADjAQFLS83FnApg3L7yDduyx4SABC5oSQf/+/WtHaxYVFTFgwIA6++Tm5lJSUkJNTQ01NTWUlJSQm5tLRkYGhx12GEuWLCEIAiZNmlR7/O7O+9FHH7Fw4cIvbdu4cSMbNmyo/bmkpIQOHTo02veWpERhSYvY7Nlw3nmQlQULFsD2V3WkyI0ZM4a5c+eSnZ3N3LlzGTNmDADl5eUMHz4cgFatWnHzzTfTpUsXunTpwtixY2sfU06YMIHhw4eTlZXFiSeeSJ8+fXZ7XoDnnnuOXr168fWvf7122+rVq/nBD37AKaecwumnn07fvn3p3bt3U/0aJCkysZ29O9LEIg8QlRkzYNAgaN8+XEngyCOjTiSljs6dOztdh6TGUL8XbBuAd9Ii8swzcOGF0KlT+A6aBU2SJO3IkhaBp5+GvLxwDc65c6Fly6gTSZKkRGNJa2JFRXDxxfCDH8CcOXD44VEnkiRJiciS1oQmToShQ6FHD5g1K5ywVpIkaWcsaU3kwQfh5z+H3NxwwMAhh0SdSJIkJTJLWhO45x644gro3x+mTYOvfS3qRJIkKdFZ0hrZnXfCtdeGIzmnTIGDDoo6kSRJSgaWtEZ0660wZgxcdBFMngwtWkSdSJIkJQtLWiMIArjpJhg7FoYMgccfhwNcJVWSJO0Fq0MDCwK44Qb43e9g+HB4+GFoZhWWJEl7yfrQgIIArr46LGiXX25BkyRJ+84K0UC2bQuL2X33wTXXwAMPWNAkSdK+s0Y0gK1bwznQHnoIfvlLuPtuiDXZ8quSJCkVWdL20+efwyWXwKOPhgMFxo+3oEmSpP3nwIH9sGVLuA7nn/8Mt90GN94YdSJJkpQqLGn76LPPIC8PnnsO7roLfvGLqBNJkqRUYknbB59+CoMGwQsvwL33wujRUSeSJEmpxpK2lzZtgoEDoaQEJkyAyy6LOpEkSUpFlrS9sHEj9OsHpaXwyCPws59FnUiSJKUqS1o9bdgAffvC4sUwaVI4YECSJKmxWNLq4aOPoE8fKCuDp56CH/846kSSJCnVWdL2oLoacnPhtddgyhQ4//yoE0mSpHRgSduNdesgHocVK2DqVDjvvKgTSZKkdGFJ24XVqyEnB956C2bMCO+mSZIkNRVL2k588EFY0N57D2bOhB49ok4kSZLSjSXtK95/Pyxlq1ZBcTGcdVbUiSRJUjqypO3gnXfCgvbhh+FktV27Rp1IkiSlK0vadm++GRa0DRtg3jzo0iXqRJIkKZ1Z0oB//jN8B23zZliwADp1ijqRJElKd2lf0pYvDwtaEITLPXXoEHUiSZIkaBZ1gCi99hp06wbNmsHChRY0SZKUONK2pC1dCt27w8EHhwXtpJOiTiRJkvR/0rKkLVkSPuI8/HBYtAiys6NOJEmS9GVpV9Jeegl69YJvfjMsaMcfH3UiSZKkutKqpJWWQu/e0KZN+Ijz29+OOpEkSdLOpU1JmzsXzj0Xjj02LGvHHBN1IkmSpF1Li5I2axb06xe+e1ZaCt/6VtSJJEmSdi/lS9q0aTBwILRvD/Pnw1FHRZ1IkiRpz1K6pE2ZAoMGwamnwosvwpFHRp1IkiSpflK2pD31FOTlwRlnhIulf+MbUSeSJEmqv5QsaY89BhdfDGefDbNnh/OhSZIkJZOUK2mFhTB0KPTsCTNnwqGHRp1IkiRp76VUSXvgAbj00nCqjRkz4JBDok4kSZK0b1KmpP3+93DllTBgAEydGq7JKUmSlKxSoqSNHw/XXReO5JwyBQ46KOpEkiRJ+yepS1oQwLhx8Otfw+DB4YjOAw+MOpUkSdL+OyDqAPsqCODGG8O7aJdcAhMnQvPmUaeSJElqGElZ0oIArr8e7r4bRoyACROgWVLfE5QkSfqypCtpQQBXXQX33w9XXAH33QexWNSpJEmSGlZS3X/atg0uuywsaNdea0GTJEmpK2lK2tatMGxYOFntr34Fv/udBU2SJKWupHjc+fnnkJ8fjt78zW9g7FgLmiRJSm0JX9K2bIGf/CSc/+yOO8K7aJIkSakuoUva5s2QlwfTpoWPN6+7LupEkiRJTSNhS9qnn8KFF8KsWeEAgSuvjDqRJElS00nIkvbJJzBwIMydCw8/HM6FJkmSlE4SrqR9/DH06wcLF8Kjj8LQoVEnkiRJanoJVdLWr4e+feEvf4HHHw8HDEiSSR7y3gAAA49JREFUJKWjhClp//kP9O4N5eUweTIMGhR1IkmSpOgkxGS21dXQsye8+io884wFTUoE1dXVxONxsrOzicfj1NTU7HS/oqIisrOzyc7OpqioqHb70qVL6dixI1lZWYwePZogCACYMmUK7du3p1mzZpSXl3/pXOPHjycrK4t27doxZ86c2u3FxcW0a9eOrKwsCgoKGuHbSlLiibykrV0L3bvDP/4Bzz0XDhiQFL2CggJycnKoqKggJydnp+WourqacePG8fLLL1NWVsa4ceNqy9zIkSMpLCykoqKCiooKiouLAejQoQNTp07l7LPP/tK5VqxYweTJk1m+fDnFxcVcfvnlbN26la1btzJq1Chmz57NihUrePrpp1mxYkXj/wIkKWKRl7Tu3eFf/4IZM8L30SQlhunTp5Ofnw9Afn4+06ZNq7PPnDlziMfjtGrVipYtWxKPxykuLqaqqor169fTtWtXYrEYQ4YMqT3+v/7rv2jXrt1Or5eXl8dBBx3E8ccfT1ZWFmVlZZSVlZGVlcUJJ5xAixYtyMvLY/r06Y375SUpAURe0t5+O5wLrVevqJNI2tHq1avJyMgAICMjgzVr1tTZp7KykrZt29Z+zszMpLKyksrKSjIzM+ts353dnWtn2yUp1cW+eE9EUvqJxWLzgG/t5I9uBIqCIPjGDvvWBEHQ8ivHXw8cFATBbds/3wx8AiwCxgdB0HP79rOAG4Ig6LfDsaXAL4IgKN/++UHgr0EQPLH98yPALML/mMwNgmD49u0/BU4PgqDOFNexWGwE8MXMigcHQdBhL38lkpQwEmZ0p6Sm90WJ2plYLLY6FotlBEFQFYvFMoC6t9JgJdBth8+ZQOn27Zlf2f7BHuKsBNru8HnHY3a1/UuCICgECvdwHUlKCpE/7pSUsGYA+dt/zgd29iLYHKBXLBZrGYvFWgK9gDlBEFQBG2Kx2BmxWCwGDNnF8V+9Xl4sFjsoFosdD2QDZcArQHYsFjs+Fou1APK27ytJKc2SJmlXCoB4LBarAOLbPxOLxTrHYrGJAEEQVAO3EhapV4Bbtm8DGAlMBN4E3gJmbz/+/FgsthLoCsyMxWJztp9rOfBnYAVQDIwKgmBrEASfA1cQFsL/Bf68fV9JSmm+kyZJkpSAvJMmSZKUgCxpkiRJCciSJkmSlIAsaZIkSQnIkiZJkpSALGmSJEkJyJImSZKUgCxpkiRJCej/A07cpWGkIqEnAAAAAElFTkSuQmCC\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0x11547e3f5c0>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Ponto fixo de g(x) = 3.14159265359+0.5*sin(x/2), para p0 = 0\n",
      "n\t p\t         |p - p0|\t |p - p0|/|p|\n",
      "1 \t 3.1415926536 \t 3.1415926536 \t 1.0000000000\n",
      "2 \t 3.6415926536 \t 0.5000000000 \t 0.1373025617\n",
      "3 \t 3.6260488644 \t 0.0155437891 \t 0.0042867015\n",
      "o ponto fixo da função g(x), dentro dos parâmetros de tolerância, é 3.6260488644453095\n",
      "O método foi encerrado após 3 iterações.\n",
      "\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "import numpy\n",
    "\n",
    "#Método do Ponto Fixo - Com gráfico\n",
    "\n",
    "import sympy as sp\n",
    "import matplotlib.pyplot as plt\n",
    "\n",
    "def ponto_fixo (g, p0, TOL, TOL_rel, N0): # Função método do Ponto Fixo\n",
    "    #Argumentos: \n",
    "    #g: função g(x) a ser encontrado o seu ponto fixo: g(p) = p\n",
    "    # TOL: Tolerância para o erro absoluto |p - p0|\n",
    "    # TOL_rel: Tolerância para o erro relativo |p - p0|/|p|\n",
    "    # N0: número máximo de iterações\n",
    "    \n",
    "    x = sp.symbols('x') # variavel da função sympy\n",
    "    gx = sp.sympify(g) # Transforma a string em uma função sympy\n",
    "    \n",
    "    \n",
    "    # Plota o gráfico da função g(x) e o gráfico y = x\n",
    "    plt.rcParams['figure.figsize'] = 10, 10 # Tamanho da figura\n",
    "    graf1 = sp.plotting.plot(gx, show=sp.false, line_color='green', ylabel=\"g(x)\") # Plota o gráfico da função g(x)\n",
    "    graf1.append(sp.plotting.plot(x, show=sp.false, line_color='blue')[0]) # Plota o gráfico y = x\n",
    "    graf1.xlim = [p0 - (2*abs(p0)), p0 + (2*abs(p0))] # limites do eixo x: 2*p0 acima e abaixo do ponto p0\n",
    "    graf1.ylim = [p0 - (2*abs(p0)), p0 + (2*abs(p0))] # limites do eixo y: 2*p0 acima e abaixo do ponto p0\n",
    "    graf1.legend = sp.true # Mostrar legenda\n",
    "    graf1.show()\n",
    "  \n",
    "    print ('Ponto fixo de g(x) = ' + g + ', para p0 = ' + str(p0) + '\\nn\\t p\\t         |p - p0|\\t |p - p0|/|p|')\n",
    "    \n",
    "    for n in range (1, N0+1):\n",
    "        p = float(gx.subs(x, p0)) # Calcula pn\n",
    "        erro = abs(p - p0) # Calcula o erro absoluto |p - p0|\n",
    "        erro_rel = erro / abs(p) # Calcula o erro relativo |p - p0|/|p|\n",
    "        print (n, '\\t', \"{:.10f}\".format(p), '\\t', \"{:.10f}\".format(erro), '\\t', \"{:.10f}\".format(erro_rel))\n",
    "        if (erro < TOL or erro_rel < TOL_rel): # Testa os erros\n",
    "            print (\"o ponto fixo da função g(x), dentro dos parâmetros de tolerância, é\", p)\n",
    "            break\n",
    "        if (erro_rel > 1): # Se o erro relativo subir acima de 100%, o programa encerra por precaução\n",
    "            print (\"O método falhou!\")\n",
    "            break\n",
    "        p0 = p\n",
    "    print (\"O método foi encerrado após\", n, \"iterações.\\n\")\n",
    "\n",
    "\n",
    "# Exemplo de uso da função ponto_fixo (basta alterar aqui para usar):\n",
    "\n",
    "ponto_fixo('3.14159265359+0.5*sin(x/2)', 0, 0.01, 0.01, 30)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### (17) -  Use a fixed-point iteration method to determine a solution accurate to within 10 _2 forx = 2sin(7rx) + x = 0, for x on fl, 2]. Use po = 1."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "1\n"
     ]
    },
    {
     "data": {
      "text/plain": [
       "1.3733333333333333"
      ]
     },
     "execution_count": 14,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "from scipy import linalg, optimize\n",
    "import numpy\n",
    "import math\n",
    "\n",
    "\n",
    "def ponto_fixo(N):\n",
    "#funcao do pto. fixo  \n",
    "    g = lambda x: x - ((pow(x,3) + 4*pow(x,2) - 10)/(3*pow(x,2)+8*x))\n",
    "    #g = lambda x: - (2 * numpy.sign(math.pi * x))\n",
    "    i = 1  \n",
    "    p0 = 1.5\n",
    "    TOL = pow(10,-2)\n",
    "    while (i <= N): \n",
    "        p = g(p0)\n",
    "        if (p - p0 < TOL):  \n",
    "            print(i)\n",
    "            return p  \n",
    "        i = i +1\n",
    "        p0 = p\n",
    "    raise NameError('Num. max. de iter. excedido!');\n",
    " \n",
    "ponto_fixo(500)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Exercicio 21 - 2.2\n",
    "\n",
    "Encontre uma função $g$ definida em $[0,1]$ que não satisfaça nenhuma das hipóteses do Teorema 2.2, mas possua um único ponto fixo em $[0,1]$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Resposta\n",
    "\n",
    "Um de muitos exemplos é a função $g(x)=\\sqrt{2x-1}$, definida em $[\\frac{1}{2},1]$\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 18,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": 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/fM55wHm/+dD6o0aNqlFYWFhaeVVGCgoKqFq1auwYWgOfU/bwWWWPsnxWqWXLaDBsGFs+8QRVFizg57p1mXrCCczZf3/vtyuB8vjvVevWrdO+tTmtEpVKpfYErgcOSpIk+asStYqvm5UkyWbphlE8I0aMoHXr1rFjaA18TtnDZ5U9yuRZLV8Ojz8OV14J06eHU8Yvvzy8626ddUr3e+WRcvrvVdolKt0avB+wLTD91wJVHxiRSqXaFD+bJEk5KklgyBBo2RJOPBEWLAh7nqZOhR49LFA5Jq2N5UmS3ATctOLn6U6iJEnKG2++CRddFA7MrFYtHJp54YWw0Uaxk6mMeE6UJElrY9KkUJieey4clHnGGeHsp7p1YydTGStRiUqSpFEp55AkKbvMng1XXQUPPhguC+7YMbwDz3Oe8oaTKEmSiuOHH8IxBf/6Vzjrad99w9lPe+wRO5nKmSVKkqR0LF0Kd98N110H338PTZvCTTdB27aQSvsNXcohlihJklanqChczXLZZTBjBtSvDw8/DCecEPZAKW950pckSX/l5Zdh552ha1dYuDAs402eDCefbIGSkyhJkv7kvfegVy8YNQqqVAlHFfTqBRtvHDuZMoglSpKkFaZNg0svhSefDPucTj4Zrr4aGjaMnUwZyBIlSdJ338G118I998CyZXDYYeGk8WbNYidTBrNESZLy188/s+Xjj8Mxx8BPP8Fuu4V9T/vtFzuZsoAbyyVJ+Wf5cnjkEdhmG7YZMCBcEPz00/D22xYopc0SJUnKLyNHwk47wSmnwC+/8FmPHvDJJ3D00Z73pGKxREmS8sPHH0ObNnDIIfD55+HddlOnMvPww6Fy5djplIXcEyVJym1z5oQLgR96KByc2bVrOHV8881jJ1OWs0RJknLTokVw221ho/jPP8P++0Pv3uHwTKkUWKIkSbll+fJwLcvll8M330CTJnDrrdCunXueVKrcEyVJyg1JAsOHQ8uWcPrpoUzddRd89BG0b2+BUqmzREmSst/EidC6ddg4PnUqXHJJeD3zTDeNq8y4nCdJyl5ffx2W7R55JPz8pJPCyeMNGkSNpfxgiZIkZZ9Fi8KG8d69YckSOPDA8M877hg7mfKIJUqSlD2WL4f+/cMlwd98A9tvHzaNt2njnieVO/dESZKyw5gxsMsucOqpoUzdc0/YC9W2rQVKUViiJEmZbepUOOooOOAA+PRTuOgimDIFuneHSi6oKB7/7ZMkZaaFC8PJ4nfcAcuWQceOYR/UllvGTiYBlihJUqYpLIT77w9XtcybFy4L/te/YN99YyeTfsflPElS5hgxIhyW2aNHON/p4Ydh/HgLlDKSJUqSFN9nn4UN4oceCl98Ec5+mjwZTj4ZKvhXlTKTy3mSpHi+/x6uugruvju8465LF7jxRg/LVFawREmSyt8vv0C/fnDNNWED+R57wL//DbvvHjuZlDZnpJKk8pMkMGQING0K550H1avDE0/Am29aoJR1LFGSpPLxySdwyCFwxBEwe3Y4vuDzz6FzZw/LVFZyOU+SVLYWLIArr4S77gr7nk4+GW64AerWjZ1MWiuWKElS2Vi+PJz3dNllYQP5HnuEgzN33TV2MqlUuJwnSSp948bBzjvDmWfCOuvAgAHwxhsWKOUUS5QkqfTMnAnHHgv77RfOfurVK+x7OuEEz3tSznE5T5K09n7+GW69FW6+GZYsgcMPh9tug623jp1MKjOWKElSySUJDB4MF1wQplDbbRfOezrkkNjJpDLnbFWSVDIffggHHACdOsEPP4RLgidOtEApb1iiJEnFM29euCB4xx3DBvLTT4cpU+Dcc8OlwVKecDlPkpSewsJwx92VV4azn1q1CkcW7Lhj7GRSFE6iJElrNmZMKEvnnAPrrReuahk3zgKlvGaJkiT9ta+/huOOC3ufpkyByy+HSZO8qkXC5TxJ0qr88kt4l90118DixdC+ffj5llvGTiZlDEuUJOn3Ro6Es88Oh2RutRX06QOHHRY7lZRxXM6TJAUzZ8LRR4cjCmbOhGuvhY8/tkBJf8FJlCTlu6VLoXdvuP76cNr4UUfB7bfD5pvHTiZlNEuUJOWzYcPCO+6mToXGjaFvXw/LlNLkcp4k5aPp0+GII6BtW5gzB266CT76yAIlFYOTKEnKJ0uWwC23hNJUUADHHhuW8urXj51MyjqWKEnKB0kCL7wQrmaZPh223z4s3R14YOxkUtZyOU+Sct0XX0C7dmH5bt48uO02mDDBAiWtJSdRkpSrli4NS3c33BCW7rp0gVtvhbp1YyeTcoIlSpJy0ciR0LNnuKplu+3CxcH77Rc7lZRTXM6TpFwye3a41+6QQ2DWrLCBfMIEC5RUBpxESVIuKCyEfv3CBcE//RT2P/Xp44GZUhmyRElStnv7bTjzzDBx2nxzeOyxcGGwpDLlcp4kZav586F7d9hrL/jkE/i//wuvFiipXDiJkqRskyTQvz9ceGE4smD//eGuu8IGcknlxkmUJGWTjz+GffeFU06BChVg4EB45RULlBSBJUqSssGiRXDRRbDjjvDGG9CjB0yaBF27QioVO52Ul1zOk6RM99xzcM458NVXsPPO4cynXXeNnUrKe06iJClTffUVdOgARx4JP/4Id94J77xjgZIyhJMoSco0y5eHwnTZZWEZ79hj4V//8roWKcNYoiQpk3zwAXTrBu++G858euopaNMmdipJq+ByniRlgkWL4PzzYZddQpG68MJw5pMFSspYTqIkKbahQ8O77WbODPud7rsPWraMnUrSGjiJkqRY5syBTp2gXTtYsAD69oW33rJASVnCSZQklbeiIrj3XujVK7zr7sgj4Y47oH792MkkFYMlSpLK08cfh43jb70VStOAAXDEEbFTSSoBl/MkqTwsWQKXXBJOHH/nHfjHP+DTTy1QUhZzEiVJZW3kSDjjDJg2LZSo++4L78KTlNWcRElSWZk/H04+GQ45BL75Bm67Df77XwuUlCOcRElSaUsSNh03Dk48EebOhdat4Z57oFGj2MkklSJLlCSVpq+/hh49aDlkCNSsCQMHQpcukErFTiaplFmiJKk0FBXB/ffDRRfBjz8yZ//9qTtoEGyySexkksqIJUqS1tbkyeHYgrFjw7EFjz3Gh5UrU9cCJeU0N5ZLUkktWwY33QTNm4cCdeaZ4b67du1iJ5NUDpxESVJJvP8+nHZauCy4cWN44AHYZ5/YqSSVIydRklQcS5aE61p22w0++igcoDlxogVKykNOoiQpXWPHwumnw5QpsPPO8OCD0KJF7FSSInESJUlr8sMP0L077L8/zJoFt94Kb79tgZLynJMoSVqd4cPD9GnWLDjwwHBly1ZbxU4lKQM4iZKkVVm4EP7+d2jTJkyi7rsPRo2yQElayUmUJP3Riy+G5bvZs8OVLffdBw0bxk4lKcM4iZKkFebPD/fdtW8PixeHjePDhlmgJK2SkyhJAnj+eTjjDPjmGzjsMLj3Xthss9ipJGUwJ1GS8tu8eXD88dChAxQUQP/+8MILFihJa+QkSlL+euaZcFXL3Llw+OFw991Qr17sVJKyhJMoSfnnu+/g2GOhY0coLITHHoPnnrNASSoWJ1GS8svTT0OPHmEZ78gj4a67oE6d2KkkZSEnUZLyw9y5cPTR0KlT+PmTT8J//mOBklRiTqIk5b5nngnnPs2bF4pUv36wySaxU0nKck6iJOWuBQvghBPC3qeiojB9evppC5SkUuEkSlJuGj4cTj01nDrerh3cf79Ld5JKlZMoSbnlp5/CoZlt2oR/fughGDLEAiWp1DmJkpQ7xo2Dk0+G6dPhwANDgdp889ipJOUoJ1GSst+SJXDeebD//uHalr59YeRIC5SkMuUkSlJ2Gz8+XBo8aRLsuWe4tmWbbWKnkpQHnERJyk6//AKXXx6K07RpcNNN8NprFihJ5cZJlKTs89FHYfo0YQK0bAkDBkCzZrFTScozTqIkZY/ly+Hmm2HnnUORuvxyeOcdC5SkKJxEScoOM2aE6dNrr8G224bp0667xk4lKY85iZKU2ZIEBg6E5s1DgTrrLHjvPQuUpOicREnKXPPnh4Mzn34a6tYNr61bx04lSYAlSlKmGjkyHJw5ezYcdRTcdx/UrBk7lSSt5HKepMyyZAmcey4ccki4tuXhh2HwYAuUpIyT9iQqlUq9DNQBioCfgLOTJJlQVsEk5aEJE6BLF/j0U9h777AXaostYqeSpFUqziSqU5IkzZMkaQncBjxURpkk5ZsVRxfsthtMngzXXw9jx1qgJGW0tCdRSZIs/M1PaxAmUpK0dr78MhxdMG5cOLrg0UfDOVCSlOGKtbE8lUoNAA749aeHln4cSXkjSUJhOuss+PHH8HrzzVCtWuxkkpSWVJIkxf+iVOok4NgkSdr+4ePnAef95kPrjxo1qkZhYeHapVSZKygooGrVqrFjaA1y5TlVWrSI7fv2pe7YsSzdaCM+Pv985u2yS+xYpSpXnlU+8Fllj/J4Vq1bt06l+7klKlEAqVRqCVA/SZLvV/M5s5Ik2axE30DlasSIEbT2/J2MlxPP6Y03wubxL7+EDh3g/vuhVq3YqUpdTjyrPOGzyh7l9KzSLlFpbSxPpVIbpFKper/5+ZHA98D84meTlJcKC+Hqq2HffWHuXLj3XnjmmZwsUJLyQ7p7omoA/0mlUusSNpR/B7RLSjrGkpRfvvwSunaF118P17c88QRsv33sVJK0VtIqUUmSfAXsVsZZJOWip56Cbt3ghx/CIZo33gjuP5GUA7z2RVLZWLQI/vEPeOghqF0bhg6Ftm3X/HWSlCUsUZJK33vvwfHHh4MzDzkE+veHOnVip5KkUuXdeZJKT1ER9O4Ne+4J06fD7bfDsGEWKEk5yUmUpNIxZw6cdBKMHAlNmoTN4zvuGDuVJJUZJ1GS1t7QoeFddyNHwmmnheU8C5SkHGeJklRyv/wC550H7dqFc6CefjocnrneerGTSVKZczlPUslMmwadO8P48bDXXmH5rmHD2Kkkqdw4iZJUfIMHh+W68eOhVy8YM8YCJSnvOImSlL6CgrB8d/fd4eyn4cPBO8ck5SlLlKT0fP45HHssTJwIBxwAjz0GdevGTiVJ0bicJ2nNBg6EnXeGjz4KlwiPHGmBkpT3nERJ+muLF8NZZ8Ejj0C9euEog/32i51KkjKCJUrSqn30UVi+++wzaNMmXN1Su3bsVJKUMVzOk/R7SRLOetptN5gyBW65BV580QIlSX/gJErS/yxaBN26/e/MpyefDPfgSZL+xBIlKfjkEzj6aJg0CY44Ah5+GDbaKHYqScpYLudJgkcf/d/y3W23wbPPWqAkaQ2cREn5rKAA/vEPuO++8O67QYOgVavYqSQpK1iipHw1bVpYvvvgA/jb38LhmZtsEjuVJGUNl/OkfPT887DTTjBhAlxxRbi+xQIlScXiJErKJ8uWwSWXQO/eULMmDBvm3XeSVEKWKClffP01dO4Mr78eji0YNAgaNIidSpKylst5Uj4YPRp23DEUqHPPhTFjLFCStJYsUVIuKyqC666Dgw8O78QbPBj+9S9YZ53YySQp67mcJ+WqhQvhhBPClS3Nm4cCtc02sVNJUs5wEiXloo8+gl13DQXqhBPgrbcsUJJUyixRUq558knYYw+YMQP69oX+/aFatdipJCnnWKKkXLFsGZx3Hhx3HNSoETaPn3UWpFKxk0lSTnJPlJQLvv0WOnWCcePCtS1PPQV168ZOJUk5zUmUlO3efjucPj5uHJxzDrzyigVKksqBJUrKVkkCd98N++4LCxbAwIHQpw9Urhw7mSTlBZfzpGy0ZAn06AGPPAJbbgnPPAMtWsROJUl5xRIlZZsZM6BjR3j/fWjTBh57DDbaKHYqSco7LudJWWTjiRNhl11CgbriinAOlAVKkqJwEiVlgySBO+9k5//7P1hvPXj+eTj88NipJCmvWaKkTLd0adj/9NBDLKlXj/VGjYLttoudSpLyniVKymRz5oT9T2+9Ba1b8/Zpp3GQBUqSMoJ7oqRM9d//hv1Pb70FF14IQ4dSWL167FSSpF85iZIy0YAB0K1buLLl0UehS5fYiSRJf+AkSsokhYXh/ruTToLateH11y1QkpShnEQO4fACAAAeBElEQVRJmWL+fDj2WBg1Ktx/N3gwbLpp7FSSpL/gJErKBB9/DLvuGgpU9+4werQFSpIynCVKiu3FF2HPPWHmzHAX3j33wDrrxE4lSVoDS5QUS5LAbbeFQzOrVAlTqDPOiJ1KkpQm90RJMfzyC5x5Jjz0EGy/PbzwQrhIWJKUNSxRUnmbNy8coDluHBx6KDz5JNSoETuVJKmYXM6TytOnn8Luu4cCde65YQJlgZKkrOQkSiovw4eHIwx+/jlsHu/ePXYiSdJacBIllbUkgTvugMMOg4oVYcQIC5Qk5QAnUVJZWrYMzj4b7r0XGjcOxxlss03sVJKkUmCJksrK/PlwzDHwyivwt7/BU0/BRhvFTiVJKiUu50ll4YsvwgGar7wCPXvCSy9ZoCQpx1iipNL21luwxx4wdWrYC3XnnVC5cuxUkqRS5nKeVJqefhpOOCFsIH/uOWjfPnYiSVIZcRIllYYkgVtugU6dwrLduHEWKEnKcU6ipLVVWAhnnRXegde0KQwdCg0bxk4lSSpjlihpbfz0U5g+DR8OBx8clvM8gVyS8oLLeVJJzZoFrVqFAnXaaWECZYGSpLxhiZJKYsKEcAfehx/CDTfAfff5DjxJyjMu50nF9dJL4Q68ZcvgiSegc+fYiSRJETiJkorj/vvDu+6qVIHRoy1QkpTHLFFSOpIErroKunWDLbYIB2ruvXfsVJKkiFzOk9aksBDOOAMefBB22SVsIN9kk9ipJEmRWaKk1Vm8OOx/GjoUDj00HGGw/vqxU0mSMoDLedJf+e47OOigUKBOOgmGDLFASZJWskRJqzJtWtjz9M47cMkl8PDDHmEgSfodl/OkP3r/fWjbFubOhX79oEeP2IkkSRnIEiX91ssvQ8eO4Qyo//wHjjwydiJJUoZyOU9aYeBAOOwwqFQJRo2yQEmSVssSJQH07g0nngh168Ibb4Q78SRJWg1LlPJbkkCvXnDhhbDDDvDmm7D99rFTSZKygHuilL+WLw+HaD7wAOyxRzjKYOONY6eSJGUJS5Ty09Kl0KVL2Dx+yCHwzDOw3nqxU0mSsojLeco/ixZBu3ahQHXqBC+8YIGSJBWbJUr55fvvwynko0ZB9+7w+OOwzjqxU0mSspAlSvlj1izYZx/473/h0kvh7ruhYsXYqSRJWco9UcoPkyfDwQfDzJlw++3wz3/GTiRJynKWKOW+99+HQw+F+fPhkUfCZcKSJK0lS5Ry2+uvh1PIly4N78A7/PDYiSRJOcISpdw1ahQccQRUqADDh8P++8dOJEnKIW4sV2564YVwjME668DIkRYoSVKps0Qp9zz1FBx1FFSvDq++Gk4jlySplFmilFv694fjjoPatWHsWGjZMnYiSVKOskQpd9x9N5x8MtSvD+PGeZGwJKlMWaKUG267DXr0gK23htdeC6+SJJUhS5SyW5LA1VfDBRfADjuECVTDhrFTSZLygEccKHslCfTqBbfcAjvtBCNGQK1asVNJkvKEJUrZKUnC1S19+sCee8JLL8GGG8ZOJUnKI5YoZZ8kgX/8A/r2DRcKDx0ajjOQJKkcWaKUXZIEzj4b+vWD/faDF1+E9dePnUqSlIfcWK7sUVQEPXuGAnXAAWECZYGSJEXiJErZoagIzjwT7rsPDjoIhgyBatVip5Ik5TFLlDJfURF07w4PPAAHHwzPPw/rrhs7lSQpz7mcp8xWVASnnx4KVOvWFihJUsZwEqXMtXw5nHpquA+vTRt45hmoWjV2KkmSACdRylRFRf8rUIcdBs8+a4GSJGUUS5Qyz4o9UP37Q7t28J//QJUqsVNJkvQ7lihllhXnQK3YAzV4sAVKkpSRLFHKHEkC550Hd90FBx4YlvAsUJKkDGWJUmZYcZnwv/8drnIZMsR34UmSMpolSpnhyivhllvCZcJDh8J668VOJEnSalmiFN/118O118Iuu8CwYV4mLEnKCpYoxXXrrXDZZdCyJYwYATVqxE4kSVJaLFGKp08fuOgiaNoURo6EjTeOnUiSpLRZohTHAw/AuefCttvCqFFQq1bsRJIkFUtaJSqVSlVNpVLPpVKpyalUakIqlRqeSqUalW005axBg6BbN9hySxg9GjbdNHYiSZKKrTiTqPuAJkmStARe/PXnUvG89BJ07Qp164YJVL16sRNJklQiaZWoJEkKkiR5KUmS5NcPvQ1sWXaxlJPGjYOOHcPm8ZEjYYstYieSJKnEUv/rRcX4olRqAPB9kiT//MPHzwPO+82H1h81alSNwsLCtUupMldQUEDVMrzgd4MpU9j14osBGH/TTfzYuHGZfa9cVtbPSaXHZ5U9fFbZozyeVevWrVPpfm6xS1QqlboEaA8clCTJz2v43FlJkmxWrG+gKEaMGEHr1q3L5jf/9FPYd19YvBiGD4f99iub75MHyvQ5qVT5rLKHzyp7lNOzSrtEVSrW75pKXQAcBfxtTQVKAmD6dDj4YPjhB3juOQuUJClnpF2ifl2qO45QoBaWXSTljDlzQoGaMwcefxwOOyx2IkmSSk1aJSqVStUHbgOmAa+mUimApUmS7F6G2ZTN5s+HQw6BL76Ae++Fzp1jJ5IkqVSlVaKSJJlFMdYIled+/hnat4ePPw6XCnfrFjuRJEmlzhPLVboKC8PU6c034YIL4MILYyeSJKlMWKJUepIEzjgDXnghHKh5882xE0mSVGYsUSo9V1wBDz4IrVvDQw9BBf/1kiTlLv+WU+m46y647jrYdVcYPBgqV46dSJKkMmWJ0tobPBjOOgu22QaGDoX114+dSJKkMmeJ0toZMwa6dIFNN4URI6B27diJJEkqF5YoldzEiXDEEVC1arjOxQuFJUl5pFjXvkgrffUVtG0LBQVhAtWiRexEkiSVK0uUiu+HH8IVLrNnw5NPwv77x04kSVK5czlPxbNsGRxzDHz0Edx0Exx7bOxEkiRFYYlS+lYcpjlyZLjK5aKLYieSJCkaS5TSd8MN4RDNQw+Ffv0g5XWKkqT8ZYlSeh57DC67DFq2hKeegkpup5Mk5TdLlNZs7Fj4+9+hfn148UWoXj12IkmSorNEafU++ww6dIAqVcJp5JttFjuRJEkZwTUZ/bW5c8NRBj/9BC+9BM2bx04kSVLGsERp1ZYuhaOOgunT4f774ZBDYieSJCmjuJynP0sS6N4d3ngDzj8fTjstdiJJkjKOJUp/1rs39O8flvJuvjl2GkmSMpIlSr/3wgtw8cWwww7w+ONQsWLsRJIkZSRLlP7no4/g+ONh441hyBDYYIPYiSRJyliWKAGwzsKF0L592FD+zDOw5ZaxI0mSlNF8d55g6VJaXnMNfPklPPgg7Ltv7ESSJGU8J1H57td34m306adw3nnhZHJJkrRGlqh816cP9O/Pd7vuCrfcEjuNJElZwxKVz159FS64AJo04cNevXwnniRJxWCJylczZ0KnTlCtGjz3HIXrrRc7kSRJWcUSlY+WLIEjj4R582DgQNh229iJJEnKOpaofJMkcMYZ8P77cMUVcMQRsRNJkpSVLFH5pm9fGDAA2rWDK6+MnUaSpKxliconY8eGYwwaN4ZHH4UKPn5JkkrKv0XzxVdfwTHHwLrrwnPPQY0asRNJkpTVPLE8H/zyCxx9NHz3HTz7LGy3XexEkiRlPSdR+eCCC+C//4VevaBDh9hpJEnKCZaoXDdoUNhMvt9+cO21sdNIkpQzLFG5bNIkOO00qFMHnnwSKrl6K0lSafFv1Vy1eHHYB/Xzz/Dii6FISZKkUmOJykUrDtT85BO48cawlCdJkkqVy3m56P77wzlQ7drBRRfFTiNJUk6yROWa996Ds8+GRo2gf38P1JQkqYz4N2wu+eEH6NQp/PPTT8PGG8fNI0lSDnNPVK5IEujeHaZNgzvvhF12iZ1IkqSc5iQqVzz0UDgTqkMH6NEjdhpJknKeJSoXfPZZ2AfVoAE8+CCkUrETSZKU8yxR2a6gADp3hqVL4bHH3AclSVI5sURluwsugA8/hCuvhH32iZ1GkqS8YYnKZs89B/36hcM0L700dhpJkvKKJSpbzZwJf/97WL579FGoWDF2IkmS8opHHGSj5cvhhBNgwQIYMgTq14+dSJKkvOMkKhv17g3jxsFZZ0H79rHTSJKUlyxR2WbCBLj8cth2W7j55thpJEnKW5aobFJQAF27htPJH30UqlWLnUiSpLzlnqhscskl8MkncN11sPPOsdNIkpTXnERli9Gj4V//gj33hIsvjp1GkqS8Z4nKBgsXwsknw3rrwcCBUMkBoiRJsfm3cTbo2RNmzYL774ettoqdRpIk4SQq8w0aBI8/DocfDqeeGjuNJEn6lSUqk337LfToAbVrhylUKhU7kSRJ+pXLeZkqSUKBmj8fnn4aNtkkdiJJkvQbTqIy1dNPwzPPwDHHwNFHx04jSZL+wBKViebODZvJa9WCO++MnUaSJK2Cy3mZ6KyzYN48ePJJl/EkScpQTqIyzeDBYSnvqKOgU6fYaSRJ0l+wRGWSefPCZvKNN4a77vLdeJIkZTCX8zLJOefAd9/BY4/BppvGTiNJklbDSVSmGDoUnngiHKp53HGx00iSpDWwRGWCRYvCMl716i7jSZKUJVzOywRXXAEzZ0LfvrDZZrHTSJKkNDiJiu2996BPH9h9dzjzzNhpJElSmixRMRUWQrduUKEC3HcfVKwYO5EkSUqTJSqmO+6A99+H88+H5s1jp5EkScVgiYrlyy/h8sthyy3DnihJkpRVLFExJEm42uXnn+Gee6BatdiJJElSMVmiYnjhBXjxxXAe1MEHx04jSZJKwBJV3pYsgXPPhfXXh969Y6eRJEkl5DlR5e2WW2D69FCg6tWLnUaSJJWQk6jyNH063HQTbL99uCdPkiRlLUtUeTr3XCgogDvvhMqVY6eRJElrwRJVXl56CYYMgWOPhQMOiJ1GkiStJUtUeSgoCMt3663nZnJJknKEG8vLw223wRdfwM03Q/36sdNIkqRS4CSqrM2eDTfcAI0bhz1RkiQpJ1iiytpll4WTyXv3hnXWiZ1GkiSVEktUWXr/fXjkETjoIGjXLnYaSZJUiixRZSVJ4Pzzwz/ffjukUnHzSJKkUmWJKivPPw9jxsCpp0Lz5rHTSJKkUmaJKgu//AIXXhjux7v22thpJElSGbBElYV+/WDqVLjkEqhTJ3YaSZJUBixRpW3BArjmGmjY0CMNJEnKYR62WdpuuQUWLoS+fWHddWOnkSRJZcRJVGmaPRv69IFmzeC442KnkSRJZcgSVZquuQaWLAknlFesGDuNJEkqQ5ao0jJlCjzwALRqBYcdFjuNJEkqY5ao0nLZZbB8Odx4owdrSpKUByxRpeG99+Cpp8LVLq1axU4jSZLKgSWqNFxySZg+3XBD7CSSJKmcWKLW1tix8PLL0KVLeFeeJEnKC5aotXXVVeGdeFddFTuJJEkqR5aotTFmTPhx4omw1Vax00iSpHJkiVobV18dplCXXho7iSRJKmeWqJJyCiVJUl6zRJWUUyhJkvKaJaoknEJJkpT3LFEl4RRKkqS8l1aJSqVSd6RSqRmpVCpJpVJNyzpURnvzzTCF6trVKZQkSXks3UnUYKAV8GUZZskON98cTie/+OLYSSRJUkSV0vmkJEnGAaTy/WLdTz6BIUPgyCNhu+1ip5EkSRGlkiRJ/5NTqRlAuyRJPv6LXz8POO83H1p/1KhRNQoLC9cqZKZo2rs3m40axdv//jc/bLtt7DilqqCggKpVq8aOoTXwOWUPn1X28Fllj/J4Vq1bt057YlSqJWoVnz8rSZLN0v4GmWzmzLAHap994JVXYqcpdSNGjKB169axY2gNfE7Zw2eVPXxW2aOcnlXaJcp356Xr9tuhsBB69YqdRJIkZQBLVDrmzYP774cdd4SDD46dRpIkZYB0jzjol0qlZgH1gVGpVGpq2cbKMHfeCT//HKZQ+b65XpIkAWmWqCRJeiZJUj9JkkpJktRJkmTrsg6WMQoKoF8/2GIL6NgxdhpJkpQhXM5bk8cfD8t555wTTimXJEnCErV6SQJ9+kD16vD3v8dOI0mSMoglanXGjIEPPwwFaoMNYqeRJEkZxBK1Ov/+d9hIfvbZsZNIkqQMY4n6K1OnwgsvQPv2XjQsSZL+xBL1V/r2DXuizj03dhJJkpSBLFGrsmgRPPwwNGsG++8fO40kScpAlqhVeeIJ+Okn6NHDwzUlSdIqWaJW5d57Yf31oUuX2EkkSVKGskT90bvvwnvvwfHHh/OhJEmSVsES9Uf33htezzgjbg5JkpTRLFG/9cMP4ZqX3XaDHXeMnUaSJGUwS9RvPfYY/PwzdO8eO4kkScpwlqgVkgTuuQdq1IBjj42dRpIkZThL1ArvvgsffQQnnADrrRc7jSRJynCWqBUeeSS8nnJK1BiSJCk7WKIACgrCAZvNmrmhXJIkpcUSBeGi4QUL4KSTPKFckiSlxRIF0L8/VKzoCeWSJCltlqg5c2D4cDj0UKhTJ3YaSZKUJSxRjz0Gy5fDySfHTiJJkrKIJWrgQNhwQ2jfPnYSSZKURfK7RH32GXz4IRx9NFSpEjuNJEnKIvldogYNCq+dO8fNIUmSsk7+lqgkgSefhE02gf32i51GkiRlmfwtUR9+CJ9/DsccA5UqxU4jSZKyTP6WqCefDK8u5UmSpBLIzxK1Yimvfn3Ya6/YaSRJUhbKzxI1fjzMmAGdOkGF/PyPQJIkrZ38bBCDB4fXTp3i5pAkSVkr/0pUksCzz8Jmm8Guu8ZOI0mSslT+lajPPoOpU+GII1zKkyRJJZZ/LeK558Jrhw5xc0iSpKyWnyWqRg0P2JQkSWslv0rU11+Hd+Yddhiss07sNJIkKYvlV4kaMiS8upQnSZLWUn6VqKFDwxUvrVvHTiJJkrJc/pSoggJ45RVo1Qo22CB2GkmSlOXyp0S99hosWQJt2sROIkmSckD+lKjhw8ProYfGzSFJknJCfpWounWhWbPYSSRJUg7IjxI1cyZ8+mmYQqVSsdNIkqQckB8lasSI8OpSniRJKiX5U6IqVICDD46dRJIk5YjcL1FFRTBmDOy0E2y0Uew0kiQpR+R+ifrkE/j+e9h//9hJJElSDsn9EjV2bHi1REmSpFKU+yVqzJiwH6pVq9hJJElSDsntEpUkYRK1445Qo0bsNJIkKYfkdon69FOYNw/22y92EkmSlGNyu0SNGRNe3Q8lSZJKWe6XqFQK9tkndhJJkpRjcrdEJQm8/jq0aAEbbhg7jSRJyjG5W6K++gq++Qb22CN2EkmSlINyt0S980543X33uDkkSVJOskRJkiSVQG6XqBo1oEmT2EkkSVIOys0StWwZvPce7LprOK1ckiSplOVmw/j4Y1iyxE3lkiSpzORmifrvf8PrbrvFzSFJknJWbpaoCRPC6047xc0hSZJyVm6WqA8/hJo1oV692EkkSVKOyr0SVVQUSlTz5uHKF0mSpDKQeyVqxgxYtCiUKEmSpDKSeyXqww/Da4sWcXNIkqSclnslauLE8OokSpIklaHcK1EffhgO2Nxhh9hJJElSDsu9EjVxYrjqpWrV2EkkSVIOy60StXgxfPEFNGsWO4kkScpxuVWipk4Nr9ttFzeHJEnKeblVoiZPDq+NG8fNIUmScp4lSpIkqQRys0Rts03cHJIkKeflXonadFOoUSN2EkmSlONyr0S5lCdJkspB7pSo77+H+fNdypMkSeUid0rUl1+G10aNosaQJEn5IXdK1FdfhdcGDeLmkCRJeSH3SlTDhnFzSJKkvJB7JcpJlCRJKge5V6Lq14+bQ5Ik5YXcKlG1asG668ZOIkmS8kBulSiX8iRJUjnJjRK1fDl8/bUlSpIklZvcKFHffguFhZYoSZJUbnKjRM2ZE17r1YubQ5Ik5Y3cKFHffRdea9eOm0OSJOWN3ChR8+aFV0uUJEkqJ7lRolZMomrViptDkiTljdwoUU6iJElSOcuNEuUkSpIklbPcKFHz5kGFCrDRRrGTSJKkPJEbJWrxYlhvvVCkJEmSykFutI4lS7wzT5IklavcKVFVq8ZOIUmS8khulKiCAidRkiSpXOVGiXISJUmSyllulCgnUZIkqZzlRolyEiVJkspZbpQoJ1GSJKmcZX+JShKPOJAkSeUu+0vUsmWhSLmcJ0mSylH2l6glS8KrkyhJklSOsr9EFRSEVydRkiSpHKVdolKp1DapVOrNVCo1OZVK/TeVSm1flsHS5iRKkiRFUJxJ1L3AfUmSNAZuAR4sm0jF5CRKkiRFkFaJSqVSmwA7AY/++qH/AFukUqlGZROrGJxESZKkCNKdRDUAZidJUgiQJEkCzAQallWwtDmJkiRJEaRCH1rDJ6VSOwMDkiTZ4TcfGw+cnyTJuN987DzgvN986foNGjSosWTFtKgMpAoLqbhkCUVVqlC0zjpl9n1yXVFRERUqZP/7DHKdzyl7+Kyyh88qe5THs5o3b97nSZJsm87npluiNgGmADWTJClMpVIpYA6wR5IkM1bzdbOSJNksvdiKqXbt2nz33XexY2gNfE7Zw2eVPXxW2aOcnlUq3U9Mq84lSTIX+ADo+uuHOgIzVlegJEmSclmlYnxud+CRVCp1CfAjcFLZRJIkScp8aZeoJEk+B/YswyyK6Mgjj4wdQWnwOWUPn1X28Fllj0x7Vu6kEwAdO3aMHUFp8DllD59V9vBZZY9Me1aWKEmSpBKwREmSJJWAJSqPFRQU0KFDBxo3bkyPHj049NBDmTFjRuxYWo1HH32UVCrFxx9/HDuKVmPp0qX069ePbbbZhh122IGuXbuu+YtU7kaMGMHOO+9Mz549adq0Kf37948dSb8655xzaNSo0Z/+9+7rr79mr732onHjxuy22258+umnEVMW7915ykHdunWjTZs2vPzyy0yZMoVu3brx8ssvx46lVXj//feZNGkSDRvGvyhAq9erVy9SqRSTJ08mlUoxZ86c2JH0B0mScPzxx/Pqq68yZ84cmjRpwrbbbstRRx1F9erVY8fLe0cffTQXXXQRrVq1+t3H77jjDs4991xOPvlkBg8ezKmnnspbb70VKaWTqLxWtWpV2rZtSzg7FfbYYw+mTZsWOZVWZenSpfTs2ZOzzjpr5fNSZlq8eDEPP/wwp5xyyspnVbdu3cip9FcWLlwIwI8//kjNmjWpUqVK5EQC2Hfffalfv/7vPjZ37lymTp26crLbsWNHpk+fHnUFxRKlle644w7at28fO4ZW4YorrqBr167UqVMndhStwRdffEHNmjV54okn2GWXXdhnn30YPXp07Fj6g1QqxVNPPcVRRx3FiSeeSKtWrejfvz/reH1Yxvrqq6+oWbMmlSqFRbRUKkXDhg2ZOXNmtEyWKAHw5JNPMmXKFK6//vrYUfQHb731FuPHj6dHjx6xoygNy5YtY9q0aTRs2JB3332XO++8k86dO3utSIYpLCzkxhtv5Pnnn2fAgAGMHj2ak046ifnz58eOpmJI5+q6smSJEr179+aNN95g2LBhVKtWLXYc/cHYsWOZNGkSW2yxBSeeeCKzZs2idevWDBs2LHY0rcLmm29OhQoVOOCAAwBo0aIFW2yxBZ988knkZPqtCRMmMHv2bPbee28Adt11V+rVq8fEiRMjJ9NfadCgAfPmzaOwsBAIBeqrr76Kuk/UEpXnbr/9dp544gluuOEGNtxww9hxtAq9evVi9uzZzJgxgwEDBlC/fn1GjBhBmzZtYkfTKtSqVYuDDjqI9957D4Avv/yS6dOn06RJk8jJ9FsNGjRg1qxZfP755wBMnTqVL774gsaNG0dOpr+yySabsNVWW/Hoo48C8J///IdGjRrRqFGjaJlSZTkKS6VSs5Ik2azMvoHWyqxZs2jQoAFbbrklANWrV6dKlSq88847kZPpr4wYMYLu3bvz4osv0rRp09hx9BemTZtGhw4dSJKEihUrcuWVV2bcdRVi5f+BXLx4Meuvvz6XXHIJnTt3jh1LQM+ePXn++ef55ptvqFWrFuuvvz5Tp07lgQce4MEHH+T7779ngw02oH///uywww6l/e3TfveOJUpA+Mu5devWsWNoDXxO2cNnlT18VtmjnJ5V2iXK5TxJkqQSsERJkiSVgCVKkiSpBCxRkiRJJWCJkiRJKgFLlCRJUglYoiRJkkrAEiVJklQCZX3Y5qQkSbYts28gSZIUSZmWKEmSpFzlcp4kSVIJWKIkSZJKwBIlSZJUApYoSZKkErBESZIklYAlSpIkqQQsUZIkSSVgiZIkSSoBS5QkSVIJ/D+j7ZAegci6uwAAAABJRU5ErkJggg==\n",
      "text/plain": [
       "<Figure size 720x720 with 1 Axes>"
      ]
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    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Informe o chute para X0:0\n",
      "Informe a tolerancia(formato 0.xxxx1):0.01\n",
      "Informe o número maximo de iterações:10\n",
      "F( 0.0 )= -3.0\n",
      "1 ª Previsão:  (6.123233995736766e-17-1j)\n",
      "(6.123233995736766e-17-1j)\n"
     ]
    },
    {
     "ename": "TypeError",
     "evalue": "can't convert complex to float",
     "output_type": "error",
     "traceback": [
      "\u001b[1;31m---------------------------------------------------------------------------\u001b[0m",
      "\u001b[1;31mTypeError\u001b[0m                                 Traceback (most recent call last)",
      "\u001b[1;32m<ipython-input-18-6e1d4b9cec3e>\u001b[0m in \u001b[0;36m<module>\u001b[1;34m()\u001b[0m\n\u001b[0;32m     70\u001b[0m \u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0;32m     71\u001b[0m \u001b[0mprint\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;34m\"F(\"\u001b[0m\u001b[1;33m,\u001b[0m\u001b[0mx0\u001b[0m\u001b[1;33m,\u001b[0m\u001b[1;34m\")=\"\u001b[0m\u001b[1;33m,\u001b[0m\u001b[0mfuncao1\u001b[0m\u001b[1;33m(\u001b[0m\u001b[0mx0\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[1;32m---> 72\u001b[1;33m \u001b[0mmetodoDoPontoFixo\u001b[0m\u001b[1;33m(\u001b[0m\u001b[0mlim1\u001b[0m\u001b[1;33m,\u001b[0m \u001b[0mlim2\u001b[0m\u001b[1;33m,\u001b[0m \u001b[0mx0\u001b[0m\u001b[1;33m,\u001b[0m \u001b[0mtol\u001b[0m\u001b[1;33m,\u001b[0m \u001b[0mnMax\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0m",
      "\u001b[1;32m<ipython-input-18-6e1d4b9cec3e>\u001b[0m in \u001b[0;36mmetodoDoPontoFixo\u001b[1;34m(pos1, pos2, x0, tol, nMax)\u001b[0m\n\u001b[0;32m     61\u001b[0m             \u001b[0mprint\u001b[0m\u001b[1;33m(\u001b[0m\u001b[0mi\u001b[0m\u001b[1;33m+\u001b[0m\u001b[1;36m1\u001b[0m\u001b[1;33m,\u001b[0m\u001b[1;34m\"ª Previsão: \"\u001b[0m\u001b[1;33m,\u001b[0m\u001b[0mfuncao2\u001b[0m\u001b[1;33m(\u001b[0m\u001b[0mx0\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0;32m     62\u001b[0m             \u001b[0mprint\u001b[0m\u001b[1;33m(\u001b[0m\u001b[0mp\u001b[0m\u001b[1;33m-\u001b[0m\u001b[0mx0\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[1;32m---> 63\u001b[1;33m             \u001b[1;32mif\u001b[0m\u001b[1;33m(\u001b[0m\u001b[0mmath\u001b[0m\u001b[1;33m.\u001b[0m\u001b[0mfabs\u001b[0m\u001b[1;33m(\u001b[0m\u001b[0mp\u001b[0m\u001b[1;33m-\u001b[0m\u001b[0mx0\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m<\u001b[0m\u001b[0mtol\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m:\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0m\u001b[0;32m     64\u001b[0m                 \u001b[1;32mreturn\u001b[0m \u001b[0mp\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0;32m     65\u001b[0m             \u001b[0mi\u001b[0m\u001b[1;33m+=\u001b[0m\u001b[1;36m1\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n",
      "\u001b[1;31mTypeError\u001b[0m: can't convert complex to float"
     ]
    }
   ],
   "source": [
    "import matplotlib.pyplot as plt\n",
    "import math\n",
    "\n",
    "#######################\n",
    "##  GRAFICO DE G(X)  ##\n",
    "#######################\n",
    "\n",
    "#variáveis utilizadas na plotagem do gráfico\n",
    "def funcao(x):\n",
    "    return (2*x-1)**(1/2)\n",
    "\n",
    "#declaração das variáveis\n",
    "j=0\n",
    "lim1 = 0.5 #limite inferior\n",
    "lim2 = 10 #limite superior\n",
    "\n",
    "#variáveis utilizadas na plotagem do gráfico\n",
    "i=lim1 \n",
    "valoresFuncao = []\n",
    "valoresX = []\n",
    "\n",
    "#Construção do gráfico da função para os intervalos designados\n",
    "while i<lim2:\n",
    "    valoresX.append(i)\n",
    "    valoresFuncao.append(funcao(i))\n",
    "    i = i + 0.01 #aumentar o número de pontos, para definir melhor a tragetória do gráfico\n",
    "\n",
    "#excução da plotagem\n",
    "#configuração dos eixos retirado de:http://www.scipy-lectures.org/intro/matplotlib/matplotlib.html\n",
    "plt.figure(figsize=(9,9), dpi=80)\n",
    "ax = plt.gca()\n",
    "\n",
    "ax.spines['right'].set_color('none')\n",
    "ax.spines['top'].set_color('none')\n",
    "ax.xaxis.set_ticks_position('bottom')\n",
    "ax.spines['bottom'].set_position(('data',0))\n",
    "ax.yaxis.set_ticks_position('left')\n",
    "ax.spines['left'].set_position(('data',0))\n",
    "\n",
    "\n",
    "plt.grid(True, which='both')\n",
    "plt.plot(valoresX, valoresFuncao, color=\"red\", linestyle=\"-\", label=\"f(x)=(2x-1)^(1/2)\")\n",
    "plt.legend(loc='upper left')\n",
    "\n",
    "plt.show()\n",
    "\n",
    "### INÍCIO DA APLICAÇÃO DO MÉTODO DO PONTO FIXO #####\n",
    "x0 = float(input(\"Informe o chute para X0:\")) #chute para x0\n",
    "tol  = float(input(\"Informe a tolerancia(formato 0.xxxx1):\"))#tolerancia\n",
    "nMax = float(input(\"Informe o número maximo de iterações:\"))#iteração\n",
    "\n",
    "def funcao2(x):\n",
    "    return (1/((2*x-1)**(1/2)))\n",
    "\n",
    "def metodoDoPontoFixo(pos1, pos2, x0, tol, nMax):    \n",
    "    if(funcao1(pos1)*funcao1(pos2)<0):\n",
    "        i=0\n",
    "        while i<=nMax:\n",
    "            p = funcao2(x0)\n",
    "            \n",
    "            print(i+1,\"ª Previsão: \",funcao2(x0))\n",
    "            print(p-x0)\n",
    "            if(math.fabs(p-x0)<tol):\n",
    "                return p\n",
    "            i+=1\n",
    "            x0 = p\n",
    "        print(\"O procedimento não foi bem sucedido!\")\n",
    "    else:\n",
    "        print(\"Não possui raiz\")\n",
    "    \n",
    "print(\"F(\",x0,\")=\",funcao1(x0))\n",
    "metodoDoPontoFixo(lim1, lim2, x0, tol, nMax)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Exercicio 23 - 2.2\n",
    "\n",
    "a) Mostre que se $A$ for qualquer número positivo, então a sequência definida por\n",
    "\n",
    "<center>$ x_{n}=\\frac{1}{2}x_{n-1}+\\frac{A}{2x_{n-1}},$ para $n \\geq 1, $</center>\n",
    "\n",
    "convergirá para $\\sqrt{A}$, sempre que $x_{0} > 0$\n",
    "\n",
    "b) O que ocorre se $x_{0} < 0$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Resposta\n",
    "\n",
    "Tomamos:\n",
    "\n",
    "$$g(x) = \\frac{x}{2} + \\frac{A}{2x}$$.\n",
    "\n",
    "Observe que $g(\\sqrt{A})=\\sqrt{A}$. Pois:\n",
    "\n",
    "<p>$$g(\\sqrt{A})=\\frac{\\sqrt{A}}{2} + \\frac{A}{2\\sqrt{A}}$$</p>\n",
    "<p>$$g(\\sqrt{A})=\\frac{4{A}}{4\\sqrt{A}}$$</p>\n",
    "<p>$$g(\\sqrt{A})=\\sqrt{A}$$</p>\n",
    "\n",
    "\n",
    "Observe que a $g^{'}(x) = \\frac{1}{2} - \\frac{A}{2x^{2}}$ se $x \\neq 0$, e consequentemente $g^{'}(x) > 0$ se $x>\\sqrt(A)$\n",
    "\n",
    "Teremos que compravar para as seguintes situações $x_0=\\sqrt{A}$, e $x_0>A$. Assim:\n",
    "\n",
    "i) Se $x_{0}=\\sqrt{A}$, então $x_{n} = \\sqrt{A}$, $\\forall$ $n$ e $$\\lim_{n \\to \\infty } x_{n}= \\sqrt{A}$$\n",
    "\n",
    "ii) Se $x_{0}>\\sqrt{A}$, então:\n",
    "\n",
    "<center>$x_{1} - \\sqrt{A} = g(x_{0}) - g(\\sqrt{A})$, implica que</center>\n",
    "\n",
    "$$g^{'}(\\xi)(x_{0}-\\sqrt(A))>0$$\n",
    "\n",
    "além disso:\n",
    "\n",
    "$$ x_{1}=\\frac{x_{0}}{2}+\\frac{A}{2x_{0}} < \\frac{x_{0}}{2} + \\frac{A}{2\\sqrt{A}}$$\n",
    "\n",
    "<b>finalizando....<b>\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**(25) Substitua, no teorema 2.4 a consideração de que \"um número positivo $k<1$ existe, com $|g'(x)|\\leqslant k$ \" por \"g satisfaz a condição de Lipschitz no intervalo $[a,b]$ com a constante de Lipschit $L<1$\". (Veja o exercício 28 da seção 1.1). Mostre que as conclusões deste teorema continuam válidas.**\n",
    "\n",
    "Diz-se que uma função $g(x)$ satisfaz a condição de Lipschitz em um intervalo $[a,b]$ com a constante L quando para todo $x,y \\in [a,b]$ o valor de |$g(x)-g(y)$|$\\leqslant L$|$x-y$|.\n",
    "\n",
    "Primeiramente é necessário observar que estamos tratando de módulos, logo, o valor da constante L será maior que 0. Além disso, como o próprio enunciado da condição de Lipschitz garante $g(x)$ é contínua no intervalo $[a,b]$, e x e y pertencem a esse mesmo intervalo, pode-se então considerar que $y=x+\\Delta x$, e reorganizar a função, de forma a obter:\n",
    "\n",
    "$$|g(x)-g(x+\\Delta x)|\\leqslant L*|x-(x+\\Delta x)|$$\n",
    "\n",
    "$$|g(x)-g(x+\\Delta x)|\\leqslant L*|-\\Delta x|$$\n",
    "\n",
    "$$|\\frac{g(x)-g(x+\\Delta x)}{-\\Delta x}|\\leqslant L$$\n",
    "\n",
    "$$|\\frac{g(x+\\Delta x)-g(x)}{\\Delta x}|\\leqslant L$$\n",
    "\n",
    "Observa-se que, à medida que o valor de y se aproxima de x, o valor de $\\Delta x$ aproxima-se de zero, e o termo $|\\frac{g(x+\\Delta x)-g(x)}{\\Delta x}|$ aproxima-se de $|g'(x)|$.\n",
    "\n",
    "Logo, dizer que \"*g satisfaz a condição de Lipschitz no intervalo $[a,b]$ com a constante de Lipschit $L<1$*\" equivale à condição imposta pelo teorema 2.4: \"*um número positivo $k<1$ existe, com $|g'(x)|\\leqslant k$* \" e, portanto, suas conclusões seguem válidas e aplicáveis mesmo com a troca de condições."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Referências Bibliográficas**\n",
    "\n",
    "BURDEN, R.L.;FAIRES,D.J.;BURDEN, A.M. **Numerical Analysis**. 10 ed. Boston, MA: Cengage Learning, 2014, cap. 2, p.47-66. ISBN 978-1-305-25366-7"
   ]
  }
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