Exercícios Resolvidos - Capítulo 2.ipynb

{
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   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Conjunto de exercícios 2.1 #"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**(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": "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": [
    "**(5) Utilize o método da bissecção para determinar as soluções com precisão de $10^{-5}$ para os problemas a seguir.**\n",
    "\n",
    "**a) $x-2^{-x} = 0$, para $0 \\leqslant x \\leqslant  1$**\n",
    "\n",
    "**b) $e^{x} - x^{2} + 3x - 2 = 0$, para $0 \\leqslant x \\leqslant  1$**\n",
    "\n",
    "**c) $2x$ $cos(2x) - {(x + 1)}^2 = 0$, para $-3 \\leqslant x \\leqslant  -2$ e $-1 \\leqslant x \\leqslant  0$**\n",
    "\n",
    "**d) $x$ $cosx - 2x^{2} + 3x - 1 = 0$, para $0,2 \\leqslant x \\leqslant  0,3$ e $1,2 \\leqslant x \\leqslant  1,3$**\n"
   ]
  },
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   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Exercícios 2.1\n",
      "\n",
      "Questão 5.a) x-2**(-x), para 0 <= x <= 1\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 -0.2071067812 \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 0.1553964425 \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 -0.0234197773 \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 0.0665710940 \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 0.0217245214 \t 0.0312500000 \t 0.0476190476   f(a)*f(p)<0.0, b=p\n",
      "6 \t 0.6250000000 \t 0.6562500000 \t 0.6406250000 \t -0.0008100080 \t 0.0156250000 \t 0.0243902439   f(a)*f(p)>0.0, a=p\n",
      "7 \t 0.6406250000 \t 0.6562500000 \t 0.6484375000 \t 0.0104666108 \t 0.0078125000 \t 0.0120481928   f(a)*f(p)<0.0, b=p\n",
      "8 \t 0.6406250000 \t 0.6484375000 \t 0.6445312500 \t 0.0048306462 \t 0.0039062500 \t 0.0060606061   f(a)*f(p)<0.0, b=p\n",
      "9 \t 0.6406250000 \t 0.6445312500 \t 0.6425781250 \t 0.0020109061 \t 0.0019531250 \t 0.0030395137   f(a)*f(p)<0.0, b=p\n",
      "10 \t 0.6406250000 \t 0.6425781250 \t 0.6416015625 \t 0.0006005959 \t 0.0009765625 \t 0.0015220700   f(a)*f(p)<0.0, b=p\n",
      "11 \t 0.6406250000 \t 0.6416015625 \t 0.6411132812 \t -0.0001046693 \t 0.0004882812 \t 0.0007616146   f(a)*f(p)>0.0, a=p\n",
      "12 \t 0.6411132812 \t 0.6416015625 \t 0.6413574219 \t 0.0002479724 \t 0.0002441406 \t 0.0003806624   f(a)*f(p)<0.0, b=p\n",
      "13 \t 0.6411132812 \t 0.6413574219 \t 0.6412353516 \t 0.0000716538 \t 0.0001220703 \t 0.0001903674   f(a)*f(p)<0.0, b=p\n",
      "14 \t 0.6411132812 \t 0.6412353516 \t 0.6411743164 \t -0.0000165072 \t 0.0000610352 \t 0.0000951928   f(a)*f(p)>0.0, a=p\n",
      "15 \t 0.6411743164 \t 0.6412353516 \t 0.6412048340 \t 0.0000275735 \t 0.0000305176 \t 0.0000475941   f(a)*f(p)<0.0, b=p\n",
      "16 \t 0.6411743164 \t 0.6412048340 \t 0.6411895752 \t 0.0000055332 \t 0.0000152588 \t 0.0000237976   f(a)*f(p)<0.0, b=p\n",
      "17 \t 0.6411743164 \t 0.6411895752 \t 0.6411819458 \t -0.0000054870 \t 0.0000076294 \t 0.0000118990\n",
      "A raiz da função, dentro dos parâmetros de tolerância, é 0.6411819458007812 \n",
      "\n",
      "Questão 5.b) exp(x)-(x**2)+3*x-2, para 0 <= x <= 1\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 0.8987212707 \t 0.5000000000 \t 1.0000000000   f(a)*f(p)<0.0, b=p\n",
      "2 \t 0.0000000000 \t 0.5000000000 \t 0.2500000000 \t -0.0284745833 \t 0.2500000000 \t 1.0000000000   f(a)*f(p)>0.0, a=p\n",
      "3 \t 0.2500000000 \t 0.5000000000 \t 0.3750000000 \t 0.4393664146 \t 0.1250000000 \t 0.3333333333   f(a)*f(p)<0.0, b=p\n",
      "4 \t 0.2500000000 \t 0.3750000000 \t 0.3125000000 \t 0.2066816912 \t 0.0625000000 \t 0.2000000000   f(a)*f(p)<0.0, b=p\n",
      "5 \t 0.2500000000 \t 0.3125000000 \t 0.2812500000 \t 0.0894331962 \t 0.0312500000 \t 0.1111111111   f(a)*f(p)<0.0, b=p\n",
      "6 \t 0.2500000000 \t 0.2812500000 \t 0.2656250000 \t 0.0305642341 \t 0.0156250000 \t 0.0588235294   f(a)*f(p)<0.0, b=p\n",
      "7 \t 0.2500000000 \t 0.2656250000 \t 0.2578125000 \t 0.0010663677 \t 0.0078125000 \t 0.0303030303   f(a)*f(p)<0.0, b=p\n",
      "8 \t 0.2500000000 \t 0.2578125000 \t 0.2539062500 \t -0.0136986837 \t 0.0039062500 \t 0.0153846154   f(a)*f(p)>0.0, a=p\n",
      "9 \t 0.2539062500 \t 0.2578125000 \t 0.2558593750 \t -0.0063148068 \t 0.0019531250 \t 0.0076335878   f(a)*f(p)>0.0, a=p\n",
      "10 \t 0.2558593750 \t 0.2578125000 \t 0.2568359375 \t -0.0026238823 \t 0.0009765625 \t 0.0038022814   f(a)*f(p)>0.0, a=p\n",
      "11 \t 0.2568359375 \t 0.2578125000 \t 0.2573242188 \t -0.0007786731 \t 0.0004882812 \t 0.0018975332   f(a)*f(p)>0.0, a=p\n",
      "12 \t 0.2573242188 \t 0.2578125000 \t 0.2575683594 \t 0.0001438683 \t 0.0002441406 \t 0.0009478673   f(a)*f(p)<0.0, b=p\n",
      "13 \t 0.2573242188 \t 0.2575683594 \t 0.2574462891 \t -0.0003173971 \t 0.0001220703 \t 0.0004741584   f(a)*f(p)>0.0, a=p\n",
      "14 \t 0.2574462891 \t 0.2575683594 \t 0.2575073242 \t -0.0000867631 \t 0.0000610352 \t 0.0002370230   f(a)*f(p)>0.0, a=p\n",
      "15 \t 0.2575073242 \t 0.2575683594 \t 0.2575378418 \t 0.0000285530 \t 0.0000305176 \t 0.0001184975   f(a)*f(p)<0.0, b=p\n",
      "16 \t 0.2575073242 \t 0.2575378418 \t 0.2575225830 \t -0.0000291050 \t 0.0000152588 \t 0.0000592522   f(a)*f(p)>0.0, a=p\n",
      "17 \t 0.2575225830 \t 0.2575378418 \t 0.2575302124 \t -0.0000002760 \t 0.0000076294 \t 0.0000296252\n",
      "A raiz da função, dentro dos parâmetros de tolerância, é 0.25753021240234375 \n",
      "\n",
      "Questão 5.c) parte 1: 2*x*cos(2*x)-(x+1)**2, para -3 <= x <= -2\n",
      "n\t a\t         b\t         p\t         f(p)\t         (b-a)/2\t e. relat. p\tMundança\n",
      "1 \t -3.0000000000 \t -2.0000000000 \t -2.5000000000 \t -3.6683109273 \t 0.5000000000 \t 1.0000000000   f(a)*f(p)>0.0, a=p\n",
      "2 \t -2.5000000000 \t -2.0000000000 \t -2.2500000000 \t -0.6139189026 \t 0.2500000000 \t 0.1111111111   f(a)*f(p)>0.0, a=p\n",
      "3 \t -2.2500000000 \t -2.0000000000 \t -2.1250000000 \t 0.6302468321 \t 0.1250000000 \t 0.0588235294   f(a)*f(p)<0.0, b=p\n",
      "4 \t -2.2500000000 \t -2.1250000000 \t -2.1875000000 \t 0.0380755317 \t 0.0625000000 \t 0.0285714286   f(a)*f(p)<0.0, b=p\n",
      "5 \t -2.2500000000 \t -2.1875000000 \t -2.2187500000 \t -0.2808361755 \t 0.0312500000 \t 0.0140845070   f(a)*f(p)>0.0, a=p\n",
      "6 \t -2.2187500000 \t -2.1875000000 \t -2.2031250000 \t -0.1195568145 \t 0.0156250000 \t 0.0070921986   f(a)*f(p)>0.0, a=p\n",
      "7 \t -2.2031250000 \t -2.1875000000 \t -2.1953125000 \t -0.0402785142 \t 0.0078125000 \t 0.0035587189   f(a)*f(p)>0.0, a=p\n",
      "8 \t -2.1953125000 \t -2.1875000000 \t -2.1914062500 \t -0.0009851950 \t 0.0039062500 \t 0.0017825312   f(a)*f(p)>0.0, a=p\n",
      "9 \t -2.1914062500 \t -2.1875000000 \t -2.1894531250 \t 0.0185743370 \t 0.0019531250 \t 0.0008920607   f(a)*f(p)<0.0, b=p\n",
      "10 \t -2.1914062500 \t -2.1894531250 \t -2.1904296875 \t 0.0088018514 \t 0.0009765625 \t 0.0004458315   f(a)*f(p)<0.0, b=p\n",
      "11 \t -2.1914062500 \t -2.1904296875 \t -2.1909179688 \t 0.0039101468 \t 0.0004882812 \t 0.0002228661   f(a)*f(p)<0.0, b=p\n",
      "12 \t -2.1914062500 \t -2.1909179688 \t -2.1911621094 \t 0.0014629304 \t 0.0002441406 \t 0.0001114206   f(a)*f(p)<0.0, b=p\n",
      "13 \t -2.1914062500 \t -2.1911621094 \t -2.1912841797 \t 0.0002389813 \t 0.0001220703 \t 0.0000557072   f(a)*f(p)<0.0, b=p\n",
      "14 \t -2.1914062500 \t -2.1912841797 \t -2.1913452148 \t -0.0003730784 \t 0.0000610352 \t 0.0000278528   f(a)*f(p)>0.0, a=p\n",
      "15 \t -2.1913452148 \t -2.1912841797 \t -2.1913146973 \t -0.0000670414 \t 0.0000305176 \t 0.0000139266   f(a)*f(p)>0.0, a=p\n",
      "16 \t -2.1913146973 \t -2.1912841797 \t -2.1912994385 \t 0.0000859717 \t 0.0000152588 \t 0.0000069634   f(a)*f(p)<0.0, b=p\n",
      "17 \t -2.1913146973 \t -2.1912994385 \t -2.1913070679 \t 0.0000094656 \t 0.0000076294 \t 0.0000034817\n",
      "A raiz da função, dentro dos parâmetros de tolerância, é -2.1913070678710938 \n",
      "\n",
      "Questão 5.c) parte 2: 2*x*cos(2*x)-(x+1)**2, para -1 <= x <= 0\n",
      "n\t a\t         b\t         p\t         f(p)\t         (b-a)/2\t e. relat. p\tMundança\n",
      "1 \t -1.0000000000 \t 0.0000000000 \t -0.5000000000 \t -0.7903023059 \t 0.5000000000 \t 1.0000000000   f(a)*f(p)<0.0, b=p\n",
      "2 \t -1.0000000000 \t -0.5000000000 \t -0.7500000000 \t -0.1686058025 \t 0.2500000000 \t 0.3333333333   f(a)*f(p)<0.0, b=p\n",
      "3 \t -1.0000000000 \t -0.7500000000 \t -0.8750000000 \t 0.2963055974 \t 0.1250000000 \t 0.1428571429   f(a)*f(p)>0.0, a=p\n",
      "4 \t -0.8750000000 \t -0.7500000000 \t -0.8125000000 \t 0.0528815944 \t 0.0625000000 \t 0.0769230769   f(a)*f(p)>0.0, a=p\n",
      "5 \t -0.8125000000 \t -0.7500000000 \t -0.7812500000 \t -0.0608144244 \t 0.0312500000 \t 0.0400000000   f(a)*f(p)<0.0, b=p\n",
      "6 \t -0.8125000000 \t -0.7812500000 \t -0.7968750000 \t -0.0046805612 \t 0.0156250000 \t 0.0196078431   f(a)*f(p)<0.0, b=p\n",
      "7 \t -0.8125000000 \t -0.7968750000 \t -0.8046875000 \t 0.0239251797 \t 0.0078125000 \t 0.0097087379   f(a)*f(p)>0.0, a=p\n",
      "8 \t -0.8046875000 \t -0.7968750000 \t -0.8007812500 \t 0.0095780659 \t 0.0039062500 \t 0.0048780488   f(a)*f(p)>0.0, a=p\n",
      "9 \t -0.8007812500 \t -0.7968750000 \t -0.7988281250 \t 0.0024376411 \t 0.0019531250 \t 0.0024449878   f(a)*f(p)>0.0, a=p\n",
      "10 \t -0.7988281250 \t -0.7968750000 \t -0.7978515625 \t -0.0011242441 \t 0.0009765625 \t 0.0012239902   f(a)*f(p)<0.0, b=p\n",
      "11 \t -0.7988281250 \t -0.7978515625 \t -0.7983398438 \t 0.0006560033 \t 0.0004882812 \t 0.0006116208   f(a)*f(p)>0.0, a=p\n",
      "12 \t -0.7983398438 \t -0.7978515625 \t -0.7980957031 \t -0.0002342943 \t 0.0002441406 \t 0.0003059039   f(a)*f(p)<0.0, b=p\n",
      "13 \t -0.7983398438 \t -0.7980957031 \t -0.7982177734 \t 0.0002108110 \t 0.0001220703 \t 0.0001529286   f(a)*f(p)>0.0, a=p\n",
      "14 \t -0.7982177734 \t -0.7980957031 \t -0.7981567383 \t -0.0000117525 \t 0.0000610352 \t 0.0000764701   f(a)*f(p)<0.0, b=p\n",
      "15 \t -0.7982177734 \t -0.7981567383 \t -0.7981872559 \t 0.0000995265 \t 0.0000305176 \t 0.0000382336   f(a)*f(p)>0.0, a=p\n",
      "16 \t -0.7981872559 \t -0.7981567383 \t -0.7981719971 \t 0.0000438863 \t 0.0000152588 \t 0.0000191172   f(a)*f(p)>0.0, a=p\n",
      "17 \t -0.7981719971 \t -0.7981567383 \t -0.7981643677 \t 0.0000160667 \t 0.0000076294 \t 0.0000095587\n",
      "A raiz da função, dentro dos parâmetros de tolerância, é -0.7981643676757812 \n",
      "\n",
      "Questão 5.d) parte 1: x*cos(x)-2*(x**2)+(3*x)-1, para 0.2 <= x <= 0.3\n",
      "n\t a\t         b\t         p\t         f(p)\t         (b-a)/2\t e. relat. p\tMundança\n"
     ]
    },
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "1 \t 0.2000000000 \t 0.3000000000 \t 0.2500000000 \t -0.1327718946 \t 0.0500000000 \t 1.0000000000   f(a)*f(p)>0.0, a=p\n",
      "2 \t 0.2500000000 \t 0.3000000000 \t 0.2750000000 \t -0.0615830707 \t 0.0250000000 \t 0.0909090909   f(a)*f(p)>0.0, a=p\n",
      "3 \t 0.2750000000 \t 0.3000000000 \t 0.2875000000 \t -0.0271127188 \t 0.0125000000 \t 0.0434782609   f(a)*f(p)>0.0, a=p\n",
      "4 \t 0.2875000000 \t 0.3000000000 \t 0.2937500000 \t -0.0101609593 \t 0.0062500000 \t 0.0212765957   f(a)*f(p)>0.0, a=p\n",
      "5 \t 0.2937500000 \t 0.3000000000 \t 0.2968750000 \t -0.0017562321 \t 0.0031250000 \t 0.0105263158   f(a)*f(p)>0.0, a=p\n",
      "6 \t 0.2968750000 \t 0.3000000000 \t 0.2984375000 \t 0.0024283062 \t 0.0015625000 \t 0.0052356021   f(a)*f(p)<0.0, b=p\n",
      "7 \t 0.2968750000 \t 0.2984375000 \t 0.2976562500 \t 0.0003375236 \t 0.0007812500 \t 0.0026246719   f(a)*f(p)<0.0, b=p\n",
      "8 \t 0.2968750000 \t 0.2976562500 \t 0.2972656250 \t -0.0007089827 \t 0.0003906250 \t 0.0013140604   f(a)*f(p)>0.0, a=p\n",
      "9 \t 0.2972656250 \t 0.2976562500 \t 0.2974609375 \t -0.0001856366 \t 0.0001953125 \t 0.0006565988   f(a)*f(p)>0.0, a=p\n",
      "10 \t 0.2974609375 \t 0.2976562500 \t 0.2975585938 \t 0.0000759667 \t 0.0000976562 \t 0.0003281917   f(a)*f(p)<0.0, b=p\n",
      "11 \t 0.2974609375 \t 0.2975585938 \t 0.2975097656 \t -0.0000548292 \t 0.0000488281 \t 0.0001641228   f(a)*f(p)>0.0, a=p\n",
      "12 \t 0.2975097656 \t 0.2975585938 \t 0.2975341797 \t 0.0000105702 \t 0.0000244141 \t 0.0000820546   f(a)*f(p)<0.0, b=p\n",
      "13 \t 0.2975097656 \t 0.2975341797 \t 0.2975219727 \t -0.0000221291 \t 0.0000122070 \t 0.0000410290   f(a)*f(p)>0.0, a=p\n",
      "14 \t 0.2975219727 \t 0.2975341797 \t 0.2975280762 \t -0.0000057793 \t 0.0000061035 \t 0.0000205141\n",
      "A raiz da função, dentro dos parâmetros de tolerância, é 0.29752807617187504 \n",
      "\n",
      "Questão 5.d) parte 2: x*cos(x)-2*(x**2)+(3*x)-1, para 1.2 <= x <= 1.3\n",
      "n\t a\t         b\t         p\t         f(p)\t         (b-a)/2\t e. relat. p\tMundança\n",
      "1 \t 1.2000000000 \t 1.3000000000 \t 1.2500000000 \t 0.0191529530 \t 0.0500000000 \t 1.0000000000   f(a)*f(p)>0.0, a=p\n",
      "2 \t 1.2500000000 \t 1.3000000000 \t 1.2750000000 \t -0.0545853516 \t 0.0250000000 \t 0.0196078431   f(a)*f(p)<0.0, b=p\n",
      "3 \t 1.2500000000 \t 1.2750000000 \t 1.2625000000 \t -0.0172248917 \t 0.0125000000 \t 0.0099009901   f(a)*f(p)<0.0, b=p\n",
      "4 \t 1.2500000000 \t 1.2625000000 \t 1.2562500000 \t 0.0010868925 \t 0.0062500000 \t 0.0049751244   f(a)*f(p)>0.0, a=p\n",
      "5 \t 1.2562500000 \t 1.2625000000 \t 1.2593750000 \t -0.0080382883 \t 0.0031250000 \t 0.0024813896   f(a)*f(p)<0.0, b=p\n",
      "6 \t 1.2562500000 \t 1.2593750000 \t 1.2578125000 \t -0.0034680196 \t 0.0015625000 \t 0.0012422360   f(a)*f(p)<0.0, b=p\n",
      "7 \t 1.2562500000 \t 1.2578125000 \t 1.2570312500 \t -0.0011886439 \t 0.0007812500 \t 0.0006215040   f(a)*f(p)<0.0, b=p\n",
      "8 \t 1.2562500000 \t 1.2570312500 \t 1.2566406250 \t -0.0000503958 \t 0.0003906250 \t 0.0003108486   f(a)*f(p)<0.0, b=p\n",
      "9 \t 1.2562500000 \t 1.2566406250 \t 1.2564453125 \t 0.0005183683 \t 0.0001953125 \t 0.0001554485   f(a)*f(p)>0.0, a=p\n",
      "10 \t 1.2564453125 \t 1.2566406250 \t 1.2565429688 \t 0.0002340162 \t 0.0000976563 \t 0.0000777182   f(a)*f(p)>0.0, a=p\n",
      "11 \t 1.2565429688 \t 1.2566406250 \t 1.2565917969 \t 0.0000918177 \t 0.0000488281 \t 0.0000388576   f(a)*f(p)>0.0, a=p\n",
      "12 \t 1.2565917969 \t 1.2566406250 \t 1.2566162109 \t 0.0000207128 \t 0.0000244141 \t 0.0000194284   f(a)*f(p)>0.0, a=p\n",
      "13 \t 1.2566162109 \t 1.2566406250 \t 1.2566284180 \t -0.0000148410 \t 0.0000122070 \t 0.0000097141   f(a)*f(p)<0.0, b=p\n",
      "14 \t 1.2566162109 \t 1.2566284180 \t 1.2566223145 \t 0.0000029360 \t 0.0000061035 \t 0.0000048571\n",
      "A raiz da função, dentro dos parâmetros de tolerância, é 1.256622314453125 \n",
      "\n"
     ]
    }
   ],
   "source": [
    "#Questão 5 - Exercícios 2.1\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 bisseccao (f, a, b, TOL, ep, efp, N0): # Função método da Bissecção   \n",
    "    #Argumentos: \n",
    "    #f: função f(x) a ser encontrada a raiz (em string)\n",
    "    # a e b: extremos do intervalo [a, b] contendo a raíz, com f(a) * f(b) < 0\n",
    "    # TOL: Tolerância para o tamanho do intervalo (b-a)/2 | ep: Tolerância para o erro relativo de p\n",
    "    # efp: Tolerância para o erro de f(p) | N0: Número máximo de iterações\n",
    "    x = sp.symbols('x') # variavel da função sympy\n",
    "    fx = sp.sympify(f) # Transforma a string em uma função sympy\n",
    "    ep_val=0.0\n",
    "    pn=0.0\n",
    "    print (f + ', para ' + str(a) + ' <= x <= ' + str(b) + '\\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",
    "        FP = float(fx.subs(x, p))\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 < ep):\n",
    "            print (\"\\nA raiz da função, dentro dos parâmetros de tolerância, é\", p, \"\\n\")\n",
    "            break\n",
    "        if (sgn(FP) * sgn(float(fx.subs(x,a))) > 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",
    "# Início da resolução:\n",
    "\n",
    "print (\"Exercícios 2.1\\n\")\n",
    "print (\"Questão 5.a) \", end=\"\"); bisseccao('x-2**(-x)', 0, 1, 0.00001, 0, 0, 100)\n",
    "print (\"Questão 5.b) \", end=\"\"); bisseccao('exp(x)-(x**2)+3*x-2', 0, 1, 0.00001, 0, 0, 100)\n",
    "print (\"Questão 5.c) parte 1: \", end=\"\"); bisseccao('2*x*cos(2*x)-(x+1)**2', -3, -2, 0.00001, 0, 0, 100)\n",
    "print (\"Questão 5.c) parte 2: \", end=\"\"); bisseccao('2*x*cos(2*x)-(x+1)**2', -1, 0, 0.00001, 0, 0, 100)\n",
    "print (\"Questão 5.d) parte 1: \", end=\"\"); bisseccao('x*cos(x)-2*(x**2)+(3*x)-1', 0.2, 0.3, 0.00001, 0, 0, 100)\n",
    "print (\"Questão 5.d) parte 2: \", end=\"\"); bisseccao('x*cos(x)-2*(x**2)+(3*x)-1', 1.2, 1.3, 0.00001, 0, 0, 100)"
   ]
  },
  {
   "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. Para este exemplo, foi definido o limite de 30 iterações."
   ]
  },
  {
   "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": [
    "**(9) **\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": [
    "**(11) Considere que $f(x)=(x+2)*(x+1)*x*(x-1)^3*(x-2)$. Para qual zero de f convergirá o método da bissecção quando aplicado a cada um dos seguintes intervalos?**\n",
    "\n",
    "**a) [-3,2.5]** → $- 2$\n",
    "\n",
    "**b) [-2.5,3]** → $1$\n",
    "\n",
    "**c) [-1.75,1.5]** → $0$\n",
    "\n",
    "**d) [-1.5, 1.75]** → $0$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 39,
   "metadata": {
    "scrolled": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Questão 11.a) (x+2)*((x+1)**2)*(x)*((x-1)**3)*(x-2), para -3 <= x <= 2.5\n",
      "n\t a\t         b\t         p\t         f(p)\t         Tolerância: (b-a)/2\tMundança\n",
      "1 \t -3.0000000000 \t 2.5000000000 \t -0.2500000000 \t -1.0814666748 \t 2.7500000000   \tf(a)*f(p)<0.0, b=p\n",
      "2 \t -3.0000000000 \t -0.2500000000 \t -1.6250000000 \t -15.6077608466 \t 1.3750000000   \tf(a)*f(p)<0.0, b=p\n",
      "3 \t -3.0000000000 \t -1.6250000000 \t -2.3125000000 \t 195.1313980634 \t 0.6875000000   \tf(a)*f(p)>0.0, a=p\n",
      "4 \t -2.3125000000 \t -1.6250000000 \t -1.9687500000 \t -5.9956827839 \t 0.3437500000   \tf(a)*f(p)<0.0, b=p\n",
      "5 \t -2.3125000000 \t -1.9687500000 \t -2.1406250000 \t 50.2346118775 \t 0.1718750000   \tf(a)*f(p)>0.0, a=p\n",
      "6 \t -2.1406250000 \t -1.9687500000 \t -2.0546875000 \t 14.4457201872 \t 0.0859375000   \tf(a)*f(p)>0.0, a=p\n",
      "7 \t -2.0546875000 \t -1.9687500000 \t -2.0117187500 \t 2.6444896230 \t 0.0429687500   \tf(a)*f(p)>0.0, a=p\n",
      "8 \t -2.0117187500 \t -1.9687500000 \t -1.9902343750 \t -2.0332667464 \t 0.0214843750   \tf(a)*f(p)<0.0, b=p\n",
      "9 \t -2.0117187500 \t -1.9902343750 \t -2.0009765625 \t 0.2117111254 \t 0.0107421875   \tf(a)*f(p)>0.0, a=p\n",
      "10 \t -2.0009765625 \t -1.9902343750 \t -1.9956054688 \t -0.9336803796 \t 0.0053710938   \tf(a)*f(p)<0.0, b=p\n",
      "11 \t -2.0009765625 \t -1.9956054688 \t -1.9982910156 \t -0.3667810625 \t 0.0026855469   \tf(a)*f(p)<0.0, b=p\n",
      "12 \t -2.0009765625 \t -1.9982910156 \t -1.9996337891 \t -0.0789929936 \t 0.0013427734   \tf(a)*f(p)<0.0, b=p\n",
      "13 \t -2.0009765625 \t -1.9996337891 \t -2.0003051758 \t 0.0659934409 \t 0.0006713867   \tf(a)*f(p)>0.0, a=p\n",
      "14 \t -2.0003051758 \t -1.9996337891 \t -1.9999694824 \t -0.0065910425 \t 0.0003356934   \tf(a)*f(p)<0.0, b=p\n",
      "15 \t -2.0003051758 \t -1.9999694824 \t -2.0001373291 \t 0.0296783651 \t 0.0001678467   \tf(a)*f(p)>0.0, a=p\n",
      "16 \t -2.0001373291 \t -1.9999694824 \t -2.0000534058 \t 0.0115379550 \t 0.0000839233   \tf(a)*f(p)>0.0, a=p\n",
      "17 \t -2.0000534058 \t -1.9999694824 \t -2.0000114441 \t 0.0024720299 \t 0.0000419617   \tf(a)*f(p)>0.0, a=p\n",
      "18 \t -2.0000114441 \t -1.9999694824 \t -1.9999904633 \t -0.0020598629 \t 0.0000209808   \tf(a)*f(p)<0.0, b=p\n",
      "19 \t -2.0000114441 \t -1.9999904633 \t -2.0000009537 \t 0.0002059944 \t 0.0000104904   \tf(a)*f(p)>0.0, a=p\n",
      "20 \t -2.0000009537 \t -1.9999904633 \t -1.9999957085 \t -0.0009269565 \t 0.0000052452\n",
      "A raiz da função, dentro dos parâmetros de tolerância, é -1.9999957084655762 \n",
      "\n",
      "Questão 11.b) (x+2)*((x+1)**2)*(x)*((x-1)**3)*(x-2), para -2.5 <= x <= 3\n",
      "n\t a\t         b\t         p\t         f(p)\t         Tolerância: (b-a)/2\tMundança\n",
      "1 \t -2.5000000000 \t 3.0000000000 \t 0.2500000000 \t 0.6488800049 \t 2.7500000000   \tf(a)*f(p)>0.0, a=p\n",
      "2 \t 0.2500000000 \t 3.0000000000 \t 1.6250000000 \t -3.7161335349 \t 1.3750000000   \tf(a)*f(p)<0.0, b=p\n",
      "3 \t 0.2500000000 \t 1.6250000000 \t 0.9375000000 \t 0.0026816467 \t 0.6875000000   \tf(a)*f(p)>0.0, a=p\n",
      "4 \t 0.9375000000 \t 1.6250000000 \t 1.2812500000 \t -0.3498439870 \t 0.3437500000   \tf(a)*f(p)<0.0, b=p\n",
      "5 \t 0.9375000000 \t 1.2812500000 \t 1.1093750000 \t -0.0178857908 \t 0.1718750000   \tf(a)*f(p)<0.0, b=p\n",
      "6 \t 0.9375000000 \t 1.1093750000 \t 1.0234375000 \t -0.0001592853 \t 0.0859375000   \tf(a)*f(p)<0.0, b=p\n",
      "7 \t 0.9375000000 \t 1.0234375000 \t 0.9804687500 \t 0.0000870653 \t 0.0429687500   \tf(a)*f(p)>0.0, a=p\n",
      "8 \t 0.9804687500 \t 1.0234375000 \t 1.0019531250 \t -0.0000000896 \t 0.0214843750   \tf(a)*f(p)<0.0, b=p\n",
      "9 \t 0.9804687500 \t 1.0019531250 \t 0.9912109375 \t 0.0000080515 \t 0.0107421875   \tf(a)*f(p)>0.0, a=p\n",
      "10 \t 0.9912109375 \t 1.0019531250 \t 0.9965820312 \t 0.0000004770 \t 0.0053710938   \tf(a)*f(p)>0.0, a=p\n",
      "11 \t 0.9965820312 \t 1.0019531250 \t 0.9992675781 \t 0.0000000047 \t 0.0026855469   \tf(a)*f(p)>0.0, a=p\n",
      "12 \t 0.9992675781 \t 1.0019531250 \t 1.0006103516 \t -0.0000000027 \t 0.0013427734   \tf(a)*f(p)<0.0, b=p\n",
      "13 \t 0.9992675781 \t 1.0006103516 \t 0.9999389648 \t 0.0000000000 \t 0.0006713867   \tf(a)*f(p)>0.0, a=p\n",
      "14 \t 0.9999389648 \t 1.0006103516 \t 1.0002746582 \t -0.0000000002 \t 0.0003356934   \tf(a)*f(p)<0.0, b=p\n",
      "15 \t 0.9999389648 \t 1.0002746582 \t 1.0001068115 \t -0.0000000000 \t 0.0001678467   \tf(a)*f(p)<0.0, b=p\n",
      "16 \t 0.9999389648 \t 1.0001068115 \t 1.0000228882 \t -0.0000000000 \t 0.0000839233   \tf(a)*f(p)<0.0, b=p\n",
      "17 \t 0.9999389648 \t 1.0000228882 \t 0.9999809265 \t 0.0000000000 \t 0.0000419617   \tf(a)*f(p)>0.0, a=p\n",
      "18 \t 0.9999809265 \t 1.0000228882 \t 1.0000019073 \t -0.0000000000 \t 0.0000209808   \tf(a)*f(p)<0.0, b=p\n",
      "19 \t 0.9999809265 \t 1.0000019073 \t 0.9999914169 \t 0.0000000000 \t 0.0000104904   \tf(a)*f(p)>0.0, a=p\n",
      "20 \t 0.9999914169 \t 1.0000019073 \t 0.9999966621 \t 0.0000000000 \t 0.0000052452\n",
      "A raiz da função, dentro dos parâmetros de tolerância, é 0.9999966621398926 \n",
      "\n",
      "Questão 11.c) (x+2)*((x+1)**2)*(x)*((x-1)**3)*(x-2), para -1.75 <= x <= 1.5\n",
      "n\t a\t         b\t         p\t         f(p)\t         Tolerância: (b-a)/2\tMundança\n",
      "1 \t -1.7500000000 \t 1.5000000000 \t -0.1250000000 \t -0.5429300666 \t 1.6250000000   \tf(a)*f(p)>0.0, a=p\n",
      "2 \t -0.1250000000 \t 1.5000000000 \t 0.6875000000 \t 0.2107454056 \t 0.8125000000   \tf(a)*f(p)<0.0, b=p\n",
      "3 \t -0.1250000000 \t 0.6875000000 \t 0.2812500000 \t 0.6721705087 \t 0.4062500000   \tf(a)*f(p)<0.0, b=p\n",
      "4 \t -0.1250000000 \t 0.2812500000 \t 0.0781250000 \t 0.2841457609 \t 0.2031250000   \tf(a)*f(p)<0.0, b=p\n",
      "5 \t -0.1250000000 \t 0.0781250000 \t -0.0234375000 \t -0.0958287219 \t 0.1015625000   \tf(a)*f(p)>0.0, a=p\n",
      "6 \t -0.0234375000 \t 0.0781250000 \t 0.0273437500 \t 0.1062053982 \t 0.0507812500   \tf(a)*f(p)<0.0, b=p\n",
      "7 \t -0.0234375000 \t 0.0273437500 \t 0.0019531250 \t 0.0077971743 \t 0.0253906250   \tf(a)*f(p)<0.0, b=p\n",
      "8 \t -0.0234375000 \t 0.0019531250 \t -0.0107421875 \t -0.0434190531 \t 0.0126953125   \tf(a)*f(p)>0.0, a=p\n",
      "9 \t -0.0107421875 \t 0.0019531250 \t -0.0043945312 \t -0.0176546055 \t 0.0063476562   \tf(a)*f(p)>0.0, a=p\n",
      "10 \t -0.0043945312 \t 0.0019531250 \t -0.0012207031 \t -0.0048887566 \t 0.0031738281   \tf(a)*f(p)>0.0, a=p\n",
      "11 \t -0.0012207031 \t 0.0019531250 \t 0.0003662109 \t 0.0014643069 \t 0.0015869141   \tf(a)*f(p)<0.0, b=p\n",
      "12 \t -0.0012207031 \t 0.0003662109 \t -0.0004272461 \t -0.0017097138 \t 0.0007934570   \tf(a)*f(p)>0.0, a=p\n",
      "13 \t -0.0004272461 \t 0.0003662109 \t -0.0000305176 \t -0.0001220740 \t 0.0003967285   \tf(a)*f(p)>0.0, a=p\n",
      "14 \t -0.0000305176 \t 0.0003662109 \t 0.0001678467 \t 0.0006712740 \t 0.0001983643   \tf(a)*f(p)<0.0, b=p\n",
      "15 \t -0.0000305176 \t 0.0001678467 \t 0.0000686646 \t 0.0002746393 \t 0.0000991821   \tf(a)*f(p)<0.0, b=p\n",
      "16 \t -0.0000305176 \t 0.0000686646 \t 0.0000190735 \t 0.0000762925 \t 0.0000495911   \tf(a)*f(p)<0.0, b=p\n",
      "17 \t -0.0000305176 \t 0.0000190735 \t -0.0000057220 \t -0.0000228883 \t 0.0000247955   \tf(a)*f(p)>0.0, a=p\n",
      "18 \t -0.0000057220 \t 0.0000190735 \t 0.0000066757 \t 0.0000267027 \t 0.0000123978   \tf(a)*f(p)<0.0, b=p\n",
      "19 \t -0.0000057220 \t 0.0000066757 \t 0.0000004768 \t 0.0000019073 \t 0.0000061989\n",
      "A raiz da função, dentro dos parâmetros de tolerância, é 4.76837158203125e-07 \n",
      "\n",
      "Questão 11.d) (x+2)*((x+1)**2)*(x)*((x-1)**3)*(x-2), para -1.5 <= x <= 1.75\n",
      "n\t a\t         b\t         p\t         f(p)\t         Tolerância: (b-a)/2\tMundança\n",
      "1 \t -1.5000000000 \t 1.7500000000 \t 0.1250000000 \t 0.4222789407 \t 1.6250000000   \tf(a)*f(p)<0.0, b=p\n",
      "2 \t -1.5000000000 \t 0.1250000000 \t -0.6875000000 \t -1.1380251904 \t 0.8125000000   \tf(a)*f(p)>0.0, a=p\n",
      "3 \t -0.6875000000 \t 0.1250000000 \t -0.2812500000 \t -1.1982169937 \t 0.4062500000   \tf(a)*f(p)>0.0, a=p\n",
      "4 \t -0.2812500000 \t 0.1250000000 \t -0.0781250000 \t -0.3323060593 \t 0.2031250000   \tf(a)*f(p)>0.0, a=p\n",
      "5 \t -0.0781250000 \t 0.1250000000 \t 0.0234375000 \t 0.0914396201 \t 0.1015625000   \tf(a)*f(p)<0.0, b=p\n",
      "6 \t -0.0781250000 \t 0.0234375000 \t -0.0273437500 \t -0.1121767861 \t 0.0507812500   \tf(a)*f(p)>0.0, a=p\n",
      "7 \t -0.0273437500 \t 0.0234375000 \t -0.0019531250 \t -0.0078276916 \t 0.0253906250   \tf(a)*f(p)>0.0, a=p\n",
      "8 \t -0.0019531250 \t 0.0234375000 \t 0.0107421875 \t 0.0424961360 \t 0.0126953125   \tf(a)*f(p)<0.0, b=p\n",
      "9 \t -0.0019531250 \t 0.0107421875 \t 0.0043945312 \t 0.0175001169 \t 0.0063476562   \tf(a)*f(p)<0.0, b=p\n",
      "10 \t -0.0019531250 \t 0.0043945312 \t 0.0012207031 \t 0.0048768357 \t 0.0031738281   \tf(a)*f(p)<0.0, b=p\n",
      "11 \t -0.0019531250 \t 0.0012207031 \t -0.0003662109 \t -0.0014653797 \t 0.0015869141   \tf(a)*f(p)>0.0, a=p\n",
      "12 \t -0.0003662109 \t 0.0012207031 \t 0.0004272461 \t 0.0017082535 \t 0.0007934570   \tf(a)*f(p)<0.0, b=p\n",
      "13 \t -0.0003662109 \t 0.0004272461 \t 0.0000305176 \t 0.0001220666 \t 0.0003967285   \tf(a)*f(p)<0.0, b=p\n",
      "14 \t -0.0003662109 \t 0.0000305176 \t -0.0001678467 \t -0.0006714994 \t 0.0001983643   \tf(a)*f(p)>0.0, a=p\n",
      "15 \t -0.0001678467 \t 0.0000305176 \t -0.0000686646 \t -0.0002746771 \t 0.0000991821   \tf(a)*f(p)>0.0, a=p\n",
      "16 \t -0.0000686646 \t 0.0000305176 \t -0.0000190735 \t -0.0000762954 \t 0.0000495911   \tf(a)*f(p)>0.0, a=p\n",
      "17 \t -0.0000190735 \t 0.0000305176 \t 0.0000057220 \t 0.0000228881 \t 0.0000247955   \tf(a)*f(p)<0.0, b=p\n",
      "18 \t -0.0000190735 \t 0.0000057220 \t -0.0000066757 \t -0.0000267031 \t 0.0000123978   \tf(a)*f(p)>0.0, a=p\n",
      "19 \t -0.0000066757 \t 0.0000057220 \t -0.0000004768 \t -0.0000019073 \t 0.0000061989\n",
      "A raiz da função, dentro dos parâmetros de tolerância, é -4.76837158203125e-07 \n",
      "\n"
     ]
    }
   ],
   "source": [
    "\n",
    "\n",
    "#Solução \n",
    "\n",
    "def bissecao(f, a, b, TOL, N):  \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",
    "        if ((fp == 0) or ((b-a)/2 < TOL)):  \n",
    "            return p  \n",
    "        #bissecta o intervalo  \n",
    "        i = i+1  \n",
    "        if (fa * fp > 0):  \n",
    "            a = p  \n",
    "            fa = fp  \n",
    "        else:  \n",
    "            b = p  \n",
    " \n",
    "print (\"Questão 11.a) \", end=\"\"); bisseccao('(x+2)*((x+1)**2)*(x)*((x-1)**3)*(x-2)', -3, 2.5, 0.00001, 100)\n",
    "print (\"Questão 11.b) \", end=\"\"); bisseccao('(x+2)*((x+1)**2)*(x)*((x-1)**3)*(x-2)', -2.5, 3, 0.00001, 100)\n",
    "print (\"Questão 11.c) \", end=\"\"); bisseccao('(x+2)*((x+1)**2)*(x)*((x-1)**3)*(x-2)', -1.75, 1.5, 0.00001, 100)\n",
    "print (\"Questão 11.d) \", end=\"\"); bisseccao('(x+2)*((x+1)**2)*(x)*((x-1)**3)*(x-2)', -1.5, 1.75, 0.00001, 100)\n",
    "\n"
   ]
  },
  {
   "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": [
    "## Exercícios Aplicados#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**(15) Uma gamela de comprimento L tem seção transversal semicircular com raio r. Qaundo a gamela está cheia com água até uma distância h do topo, o volume V de água é: $$V = L[0,5\\pi r^{2} - r^{2}arcsen(h/r) - h(r^{2} - h^{2})^\\frac{1}{2}]$$Suponha L = 10 $ft$, r = 1 $ft$ e V = 12,4 $ft^{3}$. Determine a profundidade de água na gamela com precisão de 0.01 $ft$.**\n",
    "\n",
    "**Resposta:**\n",
    "\n",
    "$V = L[0,5\\pi r^{2} - r^{2}arcsen(h/r) - h(r^{2} - h^{2})^\\frac{1}{2}]$\n",
    "$0 = L[0,5\\pi r^{2} - r^{2}arcsen(h/r) - h(r^{2} - h^{2})^\\frac{1}{2}] - V$\n",
    "\n",
    "Sendo L = 10 $ft$; r = 1 $ft$; 12,4 $ft^{3}$:\n",
    "\n",
    "$10[0,5\\pi - arcsin(h) - h(1 - h^{2})^{0,5}] - 12,4 = 0$\n",
    "\n",
    "Determinar o valor de h (raiz da função acima) com precisão de 0,01 $ft$ e então calcular profundidade da água = r - h"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "10.0*(0.5*pi - asin(x) - x*(1 - x**2)**0.5) - 12.4, para 0 <= x <= 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 -6.2581515070 \t 0.5000000000 \t 1.0000000000   f(a)*f(p)<0.0, b=p\n",
      "2 \t 0.0000000000 \t 0.5000000000 \t 0.2500000000 \t -1.6394538749 \t 0.2500000000 \t 1.0000000000   f(a)*f(p)<0.0, b=p\n",
      "3 \t 0.0000000000 \t 0.2500000000 \t 0.1250000000 \t 0.8144890292 \t 0.1250000000 \t 1.0000000000   f(a)*f(p)>0.0, a=p\n",
      "4 \t 0.1250000000 \t 0.2500000000 \t 0.1875000000 \t -0.4199467241 \t 0.0625000000 \t 0.3333333333   f(a)*f(p)<0.0, b=p\n",
      "5 \t 0.1250000000 \t 0.1875000000 \t 0.1562500000 \t 0.1957259025 \t 0.0312500000 \t 0.2000000000   f(a)*f(p)>0.0, a=p\n",
      "6 \t 0.1562500000 \t 0.1875000000 \t 0.1718750000 \t -0.1125363938 \t 0.0156250000 \t 0.0909090909   f(a)*f(p)<0.0, b=p\n",
      "7 \t 0.1562500000 \t 0.1718750000 \t 0.1640625000 \t 0.0414932414 \t 0.0078125000 \t 0.0476190476\n",
      "\n",
      "A raiz da função, dentro dos parâmetros de tolerância, é 0.1640625 ~= 0.164 \n",
      "\n",
      "O valor de h é 0.164 ft.\n",
      "Profundidade da água = r - h = 1 - 0.164 = 0.836 ft\n"
     ]
    }
   ],
   "source": [
    "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 bisseccao (f, a, b, TOL, ep, efp, N0): # Função método da Bissecção\n",
    "    #Argumentos: \n",
    "    #f: função f(x) a ser encontrada a raiz (em string)\n",
    "    # a e b: extremos do intervalo [a, b] contendo a raíz, com f(a) * f(b) < 0\n",
    "    # TOL: Tolerância para o tamanho do intervalo (b-a)/2 | ep: Tolerância para o erro relativo de p\n",
    "    # efp: Tolerância para o erro de f(p) | N0: Número máximo de iterações\n",
    "    x = sp.symbols('x') # variavel da função sympy\n",
    "    fx = sp.sympify(f) # Transforma a string em uma função sympy\n",
    "    ep_val=0.0\n",
    "    pn=0.0\n",
    "    print (f + ', para ' + str(a) + ' <= x <= ' + str(b) + '\\n\\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",
    "        FP = float(fx.subs(x, p))\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 < ep):\n",
    "            print (\"\\n\\nA raiz da função, dentro dos parâmetros de tolerância, é\", p, \"~=\", \"{:.3f}\".format(p), \"\\n\")\n",
    "            break\n",
    "        if (sgn(FP) * sgn(float(fx.subs(x,a))) > 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",
    "    return p\n",
    "\n",
    "# Início da resolução:\n",
    " \n",
    "h = bisseccao('10.0*(0.5*pi - asin(x) - x*(1 - x**2)**0.5) - 12.4', 0, 1, 0.01, 0, 0, 100)\n",
    "print (\"O valor de h é \" + \"{:.3f}\".format(h) + \" ft.\\nProfundidade da água = r - h = 1 - \" + \"{:.3f}\".format(h) + \" = \" + \"{:.3f}\".format(1-h) + \" ft\")"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Exercícios Teóricos#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**(17) Utilize o Teorema 2.1 para determinar um limitante para o número de iterações necessárias para obter uma aproximação com precisão de $10^{-4}$ da solução $x^{3} - x - 1 = 0$ dentro do intervalo [1, 2]. Determine uma aproximação da raiz com essa ordem de precisão.**\n",
    "\n",
    "**Resposta:**\n",
    "\n",
    "$$\\left|{p}_{N}-{p}\\right| \\leqslant  \\frac{b-a}{2^{N}}\\text{, sendo a=1 e b=2, então:}$$\n",
    "$$\\left|{p}_{N}-{p}\\right| \\leqslant  \\frac{2-1}{2^{N}}$$\n",
    "$$\\left|{p}_{N}-{p}\\right| \\leqslant  \\frac{1}{2^{N}}\\text{, para a precisão de } 10^{-4}:$$\n",
    "$$\\frac{1}{2^{N}} = 2^{-N} < 10^{-4}$$\n",
    "Aplicando ${log}_{10}$ em ambos os lados:\n",
    "$${log}_{10}(2^{-N}) < {log}_{10}(10^{-4})$$\n",
    "$$-N{log}_{10}(2) < -4$$\n",
    "$$-N < \\frac{-4}{{log}_{10}(2)}\\text{, multiplicando ambos os lados por -1}$$\n",
    "$$N>\\frac{4}{{log}_{10}(2)} \\approx 13,29$$\n",
    "\n",
    "Desta forma, um número de iterações $N \\geqslant  14$ garante uma aproximação com precisão de $10^{-4}.$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      "n\t a\t         b\t         p\t         f(p)\t         (b-a)/2\t e. relat. p\tmundança\n",
      "1 \t 1.0000000000 \t 2.0000000000 \t 1.5000000000 \t 0.8750000000 \t 0.5000000000 \t 1.0000000000   f(a)*f(p)<0.0, b=p\n",
      "2 \t 1.0000000000 \t 1.5000000000 \t 1.2500000000 \t -0.2968750000 \t 0.2500000000 \t 0.2000000000   f(a)*f(p)>0.0, a=p\n",
      "3 \t 1.2500000000 \t 1.5000000000 \t 1.3750000000 \t 0.2246093750 \t 0.1250000000 \t 0.0909090909   f(a)*f(p)<0.0, b=p\n",
      "4 \t 1.2500000000 \t 1.3750000000 \t 1.3125000000 \t -0.0515136719 \t 0.0625000000 \t 0.0476190476   f(a)*f(p)>0.0, a=p\n",
      "5 \t 1.3125000000 \t 1.3750000000 \t 1.3437500000 \t 0.0826110840 \t 0.0312500000 \t 0.0232558140   f(a)*f(p)<0.0, b=p\n",
      "6 \t 1.3125000000 \t 1.3437500000 \t 1.3281250000 \t 0.0145759583 \t 0.0156250000 \t 0.0117647059   f(a)*f(p)<0.0, b=p\n",
      "7 \t 1.3125000000 \t 1.3281250000 \t 1.3203125000 \t -0.0187106133 \t 0.0078125000 \t 0.0059171598   f(a)*f(p)>0.0, a=p\n",
      "8 \t 1.3203125000 \t 1.3281250000 \t 1.3242187500 \t -0.0021279454 \t 0.0039062500 \t 0.0029498525   f(a)*f(p)>0.0, a=p\n",
      "9 \t 1.3242187500 \t 1.3281250000 \t 1.3261718750 \t 0.0062088296 \t 0.0019531250 \t 0.0014727541   f(a)*f(p)<0.0, b=p\n",
      "10 \t 1.3242187500 \t 1.3261718750 \t 1.3251953125 \t 0.0020366507 \t 0.0009765625 \t 0.0007369197   f(a)*f(p)<0.0, b=p\n",
      "11 \t 1.3242187500 \t 1.3251953125 \t 1.3247070312 \t -0.0000465949 \t 0.0004882812 \t 0.0003685957   f(a)*f(p)>0.0, a=p\n",
      "12 \t 1.3247070312 \t 1.3251953125 \t 1.3249511719 \t 0.0009947910 \t 0.0002441406 \t 0.0001842639   f(a)*f(p)<0.0, b=p\n",
      "13 \t 1.3247070312 \t 1.3249511719 \t 1.3248291016 \t 0.0004740388 \t 0.0001220703 \t 0.0000921404   f(a)*f(p)<0.0, b=p\n",
      "14 \t 1.3247070312 \t 1.3248291016 \t 1.3247680664 \t 0.0002137072 \t 0.0000610352 \t 0.0000460723\n",
      "\n",
      "A raiz da função, dentro dos parâmetros de tolerância, é 1.32476806640625\n",
      "\n",
      "O método da bissecção foi encerrado após 14 iterações.\n"
     ]
    }
   ],
   "source": [
    "#Questão 17 - Exercícios 2.1\n",
    "\n",
    "def f(x): # Função da questão\n",
    "    return float(x**3 - x - 1)\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",
    "# Insira as tolerâncias aqui\n",
    "TOL = 0.0001 # Tolerância para o tamanho do intervalo (b-a)/2\n",
    "ep = 0.0 # Tolerância para o erro relativo de p\n",
    "efp = 0.0 # Tolerância para o erro relativo de f(p)\n",
    "N0 = 20 # Número máximo de iterações\n",
    "\n",
    "# Obtém a e b (f(a) e f(b) devem ter sinais opostos!)\n",
    "a = 1.0\n",
    "b = 2.0\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",
    "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",
    "    FP = f(p)\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 < ep):\n",
    "        print (\"\\n\\nA raiz da função, dentro dos parâmetros de tolerância, é\", p)\n",
    "        break\n",
    "    if (sgn(FP) * sgn(f(a)) > 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",
    "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": [
    "**(21) A função definida por $f(x) = sen (pi*x)$ tem zeros em cada inteiro. Mostre que quando $-1 < a < 0 e 2 < b < 3$, o método de Bissecção converge para:**\n",
    "\n",
    "**a) 0, se a a + b <2**\n",
    "\n",
    "**b) 2, se a + b > 2**\n",
    "\n",
    "**c) 1, se a + b = 2**"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 62,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Questão 21.a) sin(pi*x), para -1 <= x <= 0\n",
      "n\t a\t         b\t         p\t         f(p)\t         Tolerância: (b-a)/2\tMundança\n",
      "1 \t -1.0000000000 \t 0.0000000000 \t -0.5000000000 \t -1.0000000000 \t 0.5000000000   \tf(a)*f(p)<0.0, b=p\n",
      "2 \t -1.0000000000 \t -0.5000000000 \t -0.7500000000 \t -0.7071067812 \t 0.2500000000   \tf(a)*f(p)<0.0, b=p\n",
      "3 \t -1.0000000000 \t -0.7500000000 \t -0.8750000000 \t -0.3826834324 \t 0.1250000000   \tf(a)*f(p)<0.0, b=p\n",
      "4 \t -1.0000000000 \t -0.8750000000 \t -0.9375000000 \t -0.1950903220 \t 0.0625000000   \tf(a)*f(p)<0.0, b=p\n",
      "5 \t -1.0000000000 \t -0.9375000000 \t -0.9687500000 \t -0.0980171403 \t 0.0312500000   \tf(a)*f(p)<0.0, b=p\n",
      "6 \t -1.0000000000 \t -0.9687500000 \t -0.9843750000 \t -0.0490676743 \t 0.0156250000   \tf(a)*f(p)<0.0, b=p\n",
      "7 \t -1.0000000000 \t -0.9843750000 \t -0.9921875000 \t -0.0245412285 \t 0.0078125000   \tf(a)*f(p)<0.0, b=p\n",
      "8 \t -1.0000000000 \t -0.9921875000 \t -0.9960937500 \t -0.0122715383 \t 0.0039062500   \tf(a)*f(p)<0.0, b=p\n",
      "9 \t -1.0000000000 \t -0.9960937500 \t -0.9980468750 \t -0.0061358846 \t 0.0019531250   \tf(a)*f(p)<0.0, b=p\n",
      "10 \t -1.0000000000 \t -0.9980468750 \t -0.9990234375 \t -0.0030679568 \t 0.0009765625   \tf(a)*f(p)<0.0, b=p\n",
      "11 \t -1.0000000000 \t -0.9990234375 \t -0.9995117188 \t -0.0015339802 \t 0.0004882812   \tf(a)*f(p)<0.0, b=p\n",
      "12 \t -1.0000000000 \t -0.9995117188 \t -0.9997558594 \t -0.0007669903 \t 0.0002441406   \tf(a)*f(p)<0.0, b=p\n",
      "13 \t -1.0000000000 \t -0.9997558594 \t -0.9998779297 \t -0.0003834952 \t 0.0001220703   \tf(a)*f(p)<0.0, b=p\n",
      "14 \t -1.0000000000 \t -0.9998779297 \t -0.9999389648 \t -0.0001917476 \t 0.0000610352   \tf(a)*f(p)<0.0, b=p\n",
      "15 \t -1.0000000000 \t -0.9999389648 \t -0.9999694824 \t -0.0000958738 \t 0.0000305176   \tf(a)*f(p)<0.0, b=p\n",
      "16 \t -1.0000000000 \t -0.9999694824 \t -0.9999847412 \t -0.0000479369 \t 0.0000152588   \tf(a)*f(p)<0.0, b=p\n",
      "17 \t -1.0000000000 \t -0.9999847412 \t -0.9999923706 \t -0.0000239684 \t 0.0000076294\n",
      "A raiz da função, dentro dos parâmetros de tolerância, é -0.9999923706054688 \n",
      "\n",
      "Questão 21.b) sin(pi*x), para 0 <= x <= 4\n",
      "n\t a\t         b\t         p\t         f(p)\t         Tolerância: (b-a)/2\tMundança\n",
      "1 \t 0.0000000000 \t 4.0000000000 \t 2.0000000000 \t 0.0000000000 \t 2.0000000000\n",
      "A raiz da função, dentro dos parâmetros de tolerância, é 2.0 \n",
      "\n",
      "Questão 21.c) sin(pi*x), para 0 <= x <= 2\n",
      "n\t a\t         b\t         p\t         f(p)\t         Tolerância: (b-a)/2\tMundança\n",
      "1 \t 0.0000000000 \t 2.0000000000 \t 1.0000000000 \t 0.0000000000 \t 1.0000000000\n",
      "A raiz da função, dentro dos parâmetros de tolerância, é 1.0 \n",
      "\n"
     ]
    }
   ],
   "source": [
    "\n",
    "def bissecao(f, a, b, TOL, N):  \n",
    "    i = 1  \n",
    "    fa = f(a)  \n",
    "    while (i <= N):  \n",
    "        #iteracao da bisseção  \n",
    "        p = a + (b-a)/2  \n",
    "        fp = f(p)  \n",
    "        #condição de parada  \n",
    "        if ((fp == 0) or ((b-a)/2 < TOL)):  \n",
    "            return p  \n",
    "        #bissecta o intervalo  \n",
    "        i = i+1  \n",
    "        if (fa * fp > 0):  \n",
    "            a = p  \n",
    "            fa = fp  \n",
    "        else:  \n",
    "            b = p  \n",
    " \n",
    "print (\"Questão 21.a) \", end=\"\"); bisseccao('sin(pi*x)', -1, 0, 0.00001, 100)\n",
    "print (\"Questão 21.b) \", end=\"\"); bisseccao('sin(pi*x)', 0, 4  , 0.00001, 100)\n",
    "print (\"Questão 21.c) \", end=\"\"); bisseccao('sin(pi*x)', 0, 2, 0.00001, 100)\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Questões de Discussão#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**(1) Derive uma função $f$ para a qual o método da bissecção converge para um valor que não é um zero de $f$**\n",
    "\n",
    "Se uma função f for derivada e pelo método da bissecção converge para um valor não igual a 0 de $f$, então pode-se dizer que a função não é possivelmente contínua no intervalo acima mencionado ou analisado. Ou que os sinais do intervalo não diferem uns dos outros."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**(3) O método da bissecção é sensível ao valor inicial? Justifique sua resposta.**\n",
    "\n",
    "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": [
    "# Conjunto de Exercícios 2.2 #"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**(1) Use algebraic manipulation to show that each of the following functions has a fixed point at p precisely when /(p) = 0. where /(x) = x 4 + 2x 2 — x - 3."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "\n",
    "\n",
    "\n",
    "\n",
    "(a) $f(x) = 0 = x^4+ 2x^2−x−3$ ⇒ $x^4 = 3 + x−2x^2$ ⇒ $x = (3 + x−2x^2)^\\frac{1}{4}  = g1(x)$.\n",
    "\n",
    "(b) $f(x) = 0 = x^4+ 2x^2−x−3$ ⇒  $x^2=\\frac{(3 + x−x^4)}{2}$ ⇒ $x = (\\frac{3 + x−x^4}{2})^\\frac{1}{2}  = g2(x)$\n",
    "\n",
    "(c) $x^4+ 2x^2 −x −3 = 0$ ⇒ $x^4+ 2x^2 = x+ 3$ ⇒ $x^2(x^2+ 2)$ = $x+ 3$\n",
    "\n",
    "Como $(x^2+ 2) != 0$ para todo x∈R, dividimos a igualdade por $(x^2+ 2)$ e obtemos:\n",
    "$x^2=x+ 3x^2+ 2 $⇒ $x=(\\frac{x+ 3}{x^2 + 2})^\\frac{1}{2} =g3(x) $\n",
    "\n",
    "\n",
    "\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**(3) Considere que $f(x)=x^3-2x+1$. Para resolver $f(x)=0$, os quatro problemas de ponto fixo são propostos. Derive cada um dos métodos de ponto fixo, e calcule $p_1$, $p_2$, $p_3$, $p_4$. Qual dos métodos parece ser mais apropriado?**\n",
    "\n",
    "**a) $x=\\frac{1}{2}(x^3 +1)$,   $p_0=\\frac{1}{2}$**\n",
    "\n",
    "**b) $x=\\frac{2}{x}-\\frac{1}{x^2}$,  $p_0=\\frac{1}{2}$**\n",
    "\n",
    "**c) $x=\\sqrt{2-\\frac{1}{x}}$, $p_0=\\frac{1}{2}$**\n",
    "\n",
    "**d) $x=-\\sqrt[3]{1-2x}$, $p_0=\\frac{1}{2}$**"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 95,
   "metadata": {},
   "outputs": [],
   "source": [
    "import sympy as sp\n",
    "import math\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",
    "    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) >0: # 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) >0: # Se o erro relativo subir acima de 100%, o programa encerra por precaução\n",
    "            print (\"O método falhou!\")\n",
    "            break\n",
    "        p0 = p\n",
    "  \n",
    "\n",
    "print (\"O método foi encerrado após\", n, \"iterações.\\n\"); ponto_fixo('1/2*((x**3)+1)', 0.5, 0.0000000001, 0.0000000001, 30)\n",
    "print (\"O método foi encerrado após\", n, \"iterações.\\n\"); ponto_fixo ('2/x-1/(x**2)', 0.5, 0.0000000001, 0.0000000001, 30)\n",
    "print (\"O método foi encerrado após\", n, \"iterações.\\n\"); ponto_fixo('((2-(1/x)))**(1/2)', 0.5, 0.0000000001, 0.0000000001, 30)\n",
    "print (\"O método foi encerrado após\", n, \"iterações.\\n\"); ponto_fixo ('-(1-2x)**(1/3)', 0.5, 0.0000000001, 0.0000000001, 30)\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**(5) Os quatro métodos a seguir são propostos para o cálculo de $\\sqrt[3]{21}$. Classifique-os de acordo com sua velocidade de convergência aparente, supondo $p_0=1$.**\n",
    "\n",
    "**a) $p_n = \\frac{20 p_{n-1} + 21/p{n-1}^2}{21}$**\n",
    "\n",
    "**b) $p_n =p_{n-1} - \\frac{p_{n-1}^3-21}{3p_{n-1}^2}$**\n",
    "\n",
    "**c) $p_n = p_{n-1} - \\frac{p_{n-1}^4-21 p_{n-1}}{p_{n-1}^2 - 21}$**\n",
    "\n",
    "**d) $p_n = ( \\frac{21}{p_{n-1}})$**\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**a.** <p style=\"text-align: center;\">$p_{n}=((20*p_{n-1} + (21/(p_{n-1})^2))/21)$\n",
    "\n",
    "<p style=\"text-align: center;\">$g(x)= ((20*x + (21/(x)^2))/21)$\n",
    "\n",
    "<p style=\"text-align: center;\">$g'(x)=(20/21)-(2/(x^2))=0.857$\n",
    "\n",
    "\n",
    "**b.** <p style=\"text-align: center;\"> $p_{n}=p_{n-1} - ((p_{n-1})^3 - 21) /3*p_{n-1}^2)$\n",
    "\n",
    "\n",
    "<p style=\"text-align: center;\">$g'(x)=(2/3)-(7/(x^3)=0.333$\n",
    "\n",
    "**c.** <p style=\"text-align: center;\"> $p_{n}=p_{n-1} - (((p_{n-1)^4)- 21*p_{n-1})/((p_{n-1})^2) -21$ \n",
    "\n",
    "<p style=\"text-align: center;\">$g'(x)=5.7506 > 1$\n",
    "\n",
    "**d.**<p style=\"text-align: center;\"> $p_{n}= (21/p_{n-1})^(1/2)$\n",
    "\n",
    "<p style=\"text-align: center;\">$g'(x)=-0.5$ <p>\n",
    "\n",
    "Então  sua velocidad de convergencia: 1)b, 2)d, 3)a. C no converge.\n",
    "\n"
   ]
  },
  {
   "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": {
    "scrolled": false
   },
   "outputs": [
    {
     "data": {
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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": 1,
   "metadata": {
    "scrolled": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "<matplotlib.figure.Figure at 0x2b331905be0>"
      ]
     },
     "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.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": [
    "**(15) Determine todos os zeros de $f(x) = x^{2}+10cos(x)$ utilizando a iteração de ponto fixo para uma função de iteração g adequada. Determine os zeros com precisão $10^{-4}$.**\n",
    "\n",
    "**Resposta:**\n",
    "\n",
    "funções iteração: $g1(x) = arccos(0,1x^{2})$ = para $p0$ = 1 e -1 e $g2(x) = (2x^{2} - 10cos(x))/(3x)$ = para $p0$ = 3 e -3"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": 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\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0xaaafef0>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Raiz 1:\n",
      "\n",
      "Ponto fixo de g(x) = acos((-0.1)*(x**2)), para p0 = 1\n",
      "n\t p\t         |p - p0|\n",
      "1 \t 1.6709637480 \t 0.6709637480\n",
      "2 \t 1.8537696849 \t 0.1828059369\n",
      "3 \t 1.9215931414 \t 0.0678234566\n",
      "4 \t 1.9490003588 \t 0.0274072173\n",
      "5 \t 1.9604415337 \t 0.0114411749\n",
      "6 \t 1.9652816584 \t 0.0048401247\n",
      "7 \t 1.9673407208 \t 0.0020590624\n",
      "8 \t 1.9682187558 \t 0.0008780350\n",
      "9 \t 1.9685935496 \t 0.0003747939\n",
      "10 \t 1.9687536013 \t 0.0001600517\n",
      "11 \t 1.9688219622 \t 0.0000683609\n",
      "o ponto fixo da função g(x), dentro dos parâmetros de tolerância, é 1.9688219621820693\n",
      "O método foi encerrado após 11 iterações.\n",
      "\n",
      "Raiz 2:\n",
      "\n",
      "Ponto fixo de g(x) = acos((-0.1)*(x**2)), para p0 = -1\n",
      "n\t p\t         |p - p0|\n",
      "1 \t 1.6709637480 \t 2.6709637480\n",
      "2 \t 1.8537696849 \t 0.1828059369\n",
      "3 \t 1.9215931414 \t 0.0678234566\n",
      "4 \t 1.9490003588 \t 0.0274072173\n",
      "5 \t 1.9604415337 \t 0.0114411749\n",
      "6 \t 1.9652816584 \t 0.0048401247\n",
      "7 \t 1.9673407208 \t 0.0020590624\n",
      "8 \t 1.9682187558 \t 0.0008780350\n",
      "9 \t 1.9685935496 \t 0.0003747939\n",
      "10 \t 1.9687536013 \t 0.0001600517\n",
      "11 \t 1.9688219622 \t 0.0000683609\n",
      "o ponto fixo da função g(x), dentro dos parâmetros de tolerância, é 1.9688219621820693\n",
      "O método foi encerrado após 11 iterações.\n",
      "\n",
      "Raiz 3:\n",
      "\n",
      "Ponto fixo de g(x) = (2*(x**2) - 10*cos(x))/(3*x), para p0 = 3\n",
      "n\t p\t         |p - p0|\n",
      "1 \t 3.0999916629 \t 0.0999916629\n",
      "2 \t 3.1410024964 \t 0.0410108335\n",
      "3 \t 3.1552337889 \t 0.0142312925\n",
      "4 \t 3.1598366543 \t 0.0046028653\n",
      "5 \t 3.1612890655 \t 0.0014524112\n",
      "6 \t 3.1617437074 \t 0.0004546419\n",
      "7 \t 3.1618856621 \t 0.0001419546\n",
      "8 \t 3.1619299500 \t 0.0000442879\n",
      "o ponto fixo da função g(x), dentro dos parâmetros de tolerância, é 3.1619299499505833\n",
      "O método foi encerrado após 8 iterações.\n",
      "\n",
      "Raiz 4:\n",
      "\n",
      "Ponto fixo de g(x) = (2*(x**2) - 10*cos(x))/(3*x), para p0 = -3\n",
      "n\t p\t         |p - p0|\n",
      "1 \t -3.0999916629 \t 0.0999916629\n",
      "2 \t -3.1410024964 \t 0.0410108335\n",
      "3 \t -3.1552337889 \t 0.0142312925\n",
      "4 \t -3.1598366543 \t 0.0046028653\n",
      "5 \t -3.1612890655 \t 0.0014524112\n",
      "6 \t -3.1617437074 \t 0.0004546419\n",
      "7 \t -3.1618856621 \t 0.0001419546\n",
      "8 \t -3.1619299500 \t 0.0000442879\n",
      "o ponto fixo da função g(x), dentro dos parâmetros de tolerância, é -3.1619299499505833\n",
      "O método foi encerrado após 8 iterações.\n",
      "\n"
     ]
    }
   ],
   "source": [
    "import sympy as sp\n",
    "import numpy as np\n",
    "import matplotlib.pyplot as plt\n",
    "\n",
    "def ponto_fixo (g, p0, TOL, N0): # Função método do Ponto Fixo\n",
    "    # Argumentos:  g: função g(x) a ser encontrado o seu ponto fixo: g(p) = p  |  TOL: Tolerância para o erro absoluto |p - p0|\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",
    "    print ('Ponto fixo de g(x) = ' + g + ', para p0 = ' + str(p0) + '\\nn\\t p\\t         |p - p0|')\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",
    "        print (n, '\\t', \"{:.10f}\".format(p), '\\t', \"{:.10f}\".format(erro))\n",
    "        if (erro < TOL): # Testa os erros\n",
    "            print (\"o ponto fixo da função g(x), dentro dos parâmetros de tolerância, é\", p)\n",
    "            break\n",
    "        p0 = p\n",
    "    print (\"O método foi encerrado após\", n, \"iterações.\\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",
    "x = np.arange(-5.0, 5, 0.2)\n",
    "plt.plot(x, (x**2)+(10*np.cos(x)), label=\"f(x)\")\n",
    "plt.plot(x, x, label=\"y = x\")\n",
    "x = np.arange(-3.0, 3, 0.2)\n",
    "plt.plot(x, np.arccos((-0.1)*(x**2)), label=\"g1(x)\")\n",
    "x = np.arange(-5.0, -1.0, 0.2)\n",
    "plt.plot(x, (2*(x**2)-10*np.cos(x))/(3*x), label=\"g2(x)\")\n",
    "plt.legend(); plt.grid(); plt.show()\n",
    "\n",
    "print(\"Raiz 1:\\n\"); ponto_fixo('acos((-0.1)*(x**2))', 1, 0.0001, 30)\n",
    "print(\"Raiz 2:\\n\"); ponto_fixo('acos((-0.1)*(x**2))', -1, 0.0001, 30)\n",
    "print(\"Raiz 3:\\n\"); ponto_fixo('(2*(x**2) - 10*cos(x))/(3*x)', 3, 0.0001, 30)\n",
    "print(\"Raiz 4:\\n\"); ponto_fixo('(2*(x**2) - 10*cos(x))/(3*x)', -3, 0.0001, 30)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**(17) Use o método da iteração de ponto fixo para determinar a solução, com precisão de $10^-2$, da equação $x = 2 sen(7 \\pi x) + x = 0$ , for $x\\in [l, 2]$. Use $p_0 = 1$.**"
   ]
  },
  {
   "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": [
    "# Exercícios Aplicados#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Aqui só a 18"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Exercícios Teóricos#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**(19) Sejam $g ∈ C^1[a,b]$, e $p \\in (a,b)$ com $ g(p)=p$ e $|g'(p) > 1|$ . Mostre que existe um $\\delta > 0$ tal que se $0 < |p_0 - p| < \\delta$, então, $|p_0 - p| < |p_1 - p|$. Desse modo, independentemente do quão próxima seja a aproximação inicial $p_0$ de $p$, a próxima iteração $p_1$ estará mais afastada, de modo que a iteração de ponto fixo não convergirá, caso $p_0≠p$.**\n",
    "\n",
    "\n",
    "Uma vez que $g'$ é contínuo em $p$ e $|g'(p)> 1|$ deixando $∈ = |g'(p)| -1$ existe um número $δ> 0$ tal que\n",
    "\n",
    "<p style=\"text-align: center;\"> $|g'(x) - g'(p)| < ∈ = |g'(p)| -1,$\n",
    "\n",
    "sempre que $0 <|x-p| <δ$ desde\n",
    "\n",
    "\n",
    "<p style=\"text-align: center;\"> $|g'(x) - g'(p)|>= |g'(p)|-|g'(x)|$,\n",
    "\n",
    "para qualquer x $0<|x-p|<δ, tenhemos\n",
    "\n",
    "<p style=\"text-align: center;\"> $|g'(x)| >= |g'(p)|-|g'(x)-g'(p)| > |g'(p)-(|g'(p)|-1) = 1$.<p>\n",
    "\n",
    "Se $p_0$ e escolhido então $0 <|p-p_0| <δ$ nós temos o teorema do valor médio que\n",
    "\n",
    "<p style=\"text-align: center;\"> $|(p_1)-p| = |g(p_0) - g(p)| = |g'(ε)||(p_0)-p|,$ <p>\n",
    "\n",
    "para ε entre $p_0$ e $p$. $0<|p-ε| < δ$ então\n",
    "\n",
    "<p style=\"text-align: center;\"> $|(p_1)-p| = |g'(ε)||(p_0)-p| > |(p_0)-p|$. <p>.\n",
    "    \n",
    "\n",
    "Isso significa que, quando uma aproximação se aproxima de $p$, mas não é igual a $p$, os termos sucessivos da seqüência se afastam de $p$. Portanto, a seqüência não pode convergir para $p$. (Burden, 2014)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**(21) Encontre uma função $g$ definida em $[0,1]$ que não satisfaça nenhuma das hipóteses do Teorema 2.3, mas possua um único ponto fixo em $[0,1]$**\n",
    "\n",
    "Um de muitos exemplos é a função $g(x)=\\sqrt{2x-1}$, definida em $[\\frac{1}{2},1]$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 18,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": 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\n",
      "text/plain": [
       "<Figure size 720x720 with 1 Axes>"
      ]
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     "name": "stdout",
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     "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": [
    "**(23)**\n",
    "\n",
    "**a) Utilize o Teorema 2.3 para mostrar que a sequência definida por:**\n",
    "\n",
    "$$x_{n} = \\frac{1}{2}x_{n-1}+\\frac{1}{x_{n-1}},\\,\\,\\,para\\,\\,\\,n\\geq1$$\n",
    "\n",
    "**converge para $\\sqrt{2}$ sempre que $x_{0} > \\sqrt{2}$.**\n",
    "\n",
    "**b) Use o fato de que $0<(x_{0}-\\sqrt{2})^2$ sempre que $x_{0}\\neq \\sqrt{2}$ para mostrar que se $0<x_{0}<\\sqrt{2}$ então $x_{1} > \\sqrt{2}$.**\n",
    "\n",
    "**c) Use os resultados das partes (a) e (b) para mostrar que a sequência em (a) converge para $\\sqrt{2}$, sempre que $x_{0} > 0$.**"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Resposta - Letra a)**\n",
    "\n",
    "Tomamos \n",
    "\n",
    "$$f(x)=\\frac{x}{2} + \\frac{1}{x}$$\n",
    "\n",
    "Quando $x=\\sqrt{2}$, \n",
    "\n",
    "$$f(\\sqrt{2})=\\sqrt{2}$$\n",
    "\n",
    "Quando $x\\neq0$, \n",
    "\n",
    "$$g^{'}(x) = \\frac{1}{2}-\\frac{1}{x^{2}}$$\n",
    "\n",
    "Se $x>\\sqrt{2}$, então $\\frac{1}{x^{2}} < \\frac{1}{\\sqrt{2}}$, assim:\n",
    "$$g^{'}(x)=\\frac{1}{2}-\\frac{1}{(\\sqrt{2})^{2}}>0$$\n",
    "\n",
    "Suponhamos que $x_{0}>\\sqrt{2}$, então\n",
    "\n",
    "$$x_{1}-\\sqrt{2} = g(x_{0}) - g(\\sqrt{2}) = g^{'}(\\xi)(x_{0}-\\sqrt{s})$$\n",
    "\n",
    "onde $\\sqrt{2}<\\xi<x$. Portanto, \n",
    "\n",
    "<center>$x_{1} - \\sqrt{2}>0$ e $x_{1}>\\sqrt{2}$.</center>\n",
    "\n",
    "Além disso:\n",
    "\n",
    "$$x_{1} = \\frac{x_{0}}{2}+\\frac{1}{x_{0}} < \\frac{x_{0}}{2}+\\frac{1}{\\sqrt{2}} = \\frac{x_{0}+\\sqrt{2}}{2}$$\n",
    "\n",
    "e $\\sqrt{2}<x_{1}<x_{0}$. Por indução, temos:\n",
    "\n",
    "$$\\sqrt{2}<x_{n+1}<x_{n}<...<x_{0}$$\n",
    "\n",
    "Portanto, $\\left \\{ x_{n} \\right \\}$ é uma sequência decrescente que tem um limite inferior e deve convergir.\n",
    "\n",
    "Então,\n",
    "\n",
    "$$p = \\lim_{n\\rightarrow \\infty}{(\\frac{x_{n+1}}{2}+\\frac{1}{x_{n-1}})} = \\frac{p}{2}+\\frac{1}{p}$$\n",
    "\n",
    "Portanto,\n",
    "\n",
    "$$p = \\frac{p}{2}+\\frac{1}{p}$$\n",
    "\n",
    "Observe que podemos afirmar que $p=\\pm\\sqrt{2}$, quando $x_{n}>\\sqrt{2},$  $\\forall$ $n$\n",
    "\n",
    "$$\\lim_{n\\rightarrow \\infty}{x_{n}} = \\sqrt{2}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Resposta - Letra b)**\n",
    "\n",
    "Já temos que:\n",
    "\n",
    "$$0<(x_{0}-\\sqrt{2})^2$$\n",
    "\n",
    "e,\n",
    "\n",
    "$$(x_{0}-\\sqrt{2})^2 = x_{0}^{2}-2x_{0}\\sqrt{2}+2$$\n",
    "\n",
    "Logo\n",
    "\n",
    "$$0 < x_{0}^{2}-2x_{0}\\sqrt{2}+2$$\n",
    "\n",
    "E,\n",
    "\n",
    "$$2x_{0}\\sqrt{2}<x_{0}^{2}+2$$\n",
    "\n",
    "Assim,\n",
    "\n",
    "$$\\sqrt{2}<\\frac{x_{0}^{2}}{2x_{0}}+\\frac{2}{2x_{0}} \\Rightarrow \\sqrt{2}<\\frac{x_{0}}{2}+\\frac{1}{x_{0}}=x_{1}$$\n",
    "\n",
    "Portanto,\n",
    "\n",
    "$$x_{1} > \\sqrt{2}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Resposta - Letra c)**\n",
    "\n",
    "Quando analisado o resultado de (A) vemos que a $\\left \\{ x_{n} \\right \\}$ convergirá para $$\\lim_{n\\rightarrow \\infty}{x_{n}} = \\sqrt{2}$$\n",
    "sempre que $0<x_{0}<\\sqrt{2}$ e $\\sqrt{2}<x_{1}$(Já demostrado em B)\n",
    "\n",
    "Se $x_{0} = \\sqrt{2}$, implica que $x_{}=\\sqrt{2}$ para todo $n$, e\n",
    "\n",
    "$$\\lim_{n\\rightarrow \\infty}{x_{n}} = \\sqrt{2}$$\n",
    "\n",
    "Se $x_{0} > \\sqrt{2}$, remete-nos ao já demonstrado em A e\n",
    "\n",
    "$$\\lim_{n\\rightarrow \\infty}{x_{n}} = \\sqrt{2}$$"
   ]
  },
  {
   "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": [
    "# Questões de Discussão#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**(1) Forneça uma visão geral de como a teoria do caos e a iteração do ponto fixo estão relacionadas.**\n",
    "\n",
    "A teoria do caos estuda comportamentos aparentemente aleatórios, com causas puramente deterministas. A convergência do método do ponto fixo segue um comportamento que pode ser considerado caótico em alguns casos, difícil de prever, divergindo ou convergindo depedendo da função iteração escolhida. Os fractais são figuras que não seguem a geometria clássica euclidiana, possuíndo uma construção irregular e aleatória com autossimilaridade das formas e características. A Geometria Fractal é considerada a geometria da Teoria do Caos. Aproveitando a aleatoriedade da convergência do método iterativo do Ponto Fixo, ele pode ser utilizado para a construção de algoritmos que gerem fractais."
   ]
  }
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