{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Chapter 6 - Roots of non linear equations"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example No. 6_01 Pg No. 126"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The largest possible root is  x1 = [[4]]\n",
      "No root can be larger than the value = [[4]]\n",
      "\n",
      "All real roots lie in the interval (-3.741657,3.741657)\n",
      "\n",
      "We can use these two points as initial guesses for the bracketing methods and one of them for open end methods\n"
     ]
    }
   ],
   "source": [
    "from numpy import mat,shape\n",
    "from math import sqrt\n",
    "\n",
    "#Possible Initial guess values for roots\n",
    "\n",
    "A = mat([[ 2],[-8] ,[2],[12]]) # Coefficients of x terms in the decreasing order of power\n",
    "n = shape(A)#\n",
    "x1 = -A[1]/A[0]#\n",
    "print 'The largest possible root is  x1 =',x1\n",
    "print 'No root can be larger than the value =',x1\n",
    "\n",
    "x = sqrt((A[1]/A[0])**2 - 2*(A[2]/A[0])**2)\n",
    "\n",
    "print '\\nAll real roots lie in the interval (-%f,%f)\\n'%(x,x)\n",
    "print 'We can use these two points as initial guesses for the bracketing methods and one of them for open end methods'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example No. 6_03 Pg No. "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "p(4) = 1\n",
      "\n",
      " p(3)= -2\n",
      "\n",
      "\n",
      " p(2)= 1\n",
      "\n",
      "\n",
      " f(2) = p(1) = 0\n"
     ]
    }
   ],
   "source": [
    "from numpy import mat,shape\n",
    "from math import sqrt\n",
    "#Evaluating Polynomial using Horner's rule\n",
    "\n",
    "#Coefficients of x terms in the increasing order of power\n",
    "A = mat([[6],[1],[-4],[1]])\n",
    "x = 2\n",
    "N=shape(A)\n",
    "n,c = N[0],N[1]\n",
    "p=[0,0,0]\n",
    "p.append(A[n-1])\n",
    "print 'p(4) =',p[n-1][0,0]\n",
    "for i in range(1,n-1):\n",
    "    p[n-i]= p[n-i+1-1]*x + A[n-i-1]\n",
    "    print '\\n p(%d)= %d\\n'%(n-i,p[n-i])\n",
    "\n",
    "print '\\n f(%d) = p(1) = %d'%(x,p[0])\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example No. 6_04 Pg No. 132"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "First finding the interval that contains a root,this can be done by using Eq 6.10\n",
      "\n",
      " Both the roots lie in the interval (-6,6) \n",
      "\n",
      "\n",
      " The root lies in the interval (5,5)\n",
      "\n"
     ]
    }
   ],
   "source": [
    "from numpy import poly1d,polyval, sqrt\n",
    "\n",
    "#Root of a Equation Using Bisection Method\n",
    "\n",
    "#Coefficients in increasing order of power of x starting from 0\n",
    "A = [-10 ,-4, 1]#\n",
    "print 'First finding the interval that contains a root,this can be done by using Eq 6.10'\n",
    "xmax = sqrt((A[1]/A[2])**2 - 2*(A[0]/A[2]))\n",
    "print '\\n Both the roots lie in the interval (-%d,%d) \\n'%(xmax,xmax)\n",
    "x = range(-6,7)\n",
    "p= poly1d(A)# p = poly(A,'x'%('c'\n",
    "\n",
    "fx = [p(xx) for xx in x]#\n",
    "for i in range(0,12):\n",
    "    if fx[i]*fx[i] < 0:\n",
    "        break \n",
    "    \n",
    "\n",
    "print '\\n The root lies in the interval (%d,%d)\\n'%(x[i],x[i])\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example No. 6_05 Pg No. 139"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Iteration No. 1 \n",
      "\n",
      "x0 = 1.000000 \n",
      "\n",
      "Iteration No. 2 \n",
      "\n",
      "x0 = 1.000000 \n",
      "\n",
      "Iteration No. 3 \n",
      "\n",
      "x0 = 1.000000 \n",
      "\n",
      "Iteration No. 4 \n",
      "\n",
      "x0 = 1.000000 \n",
      "\n",
      "Iteration No. 5 \n",
      "\n",
      "x0 = 1.000000 \n",
      "\n",
      "Iteration No. 6 \n",
      "\n",
      "x0 = 1.000000 \n",
      "\n",
      "Iteration No. 7 \n",
      "\n",
      "x0 = 1.000000 \n",
      "\n",
      "Iteration No. 8 \n",
      "\n",
      "x0 = 1.000000 \n",
      "\n",
      "Iteration No. 9 \n",
      "\n",
      "x0 = 1.000000 \n",
      "\n",
      "Iteration No. 10 \n",
      "\n",
      "x0 = 1.000000 \n",
      "\n",
      "Iteration No. 11 \n",
      "\n",
      "x0 = 1.000000 \n",
      "\n",
      "Iteration No. 12 \n",
      "\n",
      "x0 = 1.000000 \n",
      "\n",
      "Iteration No. 13 \n",
      "\n",
      "x0 = 1.000000 \n",
      "\n",
      "Iteration No. 14 \n",
      "\n",
      "x0 = 1.000000 \n",
      "\n",
      "Iteration No. 15 \n",
      "\n",
      "x0 = 1.000000 \n",
      "\n"
     ]
    }
   ],
   "source": [
    "from numpy import poly1d,polyval\n",
    "\n",
    "#False Position Method\n",
    "\n",
    "#Coefficients of polynomial in increasing order of power of x\n",
    "A = [-2 , -1,  1]\n",
    "x1 = 1 #\n",
    "x2 = 3 #\n",
    "fx = poly1d(A)\n",
    "for i in range(1,16):\n",
    "    print 'Iteration No. %d \\n'%(i)\n",
    "    fx1 = fx(x1)\n",
    "    fx2 = fx(x2)\n",
    "    x0 = x1 - fx1*(x2-x1)/(fx2-fx1) \n",
    "    print 'x0 = %f \\n'%(x0)#\n",
    "    fx0 = fx(x0)#\n",
    "    if fx1*fx0 < 0:\n",
    "        x2 = x0 \n",
    "    else:\n",
    "        x1 = x0 #\n",
    "      "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example No. 6_07 Pg No. 147"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "x2 = 1.000000\n",
      "\n",
      "Since f(1.000000) = 0, the root closer to the point x = 0 is 1.000000 \n",
      "\n"
     ]
    }
   ],
   "source": [
    "from numpy import poly1d,polyval,polyder\n",
    "#Root of the Equation using Newton Raphson Method\n",
    "\n",
    "#Coefficients of polynomial in increasing order of power of x\n",
    "A = [ 2,  -3,  1]#\n",
    "fx = poly1d(A)\n",
    "dfx = polyder(fx)\n",
    "\n",
    "x=[0]\n",
    "f=[]\n",
    "df=[]\n",
    "for i in range(1,11):\n",
    "    f.append(fx(x[i-1]))\n",
    "    if f[i-1] != 0:\n",
    "        df.append(dfx(x[i-1]))\n",
    "        x.append(x[i-1] - f[i-1]/df[i-1])\n",
    "        print 'x%d = %f\\n'%(i+1,x[i])#    \n",
    "    else:\n",
    "        print 'Since f(%f) = 0, the root closer to the point x = 0 is %f \\n'%(x[i-1],x[i-1] )\n",
    "        break\n",
    "    "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example No. 6_08 Pg No. 151"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "x2 = 3.342105\n",
      "\n",
      "x3 = 2.248631\n",
      "\n",
      "x4 = 1.535268\n",
      "\n",
      "x5 = 1.079139\n",
      "\n",
      "x6 = 0.797330\n",
      "\n",
      "x7 = 0.632716\n",
      "\n",
      "From the results we can see that number of correct digits approximately doubles with each iteration\n"
     ]
    }
   ],
   "source": [
    "from numpy import poly1d,polyval,polyder\n",
    "\n",
    "#Root of the Equation using Newton Raphson Method\n",
    "\n",
    "#Coefficients of polynomial in increasing order of power of x\n",
    "A = [ 6,  1 , -4 , 1 ]\n",
    "fx = poly1d(A)\n",
    "dfx = polyder(fx)\n",
    "f=[];df=[]\n",
    "x = [5.0] #\n",
    "for i in range(1,7):\n",
    "    f.append(fx(x[i-1]))\n",
    "    if f[i-1] != 0:\n",
    "        df.append(dfx(x[i-1]))\n",
    "        x.append(x[i-1] - f[i-1]/df[i-1])\n",
    "        print 'x%d = %f\\n'%(i+1,x[i])\n",
    "    \n",
    "\n",
    "print 'From the results we can see that number of correct digits approximately doubles with each iteration'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example No. 6_09 Pg No. 153"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " For Iteration No. 1\n",
      "\n",
      "\n",
      " x1 = 4.000000\n",
      " x2 = 2.000000 \n",
      " fx1 = -175.000000 \n",
      " fx2 = -47.000000 \n",
      " x3 = 1.265625 \n",
      "\n",
      "\n",
      " For Iteration No. 2\n",
      "\n",
      "\n",
      " x1 = 2.000000\n",
      " x2 = 1.265625 \n",
      " fx1 = -47.000000 \n",
      " fx2 = -20.080566 \n",
      " x3 = 0.717818 \n",
      "\n",
      "\n",
      " For Iteration No. 3\n",
      "\n",
      "\n",
      " x1 = 1.265625\n",
      " x2 = 0.717818 \n",
      " fx1 = -20.080566 \n",
      " fx2 = -7.023891 \n",
      " x3 = 0.423122 \n",
      "\n",
      "\n",
      " For Iteration No. 4\n",
      "\n",
      "\n",
      " x1 = 0.717818\n",
      " x2 = 0.423122 \n",
      " fx1 = -7.023891 \n",
      " fx2 = -2.482815 \n",
      " x3 = 0.261999 \n",
      "\n",
      "\n",
      " For Iteration No. 5\n",
      "\n",
      "\n",
      " x1 = 0.423122\n",
      " x2 = 0.261999 \n",
      " fx1 = -2.482815 \n",
      " fx2 = -0.734430 \n",
      " x3 = 0.194317 \n",
      "\n",
      "\n",
      " For Iteration No. 6\n",
      "\n",
      "\n",
      " x1 = 0.261999\n",
      " x2 = 0.194317 \n",
      " fx1 = -0.734430 \n",
      " fx2 = -0.154860 \n",
      " x3 = 0.176233 \n",
      "\n",
      "This can be still continued further for accuracy\n"
     ]
    }
   ],
   "source": [
    "from __future__ import division\n",
    "from numpy import poly1d\n",
    "#Root of the equation using SECANT Method\n",
    "\n",
    "#Coefficients of polynomial in increasing order of power of x\n",
    "A = [ -10,  -4,  1]\n",
    "x1 = 4 #\n",
    "x2 = 2 #\n",
    "fx = poly1d(A)\n",
    "for i in range(1,7):\n",
    "    print '\\n For Iteration No. %d\\n'%(i)\n",
    "    fx1 = fx(x1)\n",
    "    fx2 = fx(x2)\n",
    "    x3 = x2 - fx2*(x2-x1)/(fx2-fx1) #\n",
    "    print '\\n x1 = %f\\n x2 = %f \\n fx1 = %f \\n fx2 = %f \\n x3 = %f \\n'%(x1,x2,fx1,fx2,x3) #\n",
    "    x1 = x2#\n",
    "    x2 = x3#\n",
    "\n",
    "print 'This can be still continued further for accuracy'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example No. 6_11 Pg No. 161"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " x1 = -1.000000\n",
      "\n",
      "\n",
      " x2 = 1.000000\n",
      "\n",
      "\n",
      " x3 = 1.000000\n",
      "\n",
      "\n",
      "1.000000 is root of the equation,since x3 - x2 = 0 \n",
      "\n",
      "\n",
      "x1 = 1.000000\n",
      "\n",
      "\n",
      "x2 = 1.000000\n",
      "\n",
      "\n",
      " 1.000000 is root of the equation,since x2 - x1 = 0\n"
     ]
    }
   ],
   "source": [
    "from numpy import poly1d\n",
    "from __future__ import division\n",
    "#Fixed point method\n",
    "\n",
    "#Coefficients of polynomial in increasing order of power of x\n",
    "A = [ -2,  1,  1 ]#\n",
    "B = [ 2, 0, -1 ]#\n",
    "gx = poly1d(B)\n",
    "x = [0] ##initial guess x0 = 0\n",
    "for i in range(2,11):\n",
    "    x.append (gx(x[i-2]))\n",
    "    print '\\n x%d = %f\\n'%(i-1,x[i-1])\n",
    "    if (x[i-1]-x[(i-2)]) == 0:\n",
    "        print '\\n%f is root of the equation,since x%d - x%d = 0 \\n'%(x[i-1],i-1,i-2)\n",
    "        break\n",
    "    \n",
    "\n",
    "#Changing initial guess x0 = -1\n",
    "x[0] = -1 #\n",
    "for i in range(2,11):\n",
    "    x[i-1]= gx(x[i-2])\n",
    "    print '\\nx%d = %f\\n'%(i-1,x[i-1])\n",
    "    if (x[i-1]-x[i-2]) == 0:\n",
    "        print '\\n %f is root of the equation,since x%d - x%d = 0'%(x[i-1],i-1,i-2)\n",
    "        break"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example No. 6_12 Pg No. 162"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " x0 = 1.000000 \n",
      "\n",
      " x1 = 2.500000 \n",
      "\n",
      " x2 = 1.666667 \n",
      "\n",
      " x3 = 1.250000 \n",
      "\n",
      " x4 = 1.000000 \n",
      "\n",
      "\n",
      " x0 = 0.000000 \n",
      "\n",
      " x1 = 1.000000 \n",
      "\n",
      " x2 = 7.000000 \n",
      "\n",
      " x3 = 15.000000 \n",
      "\n",
      " x4 = 25.000000 \n",
      "\n",
      "\n",
      " x0 = 1.000000 \n",
      "\n",
      " x1 = 2.250000 \n",
      "\n",
      " x2 = 2.333333 \n",
      "\n",
      " x3 = 2.625000 \n",
      "\n",
      " x4 = 3.000000 \n",
      "\n",
      " x5 = 3.416667 \n",
      "\n",
      " x6 = 3.857143 \n",
      "\n"
     ]
    }
   ],
   "source": [
    "#Fixed point method\n",
    "\n",
    "A = [ -5,  0,  1 ]#\n",
    "def g(x):\n",
    "    x = 5.0/x\n",
    "    return x\n",
    "x = [1] #\n",
    "print '\\n x0 = %f \\n'%(x[0])\n",
    "for i in range(2,6):\n",
    "    x.append(g(i))\n",
    "    print ' x%d = %f \\n'%(i-1,x[i-1])\n",
    "    \n",
    "\n",
    "#Defining g(x) in different way\n",
    "def g(x):\n",
    "    x = x**2 + x - 5\n",
    "    return x\n",
    "x=[0]\n",
    "print '\\n x0 = %f \\n'%(x[0])\n",
    "for i in range(2,6):\n",
    "    x.append(g(i))\n",
    "    print ' x%d = %f \\n'%(i-1,x[i-1])\n",
    "\n",
    "#Third form of g(x)\n",
    "def g(x):\n",
    "    x = (x + 5/x)/2\n",
    "    return x\n",
    "x=[1]\n",
    "print '\\n x0 = %f \\n'%(x[0])\n",
    "for i in range(2,8):\n",
    "    x.append(g(i))\n",
    "    print ' x%d = %f \\n'%(i-1,x[i-1])"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example No. 6_13 Pg No. 169"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " x**2 - y**2 = 3 \n",
      " x**2 + x*y \n",
      "\n",
      "\n",
      " x0 = 1.000000 \n",
      " y0 = 1.000000 \n",
      "\n",
      "\n",
      " x1 = 2.500000 \n",
      " y1 = 5.000000 \n",
      "\n",
      "\n",
      " x2 = 2.750000 \n",
      " y2 = 1.000000 \n",
      "\n",
      "\n",
      " x3 = 3.500000 \n",
      " y3 = -1.000000 \n",
      "\n"
     ]
    }
   ],
   "source": [
    "#Solving System of non-linear equations using FIXED POINT METHOD\n",
    "\n",
    "print ' x**2 - y**2 = 3 \\n x**2 + x*y \\n'\n",
    "def f(x,y):\n",
    "    x = y + 3/(x+y)\n",
    "    return x\n",
    "def g(x):\n",
    "    y = (6-x**2)/x\n",
    "    return y\n",
    "x=[1]\n",
    "y=[1]\n",
    "print '\\n x0 = %f \\n y0 = %f \\n'%(x[0],y[0])\n",
    "for i in range(2,5):\n",
    "    x.append(f((i-1),(i-1)))\n",
    "    y.append(g((i-1)))\n",
    "    print '\\n x%d = %f \\n y%d = %f \\n'%(i-1,x[i-1],i-1,y[i-1])"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example No. 6_14 Pg No. 172"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "x**2 + x*y = 6 \n",
      " x**2 - y**2 = 3 \n",
      "\n",
      "\n",
      " x1 = 0.875000 \n",
      " y1 = -0.625000 \n",
      "\n",
      "\n",
      " x2 = -0.437500 \n",
      " y2 = -2.687500 \n",
      "\n"
     ]
    }
   ],
   "source": [
    "#Solving System of Non-linear equations using Newton Raphson Method\n",
    "\n",
    "\n",
    "print 'x**2 + x*y = 6 \\n x**2 - y**2 = 3 \\n'#\n",
    "def F(x,y):\n",
    "    f = x**2 + x*y - 6\n",
    "    return f\n",
    "def G(x,y):\n",
    "    g = x**2 - y**2 -3\n",
    "    return g\n",
    "def dFx(x,y):\n",
    "    f1 = 2*x + y\n",
    "    return f1\n",
    "def dFy(x,y):\n",
    "    f2 = y\n",
    "    return f2\n",
    "def dGx(x,y):\n",
    "    g1 = 2*x\n",
    "    return g1\n",
    "def dGy(x,y):\n",
    "    g2 = -2*y\n",
    "    return g2\n",
    "x=[1]\n",
    "y=[1]\n",
    "\n",
    "for i in range(2,4):\n",
    "    Fval = F(i,i)\n",
    "    Gval = G(i,i)\n",
    "    f1 = dFx(i-1,i-1)\n",
    "    f2 = dFy(i-1,i-1)\n",
    "    g1 = dGx(i-1,i-1)\n",
    "    g2 = dGy(i-1,i-1)\n",
    "    D =  f1*g2 - f2*g1 \n",
    "    \n",
    "    x.append(x[i-2] - (Fval*g2 - Gval*f2)/D )\n",
    "    y.append(y[i-2] - (Gval*f1 - Fval*g1)/D )\n",
    "    print '\\n x%d = %f \\n y%d = %f \\n'%(i-1,x[i-1],i-1,y[i-1])    "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example No. 6_15 Pg No. 176"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "b3 = 0.000000\n",
      "\n",
      "b2 = 0.000000\n",
      "\n",
      "b1 = 0.000000\n",
      "\n",
      "Thus the polynomial is\n",
      "    2\n",
      "15 x - 7 x + 1\n"
     ]
    }
   ],
   "source": [
    "from __future__ import division\n",
    "from numpy import poly1d, arange\n",
    "#Synthetic Division\n",
    "\n",
    "a = [-9 ,15 ,-7, 1]\n",
    "b=[0,0,0,0]\n",
    "for i in arange(3,0,-1):\n",
    "    b[i]= a[i] + b[i]*3\n",
    "    print 'b%d = %f\\n'%(i,b[i-1])\n",
    "print 'Thus the polynomial is'    \n",
    "print poly1d(b)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example No. 6_16 Pg No. 187"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "b3 = 1.000000 \n",
      "\n",
      "b2 = 1.800000 \n",
      "\n",
      "b1 = 3.240000 \n",
      "\n",
      "b0 = 14.032000 \n",
      "\n",
      "c3 = 1.000000 \n",
      "\n",
      "c2 = 1.800000 \n",
      "\n",
      "c1 = 3.240000 \n",
      "\n",
      "c0 = 14.032000 \n",
      "\n",
      "\n",
      " D = 4.240000 \n",
      " du = -0.694340 \n",
      " dv = -5.250566 \n",
      " u = 1.105660\n",
      " v = -1.694340 \n",
      "\n"
     ]
    }
   ],
   "source": [
    "#Quadratic factor of a polynomial using Bairstow's Method\n",
    "\n",
    "a = [ 10, 1 ,0 ,1]#\n",
    "n = len(a)#\n",
    "u = 1.8 #\n",
    "v = -1 #\n",
    "b=[];c=[]\n",
    "for nn in range(n):\n",
    "    b.append(0)\n",
    "    c.append(0)\n",
    "b[n-1] = a[n-1]\n",
    "b[n-2] = a[n-2] + u*b[n-1]\n",
    "c[n-1] = 0 \n",
    "c[n-2] = b[n-1]\n",
    "\n",
    "\n",
    "for i in range(n-2,0,-1):\n",
    "    b[i-1]= a[i-1]+ u*b[i] + v*b[i+1]\n",
    "    c[i-1]= b[i] + u*c[i] + v*c[i+1] \n",
    "\n",
    "\n",
    "for i in range(n,0,-1):\n",
    "    print 'b%d = %f \\n'%(i-1,b[i-1])\n",
    "\n",
    "for i in range(n,0,-1):\n",
    "    print 'c%d = %f \\n'%(i-1,b[i-1])\n",
    "\n",
    "\n",
    "D = c[1]*c[1] - c[0]*c[2]\n",
    "du = -1*(b[1]*c[1] - c[0]*c[2])/D \n",
    "dv = -1*(b[0]*c[1] - b[1]*c[0])/D \n",
    "u = u + du #\n",
    "v = v + du #\n",
    "print '\\n D = %f \\n du = %f \\n dv = %f \\n u = %f\\n v = %f \\n'%(D,du,dv,u,v)\n",
    "    "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example No. 6_17 Pg No. 197"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " x1 = 0.000000\n",
      " x2 = 1.000000\n",
      " x3 = 2.000000\n",
      " f1 = -20.000000\n",
      " f2 = -7.000000\n",
      " f3 = 16.000000\n",
      " h1 = -2.000000\n",
      " h2 = -1.000000\n",
      " d1 = -36.000000\n",
      " d2 = -23.000000\n",
      " a0 = 16.000000\n",
      " a1 = 28.000000\n",
      " a2 = 5.000000\n",
      " h4 = -0.645934\n",
      " x4 = 1.354066\n",
      " \n",
      "\n",
      " x1 = 1.000000\n",
      " x2 = 2.000000\n",
      " x3 = 1.354066\n",
      " f1 = -7.000000\n",
      " f2 = 16.000000\n",
      " f3 = -0.309679\n",
      " h1 = -0.354066\n",
      " h2 = 0.645934\n",
      " d1 = -6.690321\n",
      " d2 = 16.309679\n",
      " a0 = -0.309679\n",
      " a1 = 21.145451\n",
      " a2 = 6.354066\n",
      " h4 = 0.014581\n",
      " x4 = 1.368647\n",
      " \n",
      "\n",
      " x1 = 2.000000\n",
      " x2 = 1.354066\n",
      " x3 = 1.368647\n",
      " f1 = 16.000000\n",
      " f2 = -0.309679\n",
      " f3 = -0.003394\n",
      " h1 = 0.631353\n",
      " h2 = -0.014581\n",
      " d1 = 16.003394\n",
      " d2 = -0.306286\n",
      " a0 = -0.003394\n",
      " a1 = 21.103381\n",
      " a2 = 6.722713\n",
      " h4 = 0.000161\n",
      " x4 = 1.368808\n",
      " \n",
      "\n",
      " x1 = 1.354066\n",
      " x2 = 1.368647\n",
      " x3 = 1.368808\n",
      " f1 = -0.309679\n",
      " f2 = -0.003394\n",
      " f3 = -0.000001\n",
      " h1 = -0.014742\n",
      " h2 = -0.000161\n",
      " d1 = -0.309678\n",
      " d2 = -0.003392\n",
      " a0 = -0.000001\n",
      " a1 = 21.096136\n",
      " a2 = 6.091521\n",
      " h4 = 0.000000\n",
      " x4 = 1.368808\n",
      " \n",
      "root of the polynomial is x4 = 1.368808\n"
     ]
    }
   ],
   "source": [
    "from math import sqrt\n",
    "#Solving Leonard's equation using MULLER'S Method\n",
    "\n",
    "def f(x):\n",
    "    y = x**3 + 2*x**2 + 10*x - 20\n",
    "    return y\n",
    "x1 = 0 #\n",
    "x2 = 1 #\n",
    "x3 = 2 #\n",
    "for i in range(1,11):\n",
    "    f1 = f(x1)\n",
    "    f2 = f(x2)\n",
    "    f3 = f(x3)\n",
    "    h1 = x1-x3 #\n",
    "    h2 = x2-x3 #\n",
    "    d1 = f1 - f3 #\n",
    "    d2 = f2 - f3 #\n",
    "    D = h1*h2*(h1-h2)#\n",
    "    a0 = f3 #\n",
    "    a1 = (d2*h1**2 - d1*h2**2)/D #\n",
    "    a2 = (d1*h2 - d2*h1)/D #\n",
    "    if abs(-2*a0/( a1 + sqrt( a1**2 - 4*a0*a2 ) )) < abs( -2*a0/( a1 - sqrt( a1**2 - 4*a0*a2 ) )):\n",
    "        h4 = -2*a0/(a1 + sqrt(a1**2 - 4*a0*a2))#\n",
    "    else:\n",
    "        h4 = -2*a0/(a1 - sqrt(a1**2 - 4*a0*a2))\n",
    "    \n",
    "    x4 = x3 + h4 #\n",
    "    print '\\n x1 = %f\\n x2 = %f\\n x3 = %f\\n f1 = %f\\n f2 = %f\\n f3 = %f\\n h1 = %f\\n h2 = %f\\n d1 = %f\\n d2 = %f\\n a0 = %f\\n a1 = %f\\n a2 = %f\\n h4 = %f\\n x4 = %f\\n '%(x1,x2,x3,f1,f2,f3,h1,h2,d1,d2,a0,a1,a2,h4,x4) #\n",
    "    relerr = abs((x4-x3)/x4)#\n",
    "    if relerr <= 0.00001:\n",
    "        print 'root of the polynomial is x4 = %f'%(x4)\n",
    "        break\n",
    "    \n",
    "    x1 = x2 #\n",
    "    x2 = x3 #\n",
    "    x3 = x4 #"
   ]
  }
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