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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Chapter 24: Numerical Methods"
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"## Example 24.1, page no. 636"
]
},
{
"cell_type": "code",
"execution_count": 9,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Finding roots of this equation by bisection method\n",
"f(2) is -ve and f(3) is +ve so root lies between 2 and 3\n",
"The root is: 2.6875\n"
]
}
],
"source": [
"import numpy\n",
"\n",
"x = numpy.poly([0])\n",
"p = x**3-4*x-9\n",
"print \"Finding roots of this equation by bisection method\"\n",
"print 'f(2) is -ve and f(3) is +ve so root lies between 2 and 3'\n",
"l = 2.\n",
"m = 3.\n",
"def f(x):\n",
" y = x**3-4*x-9\n",
" return y\n",
"for i in range(1,5):\n",
" k = 1.0/2.*(l+m)\n",
" if(f(k)<0):\n",
" l = k\n",
" else:\n",
" m = k\n",
"print \"The root is: \", k"
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"## Example 24.3, page no. 638"
]
},
{
"cell_type": "code",
"execution_count": 17,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"f(x)=xeˆx−cos(x)\n",
"We are required to find the roots of f(x) by the method of false position \n",
"f(0)=−ve and f(1)=+ve so s root lie between 0 and 1 \n",
"finding the roots by false position method \n",
"The root of the equation is :\n",
"0.517747878322\n"
]
}
],
"source": [
"import math\n",
"\n",
"print \"f(x)=xeˆx−cos(x)\"\n",
"def f(x):\n",
" y = x*math.e**(x)-math.cos(x)\n",
" return y\n",
"print \"We are required to find the roots of f(x) by the method of false position \"\n",
"print \"f(0)=−ve and f(1)=+ve so s root lie between 0 and 1 \"\n",
"print \"finding the roots by false position method \"\n",
"l = 0.0\n",
"m = 1.0\n",
"for i in range(1,11):\n",
" k = l-(m-l)*f(l)/(f(m)-f(l))\n",
" if(f(k)<0):\n",
" l = k\n",
" else:\n",
" m = k\n",
"print \"The root of the equation is :\"\n",
"print k"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 24.4, page no. 638"
]
},
{
"cell_type": "code",
"execution_count": 16,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"f(x) = x∗math.log(x)−1.2\n",
"We are required to find the roots of f(x) by the method of false position \n",
"f(2)=−ve and f(3)=+ve so s root lie between 2 and 3 \n",
"finding the roots by false position method \n",
"The root of the equation is : 2.74063625664\n"
]
}
],
"source": [
"import math\n",
"\n",
"print \"f(x) = x∗math.log(x)−1.2\"\n",
"def f(x):\n",
" y = x*math.log10(x)-1.2\n",
" return y\n",
"print \"We are required to find the roots of f(x) by the method of false position \"\n",
"print \"f(2)=−ve and f(3)=+ve so s root lie between 2 and 3 \"\n",
"print \"finding the roots by false position method \"\n",
"l = 2.\n",
"m = 3.\n",
"for i in range(1,4):\n",
" k = l-(m-l)*f(l)/(f(m)-f(l))\n",
" if(f(k)<0):\n",
" l = k\n",
" else:\n",
" m = k\n",
"print \"The root of the equation is : \",k"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 24.5, page no. 639"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"To find the roots of f(x) = 3x−cos(x)−1 by newtons method \n",
"f(0)=−ve and f(1) is +ve so a root lies between 0 and 1 \n",
"Let us take x0 =0.6 as the root is closer to 1 \n",
"Root is given by r=x0−f(xn)/der(f(xn))\n",
"Approximated root in each steps are \n",
"0.607290551153\n",
"0.607096741973\n",
"0.607101775605\n"
]
}
],
"source": [
"import math,numpy\n",
"from scipy.misc import derivative\n",
"\n",
"print \"To find the roots of f(x) = 3x−cos(x)−1 by newtons method \"\n",
"print \"f(0)=−ve and f(1) is +ve so a root lies between 0 and 1 \"\n",
"l = 0\n",
"m = 1\n",
"def f(x):\n",
" y = 3*x-math.cos(x)-1\n",
" return y\n",
"x0 = 0.6\n",
"print \"Let us take x0 =0.6 as the root is closer to 1 \"\n",
"print \"Root is given by r=x0−f(xn)/der(f(xn))\"\n",
"print \"Approximated root in each steps are \"\n",
"for i in range(1,4):\n",
" k = x0-f(x0)/derivative(f,x0)\n",
" print k\n",
" x0 = k"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 24.6, page no. 640"
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"To find square root of 28 by newtons method let x=sqrt(28) ie xˆ2−28=0 \n",
"To find the roots by newtons method\n",
"f(5)=−ve and f(6) is +ve so a root lies between 5 and 6 \n",
"Let us take x0 = 5.5 \n",
"Root is given by rn=xn−f(xn)/der(f(xn))\n",
"Approximated root in each steps are\n",
"5.29545454545\n",
"5.29150409676\n",
"5.29150262213\n",
"5.29150262213\n"
]
}
],
"source": [
"from scipy.misc import derivative\n",
"\n",
"print \"To find square root of 28 by newtons method let x=sqrt(28) ie xˆ2−28=0 \"\n",
"def f(x):\n",
" y = x**2-28\n",
" return y\n",
"print \"To find the roots by newtons method\"\n",
"print \"f(5)=−ve and f(6) is +ve so a root lies between 5 and 6 \"\n",
"l = 5\n",
"m = 6\n",
"print \"Let us take x0 = 5.5 \"\n",
"print \"Root is given by rn=xn−f(xn)/der(f(xn))\"\n",
"print \"Approximated root in each steps are\"\n",
"x0 = 5.5\n",
"for i in range(1,5):\n",
" k = x0-f(x0)/derivative(f,x0)\n",
" print k\n",
" x0 = k"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 24.7, page no. 641"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"To find square root of 28 by newtons method let x=sqrt(28) ie xˆ2−28=0 \n",
"To find the roots by newtons method\n",
"f(5)=−ve and f(6) is +ve so a root lies between 5 and 6 \n",
"Let us take x0 = 5.5 \n",
"Root is given by rn=xn−f(xn)/der(f(xn))\n",
"Approximated root in each steps are\n",
"5.29545454545\n",
"5.29150409676\n",
"5.29150262213\n",
"5.29150262213\n"
]
}
],
"source": [
"from scipy.misc import derivative\n",
"\n",
"print \"To find square root of 28 by newtons method let x=sqrt(28) ie xˆ2−28=0 \"\n",
"def f(x):\n",
" y = x**2-28\n",
" return y\n",
"print \"To find the roots by newtons method\"\n",
"print \"f(5)=−ve and f(6) is +ve so a root lies between 5 and 6 \"\n",
"l = 5\n",
"m = 6\n",
"print \"Let us take x0 = 5.5 \"\n",
"print \"Root is given by rn=xn−f(xn)/der(f(xn))\"\n",
"print \"Approximated root in each steps are\"\n",
"x0 = 5.5\n",
"for i in range(1,5):\n",
" k = x0-f(x0)/derivative(f,x0)\n",
" print k\n",
" x0 = k"
]
}
],
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|
