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authorTrupti Kini2016-06-22 23:30:24 +0600
committerTrupti Kini2016-06-22 23:30:24 +0600
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Added(A)/Deleted(D) following books
A Higher_Engineering_Mathematics_by_B._S._Grewal/chapter10_2.ipynb A Higher_Engineering_Mathematics_by_B._S._Grewal/chapter13_2.ipynb A Higher_Engineering_Mathematics_by_B._S._Grewal/chapter1_2.ipynb A Higher_Engineering_Mathematics_by_B._S._Grewal/chapter21_2.ipynb A Higher_Engineering_Mathematics_by_B._S._Grewal/chapter22_2.ipynb A Higher_Engineering_Mathematics_by_B._S._Grewal/chapter23_2.ipynb A Higher_Engineering_Mathematics_by_B._S._Grewal/chapter24_2.ipynb A Higher_Engineering_Mathematics_by_B._S._Grewal/chapter26_2.ipynb A Higher_Engineering_Mathematics_by_B._S._Grewal/chapter27_2.ipynb A Higher_Engineering_Mathematics_by_B._S._Grewal/chapter28_2.ipynb A Higher_Engineering_Mathematics_by_B._S._Grewal/chapter2_2.ipynb A Higher_Engineering_Mathematics_by_B._S._Grewal/chapter34_2.ipynb A Higher_Engineering_Mathematics_by_B._S._Grewal/chapter35_2.ipynb A Higher_Engineering_Mathematics_by_B._S._Grewal/chapter4_2.ipynb A Higher_Engineering_Mathematics_by_B._S._Grewal/chapter5_2.ipynb A Higher_Engineering_Mathematics_by_B._S._Grewal/chapter6_2.ipynb A Higher_Engineering_Mathematics_by_B._S._Grewal/chapter9_2.ipynb A Higher_Engineering_Mathematics_by_B._S._Grewal/screenshots/screen1_2.png A Higher_Engineering_Mathematics_by_B._S._Grewal/screenshots/screen2_2.png A Higher_Engineering_Mathematics_by_B._S._Grewal/screenshots/screen3_2.png A Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter1.ipynb A Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter10.ipynb A Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter2.ipynb A Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter3.ipynb A Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter4.ipynb A Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter5.ipynb A Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter6.ipynb A Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter7.ipynb A Linear_Algebra_by_K._Hoffman_and_R._Kunze/Chapter8.ipynb A Linear_Algebra_by_K._Hoffman_and_R._Kunze/screenshots/InverseMatrix.png A Linear_Algebra_by_K._Hoffman_and_R._Kunze/screenshots/determinant.png A Linear_Algebra_by_K._Hoffman_and_R._Kunze/screenshots/scalorpolynomial.png
Diffstat (limited to 'Higher_Engineering_Mathematics_by_B._S._Grewal')
-rw-r--r--Higher_Engineering_Mathematics_by_B._S._Grewal/chapter10_2.ipynb720
-rw-r--r--Higher_Engineering_Mathematics_by_B._S._Grewal/chapter13_2.ipynb600
-rw-r--r--Higher_Engineering_Mathematics_by_B._S._Grewal/chapter1_2.ipynb737
-rw-r--r--Higher_Engineering_Mathematics_by_B._S._Grewal/chapter21_2.ipynb1304
-rw-r--r--Higher_Engineering_Mathematics_by_B._S._Grewal/chapter22_2.ipynb244
-rw-r--r--Higher_Engineering_Mathematics_by_B._S._Grewal/chapter23_2.ipynb762
-rw-r--r--Higher_Engineering_Mathematics_by_B._S._Grewal/chapter24_2.ipynb331
-rw-r--r--Higher_Engineering_Mathematics_by_B._S._Grewal/chapter26_2.ipynb522
-rw-r--r--Higher_Engineering_Mathematics_by_B._S._Grewal/chapter27_2.ipynb1084
-rw-r--r--Higher_Engineering_Mathematics_by_B._S._Grewal/chapter28_2.ipynb437
-rw-r--r--Higher_Engineering_Mathematics_by_B._S._Grewal/chapter2_2.ipynb1360
-rw-r--r--Higher_Engineering_Mathematics_by_B._S._Grewal/chapter34_2.ipynb1414
-rw-r--r--Higher_Engineering_Mathematics_by_B._S._Grewal/chapter35_2.ipynb536
-rw-r--r--Higher_Engineering_Mathematics_by_B._S._Grewal/chapter4_2.ipynb1208
-rw-r--r--Higher_Engineering_Mathematics_by_B._S._Grewal/chapter5_2.ipynb317
-rw-r--r--Higher_Engineering_Mathematics_by_B._S._Grewal/chapter6_2.ipynb743
-rw-r--r--Higher_Engineering_Mathematics_by_B._S._Grewal/chapter9_2.ipynb476
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+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 10 : Fourier Series"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 10.1, page no. 327"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 20,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "finding the fourier series of given function\n",
+ "enter the no of terms up to each of sin or cos terms in the expansion : 3\n",
+ "0.158857730350203*sin(x) + 0.127086184280162*sin(2*x) + 0.0953146382101218*sin(3*x) + 0.158857730350203*cos(x) + 0.0635430921400812*cos(2*x) + 0.0317715460700406*cos(3*x) + 0.158857730350203\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy,math,numpy\n",
+ "\n",
+ "print \"finding the fourier series of given function\"\n",
+ "x = sympy.Symbol('x')\n",
+ "ao = 1/math.pi*sympy.integrate(sympy.exp(-1*x),(x,0,2*math.pi))\n",
+ "s = ao/2\n",
+ "n = int(raw_input(\"enter the no of terms up to each of sin or cos terms in the expansion : \"))\n",
+ "for i in range (1,n+1):\n",
+ " ai = 1/math.pi*sympy.integrate(sympy.exp(-x)*sympy.cos(i*x),(x,0,2*math.pi))\n",
+ " bi = 1/math.pi*sympy.integrate(sympy.exp(-x)*sympy.sin(i*x),(x,0,2*math.pi))\n",
+ " s = s+float(ai)*sympy.cos(i*x)+float(bi)*sympy.sin(i*x)\n",
+ "print s"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 10.2, page no. 328"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 11,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "To find the fourier transform of given function\n",
+ "Piecewise((2, s == 0), (1.0*I*exp(-1.0*I*s)/s - 1.0*I*exp(1.0*I*s)/s, True))\n",
+ "1.57079632679490\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,math,sympy\n",
+ "\n",
+ "print \"To find the fourier transform of given function\"\n",
+ "x = sympy.Symbol('x')\n",
+ "s = sympy.Symbol('s')\n",
+ "F = sympy.integrate(sympy.exp(1j*s*x),(x,-1,1))\n",
+ "print F\n",
+ "F1 = sympy.integrate(sympy.sin(x)/x,(x,0,numpy.inf))\n",
+ "print F1"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 10.3, page no. 328"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 13,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "finding the fourier series of given function\n",
+ "enter the no of terms up to each of sin or cos terms in the expansion : 3\n",
+ "3.0*sin(x) - 0.5*sin(2*x) + 1.0*sin(3*x) - 0.636619772367581*cos(x) - 0.0707355302630646*cos(3*x) - 0.785398163397448\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy,math\n",
+ "\n",
+ "print \"finding the fourier series of given function\"\n",
+ "x = sympy.Symbol('x')\n",
+ "ao = 1/math.pi*(sympy.integrate(-1*math.pi*x**0,(x,-math.pi,0))+sympy.integrate(x,(x,0,math.pi)))\n",
+ "s = ao/2\n",
+ "n = int(raw_input(\"enter the no of terms up to each of sin or cos terms in the expansion : \"))\n",
+ "for i in range (1,n+1):\n",
+ " ai = 1/math.pi*(sympy.integrate(-1*math.pi*sympy.cos(i*x),(x,-1*math.pi,0))+sympy.integrate(x*sympy.cos(i*x),(x,0,math.pi)))\n",
+ " bi = 1/math.pi*(sympy.integrate(-1*math.pi*x**0*sympy.sin(i*x),(x,-1*math.pi,0))+sympy.integrate(x*sympy.sin(i*x),(x,0,math.pi)))\n",
+ " s = s+float(ai)*sympy.cos(i*x)+float(bi)*sympy.sin(i*x)\n",
+ "print s"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 10.4, page no. 329"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 15,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "finding the fourier series of given function\n",
+ "enter the no of terms up to each of sin or cos terms in the expansion :3\n",
+ "(exp(l) - exp(-l))/(2*l)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy,math\n",
+ "\n",
+ "print \"finding the fourier series of given function\"\n",
+ "x = sympy.Symbol('x')\n",
+ "l = sympy.Symbol('l')\n",
+ "ao = 1/l*sympy.integrate(sympy.exp(-1*x),(x,-l,l))\n",
+ "s = ao/2\n",
+ "n = int(raw_input(\"enter the no of terms up to each of sin or cos terms in the expansion :\"))\n",
+ "for i in range (1,n+1):\n",
+ " ai = 1/l*sympy.integrate(sympy.exp(-x)*sympy.cos(i*math.pi*x/l),(x,-l,l))\n",
+ " bi = 1/l*sympy.integrate(sympy.exp(-x)*sympy.sin(i*math.pi*x/l),(x,-l,l))\n",
+ " s = s+float(ai)*sympy.cos(i*math.pi*x/l)+float(bi)*sympy.sin(i*math.pi*x/l)\n",
+ "print s"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 10.5, page no. 330"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 17,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "finding the fourier series of given function\n",
+ "enter the no of terms up to each of terms in the expansion : 3\n",
+ "2.0*sin(x) - 1.0*sin(2*x) + 0.666666666666667*sin(3*x)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math,sympy\n",
+ "\n",
+ "print \"finding the fourier series of given function\"\n",
+ "x = sympy.Symbol('x')\n",
+ "l = sympy.Symbol('l')\n",
+ "s = 0\n",
+ "n = int(raw_input(\"enter the no of terms up to each of terms in the expansion : \"))\n",
+ "for i in range(1,n+1):\n",
+ " bi = 2/math.pi*sympy.integrate(x*sympy.sin(i*x),(x,0,math.pi))\n",
+ " s = s+float(bi)*sympy.sin(i*x)\n",
+ "print s"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 10.6, page no. 332"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 19,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "finding the fourier series of given function\n",
+ "enter the no of terms up to each of sin or cos terms in the expansion :3\n",
+ "l**2/3\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math,sympy\n",
+ "\n",
+ "print \"finding the fourier series of given function\"\n",
+ "x = sympy.Symbol('x')\n",
+ "l = sympy.Symbol('l')\n",
+ "ao = 2/l*sympy.integrate(x**2,(x,0,l))\n",
+ "s = ao/2\n",
+ "n = int(raw_input(\"enter the no of terms up to each of sin or cos terms in the expansion :\"))\n",
+ "for i in range(1,n+1):\n",
+ " ai = 2/l*sympy.integrate(x**2*sympy.cos(i*math.pi*x/l),(x,0,l))\n",
+ " s = s+float(ai)*sympy.cos(i*math.pi*x/l)\n",
+ "print s"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "## Example 10.7, page no. 333"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "finding the fourier series of given function\n",
+ "enter the no of terms up to each of sin or cos terms in the expansion :2\n",
+ "7.06789929214115e-17*cos(x) + 0.424413181578387*cos(2*x) + 0.636619772367581\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy,math\n",
+ "\n",
+ "print \"finding the fourier series of given function\"\n",
+ "x = sympy.Symbol('x')\n",
+ "ao = 2/math.pi*(sympy.integrate(sympy.cos(x),(x,0,math.pi/2))+sympy.integrate(-sympy.cos(x),(x,math.pi/2,math.pi)))\n",
+ "s = ao/2\n",
+ "n = int(raw_input(\"enter the no of terms up to each of sin or cos terms in the expansion :\"))\n",
+ "for i in range(1,n+1):\n",
+ " ai = 2/math.pi*(sympy.integrate(sympy.cos(x)*sympy.cos(i*x),(x,0,math.pi/2))+sympy.integrate(-sympy.cos(x)*sympy.cos(i*x),(x,math.pi/2,math.pi)))\n",
+ " s = s+float(ai)*sympy.cos(i*x)\n",
+ "print s"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "## Example 10.8, page no. 334"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "finding the fourier series of given function\n",
+ "enter the no of terms up to each of sin or cos terms in the expansion : 3\n",
+ "0.810569469138702*cos(x) + 7.79634366503875e-17*cos(2*x) + 0.0900632743487446*cos(3*x)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy,math\n",
+ "\n",
+ "print \"finding the fourier series of given function\"\n",
+ "x = sympy.Symbol('x')\n",
+ "ao = 2/math.pi*(sympy.integrate((1-2*x/math.pi),(x,0,math.pi)))\n",
+ "s = ao/2\n",
+ "n = int(raw_input(\"enter the no of terms up to each of sin or cos terms in the expansion : \"))\n",
+ "for i in range(1,n+1):\n",
+ " ai = 2/math.pi*(sympy.integrate((1-2*x/math.pi)*sympy.cos(i*x),(x,0,math.pi)))\n",
+ " s = s+float(ai)*sympy.cos(i*x)\n",
+ "print s"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 10.9, page no. 336"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 9,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "finding the fourier series of given function\n",
+ "enter the no of terms up to each of sin or cos terms in the expansion : 3\n",
+ "0.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy,math\n",
+ "\n",
+ "print \"finding the fourier series of given function\"\n",
+ "x = sympy.Symbol('x')\n",
+ "l = sympy.Symbol('l')\n",
+ "s = 0\n",
+ "n = int(raw_input(\"enter the no of terms up to each of sin or cos terms in the expansion : \"))\n",
+ "for i in range(1,n+1):\n",
+ " bi = sympy.integrate(x*sympy.sin(i*math.pi*x/2),(x,0,2)) \n",
+ " s = s+float(bi)*sympy.sin(i*math.pi*x/2)\n",
+ "print float(s)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 10.10, page no. 337"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 10,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "finding the fourier series of given function\n",
+ "enter the no of terms up to each of sin or cos terms in the expansion : 5\n",
+ "1.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy,math\n",
+ "\n",
+ "print \"finding the fourier series of given function\"\n",
+ "x = sympy.Symbol('x')\n",
+ "ao = 2/2*(sympy.integrate(x,(x,0,2)))\n",
+ "s = ao/2\n",
+ "n = int(raw_input(\"enter the no of terms up to each of sin or cos terms in the expansion : \"))\n",
+ "for i in range(1,n+1):\n",
+ " ai = 2/2*(sympy.integrate(x*sympy.cos(i*math.pi*x/2),(x,0,2)))\n",
+ " s = s +float(ai)*sympy.cos(i*math.pi*x/2)\n",
+ "print float(s)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 10.11, page no. 338"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 13,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "finding the fourier series of given function\n",
+ "enter the no of terms up to each of sin or cos terms in the expansion : 2\n",
+ "0.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy,math\n",
+ "\n",
+ "print \"finding the fourier series of given function\"\n",
+ "x = sympy.Symbol('x')\n",
+ "ao = 0\n",
+ "s = ao\n",
+ "n = int(raw_input(\"enter the no of terms up to each of sin or cos terms in the expansion : \"))\n",
+ "for i in range(1,n+1):\n",
+ " bi = 2/1*(sympy.integrate((1/4-x)*sympy.sin(i*math.pi*x),(x,0,1/2))+sympy.integrate((x-3/4)*sympy.sin(i*math.pi*x),(x,1/2,1)))\n",
+ " s = s+float(bi)*sympy.sin(i*math.pi*x)\n",
+ "print float(s)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 10.12, page no. 339"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 15,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "finding the fourier series of given function\n",
+ "enter the no of terms up to each of sin or cos terms in the expansion : 3\n",
+ "-4.0*cos(x) + 0.999999999999999*cos(2*x) - 0.444444444444444*cos(3*x) + 3.28986813369645\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy,math\n",
+ "\n",
+ "print \"finding the fourier series of given function\"\n",
+ "x = sympy.Symbol('x')\n",
+ "ao = 1/math.pi*sympy.integrate(x**2,(x,-math.pi,math.pi))\n",
+ "s = ao/2\n",
+ "n = int(raw_input(\"enter the no of terms up to each of sin or cos terms in the expansion : \"))\n",
+ "for i in range(1,n+1):\n",
+ " ai = 1/math.pi*sympy.integrate((x**2)*sympy.cos(i*x),(x,-math.pi,math.pi))\n",
+ " bi = 1/math.pi*sympy.integrate((x**2)*sympy.sin(i*x),(x,-math.pi,math.pi))\n",
+ " s = s+float(ai)*sympy.cos(i*x)+float(bi)*sympy.sin(i*x)\n",
+ "print s"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 10.13, page no. 341"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 13,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The complex form of series is summation of f(n,x) where n varies from −%inf to %inf and f(n,x) is given by : \n",
+ "0\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy,math\n",
+ "\n",
+ "print \"The complex form of series is summation of f(n,x) where n varies from −%inf to %inf and f(n,x) is given by : \"\n",
+ "n = sympy.Symbol('n')\n",
+ "x = sympy.Symbol('x')\n",
+ "a = sympy.exp(-x)\n",
+ "b = sympy.exp(-1j*math.pi*n*x)\n",
+ "cn = 1/2*sympy.Integral(sympy.exp(-x)*sympy.exp(-1j*math.pi*n*x),(x,-sympy.oo,sympy.oo))\n",
+ "fnx = float(cn)*sympy.exp(1j*n*math.pi*x) \n",
+ "print fnx"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 10.14, page no. 342"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Practical harmoninc analysis\n",
+ "No of sin or cos term in expansion : 5\n",
+ "-0.522944001594253*sin(x) - 0.410968418886797*sin(2*x) + 0.0927272727272729*sin(3*x) + 0.111796006670355*sin(4*x) - 0.126146907496654*sin(5*x) - 0.373397459621556*cos(x) + 0.159090909090909*cos(2*x) - 0.258181818181819*cos(3*x) - 0.257272727272728*cos(4*x) - 0.546602540378445*cos(5*x) + 1.30454545454545\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy, math, numpy\n",
+ "\n",
+ "print \"Practical harmoninc analysis\"\n",
+ "x = sympy.Symbol('x')\n",
+ "xo = numpy.array([math.pi/6, math.pi/3, math.pi/2, 2*math.pi/3, 5*math.pi/6, math.pi, 7*math.pi/6, 4*math.pi/3, \\\n",
+ " 3*math.pi/2, 5*math.pi/3, 11*math.pi/6])\n",
+ "yo = numpy.array([1.10, 0.30, 0.16, 1.50, 1.30, 2.16, 1.25, 1.30, 1.52, 1.76, 2.00])\n",
+ "ao = 2*numpy.sum(yo)/len(xo)\n",
+ "s = ao/2\n",
+ "n = int(raw_input('No of sin or cos term in expansion : '))\n",
+ "for i in range(1,n+1):\n",
+ " an = 2*sum(yo*numpy.cos(i*xo))/len(yo)\n",
+ " bn = 2*sum(yo*numpy.sin(i*xo))/len(yo)\n",
+ " s = s+float(an)*sympy.cos(i*x)+float(bn)*sympy.sin(i*x)\n",
+ "print s"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "## Example 10.15, page no. 342"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 11,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Practical harmonic analysis\n",
+ "No of sin or cos term in expansion :5\n",
+ "-1.61653377487451e-16*sin(6.28318530717959*x/T) + 1.5*cos(6.28318530717959*x/T) + 0.75\n",
+ "1.5 -1.61653377487e-16\n",
+ "Direct current :\n",
+ "1.5\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math,sympy,numpy\n",
+ "\n",
+ "print \"Practical harmonic analysis\"\n",
+ "x = sympy.Symbol('x')\n",
+ "T = sympy.Symbol('T')\n",
+ "xo = numpy.array([1/6,1/3,1/2,2/3,5/6,1])\n",
+ "yo = numpy.array([1.30,1.05,1.30,-0.88,-0.25,1.98])\n",
+ "ao = 2*sum(yo)/len(xo)\n",
+ "s = ao/2\n",
+ "n = int(raw_input(\"No of sin or cos term in expansion :\"))\n",
+ "i = 1\n",
+ "an = 2*sum(yo*numpy.cos(i*xo*2*math.pi))/len(yo)\n",
+ "bn = 2*sum(yo*numpy.sin(i*xo*2*math.pi))/len(yo)\n",
+ "s = s+float(an)*sympy.cos(i*x*2*math.pi/T)+float(bn)*sympy.sin(i*x*2*math.pi/T)\n",
+ "print s\n",
+ "print \"Direct current :\"\n",
+ "i = math.sqrt(an**2+bn**2)\n",
+ "print i"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 10.16, page no. 343"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 19,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Practical harmonic analysis\n",
+ "Input xo matrix (in factor of T):1\n",
+ "No of sin or cos term in expansion :5\n",
+ "-3.67394039744206e-16*sin(6.28318530717959*x/T) + 1.5*cos(6.28318530717959*x/T) + 4.5\n",
+ "Direct current :\n",
+ "1.5\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math,sympy,numpy\n",
+ "\n",
+ "print \"Practical harmonic analysis\"\n",
+ "x = sympy.Symbol('x')\n",
+ "T = sympy.Symbol('T')\n",
+ "xo = int(raw_input(\"Input xo matrix (in factor of T):\"))\n",
+ "yo = numpy.array([1.30,1.05,1.30,-0.88,-0.25,1.98])\n",
+ "ao = 2*sum(yo)/xo \n",
+ "s = ao/2\n",
+ "n = int(raw_input(\"No of sin or cos term in expansion :\"))\n",
+ "i = 1\n",
+ "an = 2*sum(yo*numpy.cos(i*xo*2*math.pi))/len(yo)\n",
+ "bn = 2*sum(yo*numpy.sin(i*xo*2*math.pi))/len(yo)\n",
+ "s = s+float(an)*sympy.cos(i*x*2*math.pi/T)+float(bn)*sympy.sin(i*x*2*math.pi/T)\n",
+ "print s\n",
+ "print \"Direct current :\"\n",
+ "i = math.sqrt(an**2+bn**2)\n",
+ "print i"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 10.17, page no. 344"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 20,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Practical harmonic analysis\n",
+ "Input xo matrix (in factor of T):1\n",
+ "No of sin or cos term in expansion :4\n",
+ "-3.67394039744206e-16*sin(6.28318530717959*x/T) + 1.5*cos(6.28318530717959*x/T) + 4.5\n",
+ "Direct current :\n",
+ "1.5\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math,sympy,numpy\n",
+ "\n",
+ "print \"Practical harmonic analysis\"\n",
+ "x = sympy.Symbol('x')\n",
+ "T = sympy.Symbol('T')\n",
+ "xo = int(raw_input(\"Input xo matrix (in factor of T):\"))\n",
+ "yo = numpy.array([1.30,1.05,1.30,-0.88,-0.25,1.98])\n",
+ "ao = 2*sum(yo)/xo \n",
+ "s = ao/2\n",
+ "n = int(raw_input(\"No of sin or cos term in expansion :\"))\n",
+ "i = 1\n",
+ "an = 2*sum(yo*numpy.cos(i*xo*2*math.pi))/len(yo)\n",
+ "bn = 2*sum(yo*numpy.sin(i*xo*2*math.pi))/len(yo)\n",
+ "s = s+float(an)*sympy.cos(i*x*2*math.pi/T)+float(bn)*sympy.sin(i*x*2*math.pi/T)\n",
+ "print s\n",
+ "print \"Direct current :\"\n",
+ "i = math.sqrt(an**2+bn**2)\n",
+ "print i"
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.11+"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter13_2.ipynb b/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter13_2.ipynb
new file mode 100644
index 00000000..3dddd224
--- /dev/null
+++ b/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter13_2.ipynb
@@ -0,0 +1,600 @@
+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 13: Linear Differential Equations"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 13.1, page no. 398"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 26,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Solution to the given linear differential equation is given by: \n",
+ "y = c1*exp(-1.28077640640442*x) + c2*exp(0.780776406404415*x)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy, numpy\n",
+ "\n",
+ "print \"Solution to the given linear differential equation is given by: \"\n",
+ "c1 = sympy.Symbol('c1')\n",
+ "c2 = sympy.Symbol('c2')\n",
+ "x = sympy.Symbol('x')\n",
+ "m = numpy.poly([0])\n",
+ "f = m**2+m-2\n",
+ "r = numpy.roots([2, 1, -2])\n",
+ "y = c1*sympy.exp(r[0]*x)+c2*sympy.exp(r[1]*x)\n",
+ "print \"y = \", y"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 13.2, page no. 399"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 24,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Solution to the given linear differntial equation is given by: \n",
+ "y = (c1 + c2*x)*exp(x*(-1.5 + 1.5*I))\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy, numpy\n",
+ "\n",
+ "print 'Solution to the given linear differntial equation is given by: '\n",
+ "c1 = sympy.Symbol('c1')\n",
+ "c2 = sympy.Symbol('c2')\n",
+ "x = sympy.Symbol('x')\n",
+ "m = numpy.poly([0])\n",
+ "f = m**2+6*m+9;\n",
+ "r = numpy.roots([2, 6, 9])\n",
+ "y =(c1+x*c2)*sympy.exp(r[0]*x)\n",
+ "print \"y = \", y"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 13.3, page no. 401"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 40,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Solution to the given linear differntial equation is given by: \n",
+ "y = c1*exp(0.105616806777468*x) + c2*exp(0.105616806777468*x) + c3*exp(-0.877900280221601*x)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy, sympy\n",
+ "\n",
+ "print 'Solution to the given linear differntial equation is given by: '\n",
+ "c1 = sympy.Symbol('c1')\n",
+ "c2 = sympy.Symbol('c2')\n",
+ "c3 = sympy.Symbol('c3')\n",
+ "x = sympy.Symbol('x')\n",
+ "m = numpy.poly([0])\n",
+ "f = m**3+m*2+4*m+4;\n",
+ "r = numpy.roots([3, 2, 4, 4])\n",
+ "y = c1*sympy.exp(r[0].real*x)+c2*sympy.exp(r[1].real*x)+c3*sympy.exp(r[2].real*x)\n",
+ "print \"y = \", y"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 13.4, page no. 402"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 61,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Solution to the given linear differntial equation is given by: \n",
+ "y = c1*exp(-0.707106781186547*x) + c2*exp(-0.707106781186547*x) + c3*exp(0.707106781186547*x) + c4*exp(0.707106781186547*x)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy, sympy\n",
+ "\n",
+ "print 'Solution to the given linear differntial equation is given by: '\n",
+ "c1 = sympy.Symbol('c1')\n",
+ "c2 = sympy.Symbol('c2')\n",
+ "c3 = sympy.Symbol('c3')\n",
+ "c4 = sympy.Symbol('c4')\n",
+ "x = sympy.Symbol('x')\n",
+ "m = numpy.poly([0])\n",
+ "f = m**4+4;\n",
+ "r = numpy.roots([4, 0, 0, 0, 4])\n",
+ "y = c1*sympy.exp(r[0].real*x)+c2*sympy.exp(r[1].real*x)+c3*sympy.exp(r[2].real*x)+c4*sympy.exp(r[3].real*x)\n",
+ "print \"y = \", y"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 13.5, page no. 402"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 64,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Solution to the given linear differntial equation is given by: \n",
+ "y = [ 9041.93285661 22.41271075]\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "print 'Solution to the given linear differntial equation is given by: '\n",
+ "m = numpy.poly([0])\n",
+ "f = m**2+5*m+6\n",
+ "y = numpy.exp(f)/numpy.polyval(f,1)\n",
+ "print \"y = \", y"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 13.6, page no. 403"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Solution of the given linear equation is given by: \n",
+ "[ 0.25+0.96824584j 0.25-0.96824584j]\n",
+ "y = 1/f(D)∗[exp(-2x)+exp(x)-exp(-x)\n",
+ "using 1/f(D)exp(ax) = x/f1(D)∗exp(ax) if f(m)=0\n",
+ "y = x**2*exp(x)/6 + x*exp(-2*x)/9 + exp(-x)/4\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy, sympy\n",
+ "\n",
+ "print 'Solution of the given linear equation is given by: '\n",
+ "x = sympy.Symbol('x')\n",
+ "m = numpy.poly([0])\n",
+ "f =(m+2)*(m-1)**2;\n",
+ "r = numpy.roots([2, -1, 2])\n",
+ "print r\n",
+ "print 'y = 1/f(D)∗[exp(-2x)+exp(x)-exp(-x)'\n",
+ "print 'using 1/f(D)exp(ax) = x/f1(D)∗exp(ax) if f(m)=0'\n",
+ "y1 = x*sympy.exp(-2*x)/9\n",
+ "y2 = sympy.exp(-x)/4\n",
+ "y3 = x**2*sympy.exp(x)/6\n",
+ "y = y1+y2+y3\n",
+ "print \"y = \", y"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 13.7, page no. 404"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Solution of the given linear equation is given by: \n",
+ "Using the identity 1/f(D**2)∗sin(ax+b)[or cos(ax+b)]=1/f(-a**2)∗sin(ax+b)[or cos(ax+b)] this equation \n",
+ "can be redused to y = (4D+1)/65∗cos(2x-1)\n",
+ "y = -8*sin(2*x - 1)/65 + cos(2*x - 1)/65\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy, sympy\n",
+ "\n",
+ "print 'Solution of the given linear equation is given by: '\n",
+ "x = sympy.Symbol('x')\n",
+ "m = numpy.poly([0])\n",
+ "f = m**3+1;\n",
+ "print '''Using the identity 1/f(D**2)∗sin(ax+b)[or cos(ax+b)]=1/f(-a**2)∗sin(ax+b)[or cos(ax+b)] this equation \n",
+ "can be redused to''',\n",
+ "print 'y = (4D+1)/65∗cos(2x-1)'\n",
+ "y = (sympy.cos(2*x-1)+4*sympy.diff(sympy.cos(2*x-1),x))/65\n",
+ "print \"y = \", y"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 13.8, page no. 405"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Solution of the given linear equation is given by: \n",
+ "Using 1/f(D)exp(ax) = x/f1(D)∗exp(ax) if f(m)=0\n",
+ "y = x∗1/(3D**2+4)∗sin2x\n",
+ "Using this identity 1/f(D**2)∗sin(ax+b)[or cos(ax+b)]= 1/f(-a**2)∗sin(ax+b)[or cos (ax+b)] this equation \n",
+ "can be redused to y = -x/8∗sin2x\n",
+ "y = -0.113662178353\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy, sympy\n",
+ "\n",
+ "print 'Solution of the given linear equation is given by: '\n",
+ "x = sympy.Symbol('x')\n",
+ "m = numpy.poly([0])\n",
+ "f = m**3+4*m\n",
+ "print 'Using 1/f(D)exp(ax) = x/f1(D)∗exp(ax) if f(m)=0'\n",
+ "print 'y = x∗1/(3D**2+4)∗sin2x'\n",
+ "print '''Using this identity 1/f(D**2)∗sin(ax+b)[or cos(ax+b)]= 1/f(-a**2)∗sin(ax+b)[or cos (ax+b)] this equation \n",
+ "can be redused to''',\n",
+ "print 'y = -x/8∗sin2x'\n",
+ "x=1\n",
+ "y = -x*numpy.sin(2*x)/8\n",
+ "print \"y = \", y"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 13.9, page no. 406"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Solution of the given linear equation is given by: \n",
+ "y = 1/(D(D+1))[x**2+2x+4] can be written as (1−D+D**2)/D[x**2+2x+4] which is combination of\n",
+ "differentialtion and integration\n",
+ "y = x**3/3 + 4*x\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy, sympy\n",
+ "\n",
+ "print 'Solution of the given linear equation is given by: '\n",
+ "x = sympy.Symbol('x')\n",
+ "m = numpy.poly([0])\n",
+ "print '''y = 1/(D(D+1))[x**2+2x+4] can be written as (1−D+D**2)/D[x**2+2x+4] which is combination of\n",
+ "differentialtion and integration'''\n",
+ "g = x**2+2*x+4\n",
+ "f = g-sympy.diff(g,x)+sympy.diff(g,x,2)\n",
+ "y = sympy.integrate(f,x)\n",
+ "print 'y = ', y"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 13.11, page no. 406"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 13,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Solution to the given linear differntial equation is given by: \n",
+ "CF + PI\n",
+ "[-1.]\n",
+ "CF is given by: \n",
+ "(c1 + c2*x)*exp(-1.0*x)\n",
+ "−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−\n",
+ "PI = 8∗{1/(D−2)**2[exp(2x)]+{1/(D−2)**2[sin(2x)]+{1/(D−2)**2[x**2]}\n",
+ "Using identitties it reduces to: 4*x**2*exp(2*x) + 4*x + cos(2*x) + 3\n",
+ "The solution is y = 4*x**2*exp(2*x) + 4*x + (c1 + c2*x)*exp(-1.0*x) + cos(2*x) + 3\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy, sympy\n",
+ "\n",
+ "print 'Solution to the given linear differntial equation is given by: '\n",
+ "print 'CF + PI'\n",
+ "c1 = sympy.Symbol('c1')\n",
+ "c2 = sympy.Symbol('c2')\n",
+ "x = sympy.Symbol('x')\n",
+ "m = numpy.poly([0])\n",
+ "f = (m-2)**2;\n",
+ "r = numpy.roots([2, 2])\n",
+ "print r\n",
+ "print 'CF is given by: '\n",
+ "cf = (c1+c2*x)*sympy.exp(r[0]*x)\n",
+ "print cf\n",
+ "print '−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−'\n",
+ "print 'PI = 8∗{1/(D−2)**2[exp(2x)]+{1/(D−2)**2[sin(2x)]+{1/(D−2)**2[x**2]}'\n",
+ "print 'Using identitties it reduces to: ',\n",
+ "pi = 4*x**2*sympy.exp(2*x)+sympy.cos(2*x)+4*x+3\n",
+ "print pi\n",
+ "y = cf + pi ;\n",
+ "print 'The solution is y = ', y"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Exampe 13.12, page no. 407"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 19,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Solution to the given linear differntial equation is given by: \n",
+ "CF + PI\n",
+ "[-0.+1.41421356j 0.-1.41421356j]\n",
+ "CF is given by\n",
+ "c1*exp(1.4142135623731*I*x) + c2*exp(-1.4142135623731*I*x)\n",
+ "−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−\n",
+ "PI = 8∗{1/(D**2-4)[x∗sinh(x)]\n",
+ "Using identities it reduces to: -x*(exp(x) - exp(-x))/6\n",
+ "The solution is y = c1*exp(1.4142135623731*I*x) + c2*exp(-1.4142135623731*I*x) - x*(exp(x) - exp(-x))/6\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy, sympy\n",
+ "\n",
+ "print 'Solution to the given linear differntial equation is given by: '\n",
+ "print 'CF + PI'\n",
+ "c1 = sympy.Symbol('c1')\n",
+ "c2 = sympy.Symbol('c2')\n",
+ "x = sympy.Symbol('x')\n",
+ "m = numpy.poly([0])\n",
+ "f = m**2-4\n",
+ "r = numpy.roots([2, 0, 4])\n",
+ "print r\n",
+ "print 'CF is given by'\n",
+ "cf = c1*sympy.exp(r[0]*x)+c2*sympy.exp(r[1]*x)\n",
+ "print cf\n",
+ "print '−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−'\n",
+ "print 'PI = 8∗{1/(D**2-4)[x∗sinh(x)]'\n",
+ "print 'Using identities it reduces to: ',\n",
+ "pi = -x/6*(sympy.exp(x)-sympy.exp(-x))-2/18*(sympy.exp(x)+sympy.exp(-x))\n",
+ "print pi\n",
+ "y = cf + pi\n",
+ "print \"The solution is y = \", y"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 13.13, page no. 408"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 26,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Solution to the given linear differntial equation is given by: \n",
+ "CF + PI\n",
+ "[-0.+0.70710678j 0.-0.70710678j]\n",
+ "CF is given by\n",
+ "c1*exp(0.707106781186548*I*x) + c2*exp(-0.707106781186548*I*x)\n",
+ "−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−\n",
+ "PI = -1/10∗{1/(D**2-1)[x∗sin(3x)+cos(x)]\n",
+ "Using identities it reduces to: -x*sin(3*x) - cos(x)/2\n",
+ "The solution is y = c1*exp(0.707106781186548*I*x) + c2*exp(-0.707106781186548*I*x) - x*sin(3*x) - cos(x)/2\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy, sympy\n",
+ "\n",
+ "print 'Solution to the given linear differntial equation is given by: '\n",
+ "print 'CF + PI'\n",
+ "c1 = sympy.Symbol('c1')\n",
+ "c2 = sympy.Symbol('c2')\n",
+ "x = sympy.Symbol('x')\n",
+ "m = numpy.poly([0])\n",
+ "f = m**2-1\n",
+ "r = numpy.roots([2, 0, 1])\n",
+ "print r\n",
+ "print 'CF is given by'\n",
+ "cf = c1*sympy.exp(r[0]*x)+c2*sympy.exp(r[1]*x)\n",
+ "print cf\n",
+ "print '−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−'\n",
+ "print 'PI = -1/10∗{1/(D**2-1)[x∗sin(3x)+cos(x)]'\n",
+ "print 'Using identities it reduces to: ',\n",
+ "pi = -1/10*(x*sympy.sin(3*x)+3/5*sympy.cos(3*x))-sympy.cos(x)/2\n",
+ "print pi\n",
+ "y = cf + pi\n",
+ "print \"The solution is y = \", y"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 13.14, page no. 408"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 31,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Solution to the given linear differntial equation is given by: \n",
+ "CF + PI\n",
+ "(5.55111512313e-17+0.707106781187j)\n",
+ "CF is given by\n",
+ "(c1 + c2*x)*exp(5.55111512312578e-17*x) + (c3 + c4*x)*exp(-0.5*x)\n",
+ "−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−\n",
+ "PI = ∗{1/(D**4+2∗D+1)[x**2∗cos(x)]\n",
+ "Using identities it reduces to: 4*x**3*sin(x) - (x**4 - 9*x**2)*cos(x)\n",
+ "The solution is y = 4*x**3*sin(x) + (c1 + c2*x)*exp(5.55111512312578e-17*x) + (c3 + c4*x)*exp(-0.5*x) - (x**4 - 9*x**2)*cos(x)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy, sympy\n",
+ "\n",
+ "print 'Solution to the given linear differntial equation is given by: '\n",
+ "print 'CF + PI'\n",
+ "c1 = sympy.Symbol('c1')\n",
+ "c2 = sympy.Symbol('c2')\n",
+ "c3 = sympy.Symbol('c3')\n",
+ "c4 = sympy.Symbol('c4')\n",
+ "x = sympy.Symbol('x')\n",
+ "m = numpy.poly([0])\n",
+ "f =m**4+2*m**2+1\n",
+ "r = numpy.roots([4, 2, 2, 1])\n",
+ "print r[0]\n",
+ "print 'CF is given by'\n",
+ "cf = ((c1+c2*x)*sympy.exp(r[0].real*x)+(c3+c4*x)*sympy.exp(r[2].real*x))\n",
+ "print cf\n",
+ "print '−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−'\n",
+ "print 'PI = ∗{1/(D**4+2∗D+1)[x**2∗cos(x)]'\n",
+ "print 'Using identities it reduces to: ',\n",
+ "pi = -1/48*((x**4-9*x**2)*sympy.cos(x)-4*x**3*sympy.sin(x))\n",
+ "print pi\n",
+ "y = cf + pi\n",
+ "print \"The solution is y = \", y"
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.11+"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter1_2.ipynb b/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter1_2.ipynb
new file mode 100644
index 00000000..e7f8326b
--- /dev/null
+++ b/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter1_2.ipynb
@@ -0,0 +1,737 @@
+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 1: Solution of Equation & Curve Fitting"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.1, page no. 21"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The roots of above equation are: [-3. 2. 0.5]\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "x = numpy.poly([0])\n",
+ "p = 2*(x**3)+x**2-13*x+6\n",
+ "print \"The roots of above equation are: \", numpy.roots([2,1,-13,6])"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.2, page no. 21"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The roots of the equation are: [ 2.00000000+2.64575131j 2.00000000-2.64575131j -2.66666667+0.j ]\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "x = numpy.poly([0])\n",
+ "print \"The roots of the equation are: \", numpy.roots ([3, -4, 1, 88])"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "## Example 1.3, page no. 22"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The roots of above equation are: [ 6. 3. -2.]\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "x = numpy.poly([0])\n",
+ "p = x^3-7*(x^2)+36\n",
+ "print \"The roots of above equation are:\", numpy.roots([1, -7, 0, 36])"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.4, page no. 23"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The roots of above equation are: [ 5. -4. 2. -1.]\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "x = numpy.poly ([0]) \n",
+ "p = x**4-2*(x**3)-21*(x**2)+22*x+40\n",
+ "print \"The roots of above equation are:\", numpy.roots([1,-2,-21,22,40])"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.5, page no. 23"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The roots of above equation are: [ 4. 2. 1. 0.5]\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "x = numpy.poly ([0]) \n",
+ "p = 2*(x**4)-15*(x**3)+35*(x**2)-30*x+8\n",
+ "print \"The roots of above equation are:\", numpy.roots([2,-15,35,-30,8])"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.6, page no. 24"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 25,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The roots of above equation are: [ 2.87938524 0.65270364 -0.53208889]\n",
+ "let x1 = 0.6527036 x2 = -0.5320889 x3 = 2.8793852\n",
+ "So the equation whose roots are cube of the roots of above equation is (x−x1ˆ3)∗(x−x2ˆ3)∗(x−x3ˆ3)\n",
+ "(x - 23.8725770741465)*(x - 0.278066086195109)*(x + 0.150644263115026)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "import sympy\n",
+ "\n",
+ "x = numpy.poly ([0]) \n",
+ "p = x**3-3*(x**2)+1\n",
+ "ans = numpy.roots([1,-3, 0, 1])\n",
+ "print \"The roots of above equation are:\", ans\n",
+ "x = sympy.Symbol('x')\n",
+ "print \"let x1 = 0.6527036 x2 = -0.5320889 x3 = 2.8793852\"\n",
+ "print \"So the equation whose roots are cube of the roots of above equation is (x−x1ˆ3)∗(x−x2ˆ3)∗(x−x3ˆ3)\"\n",
+ "p1 = (x-x1**3)*(x-x2**3)*(x-x3**3)\n",
+ "print p1"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.7, page no. 25"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 37,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The roots of above equation are:\n",
+ "[ 4.48928857 2.28916855 -0.77845712]\n",
+ "let x1 = -0.7784571 x2 = 2.2891685 x3 = 4.4892886\n",
+ "Now, since we want equation whose sum of roots is 0. sum of roots of above equation is 6, so we will decrease\n",
+ "Value of each root by 2 i.e. x4 = x1-2\n",
+ "x4 = -2.7784571\n",
+ "x5 = 0.2891685\n",
+ "x6 = 2.4892886\n",
+ "Hence, the required equation is ( x−x4 ) ∗ ( x−x5 ) ∗ ( x−x6 ) = 0 −−>\n",
+ "(x - 2.4892886)*(x - 0.2891685)*(x + 2.7784571)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "import sympy\n",
+ "\n",
+ "x = numpy.poly ([0]) \n",
+ "x1 = numpy.poly ([0]) \n",
+ "x2 = numpy.poly ([0]) \n",
+ "x3 = numpy.poly ([0]) \n",
+ "x4 = numpy.poly ([0]) \n",
+ "x5 = numpy.poly ([0] ) \n",
+ "x6 = numpy.poly ([0]) \n",
+ "p = x**3-6*(x**2)+5*x+8\n",
+ "print \"The roots of above equation are:\"\n",
+ "print numpy.roots ([1, -6, 5, 8])\n",
+ "print \"let x1 = -0.7784571 x2 = 2.2891685 x3 = 4.4892886\"\n",
+ "x1 = -0.7784571\n",
+ "x2 = 2.2891685\n",
+ "x3 = 4.4892886\n",
+ "print \"Now, since we want equation whose sum of roots is 0. sum of roots of above equation is 6, so we will decrease\"\n",
+ "print \"Value of each root by 2 i.e. x4 = x1-2\"\n",
+ "x = sympy.Symbol('x')\n",
+ "x4 = x1-2\n",
+ "print \"x4 = \", x4\n",
+ "x5=x2-2\n",
+ "print \"x5 = \", x5\n",
+ "x6=x3-2\n",
+ "print \"x6 = \", x6\n",
+ "print \"Hence, the required equation is ( x−x4 ) ∗ ( x−x5 ) ∗ ( x−x6 ) = 0 −−>\"\n",
+ "p1 =( x-x4 )*(x-x5)*(x-x6)\n",
+ "print p1"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.8, page no. 28"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The roots of above equation are: [ 3. 2. 1. 0.5 0.33333333]\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "x = numpy.poly ([0]) \n",
+ "p = 6*(x**5)-41*(x**4)+97*(x**3)-97*(x**2)+41*x-6\n",
+ "print \"The roots of above equation are:\",numpy.roots([6,-41,97,-97,41,-6])"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.9, page no. 28"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 38,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The roots of above equation are: [ 2.00000000+0.j -1.00000000+0.j 0.83333333+0.5527708j\n",
+ " 0.83333333-0.5527708j 1.00000000+0.j 0.50000000+0.j ]\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "x = numpy.poly([0]) \n",
+ "p = 6*(x**6)-25*(x**5)+31*(x**4)-31*(x**2)+25*x-6\n",
+ "print \"The roots of above equation are:\", numpy.roots([6,-25,31,0,-31,25,-6])"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.10, page no. 29"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The roots of above equation are: [ 2.+3.46410162j 2.-3.46410162j -1.+0.j ]\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "x = numpy.poly ([0]) \n",
+ "p = x**3-3*(x**2)+12*x+16\n",
+ "print \"The roots of above equation are:\", numpy.roots([1,-3,12,16])"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.11, page no. 30"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 9,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The roots of above equation are: [ 0.28571429+0.24743583j 0.28571429-0.24743583j -0.25000000+0.j ]\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "x = numpy.poly ([0]) \n",
+ "p = 28*(x**3)-9*(x**2)+1\n",
+ "print \"The roots of above equation are:\",numpy.roots([28,-9,0,1])"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.12, page no. 31"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 11,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The roots of above equation are: [-5. +0.00000000e+00j 2. +2.90013456e-08j 2. -2.90013456e-08j]\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "x = numpy.poly ([0]) \n",
+ "p = x**3+x**2-16*x+20\n",
+ "print \"The roots of above equation are:\",numpy.roots ([1,1,-16,20])"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.13, page no. 31"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 12,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The roots of above equation are: [ 1.08727971+1.17131211j 1.08727971-1.17131211j -1.17455941+0.j ]\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "x = numpy.poly ([0]) \n",
+ "p = x**3-3*(x**2)+3\n",
+ "print \"The roots of above equation are:\",numpy.roots ([1,-1,0,3])"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.14, page no. 33"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 13,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The roots of above equation are: [ 6. 4. 3. -1.]\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "x = numpy.poly([0]) \n",
+ "p = x**4-12*(x**3)+41*(x**2)-18*x-72\n",
+ "print \"The roots of above equation are:\",numpy.roots ([1,-12,41,-18,-72])"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.15, page no. 34"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 14,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The roots of above equation are: [-2.30277564 2.61803399 1.30277564 0.38196601]\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "x = numpy.poly ([0]) \n",
+ "p = x**4-2*(x**3)-5*(x**2)+10*x-3\n",
+ "print \"The roots of above equation are:\", numpy.roots ([1,-2,-5,10,-3])"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.16, page no. 35"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 15,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The roots of above equation are: [ 3.73205081+0.j -2.00000000+1.73205081j -2.00000000-1.73205081j\n",
+ " 0.26794919+0.j ]\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "x = numpy.poly ([0]) \n",
+ "p = x**4-8*(x**2)-24*x+7\n",
+ "print \"The roots of above equation are:\",numpy.roots ([1,0,-8,-24,7])"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.17, page no. 35"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 16,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The roots of above equation are: [ 6.05932014 -1.65491082 1.59559067]\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "x = numpy.poly ([0]) \n",
+ "p = x**4-6*(x**3)-3*(x**2)+22*x-6\n",
+ "print \"The roots of above equation are:\",numpy.roots ([1,-6,-3,22.-6])"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.18, page no. 37"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 18,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "From the graph, it is clear that the point of intersection is nearly x = 1.43\n"
+ ]
+ },
+ {
+ "data": {
+ "image/png": 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+Nv1vnFPxHA4lH+Jw8mFGzR+lRUekVFMbvpRKGe3yGRb8toA3Fr7B92u/p1fT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\nWrV6jlWrKFRGoGyg6KiGLyK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+ "text/plain": [
+ "<matplotlib.figure.Figure at 0x7fa769850290>"
+ ]
+ },
+ "metadata": {},
+ "output_type": "display_data"
+ }
+ ],
+ "source": [
+ "%matplotlib inline\n",
+ "import matplotlib.pyplot as plt\n",
+ "import numpy\n",
+ "import math\n",
+ "\n",
+ "x = numpy.linspace(1 ,3 ,30)\n",
+ "y1 = 3-x\n",
+ "y2 = math.e**(x-1)\n",
+ "plt.xlabel('X axis')\n",
+ "plt.ylabel('Y axis')\n",
+ "plt.title('My Graph')\n",
+ "plt.plot(x, y1, \"o-\" )\n",
+ "plt.plot(x, y2, \"+-\" )\n",
+ "plt.legend([\"3-x\" ,\"e**(x-1)\"])\n",
+ "print \"From the graph, it is clear that the point of intersection is nearly x = 1.43\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.19, page no. 40"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 23,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "From the graph, it is clear that the point of intersection is nearly x=2.3\n"
+ ]
+ },
+ {
+ "data": {
+ "image/png": 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Bdwe/Y8raKWRkZXBepfNoUqMJ9avWp36V+t5/C9yKMtmpAFJyQnWSOS0NBg+GtWth0KCl\nvPrqPLZt+2XxuqZNn+Sll3rSu3fCGfYiZ6KgEED+fkkdzzzO3iN7mbV5Ftv3bz81ABzYjnOOxjUa\nU6VCFTKzMqkRW4Ovf/qaTm06cWmdS7m2ybXccPENlIsq/D/n2RYfa16rOb++5NcAlI8uf9aFym67\n9DZuu9R71aqasTX9Xtgstx/HM4+z58gedh/eza5Du9h9eDe7D+9m408b2fDjBnYc3MHhjMOcyD7B\n66tfp1ZsLTrU60BSsySa12rORTUvomrFqqfsW0Ehsk2bBkOGQP/+MHkyxMYm0LIljB8/guPHo4mJ\nyWLIEAWEc6WgEEAFv6QyszPZtm8bX/34FV/9+BWLUxez4ccNHDh+gCyXxcLtC6kRU4P29dpzR+s7\naBTXiMY1GlMjpsZpQ/nirDAZSmLKxdAorhGN4hoV2sY5x+MLHufGFjeydd9Wtvy8hWkbp7F131a2\n/ryV6jHV8wJE81rN2fyTd0RyQbUL/Dr1oSASmlJSljJu3HwyMspRsWImAwYkMXNmAmvXwowZ0Lnz\nL217905QEChhAQsKZnYBMAWoAzhggnNuXIE2icBM4Nucp6Y7554NVJ9K085DO9ny8xbGLhvLV2ne\nILD5p83Ur1qfS+pcwqW1L+W+9vdxaZ1LaV6rOc8tey6gX/JF+fIr6hdloPZtZlSuUJkr472nsPLL\ndtnsPrybf2/4N4u3e4Pr57s+Z/aW2WS7bFqe15KrG19Nu3rtaFe3Hc1rNSc66tSF0xQUQk9KylIe\nfvjUU0KLFg2jTx9YvTqB2Nggdq6MCORI4STwiHNujZlVAVaa2QLn3KYC7ZY45/oEsB8lprAvkSMn\njrBi9wqW71zOh1s+ZM3eNWRnZ3Ms8xipB1KpU7kO97W/j/6t+1O5QuUS6UuofHEHa99RFkXDag35\nQ+c/8IfOfwB+GT39cPgHVu9ZzeofVjNj0wxGfDyCvUf20ur8VrSr6w0S7eq1Iys7q0h9kcAbN27+\nKQEBICtrNMeOjSA2ViOC0hCwoOCc2wPsybl/xMw2AfWBgkEhbFIcPKkeroq/ik0/bWL5zuUs37Wc\nz3d+zrb922h9fmuuaHAFD3Z8kE4NOtEorhFPLXnK71//gfwiDmfF+TvrVa1Hvar16HXRL6tOHjx+\nkDV71vDexvd4+cuX+eHwD/x47EemrptKfPV4rm9+Pfd1uI9qFav53KdGFaUjI8P3V9Lx4/4tjy3n\nrlTmFMysEdAOWF5gkwO6mNlaYBcw1Dm3sTT65K8TWSf4YtcXLN6+mDfXvskLn71A3Sp16dSwE50a\ndOKey+6hTd02VIiucE7voy+cc3emz7B6THW6NepGt0bd8p7708I/0b1Rdz7Z8Qkfbf2Ip5c8zUW1\nLuKq+KvoGt+VrvFdqV+1PqCgUBrS0uDrrzN9bouJ0aiutAQ8KOScOpoGPOycO1Jg8yrgAufcMTO7\nDvgAaO5rP6NGjcq7n5iYSGJiYon0z1e+/5o9a1i8fTGLUxezJHUJ1WOq0ySuCakHUnmsy2NUKl/J\nr2vK6kukdBX1844pF0Nys2SSmyUD3h8Aq35YxSfffcLb69/m/pT7qV6xOl3ju3Lg+AG+O/AdF8Zd\nGICeS25mUZcuSaxdO4xvvz01zXTIkKItW11WeTwePB7POe0joAvimVl54ENgjnPuRT/abwfaO+f2\nFXg+YAvijfx4JP0u7ecNAtsX40n1UK9qPa5udDVXN76abo26UTO2JhD+GT9yqrP9+l+8fTHTN05n\nx8EdfLj1QyqVq0Rs+Vi6XdiNuy+7m8RGiVQqX6lY+xav/HUHb7zhzSxKSVnK+PEL8qWZXqsMo2IK\nqVVSzZsT+Cbws3PukULanA/86JxzZtYR+LdzrpGPdiUaFA4eP8jcb+by4dYPeX/T+9SpXIerG3uD\nQPdG3alXtZ7P1ykolF2jPKP4n27/w+ofVjNv2zzmbZvHqh9W0alBJ5Kbekcbreq0ykuF1bFydvnr\nDp5+GmUWBUBxgkIgTx9dCfQH1pnZ6pznngTiAZxzrwG3AvebWSZwDPjNub5pYb/Qvtn3DbM3z+bD\nrR/y2fef0aBaAy6udTFHTx7lzjZ3AlC/av1CAwLodFBZF2VRtK/fnvb12/PkVU9yKOMQH2//mHnb\n5nHTuzeRfjKdpKZJJDdNJv1kerC7GzKKUncgwRdx11PI/YWWmZ3JZ99/xuwts5m9ZTYHjh/g+ouu\n54aLb+CaxtfkpYbqF534w5/TQW+ve5u317/Ntn3b2LJvC43jGtPyvJYMbDeQW391a+l0NMT4qjuI\njh5Gnz7JvP226g4CLaROH5Ukf4PC4YzDDJw5kJhyMcz5Zg7x1eO5ofkN3ND8BtrXb+9zATYFBQmE\nJxc9SYf6HXj/6/dJ2ZJCi/NacFOLm7ip5U00q9nslLaRPP+QnDyc+fNPr0dNTh7B3LnPBKFHZUuo\nnT4qFcdOHuP5Zc/z7sZ3+Xb/t5zIOkHvi3pzZ+s76duirzKEJCgqRFfg5pY3c3PLmzmRdQJPqocZ\nm2bQdVJX6lSuw00tbuLmljfT+vzWER0UVHcQfsIyKBzPPM7cb+by7oZ3mbN1Dh0bdGRo56Hc1PIm\nxi0fV6Rf/pH6P6MEV/7jqkJ0BZKaJpHUNImXe73M5zs/Z8amGdz07k0A1K1Sl74X96Vt3bYRtVy1\n6g7CU9gEhRNZJ1j47ULe3fAuszbPom3dtvS7pB/jeo6jduXawe6eyCkK+7ERHRXNlfFXcjL7JFUq\nVGHv0b28tvI1ur/ZnfJR5enboi/DE4afcaHAcDB9ujfVVHUH4Sds5hRqja3FxeddTL9L+nHrr27N\nqzQtKJKH4hKZRnlGMbLbSP7z/X94a91bvLfxPVrWbkn/Vv359SW/zquTyRXKx7jqDkJLRE80/37O\n76keU92vSmKRcFIw2eFE1gnmfjOXt9a9xbxt87i68dX8ttVvub759cSUiwnZ5Ijc0YHqDkJHRAeF\ncOinSHGc6Zf/weMHmbFpBm+tf4vVP6zm5pY3k5WdxaS+k4I6/5C/9sAsk8zMJNLSEvJGBxIaFBRE\nIpQn1cOszbNYv3c9C7cvpFZsLS6rdxkPXv4gfVv0LdW++Ko9qF59GJMmJXPzzTotFEoUFETKgJEf\nj+TaptcyYeUEZm2eRc9mPbnnsnvo3ri7z1qckqbag/BRnKAQ+CNIREqUmdE1vitTbprC9oe30zW+\nK4/Me4Tm45vz52V/Zs+RPae096R6SvT9d+5U7UEkU1AQCTP55x9qxNZgcMfBrL1vLW/f/DZbf95K\ni7+14JZ/38Lcb+aSlZ1VYkEhLQ369YPUVNUeRDIFBZEw42tS2szo1LAT/+j7D3Y8soOkJkkMWzyM\npuOa8umOT9mXvu/0HRXBtGnQujXEx8PUqUk0bTrslO3e2oNrz+k9JDRoTkEkAnlSPXhSPew+vJuJ\nqyZSMboil9S+hMEdBzOw3UC/9+Or7gBUexAuNNEsIqcZ5RnFfR3u47UVr/Hqyle5pPYlPNzpYXo3\n733GiWld7yD8KSiIyGnyF7tlZGbw3sb3eGn5S+xP3+8dObQdyLJFaxk3bj57Y3dS42BD1R1ECAUF\nETmNr+I45xyf7fyMccvHkfL1R0RtaM6hBf+EVv8EzyjVHUQIBQURKbJuNzzE0mNVof1ESI+DWZNg\nR1fVHUSAMnk9BRE5N98dPgRWE1beAwlj4PYb4EQVNv90Edkuu1QK4iR06L+2SBmVW3eQ9mVD8IyC\nxaPBMxKe/wnm/x8/n7+JVq+04s01b3Ii60SwuyulREFBpAw6Y92Bi6bp8TX8s9u/eDH5Raaum0qz\ncc148fMXOXLiyCn7KelqaQk+zSmIlCFnqzvYE/M9dY9fcFrdwYrdKxj76Vg8qR7u73A/QzoOoXbl\n2iG7jLd4aaJZRPLkX966YsVMLrssicmTE86p7mDrz1v5y3/+wnsb3+O3rX5Luahy/LXnX0u+81Ii\nFBREBPC9vHX58sMYMyaZoUPPLc3Uk+ph9ubZfL7zc/6z8z+0r9eeq+Kvom+LvroAVohR9pGIADBu\n3PxTAgLAyZOjWbhwxDkHhfxXP3x8weNEWzQTVk0gPTOdxnGNuTDuwnPavwSXJppFItDhw6WzvHWl\n8pV4rsdzbB68mZqxNblswmXcO/tevjvwXYm+j5QeBQWRCDN9OqxcWTrLW+eOGM6rdB5jrhnDlsFb\nOK/SeVw24TJ+N/t3pB5IzWurTKXwoKAgEiFy6w6GDYPRo0tneeuCcwi1KtVi9DWj2TJ4C3Uq16H9\nhPYMmjWIb/d/q6AQJjTRLBIBfK1oGgrLW+9L38dfP/srf1/xdy6sfiEf/OYD4qvHl2ofyjJlH4mU\nMYXVHYSK3Os6pJ9M5/n/PE9MuRjanN+GP135J25seWOwuxfxlH0kEsHOVHcweXJoXu8gf6ZSbPlY\n7utwH88ufZa7Z9/N6j2rebTLo1SrWC24nZRTKCiIhAFfdQcffzyMMWM45xTT0lS3Sl3+1utvPNr5\nUUZ6RnLR+It4vMvjPHD5A8SWD8GoVgZpolkkDBRed7AgSD0quvyT0o1rNGbKTVNYfOdiln2/jOZ/\na87ElRPJzP4la0oT08GhoCASBkqr7iCQfFU7X1LnEt7v9z7Tfj2Nf234F796+Ve8+9W7ZLtsBYUg\nUVAQCXHTppVe3UGwdGrYiUV3LuLvvf/OXz77C+0ntGfbvm3B7laZpKAgEqJy6w6GDy+9uoNgKxdV\njl7NenFRzYt4a/1bNBvXjPs/vF+jhlKkiWaREJS/7sCbWZRAy5YwfvyIfHUHPUu97iDQ8mcrjVg8\ngjqV6/DsJ89yMvskzWs1p37V+sHtYBmgOgWREBLqdQelKfdaDQeOH+C5T57j9dWvM6TjEIZ2GUqV\nClWC3b2wUJw6BZ0+EgmSlJSlJCcPJzFxFMnJw3niiaV5V0NbvbpsBwT4ZWI6LiaOsdeOZeXvVrJ1\n31aaj2/O66teJyv7l/kUnV4qOWcdKZjZbcBc59whMxsBXAY845xbVRodzOmDRgoSUQJ5vYNIt2L3\nCh6d/yj70vfxfI/n6dmsJ08teUpXgPMhUCOFETkBoStwDfAP4JXidFBEvCKh7iBYOtTvgGeAh9FX\nj+aReY+Q9FYSe47sCXa3IoY/QSF3jHY9MNE59yFQIXBdEol8kVB3EExmRrWK1fj1r35NhagKvLby\nNdq/1p7H5j+mU0nnyJ/so11mNgG4FvizmcWguQiRYisLdQelIX+m0h8X/JETWSeYvHYy9avWp8sF\nXagQrd+uxeHPl/ttwDwgyTl3AKgBPBbQXolEoLJYd1BaYsvH8teef2XpXUuZt20erV9pzZytc4Ld\nrbBU6ESzmVXLmUuo6Wu7c25fQHt2al800SxhLVSvdxApPKmevFGDc46Ptn7EI/Me4aJaF/FC0gtc\nfN7Fwe1gkJTo9RTMLMU519vMUoHTGjnnGherl8WgoCDhouDy1gMGJDFzZoLqDoLgRNYJxi8fz58/\n/TN3tr6TEd1GEBcTB5waRCKZLrIjEkS+0kyjo4fRp08yb7+dEJLXOygLfjz6I8MXD2fW5lk80/0Z\n/rvdf/PM0mfKRAprQFJSzezuAo/LmdnIonZOJNL5SjPNyhrNsWMLFBCCqE7lOky4YQJzfjuHqeum\n0mFiB3Yc3BHsboUsfyaae5jZR2ZW38wuBT4DznqpJDO7wMw+NrMNZvaVmT1USLtxZrbVzNaaWbsi\n9l8kZGRkKM00lB3MOEj3Rt1pWqMpb6x5gzavtmHo/KFKYS3grCmpzrnbzew3wDrgKPBb59wyP/Z9\nEnjEObfGzKoAK81sgXNuU24DM+sFNHPOXWRmnfAWxV1RrL9EJIjS0uDrr5VmGsryp7A+uehJsrKz\nmLRmEvHV4+ka35VyUVofFPw7fdQceAiYAewA+ptZ5bO9zjm3xzm3Juf+EWATUHCJwz7AmzltlgNx\nZnZ+kf4CkSCbPh1at4YuXZJo0kRppuGgQnQFxl47lqV3LWXm5pm0n9CeT777JNjdCgn+hMZZwGDn\n3EIziwIeAb4EfuXvm5hZI6AdsLzApgbA9/ke7wQaAnv93bdIsORf0XTGDOjcOYGUlMhf3joS5I4Y\nWtZuycJZ0ttjAAARrUlEQVQ7FvLexvf4rxn/RfdG3Xn+2uepW6VucDsYRP4siFfdOXewwHPNnXNb\n/HoD76kjD/Csc+6DAttmA392zn2a83gh8HjBxfbMzI0c+cvcdmJiIomJif68vUhATJ/uDQj56w4k\nvB05cYRnljzDpDWTGH7VcB7s+CDlosqFVfqqx+PB4/HkPX7qqacCk5JqZq3wjgxiyKlZcM5N8eN1\n5YEPgTnOuRd9bH8V8Djn/pXz+Gugm3Nub4F2SkmVoFDdQdmzKW0TQ+YM4cejP/Jyr5dZtH1R2Kav\nBqROwcxGAd2AS4AU4DpgmXPu1rO8zvDOF/zsnHukkDa98J6a6mVmVwAvOudOm2hWUJBgUN1B2eWc\n472N7/Ho/EepGVOTBXcuoE7lOsHuVpEFaunsW4EewA/OuYFAGyDOj9ddCfQHupvZ6pzbdWZ2r5nd\nC+Cc+wj41sy+AV4DHihK50UCSXUHZdeS75awMW0jd7S+g3U/rqPxi4254Z83sHj74mB3LeD8mWhO\nd85lmVmmmVUHfgQuONuLctJWzxp0nHOD/eiDSKlT3UHZlT99tUJ0BW5ueTP3fXgfwxcP59XrX6X1\n+a2D28EA8mek8KWZ1QAmAiuA1cB/AtorkSBT3YHk1/r81iz772Xc1fYuekzpwWPzH+PIiSPB7lZA\n+PNL/gHn3H7n3KtAEjAg5zSSSESaNk11B/KL3BFDlEXxu/a/46sHvmLv0b1c8vdLmPn1zOB2LgC0\nIJ5Ijvx1B7mZRVreWgrz8faPuT/lflqc14Jx140jvno8EForsGqVVJFi8nW9A5GzycjM4PlPn+el\n5S/xp65/4uFODzP6k9Ehk8Ja0tdTmAM84JzbXhKdOxcKClKS8tcemGWSmZlEWlqC6g6k2L7Z9w0P\npDzA3qN7ubz+5bze5/VgdwkoXlA4U/bRJGCemb0JPO+cO3lOvRMJAb5qD6pXH8akSd5lKkSKY+eh\nnXRu2JkNaRv4x+p/sG7vOq5pfA3JzZJD5lSSv854+ihniYr/AZKBqfxyBTbnnHsh8N3L64dGClIi\nkpOHM3/+sz6eH8Hcuc8EoUcSaf644I/sS9/H3G1z+dt1f6Nvi75B60sgitdOAkfwLm9RFaiSc6ta\nrB6KBNnOnao9kMCKLR/LxD4TmXrTVB5f+Di3/PsWdh/eHexu+a3QoGBmPfHWJFQG2jnnRjrnnsq9\nlVoPRUpAWhr06wepqao9kMDKPV2U2CiRtfet5Vfn/Yo2r7bh1RWvku2yg9s5P5xppDAM+LVz7o/O\nuWOl1SGRkpZbdxAfD1OnJtG0qWoPJHDyzyHElIvhmauf4eMBHzNl7RQS3khgY9rGvO2heNW3M000\nJ+hEvoSz0693AJBAxYq65oGUrkvrXMqy/17Gayteo9vkbjzQ4QGeuOqJkKppyKU6BQl7BZe3fuih\nJNLTE1R3ICFp16FdDJkzhI1pG+ncsDNv3PhGwN6rpFNSRUKerxTTTz8dRlwczJiRoLoDCTlb922l\n9fmtKR9VnslrJ7P+x/X0aNKDns16hsSoQUFBwpqv5a2PHh1N584jVHcgISn/CqyNFzZmX/o+3l7/\nNldecGVwO5bDn1VSRUJWYctbnzypFFMJfTHlYphwwwSm3DiFR+Y9wm+m/YYfj/4Y1D4pKEhYO3BA\nKaYSvnJHDN0bd2fd/euIrx5Pq1da8da6twjWPKommiUs5WYWffrpUmAeu3b9cgqpadMneeklZRRJ\neFqxewV3z7qbBlUb8Or1r+atvlocgbocp0hIyV93sHVrAq+9lkxy8gi6dRtFcvIIBQQJax3qd2DF\nPSvoGt+V9hPa8/IXL+cVvZVGXYNGChI2fF3vQCSSff3T1wyaNQiA1/u8zr+++leRluVWSqpEjIK1\nB5ddlsTkyQn07w+TJ6vuQMqGFue1YOnApby64lWueuMqWtdpzcmsk5SPLh+w99RIQUKOr9qD8uWH\nMWZMMkOH6rSQlC2eVA+eVA8Hjx/kxeUvUrdyXfq26MtvLv3NWesaNFKQiOCr9uDkydEsXDhCQUHK\nnPx1DdUqVqNJjSY8tuAx6lSuQ+eGnalYrmKJvp8mmiXkHD6s5a1FfDEzBrQdwNr71rJu7zraT2jP\nF7u+KNH3UFCQkDJtGqxcqdoDEV9yRwz1qtbj/X7vMyJhBH3e6cPjCx4n/WR6ibyHgoKEhNzrHQwf\nDqNHa3lrEV/yzyGYGf0u7cf6+9ez4+AO2rzahmU7lp3ze2iiWYJu2jROW9E0JWUp48cvyLe89bWq\nPRA5g/c3vc/gOYO5peUtjLlmDFUqVCnWRLOCggSN6g5ESta+9H38Yd4fWPLdEl6/4XV6NO2h7CMJ\nTao7EAm8mrE1uavtXZzMOslvZ/y2WPvQSEECTnUHIqXPOUdUVJTWPpLQU3jdwYIg9Ugk8pkVKRbk\nUVCQgFPdgUj4UFCQgFLdgUh4UVCQgFDdgUh4UvaRlLj8dQfezKIEWraE8eNH5Ks70DUPREKRso+k\nxKSlwYMPwrp1qjsQCQVaJVVKVf7ag/37M/nuuyTuuSeBN99U3YFIuFJQkGLxVXvQoMEwEhO9p4tE\nJDxpolmKxVftwa5doxk/XrUHIuFMQUGKLC0NVq1S7YFIJFJQkCJ57z1o1QoqV1btgUgkUlAQv+TW\nHYwYAe+/Dy+/rNoDkUikiWY5q9PrDgC8k8mqPRCJLKpTkDwFl7ceMCCJmTMTdL0DkTClOgUpNl8p\nposWDaNPH1i9OkF1ByJlhOYUBPCdYpqVNZpjxxYoIIiUIQoKAkBGhlJMRURBQfBmFn39tVJMRSTA\nQcHMJpnZXjNbX8j2RDM7aGarc27DA9kfOd306dC6NXTpkkSTJkoxFSnrAj3R/AYwHphyhjZLnHN9\nAtwPKSAtDQYPhrVrYcYM6Nw5gZQUpZiKlHUBT0k1s0bAbOdcKx/bEoFHnXM3nGUfSkktQdOnewNC\n//7w9NNa0VQkUoVjSqoDupjZWmAXMNQ5tzHIfYoo+WsPzDLJzEwiLS0hZ3QQ7N6JSKgJdlBYBVzg\nnDtmZtcBHwDNfTUcNWpU3v3ExEQSExNLo39hzVftQfXqw5g0yXu6SEQii8fjwePxnNM+gnr6yEfb\n7UB759y+As/r9FExJCcPZ/78Z308P4K5c58JQo9EpDQV5/RRUFNSzex8M7Oc+x3xBql9Z3mZ+Gnn\nTtUeiEjRBPT0kZm9A3QDzjOz74GRQHkA59xrwK3A/WaWCRwDfhPI/pQVuZlFqamqPRCRotGCeBEm\n/4qmnTsv5fHHT51TaNr0SV56SammImVBOGYfSQk5ve4AIIGKFVV7ICL+00ghAuQfHajuQERyaaRQ\nBqjuQEQCSUEhjKjuQEQCTaukhhFf1zw4eHA0EyYsCFKPRCTSKCiEEdUdiEigKSiEgbQ06NdPdQci\nEngKCiFu2jTv9Q7i42Hq1CSaNtU1D0QkcDTRHKJUdyAiwaA6hSDLn2JasWImDz2URHp6guoOROSc\nFadOQUEhiHylmFauPIy4uGTeey9BdQcick7CbpXUss5XiunRo6Np2XKBAoKIBIWCQhBlZPie0jl5\nUimmIhIcCgpBdOCAUkxFJLQ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+ "text/plain": [
+ "<matplotlib.figure.Figure at 0x7fa7698ce210>"
+ ]
+ },
+ "metadata": {},
+ "output_type": "display_data"
+ }
+ ],
+ "source": [
+ "%matplotlib inline\n",
+ "import matplotlib.pyplot as plt\n",
+ "import numpy\n",
+ "import math\n",
+ "\n",
+ "x = numpy.linspace(1 ,3 ,30)\n",
+ "y1 = x\n",
+ "y2 = numpy.sin(x)+math.pi/2\n",
+ "plt.xlabel('X axis')\n",
+ "plt.ylabel('Y axis')\n",
+ "plt.title('My Graph')\n",
+ "plt.plot(x, y1, \"o-\")\n",
+ "plt.plot(x, y2, \"+-\")\n",
+ "plt.legend([\"x\", \"sin(x)+pi/2\"])\n",
+ "print \"From the graph, it is clear that the point of intersection is nearly x=2.3\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.20, page no. 41"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 29,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "From the graph, it is clear that the point of intersection is nearly x=2.3 \n"
+ ]
+ },
+ {
+ "data": {
+ "image/png": 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vVn3zTdUlS1SXLlVdtkz13HP9lx84cLquXKm6apXq6tWqH3+sOniw/7IXXjhd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\nMcYEZIHCGGNMQP8fXmj88edlNkgAAAAASUVORK5CYII=\n",
+ "text/plain": [
+ "<matplotlib.figure.Figure at 0x7fa7696dc7d0>"
+ ]
+ },
+ "metadata": {},
+ "output_type": "display_data"
+ }
+ ],
+ "source": [
+ "%matplotlib inline\n",
+ "import matplotlib.pyplot as plt\n",
+ "import numpy\n",
+ "import math\n",
+ "\n",
+ "x = numpy.linspace(0, 3, 30)\n",
+ "y1 = -1/numpy.cos(x)\n",
+ "y2 = numpy.cosh(x)\n",
+ "plt.xlabel('X axis')\n",
+ "plt.ylabel('Y axis')\n",
+ "plt.title('My Graph')\n",
+ "plt.plot(x, y1, \"o-\" )\n",
+ "plt.plot(x, y2, \"+-\" )\n",
+ "plt.legend ([\"-sec(x)\", \"cosh(x)\"])\n",
+ "print \"From the graph, it is clear that the point of intersection is nearly x=2.3 \""
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.10"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter21_2.ipynb b/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter21_2.ipynb
new file mode 100644
index 00000000..db9ef99e
--- /dev/null
+++ b/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter21_2.ipynb
@@ -0,0 +1,1304 @@
+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 21 Laplace Transform"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 21.1.1, page no. 556"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "To find the laplace of given function in t\n",
+ "(12*s/((s**2 + 1)*(s**2 + 25)), 0, True)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "from sympy.integrals import laplace_transform\n",
+ "from sympy.abc import t, s, a\n",
+ "\n",
+ "print 'To find the laplace of given function in t'\n",
+ "print laplace_transform(sympy.sin(2*t)*sympy.sin(3*t),t,s)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 21.1.2, page no. 557"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 14,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "To find the laplace of given function in t\n",
+ "(sqrt(pi)*(sqrt(2)*sin(s**2/4)*fresnelc(sqrt(2)*s/(2*sqrt(pi))) - sqrt(2)*cos(s**2/4)*fresnels(sqrt(2)*s/(2*sqrt(pi))) + cos(s**2/4 + pi/4))/2, 0, True)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "from sympy.integrals import laplace_transform\n",
+ "from sympy.abc import t, s, a\n",
+ "\n",
+ "print 'To find the laplace of given function in t'\n",
+ "print laplace_transform(sympy.cos(t**2),t,s)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 21.1.3, page no. 558"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 15,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "To find the laplace of given function in t\n",
+ "((-1)**(1/12)*pi*sqrt(s)*besseli(-1/3, 2*sqrt(3)*s**(3/2)*exp_polar(I*pi/4)/9)/18 + (-1)**(11/12)*sqrt(3)*pi*sqrt(s)*besseli(1/3, 2*sqrt(3)*s**(3/2)*exp_polar(I*pi/4)/9)/18 + (-1)**(7/12)*sqrt(3)*pi*sqrt(s)*besseli(5/3, 2*sqrt(3)*s**(3/2)*exp_polar(I*pi/4)/9)/18 - (-1)**(5/12)*pi*sqrt(s)*besseli(7/3, 2*sqrt(3)*s**(3/2)*exp_polar(I*pi/4)/9)/18 + (-1)**(1/12)*pi*sqrt(s)*besselj(-1/3, 2*sqrt(3)*s**(3/2)*exp_polar(I*pi/4)/9)/18 + (-1)**(11/12)*sqrt(3)*pi*sqrt(s)*besselj(1/3, 2*sqrt(3)*s**(3/2)*exp_polar(I*pi/4)/9)/18 + (-1)**(7/12)*sqrt(3)*pi*sqrt(s)*besselj(5/3, 2*sqrt(3)*s**(3/2)*exp_polar(I*pi/4)/9)/18 - (-1)**(5/12)*pi*sqrt(s)*besselj(7/3, 2*sqrt(3)*s**(3/2)*exp_polar(I*pi/4)/9)/18 + s**2*hyper((1,), (2/3, 5/6, 7/6, 4/3), -s**6/11664)/6 + (-1)**(1/3)*pi*besseli(2/3, 2*sqrt(3)*s**(3/2)*exp_polar(I*pi/4)/9)/(3*s) - 2*(-1)**(1/6)*sqrt(3)*pi*besseli(4/3, 2*sqrt(3)*s**(3/2)*exp_polar(I*pi/4)/9)/(9*s) - (-1)**(1/3)*pi*besselj(2/3, 2*sqrt(3)*s**(3/2)*exp_polar(I*pi/4)/9)/(3*s) + 2*(-1)**(1/6)*sqrt(3)*pi*besselj(4/3, 2*sqrt(3)*s**(3/2)*exp_polar(I*pi/4)/9)/(9*s), 0, True)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "from sympy.integrals import laplace_transform\n",
+ "from sympy.abc import t, s, a\n",
+ "\n",
+ "print 'To find the laplace of given function in t'\n",
+ "print laplace_transform((sympy.sin(t**3)),t,s)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 21.2.1, page no. 560"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 16,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "To find the laplace of given function in t\n",
+ "((2*s - 9)/(s**2 + 6*s + 34), -3, True)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "from sympy.integrals import laplace_transform\n",
+ "from sympy.abc import t, s, a\n",
+ "\n",
+ "print 'To find the laplace of given function in t'\n",
+ "f = sympy.exp(-3*t)*(2*sympy.cos(5*t)-3*sympy.sin(5*t))\n",
+ "print laplace_transform(f,t,s)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 21.2.2, page no. 560"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 17,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "To find the laplace of given function in t\n",
+ "(2/((s - 3)*((s - 3)**2 + 4)), -oo, Abs(periodic_argument(exp_polar(2*I*pi)*polar_lift(-s + 3)**2, oo)) < pi)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "from sympy.integrals import laplace_transform\n",
+ "from sympy.abc import t, s, a\n",
+ "\n",
+ "print 'To find the laplace of given function in t'\n",
+ "f = sympy.exp(3*t)*(sympy.sin(t))**2\n",
+ "print laplace_transform(f,t,s)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 21.2.3, page no. 561"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 41,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "To find the laplace of given function in t\n",
+ "InverseLaplaceTransform(exp(4*t)*sin(2*t)*cos(t), t, s, _None)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "from sympy.integrals import inverse_laplace_transform\n",
+ "from sympy.abc import t, s\n",
+ "\n",
+ "print 'To find the laplace of given function in t'\n",
+ "f = sympy.exp(4*t)*(sympy.cos(t)*sympy.sin(2*t))\n",
+ "print inverse_laplace_transform(f,t,s)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 21.4.1, page no. 563"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 26,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "To find the laplace of given function in t\n",
+ "(2*a*s/(a**2 + s**2)**2, 0, Abs(periodic_argument(polar_lift(a)**2, oo)) == 0)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "from sympy.integrals import laplace_transform\n",
+ "from sympy.abc import t, s, a\n",
+ "\n",
+ "print 'To find the laplace of given function in t'\n",
+ "f = t*sympy.sin(a*t)\n",
+ "print laplace_transform(f,t,s)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 21.4.2, page no. 564"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 27,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "To find the laplace of given function in t\n",
+ "((-a**2 + s**2)/(a**2 + s**2)**2, 0, Abs(periodic_argument(polar_lift(a)**2, oo)) == 0)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "from sympy.integrals import laplace_transform\n",
+ "from sympy.abc import t, s, a\n",
+ "\n",
+ "print 'To find the laplace of given function in t'\n",
+ "f = t*sympy.cos(a*t)\n",
+ "print laplace_transform(f,t,s)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 21.5, page no. 565"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 28,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Piecewise((u/2, s**3*u**2 == 0), (1/(s**2*u) + (-s**2*u**2 - s*u)*exp(-s*u)/(s**3*u**2), True)) + Integral(exp(-s*t), (t, u, oo))\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "\n",
+ "t = sympy.Symbol('t')\n",
+ "s = sympy.Symbol('s')\n",
+ "u = sympy.Symbol('u')\n",
+ "f = sympy.integrate(sympy.exp(-s*t)*t/u,(t,0,u))+sympy.integrate(sympy.exp(-s*t),(t,u,sympy.oo))\n",
+ "print f"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 21.7, page no. 566"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 29,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "To find the laplace of given function in t\n",
+ "(atan(a/s), 0, Abs(periodic_argument(polar_lift(a)**2, oo)) == 0)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "from sympy.integrals import laplace_transform\n",
+ "from sympy.abc import t, s, a\n",
+ "\n",
+ "print 'To find the laplace of given function in t'\n",
+ "f = sympy.sin(a*t)/t\n",
+ "print laplace_transform(f,t,s)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 21.8.1, page no. 567"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 30,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "To find the laplace of given function in t\n",
+ "((-a**2 + s**2)/(a**2 + s**2)**2, 0, Abs(periodic_argument(polar_lift(a)**2, oo)) == 0)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "from sympy.integrals import laplace_transform\n",
+ "from sympy.abc import t, s, a\n",
+ "\n",
+ "print 'To find the laplace of given function in t'\n",
+ "f = t*sympy.cos(a*t)\n",
+ "print laplace_transform(f,t,s)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 21.8.2, page no. 568"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 31,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "To find the laplace of given function in t\n",
+ "(2*a*(-a**2 + 3*s**2)/(a**2 + s**2)**3, 0, Abs(periodic_argument(polar_lift(a)**2, oo)) == 0)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "from sympy.integrals import laplace_transform\n",
+ "from sympy.abc import t, s, a\n",
+ "\n",
+ "print 'To find the laplace of given function in t'\n",
+ "f = (t**2)*sympy.sin(a*t)\n",
+ "print laplace_transform(f,t,s)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 21.8.3, page no. 569"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 32,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "To find the laplace of given function in t\n",
+ "(6/(s + 3)**4, 0, True)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "from sympy.integrals import laplace_transform\n",
+ "from sympy.abc import t, s, a\n",
+ "\n",
+ "print 'To find the laplace of given function in t'\n",
+ "f = sympy.exp(-3*t)*t**3;\n",
+ "print laplace_transform(f,t,s)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 21.8.4, page no. 570"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 33,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "To find the laplace of given function in t\n",
+ "(6*(s + 1)/((s + 1)**2 + 9)**2, -1, True)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "from sympy.integrals import laplace_transform\n",
+ "from sympy.abc import t, s, a\n",
+ "\n",
+ "print 'To find the laplace of given function in t'\n",
+ "f = sympy.exp(-t)*t*sympy.sin(3*t)\n",
+ "print laplace_transform(f,t,s)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 21.9.1, page no. 571"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 52,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "To find the laplace of given function in t\n",
+ "-MellinTransform(1, t, s) + MellinTransform(1/t, t, s)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "from sympy.integrals import mellin_transform\n",
+ "from sympy.abc import t, s, a\n",
+ "\n",
+ "print 'To find the laplace of given function in t'\n",
+ "f = (1-t)/t\n",
+ "print mellin_transform(f,t,s)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 21.9.2, page no. 572"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 49,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "To find the laplace of given function in t\n",
+ "(2**s*sqrt(pi)*a**(-s)*b**(-s)*(a*b**s - a**s*b)*gamma(s/2 - 1/2)/(4*gamma(-s/2 + 1)), (1, 2), And(Abs(periodic_argument(polar_lift(a)**2, oo)) == 0, Abs(periodic_argument(polar_lift(b)**2, oo)) == 0))\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "from sympy.integrals import mellin_transform\n",
+ "from sympy.abc import t, s, a, b\n",
+ "\n",
+ "print 'To find the laplace of given function in t'\n",
+ "f =(sympy.cos(a*t)-sympy.cos(b*t))/t\n",
+ "print mellin_transform(f,t,s)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 21.10.1, page no. 574"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 55,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "To find the given integral find the laplace of tsin(t) and put s=2\n",
+ "Integral((2*s/(s**2 + 1)**2, 0, True), s)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "from sympy.integrals import laplace_transform\n",
+ "from sympy.abc import t, s, a\n",
+ "\n",
+ "print 'To find the given integral find the laplace of tsin(t) and put s=2'\n",
+ "f = sympy.sin(t)*t\n",
+ "l = laplace_transform(f,t,s)\n",
+ "s = 2\n",
+ "print sympy.integrals.Integral(l)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 21.10.3, page no. 575"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 57,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "To find the laplace of given function in t\n",
+ "(sqrt(2)*LaplaceTransform(meijerg(((-1/2, 0, 1/4, 1/2, 3/4, 1), (1,)), ((), (-1/2, 0, 0)), 64*exp_polar(-4*I*pi)/t**4), t, s)/(8*sqrt(pi)) - 4/s - pi/(4*s) + 34*sqrt(2)/(9*s), 0, True)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "from sympy.integrals import laplace_transform\n",
+ "from sympy.abc import t, s, a\n",
+ "\n",
+ "print 'To find the laplace of given function in t'\n",
+ "f = sympy.integrate(sympy.exp(t)*sympy.sin(t)/t,(t,0,t))\n",
+ "l = laplace_transform(f,t,s)\n",
+ "print l"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 21.11.1, page no. 576"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 58,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "To find the laplace of given function in t\n",
+ "InverseLaplaceTransform(1, t, s, _None)/s - 3*InverseLaplaceTransform(1, t, s, _None)/s**2 + 4*InverseLaplaceTransform(1, t, s, _None)/s**3\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "from sympy.integrals import inverse_laplace_transform\n",
+ "from sympy.abc import t, s, a\n",
+ "\n",
+ "print 'To find the laplace of given function in t'\n",
+ "f =(s**2-3*s+4)/s**3\n",
+ "print inverse_laplace_transform(f,t,s)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 21.11.2, page no. 577"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 60,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "To find the laplace of given function in t\n",
+ "s*InverseLaplaceTransform(1, t, s, _None)/(2*s**2 - 4*s + 13) + 2*InverseLaplaceTransform(1, t, s, _None)/(2*s**2 - 4*s + 13)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "from sympy.integrals import inverse_laplace_transform\n",
+ "from sympy.abc import t, s, a\n",
+ "\n",
+ "print 'To find the laplace of given function in t'\n",
+ "f =(s+2)/(2*s**2-4*s+13)\n",
+ "il = inverse_laplace_transform(f,t,s)\n",
+ "print il"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 21.12.1, page no. 579"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 62,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "To find the laplace of given function in t\n",
+ "2*s**2*InverseLaplaceTransform(1, t, s, _None)/(s**3 - 6*s**2 + 11*s - 6) - 6*s*InverseLaplaceTransform(1, t, s, _None)/(s**3 - 6*s**2 + 11*s - 6) + 5*InverseLaplaceTransform(1, t, s, _None)/(s**3 - 6*s**2 + 11*s - 6)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "from sympy.integrals import inverse_laplace_transform\n",
+ "from sympy.abc import t, s, a\n",
+ "\n",
+ "print 'To find the laplace of given function in t'\n",
+ "f =((2*s**2-6*s+5)/(s**3-6*s**2+11*s-6))\n",
+ "il = inverse_laplace_transform(f,t,s)\n",
+ "print il"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 21.12. 3, page no. 579"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 63,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "To find the laplace of given function in t\n",
+ "4*s*InverseLaplaceTransform(1, t, s, _None)/(s*(s - 1)**2 + 2*(s - 1)**2) + 5*InverseLaplaceTransform(1, t, s, _None)/(s*(s - 1)**2 + 2*(s - 1)**2)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "from sympy.integrals import inverse_laplace_transform\n",
+ "from sympy.abc import t, s, a\n",
+ "\n",
+ "print 'To find the laplace of given function in t'\n",
+ "f =(4*s+5)/((s-1)**2*(s+2))\n",
+ "il = inverse_laplace_transform(f,t,s)\n",
+ "print il"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 21.13.1, page no. 580"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 64,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "To find the laplace of given function in t\n",
+ "5*s*InverseLaplaceTransform(1, t, s, _None)/(s**3 + s**2 + 3*s - 5) + 3*InverseLaplaceTransform(1, t, s, _None)/(s**3 + s**2 + 3*s - 5)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "from sympy.integrals import inverse_laplace_transform\n",
+ "from sympy.abc import t, s, a\n",
+ "\n",
+ "print 'To find the laplace of given function in t'\n",
+ "f =(5*s+3)/((s-1)*(s**2+2*s+5))\n",
+ "il = inverse_laplace_transform(f,t,s)\n",
+ "print il"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 21.13.2, page no. 581"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 65,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "To find the laplace of given function in t\n",
+ "s*InverseLaplaceTransform(1, t, s, _None)/(4*a**4 + s**4)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "from sympy.integrals import inverse_laplace_transform\n",
+ "from sympy.abc import t, s, a\n",
+ "\n",
+ "print 'To find the laplace of given function in t'\n",
+ "f = s/(s**4+4*a**4)\n",
+ "il = inverse_laplace_transform(f,t,s)\n",
+ "print il"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 21.14.1, page no. 583"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 66,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "To find the laplace of given function in t\n",
+ "s**2*InverseLaplaceTransform(1, t, s, _None)/(s - 2)**3\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "from sympy.integrals import inverse_laplace_transform\n",
+ "from sympy.abc import t, s, a\n",
+ "\n",
+ "print 'To find the laplace of given function in t'\n",
+ "f = s**2/(s-2)**3\n",
+ "il = inverse_laplace_transform(f,t,s)\n",
+ "print il"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 21.14.2, page no. 583"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 67,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "To find the laplace of given function in t\n",
+ "s*InverseLaplaceTransform(1, t, s, _None)/(s**2 - 4*s + 13) + 3*InverseLaplaceTransform(1, t, s, _None)/(s**2 - 4*s + 13)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "from sympy.integrals import inverse_laplace_transform\n",
+ "from sympy.abc import t, s, a\n",
+ "\n",
+ "print 'To find the laplace of given function in t'\n",
+ "f =(s+3)/((s**2-4*s+13))\n",
+ "il = inverse_laplace_transform(f,t,s)\n",
+ "print il"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 21.15.1, page no. 584"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 68,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "To find the laplace of given function in t\n",
+ "InverseLaplaceTransform(1, t, s, _None)/(a**2*s + s**3)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "from sympy.integrals import inverse_laplace_transform\n",
+ "from sympy.abc import t, s, a\n",
+ "\n",
+ "print 'To find the laplace of given function in t'\n",
+ "f =1/(s*(s**2+a**2))\n",
+ "il = inverse_laplace_transform(f,t,s)\n",
+ "print il"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 21.15.2, page no. 584"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 69,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "To find the laplace of given function in t\n",
+ "InverseLaplaceTransform(1, t, s, _None)/(s*(a + s)**3)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "from sympy.integrals import inverse_laplace_transform\n",
+ "from sympy.abc import t, s, a\n",
+ "\n",
+ "print 'To find the laplace of given function in t'\n",
+ "f =1/(s*(s+a)**3)\n",
+ "il = inverse_laplace_transform(f,t,s)\n",
+ "print il"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 21.16.1, page no. 585"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 70,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "To find the laplace of given function in t\n",
+ "s*InverseLaplaceTransform(1, t, s, _None)/(a**2 + s**2)**2\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "from sympy.integrals import inverse_laplace_transform\n",
+ "from sympy.abc import t, s, a\n",
+ "\n",
+ "print 'To find the laplace of given function in t'\n",
+ "f = s/((s**2+a**2)**2)\n",
+ "il = inverse_laplace_transform(f,t,s)\n",
+ "print il"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 21.16.2, page no. 586"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 71,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "To find the laplace of given function in t\n",
+ "s**2*InverseLaplaceTransform(1, t, s, _None)/(a**2 + s**2)**2\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "from sympy.integrals import inverse_laplace_transform\n",
+ "from sympy.abc import t, s, a\n",
+ "\n",
+ "print 'To find the laplace of given function in t'\n",
+ "f = s**2/((s**2+a**2)**2) ;\n",
+ "il = inverse_laplace_transform(f,t,s)\n",
+ "print il"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 21.16.3, page no. 587"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 72,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "To find the laplace of given function in t\n",
+ "InverseLaplaceTransform(1, t, s, _None)/(a**2 + s**2)**2\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "from sympy.integrals import inverse_laplace_transform\n",
+ "from sympy.abc import t, s, a\n",
+ "\n",
+ "print 'To find the laplace of given function in t'\n",
+ "f = 1/((s**2+a**2)**2) ;\n",
+ "il = inverse_laplace_transform(f,t,s)\n",
+ "print il"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 21.17.1, page no. 588"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 73,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "To find the laplace of given function in t\n",
+ "InverseLaplaceTransform(1, t, s, _None)/(a**2 + s**2)**2\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "from sympy.integrals import inverse_laplace_transform\n",
+ "from sympy.abc import t, s, a\n",
+ "\n",
+ "print 'To find the laplace of given function in t'\n",
+ "f = (s+2)/(s**2*(s+1)*(s-2))\n",
+ "il = inverse_laplace_transform(f,t,s)\n",
+ "print il"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 21.17.2, page no. 589"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 74,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "To find the laplace of given function in t\n",
+ "s*InverseLaplaceTransform(1, t, s, _None)/(s**2 + 4*s + 5)**2 + 2*InverseLaplaceTransform(1, t, s, _None)/(s**2 + 4*s + 5)**2\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "from sympy.integrals import inverse_laplace_transform\n",
+ "from sympy.abc import t, s, a\n",
+ "\n",
+ "print 'To find the laplace of given function in t'\n",
+ "f =(s+2)/(s**2+4*s+5)**2\n",
+ "il = inverse_laplace_transform(f,t,s)\n",
+ "print il"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 21.19.1, page no. 590"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 80,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "To find the laplace of given function in t\n",
+ "s*InverseLaplaceTransform(1, t, s, _None)/(a**2 + s**2)**2\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "from sympy.integrals import inverse_laplace_transform\n",
+ "from sympy.abc import t, s, a\n",
+ "\n",
+ "print 'To find the laplace of given function in t'\n",
+ "f = s/(s**2+a**2)**2\n",
+ "il = inverse_laplace_transform(f,t,s)\n",
+ "print il"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 21.19.2, page no. 591"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 76,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "To find the laplace of given function in t\n",
+ "s**2*InverseLaplaceTransform(1, t, s, _None)/(a**2*b**2 + a**2*s**2 + b**2*s**2 + s**4)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "from sympy.integrals import inverse_laplace_transform\n",
+ "from sympy.abc import t, s, a\n",
+ "\n",
+ "print 'To find the laplace of given function in t'\n",
+ "f = s**2/((s**2+a**2)*(s**2+b**2))\n",
+ "il = inverse_laplace_transform(f,t,s)\n",
+ "print il"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 21.28.1, page no. 598"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Laplace of the given funtion is: \n",
+ "Piecewise((1/2, s**3 == 0), (exp(-3*s)/s**2 - (-s**2 + s)*exp(-2*s)/s**3, True)) + Piecewise((1/2, s**3 == 0), (exp(-s)/s**2 + (-s**2 - s)*exp(-2*s)/s**3, True))\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "from sympy.integrals import laplace_transform\n",
+ "from sympy.abc import t, s, a\n",
+ "\n",
+ "f = sympy.integrate(sympy.exp(-s*t)*(t-1),(t ,1 ,2))+sympy.integrate(sympy.exp(-s*t)*(3-t),(t ,2 ,3))\n",
+ "print 'Laplace of the given funtion is: '\n",
+ "print f"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "## Example 21.28.2, page no. 598"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Laplace of the given funtion is: \n",
+ "Piecewise((2, s == -1), (-1/(s*exp(2)*exp(2*s) + exp(2)*exp(2*s)) + 1/(s + 1), True))\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "from sympy.integrals import laplace_transform\n",
+ "from sympy.abc import t, s, a\n",
+ "\n",
+ "f = sympy.integrate(sympy.exp(-s*t)*sympy.exp(-t),(t ,0 ,2))\n",
+ "print 'Laplace of the given funtion is: '\n",
+ "print f"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 21.34, page no. 602"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "To find the laplace transform of periodic funtion\n",
+ "Piecewise((Piecewise((0, And(s == 0, w == 0)), (1.5707963267949*exp(3.14159265358979*I*w)*sin(3.14159265358979*w) + 1.5707963267949*I*exp(3.14159265358979*I*w)*cos(3.14159265358979*w) - I*exp(3.14159265358979*I*w)*sin(3.14159265358979*w)/(2*w), s == -I*w), (1.5707963267949*exp(-3.14159265358979*I*w)*sin(3.14159265358979*w) - 1.5707963267949*I*exp(-3.14159265358979*I*w)*cos(3.14159265358979*w) + I*exp(-3.14159265358979*I*w)*sin(3.14159265358979*w)/(2*w), s == I*w), (-s*sin(3.14159265358979*w)/(s**2*exp(3.14159265358979*s) + w**2*exp(3.14159265358979*s)) - w*cos(3.14159265358979*w)/(s**2*exp(3.14159265358979*s) + w**2*exp(3.14159265358979*s)), True)), And(Or(s == -I*w, s == 0, s == I*w), Or(s == -I*w, s == I*w, w == 0))), (Piecewise((w/(s**2 + w**2), And(s == 0, w == 0)), (w/(s**2 + w**2) + 1.5707963267949*exp(3.14159265358979*I*w)*sin(3.14159265358979*w) + 1.5707963267949*I*exp(3.14159265358979*I*w)*cos(3.14159265358979*w) - I*exp(3.14159265358979*I*w)*sin(3.14159265358979*w)/(2*w), s == -I*w), (w/(s**2 + w**2) + 1.5707963267949*exp(-3.14159265358979*I*w)*sin(3.14159265358979*w) - 1.5707963267949*I*exp(-3.14159265358979*I*w)*cos(3.14159265358979*w) + I*exp(-3.14159265358979*I*w)*sin(3.14159265358979*w)/(2*w), s == I*w), (-s*sin(3.14159265358979*w)/(s**2*exp(3.14159265358979*s) + w**2*exp(3.14159265358979*s)) - w*cos(3.14159265358979*w)/(s**2*exp(3.14159265358979*s) + w**2*exp(3.14159265358979*s)) + w/(s**2 + w**2), True)), True))/(w*(-exp(-6.28318530717959*s/w) + 1))\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy, math\n",
+ "\n",
+ "print 'To find the laplace transform of periodic funtion'\n",
+ "w = sympy.Symbol('w')\n",
+ "t = sympy.Symbol('t')\n",
+ "s = sympy.Symbol('s')\n",
+ "f = 1/(1-sympy.exp(-2*math.pi*s/w))*sympy.integrate(sympy.exp(-1*s*t)*sympy.sin(w*t),(t,0,math.pi))/w\n",
+ "print f"
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.10"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter22_2.ipynb b/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter22_2.ipynb
new file mode 100644
index 00000000..d3d7419e
--- /dev/null
+++ b/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter22_2.ipynb
@@ -0,0 +1,244 @@
+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 22 : Integral Transform"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 22.1, page no. 608"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "To find the fourier sin integral\n",
+ "0.636619772367581*Integral(sin(t*u), (t, 0, oo))*Integral(sin(u*x), (u, 0, oo))\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy,math\n",
+ "\n",
+ "print \"To find the fourier sin integral\"\n",
+ "x = sympy.Symbol('x')\n",
+ "t = sympy.Symbol('t')\n",
+ "u = sympy.Symbol('u')\n",
+ "fs = 2/math.pi*sympy.integrate(sympy.sin(u*x),(u,0,sympy.oo))*(sympy.integrate(x**0*sympy.sin(u*t),(t,0,sympy.oo)))\n",
+ "print fs"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 22.2, page no. 608"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "To find the fourier transform of given function\n",
+ "Piecewise((2, s == 0), (1.0*I*exp(-1.0*I*s)/s - 1.0*I*exp(1.0*I*s)/s, True))\n",
+ "pi/2\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "\n",
+ "print \"To find the fourier transform of given function\"\n",
+ "x = sympy.Symbol('x')\n",
+ "s = sympy.Symbol('s')\n",
+ "F = sympy.integrate(sympy.exp(1j*s*x),(x,-1,1))\n",
+ "print F\n",
+ "F1 = sympy.integrate(sympy.sin(x)/x,(x,0,sympy.oo))\n",
+ "print F1"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "## Example 22.3, page no. 609"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "To find the fourier transform of given function\n",
+ "Piecewise((4/3, s**6 == 0), ((-2.0*s**4 - 2.0*I*s**3)*exp(1.0*I*s)/s**6 - (2.0*s**4 - 2.0*I*s**3)*exp(-1.0*I*s)/s**6, True))\n",
+ "Integral((x*cos(x) - sin(x))*cos(x/2)/x**3, (x, 0, +inf))\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy,numpy\n",
+ "\n",
+ "print \"To find the fourier transform of given function\"\n",
+ "x = sympy.Symbol('x')\n",
+ "s = sympy.Symbol('s')\n",
+ "F = sympy.integrate(sympy.exp(1j*s*x)*(1-x**2),(x,-1,1))\n",
+ "print F\n",
+ "F1 = sympy.integrals.Integral((x*sympy.cos(x)-sympy.sin(x))/x**3*sympy.cos(x/2),(x,0,numpy.inf))\n",
+ "print F1"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 22.4, page no. 610"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 10,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "To find the fourier sin integral\n",
+ "Piecewise((s/(s**2 + 1), Abs(periodic_argument(polar_lift(s)**2, oo)) == 0), (Integral(exp(-x)*sin(s*x), (x, 0, oo)), True))\n",
+ "Piecewise((sqrt(pi)*(-sqrt(pi)*sinh(m) + sqrt(pi)*cosh(m))/2, Abs(periodic_argument(polar_lift(m)**2, oo)) == 0), (Integral(x*sin(m*x)/(x**2 + 1), (x, 0, oo)), True))\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy,math\n",
+ "\n",
+ "print \"To find the fourier sin integral\"\n",
+ "x = sympy.Symbol('x')\n",
+ "s = sympy.Symbol('s')\n",
+ "m = sympy.Symbol('m')\n",
+ "fs = sympy.integrate(sympy.sin(s*x)*sympy.exp(-x),(x,0,sympy.oo))\n",
+ "print fs\n",
+ "f = sympy.integrate(x*sympy.sin(m*x)/(1+x**2),(x,0,sympy.oo))\n",
+ "print f"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 22.5, page no. 611"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 11,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Fourier cosine transform.\n",
+ "Piecewise((1/2, s == 0), (-sin(s)/s + cos(s)/s**2 - cos(2*s)/s**2, True)) + Piecewise((1/2, s == 0), (sin(s)/s + cos(s)/s**2 - 1/s**2, True))\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy,math\n",
+ "\n",
+ "print \"Fourier cosine transform.\"\n",
+ "x = sympy.Symbol('x')\n",
+ "s = sympy.Symbol('s')\n",
+ "f = sympy.integrate(x*sympy.cos(s*x),(x,0,1))+sympy.integrate((2-x)*sympy.cos(s*x),(x,1,2))\n",
+ "print f"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 22.6, page no. 612"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 12,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Fourier cosine transform.\n",
+ "Piecewise((s*atan(sqrt(s**2/a**2))/(a*sqrt(s**2/a**2)), Or(And(-s**2/a**2 != 1, Abs(periodic_argument(polar_lift(a)**2, oo)) == pi, Abs(periodic_argument(polar_lift(s)**2, oo)) == 0), And(Abs(periodic_argument(polar_lift(a)**2, oo)) < pi, Abs(periodic_argument(polar_lift(s)**2, oo)) == 0))), (Integral(exp(-a*x)*sin(s*x)/x, (x, 0, oo)), True))\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy,math\n",
+ "\n",
+ "print \"Fourier cosine transform.\"\n",
+ "x = sympy.Symbol('x')\n",
+ "s = sympy.Symbol('s')\n",
+ "a = sympy.Symbol('a')\n",
+ "f = sympy.integrate(sympy.exp(-a*x)/x*sympy.sin(s*x),(x,0,sympy.oo))\n",
+ "print f"
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.10"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter23_2.ipynb b/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter23_2.ipynb
new file mode 100644
index 00000000..a7ba913c
--- /dev/null
+++ b/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter23_2.ipynb
@@ -0,0 +1,762 @@
+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 23 : Statistical Methods"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 23.1, page no. 618"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 65,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The first row of A denotes the no. of students falling in the marks group starting from (5 −10)... till (40−45)\n",
+ "the second row denotes cumulative frequency (less than)\n",
+ "the third row denotes cumulative frequency(more than)\n",
+ "[[ 5. 6. 15. 10. 5. 4. 2. 2.]\n",
+ " [ 5. 11. 26. 36. 41. 45. 47. 49.]\n",
+ " [ 49. 44. 38. 23. 13. 8. 4. 2.]]\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "A = numpy.zeros([3,8])\n",
+ "print \"The first row of A denotes the no. of students falling in the marks group starting from (5 −10)... till (40−45)\"\n",
+ "A[0] = [5,6,15,10,5,4,2,2]\n",
+ "print \"the second row denotes cumulative frequency (less than)\"\n",
+ "A[1,0] = 5\n",
+ "for i in range(1,8):\n",
+ " A[1,i] = A[1,i-1]+A[0,i]\n",
+ "print \"the third row denotes cumulative frequency(more than)\"\n",
+ "A[2,0] = 49\n",
+ "for i in range (1,8):\n",
+ " A[2,i] = A[2,i-1]-A[0,i-1]\n",
+ "print A"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 23.2, page no. 619"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 63,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The first row of A represents the mid values of weekly earnings having interval of 2 in each class=x \n",
+ "The second row denotes the no. of employees or in other words frequency=f \n",
+ "Third row denotes f∗x\n",
+ "Fourth row denotes u=(x−25)/2 \n",
+ "Fifth row denotes f∗x\n",
+ "[[ 1.10000000e+01 1.30000000e+01 1.50000000e+01 1.70000000e+01\n",
+ " 1.90000000e+01 2.10000000e+01 2.30000000e+01 2.50000000e+01\n",
+ " 2.70000000e+01 2.90000000e+01 3.10000000e+01 3.30000000e+01\n",
+ " 3.50000000e+01 3.70000000e+01 3.90000000e+01 4.10000000e+01]\n",
+ " [ 3.00000000e+00 6.00000000e+00 1.00000000e+01 1.50000000e+01\n",
+ " 2.40000000e+01 4.20000000e+01 7.50000000e+01 9.00000000e+01\n",
+ " 7.90000000e+01 5.50000000e+01 3.60000000e+01 2.60000000e+01\n",
+ " 1.90000000e+01 1.30000000e+01 9.00000000e+00 7.00000000e+00]\n",
+ " [ 3.30000000e+01 7.80000000e+01 1.50000000e+02 2.55000000e+02\n",
+ " 4.56000000e+02 8.82000000e+02 1.72500000e+03 2.25000000e+03\n",
+ " 2.13300000e+03 1.59500000e+03 1.11600000e+03 8.58000000e+02\n",
+ " 6.65000000e+02 4.81000000e+02 3.51000000e+02 2.87000000e+02]\n",
+ " [ -7.00000000e+00 -6.00000000e+00 -5.00000000e+00 -4.00000000e+00\n",
+ " -3.00000000e+00 -2.00000000e+00 -1.00000000e+00 0.00000000e+00\n",
+ " 1.00000000e+00 2.00000000e+00 3.00000000e+00 4.00000000e+00\n",
+ " 5.00000000e+00 6.00000000e+00 7.00000000e+00 8.00000000e+00]\n",
+ " [ -2.10000000e+01 -3.60000000e+01 -5.00000000e+01 -6.00000000e+01\n",
+ " -7.20000000e+01 -8.40000000e+01 -7.50000000e+01 0.00000000e+00\n",
+ " 7.90000000e+01 1.10000000e+02 1.08000000e+02 1.04000000e+02\n",
+ " 9.50000000e+01 7.80000000e+01 6.30000000e+01 5.60000000e+01]]\n",
+ "Sum of all elements of third row= 13315.0\n",
+ "Sum of all elements of second row= 509.0\n",
+ "mean= 26.1591355599\n",
+ "Sum of all elements of fifth row= 295.0\n",
+ "mean by step deviation method= 26.1591355599\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "A = numpy.zeros([5,16])\n",
+ "print \"The first row of A represents the mid values of weekly earnings having interval of 2 in each class=x \"\n",
+ "A[0] = [11,13,15,17,19,21,23,25,27,29,31,33,35,37,39,41]\n",
+ "print \"The second row denotes the no. of employees or in other words frequency=f \"\n",
+ "A[1]=[3,6,10,15,24,42,75,90,79,55,36,26,19,13,9,7]\n",
+ "print \"Third row denotes f∗x\"\n",
+ "for i in range(0,16):\n",
+ " A[2,i] = A[0,i]*A[1,i]\n",
+ "print \"Fourth row denotes u=(x−25)/2 \"\n",
+ "for i in range(0,16):\n",
+ " A[3,i] = (A[0,i]-25)/2\n",
+ "print \"Fifth row denotes f∗x\"\n",
+ "for i in range(0,16):\n",
+ " A[4,i] = A[3,i]*A[1,i]\n",
+ "print A\n",
+ "b = 0\n",
+ "print \"Sum of all elements of third row=\",\n",
+ "for i in range(0,16):\n",
+ " b = b+A[2,i]\n",
+ "print b\n",
+ "f = 0\n",
+ "print \"Sum of all elements of second row=\",\n",
+ "for i in range(0,16):\n",
+ " f = f+A[1,i]\n",
+ "print f\n",
+ "print \"mean=\",b/f\n",
+ "d = 0\n",
+ "print \"Sum of all elements of fifth row= \",\n",
+ "for i in range(0,16):\n",
+ " d = d+A[4,i]\n",
+ "print d\n",
+ "print \"mean by step deviation method= \",25+(2*d/f)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 23.3, page no. 620"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 47,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The first row of A denotes the no. of students falling in the marks group startin from (5−10)... till (40−45)\n",
+ "The second row denotes cumulative frequency (less than)\n",
+ "The third row denotes cumulative frequency (more than)\n",
+ "[[ 5. 6. 15. 10. 5. 4. 2. 2.]\n",
+ " [ 5. 11. 26. 36. 41. 45. 47. 49.]\n",
+ " [ 49. 44. 38. 23. 13. 8. 4. 2.]]\n",
+ "Median falls in the class (15−20)=l+((n/2−c)∗h)/f= 19\n",
+ "Lower quartile also falls in the class (15−20)=\n",
+ "Upper quartile also falls in the class (25−30)=\n",
+ "Semiinter quartile range= 5\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "A = numpy.zeros([3,8])\n",
+ "print \"The first row of A denotes the no. of students falling in the marks group startin from (5−10)... till (40−45)\"\n",
+ "\n",
+ "A[0] = [5,6,15,10,5,4,2,2]\n",
+ "print \"The second row denotes cumulative frequency (less than)\"\n",
+ "A[1]=[5,11,26,36,41,45,47,49]\n",
+ "print \"The third row denotes cumulative frequency (more than)\"\n",
+ "A[2] = [49,44,38,23,13,8,4,2]\n",
+ "print A\n",
+ "print \"Median falls in the class (15−20)=l+((n/2−c)∗h)/f= \",15+((49/2-11)*5)/15\n",
+ "print \"Lower quartile also falls in the class (15−20)=\"\n",
+ "Q1 = 15+((49/4-11)*5)/15\n",
+ "print \"Upper quartile also falls in the class (25−30)=\"\n",
+ "Q3 =25+((3*49/4-36)*5)/5\n",
+ "print \"Semiinter quartile range= \",(Q3-Q1)/2"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 23.4, page no. 620"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 66,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The first row of A denotes the roll no. of students form 1 to 10 and that of B denotes form 11 to 20 \n",
+ "The second row of A and B denotes the corresponding marks in physics\n",
+ "The third row denotes the corresponding marks in chemistry\n",
+ "Median marks in physics=arithmetic mean of 10th and 11th student = 26\n",
+ "Median marks in chemistry=arithmetic mean of 10th and 11th student = 41\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "A = numpy.zeros([3,10])\n",
+ "print \"The first row of A denotes the roll no. of students form 1 to 10 and that of B denotes form 11 to 20 \"\n",
+ "A[0] = [1,2,3,4,5,6,7,8,9,10]\n",
+ "B[0] = [11,12,13,14,15,16,17,18,19,20]\n",
+ "print \"The second row of A and B denotes the corresponding marks in physics\"\n",
+ "A[1] = [53,54,52,32,30,60,47,46,35,28]\n",
+ "B[1] = [25,42,33,48,72,51,45,33,65,29]\n",
+ "print \"The third row denotes the corresponding marks in chemistry\"\n",
+ "A[2] = [58,55,25,32,26,85,44,80,33,72]\n",
+ "B[2] = [10,42,15,46,50,64,39,38,30,36]\n",
+ "print \"Median marks in physics=arithmetic mean of 10th and 11th student = \",(28+25)/2\n",
+ "print \"Median marks in chemistry=arithmetic mean of 10th and 11th student = \",(72+10)/2"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 23.5, page no. 621"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 9,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Let the misssing frequencies be f1 and f2 \n",
+ "Sum of given frequencies=12+30+65+25+18 = 150\n",
+ "So, f1+f2=229−c = 79\n",
+ "Median=46=40+(114.5−(12+30+f1))∗10/65)\n",
+ "f1=33.5=34 \n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Let the misssing frequencies be f1 and f2 \"\n",
+ "print \"Sum of given frequencies=12+30+65+25+18 = \",\n",
+ "c =12+30+65+25+18\n",
+ "print c\n",
+ "print \"So, f1+f2=229−c = \",229-c\n",
+ "print \"Median=46=40+(114.5−(12+30+f1))∗10/65)\"\n",
+ "print \"f1=33.5=34 \"\n",
+ "f1 = 34\n",
+ "f2 = 45"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 23.6, page no. 622"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 10,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Let the eqidistance be s, then\n",
+ "Average speed = total distance/total time taken 1800/47\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "\n",
+ "s = sympy.Symbol('s')\n",
+ "print \"Let the eqidistance be s, then\"\n",
+ "t1 = s/30\n",
+ "t2 = s/40\n",
+ "t3 = s/50\n",
+ "print \"Average speed = total distance/total time taken\",3*s/(t1+t2+t3)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 23.7, page no. 623"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 68,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The first row denotes the size of item \n",
+ "The second row denotes the corresponding frequency(f)\n",
+ "The third row denotes the corresponding deviation(d)\n",
+ "The fourth row denotes the corresponding f∗d \n",
+ "The fifth row denotes the corresponding f∗dˆ2 \n",
+ "[[ 6. 7. 8. 9. 10. 11. 12.]\n",
+ " [ 3. 6. 9. 13. 8. 5. 4.]\n",
+ " [ -3. -2. -1. 0. 1. 2. 3.]\n",
+ " [ -9. -12. -9. 0. 8. 10. 12.]\n",
+ " [ 27. 24. 9. 0. 8. 20. 36.]]\n",
+ "Sum of fourth row elements= 0.0\n",
+ "Sum of fifth row elements= 124.0\n",
+ "Sum of all frequencies = 48.0\n",
+ "mean=9+b/d = 9.0\n",
+ "Standard deviation = (c/d)ˆ0.5 = 1.60727512683\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "A = numpy.zeros([5,7])\n",
+ "print \"The first row denotes the size of item \"\n",
+ "A[0,:] = [6,7,8,9,10,11,12]\n",
+ "print \"The second row denotes the corresponding frequency(f)\"\n",
+ "A[1,:] = [3,6,9,13,8,5,4]\n",
+ "print \"The third row denotes the corresponding deviation(d)\"\n",
+ "A[2,:] = [-3,-2,-1,0,1,2,3]\n",
+ "print \"The fourth row denotes the corresponding f∗d \"\n",
+ "for i in range(0,7):\n",
+ " A[3,i] = A[1,i]*A[2,i] \n",
+ "print \"The fifth row denotes the corresponding f∗dˆ2 \"\n",
+ "for i in range(0,7):\n",
+ " A[4,i] = A[1,i]*(A[2,i]**2)\n",
+ "print A\n",
+ "b = 0\n",
+ "print \"Sum of fourth row elements= \",\n",
+ "for i in range(0,7):\n",
+ " b = b+A[3,i]\n",
+ "print b\n",
+ "c = 0\n",
+ "print \"Sum of fifth row elements= \",\n",
+ "for i in range(0,7):\n",
+ " c = c+A[4,i]\n",
+ "print c\n",
+ "d = 0\n",
+ "print \"Sum of all frequencies = \",\n",
+ "for i in range(0,7):\n",
+ " d = d+A[1,i]\n",
+ "print d\n",
+ "print \"mean=9+b/d = \",9+b/d\n",
+ "print \"Standard deviation = (c/d)ˆ0.5 = \",(c/d)**0.5"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 23.8, page no. 624"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 70,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The first row of A represents the mid values of wage classes having interval of 8 in each class=x \n",
+ "The second row denotes the no. of men or in other words frequency=f \n",
+ "Third row denotes f∗x \n",
+ "Fourth row denotes d=(x−32.5)/8 \n",
+ "Fifth row denotes f∗d \n",
+ "Sixth row denotes f∗(dˆ2)\n",
+ "[[ 8.5 16.5 24.5 32.5 40.5 48.5 56.5 64.5 72.5]\n",
+ " [ 2. 24. 21. 18. 5. 3. 5. 8. 2. ]\n",
+ " [ 17. 396. 514.5 585. 202.5 145.5 282.5 516. 145. ]\n",
+ " [ -3. -2. -1. 0. 1. 2. 3. 4. 5. ]\n",
+ " [ -6. -48. -21. 0. 5. 6. 15. 32. 10. ]\n",
+ " [ 18. 96. 21. 0. 5. 12. 45. 128. 50. ]]\n",
+ "Sum of all elements of sixth row= 375.0\n",
+ "Sum of all elements of second row= 88.0\n",
+ "mean= 4.26136363636\n",
+ "Sum of all elements of fifth row= Mean wage= 31.8636363636\n",
+ "standard deviation= 34.0402892562\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "A = numpy.zeros([6,9])\n",
+ "print \"The first row of A represents the mid values of wage classes having interval of 8 in each class=x \"\n",
+ "A[0] = [8.5,16.5,24.5,32.5,40.5,48.5,56.5,64.5,72.5]\n",
+ "print \"The second row denotes the no. of men or in other words frequency=f \"\n",
+ "A[1] = [2,24,21,18,5,3,5,8,2]\n",
+ "print \"Third row denotes f∗x \"\n",
+ "for i in range(0,9):\n",
+ " A[2,i] = A[0,i]*A[1,i] \n",
+ "print \"Fourth row denotes d=(x−32.5)/8 \"\n",
+ "for i in range(0,9):\n",
+ " A[3,i] = (A[0,i]-32.5)/8\n",
+ "print \"Fifth row denotes f∗d \"\n",
+ "for i in range(0,9):\n",
+ " A[4,i] = A[3,i]*A[1,i]\n",
+ "print \"Sixth row denotes f∗(dˆ2)\"\n",
+ "for i in range(0,9):\n",
+ " A[5,i] = A[3,i]**2*A[1,i]\n",
+ "print A\n",
+ "b = 0\n",
+ "print \"Sum of all elements of sixth row= \",\n",
+ "for i in range(0,9):\n",
+ " b = b+A[5,i]\n",
+ "print b\n",
+ "f = 0\n",
+ "print \"Sum of all elements of second row= \",\n",
+ "for i in range(0,9):\n",
+ " f = f+A[1,i]\n",
+ "print f\n",
+ "print \"mean= \",b/f\n",
+ "d = 0\n",
+ "print \"Sum of all elements of fifth row= \",\n",
+ "for i in range(0,9):\n",
+ " d = d+A[4,i]\n",
+ "print \"Mean wage= \",32.5+(8*d/f)\n",
+ "print \"standard deviation= \",8*(b/f-(d/f)**2)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 23.9, page no. 626"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 72,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The first row of A denotes the scores of A and that of B denotes that of B \n",
+ "The second row of A and B denotes the corresponding deviation \n",
+ "The third row of A and B denotes the corresponding deviation square \n",
+ "[[ 12. 115. 6. 73. 7. 19. 119. 36. 84. 29.]\n",
+ " [ -39. 64. -45. 22. -44. -32. 68. -15. 33. -22.]\n",
+ " [ 1521. 4096. 2025. 484. 1936. 1024. 4624. 225. 1089. 484.]]\n",
+ "[[ 47. 12. 16. 42. 4. 51. 37. 48. 13. 0.]\n",
+ " [ -4. -39. -35. -9. -47. 0. -14. -3. -38. -51.]\n",
+ " [ 16. 1521. 1225. 81. 2209. 0. 196. 9. 1444. 2601.]]\n",
+ "Sum of second row elements of A=b= -10.0\n",
+ "sum of second row elements of B=c= -240.0\n",
+ "Sum of third row elements of A=d= 17508.0\n",
+ "Sum of second row elements of B=e= 9302.0\n",
+ "Arithmetic mean of A= 50.0\n",
+ "Standard deviation of A= 41.8306108012\n",
+ "Arithmetic mean of B= 27.0\n",
+ "Standard deviation of A= 18.8202019118\n",
+ "Coefficient of variation of A= 83.6612216024\n",
+ "Coefficient of variation of B= 69.7044515251\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "A = numpy.zeros([3,10])\n",
+ "B = numpy.zeros([3,10])\n",
+ "print \"The first row of A denotes the scores of A and that of B denotes that of B \"\n",
+ "A[0] = [12,115,6,73,7,19,119,36,84,29]\n",
+ "B[0] = [47,12,16,42,4,51,37,48,13,0]\n",
+ "print \"The second row of A and B denotes the corresponding deviation \"\n",
+ "for i in range (0,10):\n",
+ " A[1,i] = A[0,i]-51\n",
+ " B[1,i] = B[0,i]-51\n",
+ "print \"The third row of A and B denotes the corresponding deviation square \"\n",
+ "for i in range (0,10):\n",
+ " A[2,i] = A[1,i]**2\n",
+ " B[2,i] = B[1,i]**2\n",
+ "print A\n",
+ "print B\n",
+ "b = 0\n",
+ "print \"Sum of second row elements of A=b=\",\n",
+ "for i in range (0,10):\n",
+ " b = b+A[1,i]\n",
+ "print b\n",
+ "c = 0\n",
+ "print \"sum of second row elements of B=c=\",\n",
+ "for i in range (0,10):\n",
+ " c = c+B[1,i]\n",
+ "print c\n",
+ "d = 0\n",
+ "print \"Sum of third row elements of A=d=\",\n",
+ "for i in range (0,10):\n",
+ " d = d+A[2,i]\n",
+ "print d\n",
+ "e = 0\n",
+ "print \"Sum of second row elements of B=e=\",\n",
+ "for i in range (0,10):\n",
+ " e = e+B[2,i]\n",
+ "print e\n",
+ "print \"Arithmetic mean of A= \",\n",
+ "f = 51+b/10\n",
+ "print f\n",
+ "print \"Standard deviation of A= \",\n",
+ "g = (d/10-(b/10)**2)**0.5\n",
+ "print g\n",
+ "print \"Arithmetic mean of B= \",\n",
+ "h = 51+c/10\n",
+ "print h\n",
+ "print \"Standard deviation of A= \",\n",
+ "i = (e/10-(c/10)**2)**0.5\n",
+ "print i\n",
+ "print \"Coefficient of variation of A=\",(g/f)*100\n",
+ "print \"Coefficient of variation of B=\",(i/h)*100"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 23.10, page no. 628"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 14,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "If m is the mean of entire data, then\n",
+ "116\n",
+ "If s is the standard deviation of entire data, then \n",
+ "7.74596669241\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"If m is the mean of entire data, then\"\n",
+ "m = (50*113+60*120+90*115)/(50+60+90)\n",
+ "print m\n",
+ "print \"If s is the standard deviation of entire data, then \"\n",
+ "s = (((50*6**2)+(60*7**2)+(90*8**2)+(50*3**2)+(60*4**2)+(90*1**2))/200)**0.5\n",
+ "print s"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 23.12, page no. 629"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 73,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The first row of A denotes the no. of persons falling in the weight group starting from (70−80)... till (140−150)\n",
+ "The second row denotes cumulative frequency\n",
+ "Median falls in the class (110−120) = l+((n/2−c)∗h)/f= 111\n",
+ "Lower quartile also falls in the class (90−100)= 97.8571428571\n",
+ "Upper quartile also falls in the class (120−130)= 123.444444444\n",
+ "Quartile coefficient of skewness= -0.0272952853598\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "A = numpy.zeros([2,8])\n",
+ "print \"The first row of A denotes the no. of persons falling in the weight group starting from (70−80)... till (140−150)\"\n",
+ "A[0] = [12,18,35,42,50,45,20,8]\n",
+ "print \"The second row denotes cumulative frequency\"\n",
+ "A[1,0] = 12\n",
+ "for i in range(1,8):\n",
+ " A[1,i] = A[1,i-1]+A[0,i]\n",
+ "print \"Median falls in the class (110−120) = l+((n/2−c)∗h)/f= \",\n",
+ "Q2 = 110+(8*10)/50\n",
+ "print Q2\n",
+ "print \"Lower quartile also falls in the class (90−100)= \",\n",
+ "Q1 = 90+(57.5-30)*10/35\n",
+ "print Q1\n",
+ "print \"Upper quartile also falls in the class (120−130)= \",\n",
+ "Q3 = 120+(172.5-157)*10/45\n",
+ "print Q3\n",
+ "print \"Quartile coefficient of skewness= \",(Q1+Q3-2*Q2)/(Q3-Q1)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 23.13, page no. 631"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 76,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The first row of A denotes the corresponding I.R. of students \n",
+ "The second row denotes the corresponding deviation of I.R.\n",
+ "The third row denotes the square of corresponding deviation of I.R.\n",
+ "The fourth row denotes the corresponding E.R. of students \n",
+ "The fifth row denotes the corresponding deviation of E.R.\n",
+ "The sixth row denotes the square of corresponding deviation of E.R.\n",
+ "The seventh row denotes the product of the two corresponding deviations\n",
+ "[[ 105. 104. 102. 101. 100. 99. 98. 96. 93. 92.]\n",
+ " [ 6. 5. 3. 2. 1. 0. -1. -3. -6. -7.]\n",
+ " [ 36. 25. 9. 4. 1. 0. 1. 9. 36. 49.]\n",
+ " [ 101. 103. 100. 98. 95. 96. 104. 92. 97. 94.]\n",
+ " [ 3. 5. 2. 0. -3. -2. 6. -6. -1. -4.]\n",
+ " [ 9. 25. 4. 0. 9. 4. 36. 36. 1. 16.]\n",
+ " [ 18. 25. 6. 0. -3. -0. -6. 18. 6. 28.]]\n",
+ "The sum of elements of first row=a = 990.0\n",
+ "The sum of elements of second row=b = 0.0\n",
+ "The sum of elements of third row=c = 170.0\n",
+ "The sum of elements of fourth row=d = 980.0\n",
+ "The sum of elements of fifth row=e = 0.0\n",
+ "The sum of elements of sixth row=d = 140.0\n",
+ "The sum of elements of seventh row=d = 92.0\n",
+ "Coefficient of correlation = 0.596347425668\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "A = numpy.zeros([7,10])\n",
+ "print \"The first row of A denotes the corresponding I.R. of students \"\n",
+ "A[0] = [105,104,102,101,100,99,98,96,93,92]\n",
+ "print \"The second row denotes the corresponding deviation of I.R.\"\n",
+ "for i in range(0,10):\n",
+ " A[1,i] = A[0,i]-99\n",
+ "print \"The third row denotes the square of corresponding deviation of I.R.\"\n",
+ "for i in range(0,10):\n",
+ " A[2,i] = A[1,i]**2\n",
+ "print \"The fourth row denotes the corresponding E.R. of students \"\n",
+ "A[3] = [101,103,100,98,95,96,104,92,97,94]\n",
+ "print \"The fifth row denotes the corresponding deviation of E.R.\"\n",
+ "for i in range(0,10):\n",
+ " A[4,i] = A[3,i]-98\n",
+ "print \"The sixth row denotes the square of corresponding deviation of E.R.\"\n",
+ "for i in range(0,10):\n",
+ " A[5,i] = A[4,i]**2\n",
+ "print \"The seventh row denotes the product of the two corresponding deviations\"\n",
+ "for i in range(0,10):\n",
+ " A[6,i] = A[1,i]*A[4,i]\n",
+ "print A\n",
+ "a = 0\n",
+ "print \"The sum of elements of first row=a = \",\n",
+ "for i in range(0,10):\n",
+ " a = a+A[0,i]\n",
+ "print a\n",
+ "b = 0 \n",
+ "print \"The sum of elements of second row=b = \",\n",
+ "for i in range(0,10):\n",
+ " b = b+A[1,i]\n",
+ "print b\n",
+ "c = 0\n",
+ "print \"The sum of elements of third row=c = \",\n",
+ "for i in range(0,10):\n",
+ " c = c+A[2,i]\n",
+ "print c\n",
+ "d = 0\n",
+ "print \"The sum of elements of fourth row=d = \",\n",
+ "for i in range(0,10):\n",
+ " d = d+A[3,i]\n",
+ "print d\n",
+ "e = 0\n",
+ "print \"The sum of elements of fifth row=e = \",\n",
+ "for i in range(0,10):\n",
+ " e = e+A[4,i]\n",
+ "print e\n",
+ "f = 0\n",
+ "print \"The sum of elements of sixth row=d = \",\n",
+ "for i in range(0,10):\n",
+ " f = f+A[5,i]\n",
+ "print f\n",
+ "g = 0\n",
+ "print \"The sum of elements of seventh row=d = \",\n",
+ "for i in range(0,10):\n",
+ " g = g+A[6,i]\n",
+ "print g\n",
+ "print \"Coefficient of correlation = \",g/(c*f)**0.5"
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.10"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter24_2.ipynb b/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter24_2.ipynb
new file mode 100644
index 00000000..9dd44a2c
--- /dev/null
+++ b/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter24_2.ipynb
@@ -0,0 +1,331 @@
+{
+ "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"
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.10"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter26_2.ipynb b/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter26_2.ipynb
new file mode 100644
index 00000000..dcac1bac
--- /dev/null
+++ b/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter26_2.ipynb
@@ -0,0 +1,522 @@
+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 26 : Difference Equations And Z Transform"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 26.2, page no. 667"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "yn= (-2)**n*b + 2**n*a\n",
+ "y(n+1)=yn1= (-2.0)**n*b + 2.0**n*a\n",
+ "y(n+2)=yn2= (-2.0)**n*b + 2.0**n*a\n",
+ "Eliminating a b from these equations we get :\n",
+ "The required difference equation : \n",
+ "16*yn0 - 4*yn2\n",
+ "=0\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy,numpy\n",
+ "\n",
+ "n = sympy.Symbol('n')\n",
+ "a = sympy.Symbol('a')\n",
+ "b = sympy.Symbol('b')\n",
+ "yn0 = sympy.Symbol('yn0')\n",
+ "yn1 = sympy.Symbol('yn1')\n",
+ "yn2 = sympy.Symbol('yn2')\n",
+ "yn = a*2**n+b*(-2)**n\n",
+ "print \"yn= \",yn\n",
+ "n = n+1\n",
+ "yn = yn.evalf()\n",
+ "print \"y(n+1)=yn1=\",yn\n",
+ "n = n+1\n",
+ "yn = yn.evalf()\n",
+ "print \"y(n+2)=yn2=\",yn\n",
+ "print \"Eliminating a b from these equations we get :\"\n",
+ "A = sympy.Matrix([[yn0,1,1],[yn1,2,-2],[yn2,4,4]])\n",
+ "y = A.det()\n",
+ "print \"The required difference equation : \"\n",
+ "print y\n",
+ "print \"=0\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 26.3, page no. 669"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 9,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Cumulative function is given by Eˆ3−2∗Eˆ2−5∗E+6=0 \n",
+ "[-2. 3. 1.]\n",
+ "Therefor the complete solution is : \n",
+ "un= (-2.0)**(n + 2)*c1 + 1.0**(n + 2)*c3 + 3.0**(n + 2)*c2\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy\n",
+ "\n",
+ "c1 = sympy.Symbol('c1')\n",
+ "c2 = sympy.Symbol('c2')\n",
+ "c3 = sympy.Symbol('c3')\n",
+ "print \"Cumulative function is given by Eˆ3−2∗Eˆ2−5∗E+6=0 \"\n",
+ "E = numpy.poly([0])\n",
+ "f = E**3-2*E**2-5*E+6\n",
+ "r = numpy.roots([1,-2,-5,6])\n",
+ "print r\n",
+ "print \"Therefor the complete solution is : \"\n",
+ "un = c1*(r[0])**n+c2*(r[1])**n+c3*(r[2])**n\n",
+ "print \"un=\",un"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "## Example 26.4, page no. 670"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Cumulative function is given by Eˆ2−2∗E+1=0 \n",
+ "[ 1. 1.]\n",
+ "Therefor the complete solution is : \n",
+ "un = 1.0**n*(c1 + c2*n)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy\n",
+ "\n",
+ "c1 = sympy.Symbol('c1')\n",
+ "c2 = sympy.Symbol('c2')\n",
+ "c3 = sympy.Symbol('c3')\n",
+ "n = sympy.Symbol('n')\n",
+ "print \"Cumulative function is given by Eˆ2−2∗E+1=0 \"\n",
+ "E = numpy.poly([0])\n",
+ "f = E**2-2*E+1\n",
+ "r = numpy.roots([1,-2,1])\n",
+ "print r\n",
+ "print \"Therefor the complete solution is : \"\n",
+ "un = (c1+c2*n)*(r[0])**n\n",
+ "print \"un = \",un"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 26.6, page no. 671"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "For Fibonacci Series yn2=yn1+yn0 \n",
+ "So Cumulative function is given by Eˆ2−E−1=0 \n",
+ "[ 1.61803399 -0.61803399]\n",
+ "Therefor the complete solution is : \n",
+ "un = (-0.618033988749895)**n*c2 + 1.61803398874989**n*c1\n",
+ "Now puttting n=1, y=0 and n=2, y=1 we get \n",
+ "c1=(5−sqrt(5))/10 c2=(5+sqrt(5))/10 \n",
+ "(-0.618033988749895)**n*c2 + 1.61803398874989**n*c1\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy,math\n",
+ "\n",
+ "c1 = sympy.Symbol('c1')\n",
+ "c2 = sympy.Symbol('c2')\n",
+ "c3 = sympy.Symbol('c3')\n",
+ "n = sympy.Symbol('n')\n",
+ "print \"For Fibonacci Series yn2=yn1+yn0 \"\n",
+ "print \"So Cumulative function is given by Eˆ2−E−1=0 \"\n",
+ "E = numpy.poly([0])\n",
+ "f = E**2-E-1\n",
+ "r = numpy.roots([1,-1,-1])\n",
+ "print r\n",
+ "print \"Therefor the complete solution is : \"\n",
+ "un = (c1)*(r[0])**n+c2*(r[1])**n \n",
+ "print \"un = \",un\n",
+ "print \"Now puttting n=1, y=0 and n=2, y=1 we get \"\n",
+ "print \"c1=(5−sqrt(5))/10 c2=(5+sqrt(5))/10 \"\n",
+ "c1 =(5-math.sqrt(5))/10\n",
+ "c2 =(5+math.sqrt(5))/10\n",
+ "un = un.evalf()\n",
+ "print un"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 26.7, page no. 672"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Cumulative function is given by Eˆ2−4∗E+3=0 \n",
+ "[ 3. 1.]\n",
+ "Therefor the complete solution is = cf+pi \n",
+ "cf = 1.0**n*c2 + 3.0**n*c1\n",
+ "PI=1/(Eˆ2−4E+3)[5ˆn]\n",
+ "put E=5\n",
+ "We get PI=5ˆn/8 \n",
+ "un = 1.0**n*c2 + 3.0**n*c1 + 5**n/8\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy,math\n",
+ "\n",
+ "c1 = sympy.Symbol('c1')\n",
+ "c2 = sympy.Symbol('c2')\n",
+ "c3 = sympy.Symbol('c3')\n",
+ "n = sympy.Symbol('n')\n",
+ "print \"Cumulative function is given by Eˆ2−4∗E+3=0 \"\n",
+ "E = numpy.poly([0])\n",
+ "f = E**2-4*E+3\n",
+ "r = numpy.roots([1,-4,3])\n",
+ "print r\n",
+ "print \"Therefor the complete solution is = cf+pi \"\n",
+ "cf = c1*(r[0])**n+c2*r[1]**n \n",
+ "print \"cf = \",cf\n",
+ "print \"PI=1/(Eˆ2−4E+3)[5ˆn]\"\n",
+ "print \"put E=5\"\n",
+ "print \"We get PI=5ˆn/8 \"\n",
+ "pi = 5**n/8\n",
+ "un = cf+pi\n",
+ "print \"un = \",un"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 26.8, page no. 672"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Cumulative function is given by Eˆ2−4∗E+4=0 \n",
+ "[ 2. 2.]\n",
+ "Therefor the complete solution is = cf+pi\n",
+ "cf = 2.0**n*(c1 + c2*n)\n",
+ "PI=1/(Eˆ2−4E+4)[2ˆn]\n",
+ "We get PI=n∗(n−1)/2∗2ˆ(n−2)\n",
+ "un = 2**(n - 2)*n*(n - 1)/2 + 2.0**n*(c1 + c2*n)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy,math\n",
+ "\n",
+ "c1 = sympy.Symbol('c1')\n",
+ "c2 = sympy.Symbol('c2')\n",
+ "c3 = sympy.Symbol('c3')\n",
+ "n = sympy.Symbol('n')\n",
+ "print \"Cumulative function is given by Eˆ2−4∗E+4=0 \"\n",
+ "E = numpy.poly([0])\n",
+ "f = E**2-4*E+4\n",
+ "r = numpy.roots([1,-4,4])\n",
+ "print r\n",
+ "print \"Therefor the complete solution is = cf+pi\" \n",
+ "cf = (c1+c2*n)*r[0]**n\n",
+ "print \"cf = \",cf\n",
+ "print \"PI=1/(Eˆ2−4E+4)[2ˆn]\"\n",
+ "print \"We get PI=n∗(n−1)/2∗2ˆ(n−2)\"\n",
+ "pi = n*(n-1)/math.factorial(2)*2**(n-2)\n",
+ "un = cf+pi\n",
+ "print \"un = \",un"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 26.10, page no. 674"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Cumulative function is given by Eˆ2−4=0 \n",
+ "[-2. 2.]\n",
+ "Therefor the complete solution is = cf+pi \n",
+ "CF = (-2.0)**n*(c1 + c2*n)\n",
+ " PI=1/(Eˆ2−4)[nˆ2+n−1]\n",
+ "We get PI=−nˆ2/3−7/9∗n−17/27 \n",
+ "un = (-2.0)**n*(c1 + c2*n) - n**2/3\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy,math\n",
+ "\n",
+ "c1 = sympy.Symbol('c1')\n",
+ "c2 = sympy.Symbol('c2')\n",
+ "c3 = sympy.Symbol('c3')\n",
+ "n = sympy.Symbol('n')\n",
+ "print \"Cumulative function is given by Eˆ2−4=0 \"\n",
+ "E = numpy.poly([0])\n",
+ "f = E**2-4\n",
+ "r = numpy.roots([1,0,-4]) \n",
+ "print r\n",
+ "print \"Therefor the complete solution is = cf+pi \"\n",
+ "cf = (c1+c2*n)*r[0]**n\n",
+ "print \"CF = \",cf\n",
+ "print \" PI=1/(Eˆ2−4)[nˆ2+n−1]\"\n",
+ "print \"We get PI=−nˆ2/3−7/9∗n−17/27 \"\n",
+ "pi = -n**2/3-7/9*n-17/27\n",
+ "un = cf+pi \n",
+ "print \"un = \",un"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 26.11,page no. 674"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Cumulative function is given by Eˆ2−2∗E+1=0 \n",
+ "[-2.41421356 0.41421356]\n",
+ "Therefor the complete solution is = cf+pi\n",
+ "CF = (-2.41421356237309)**n*(c1 + c2*n)\n",
+ "PI=1/(E−1)ˆ2[nˆ2∗2ˆn]\n",
+ "We get PI=2ˆn∗(nˆ2−8∗n+20)\n",
+ "un = (-2.41421356237309)**n*(c1 + c2*n) + 2**n*(n**2 - 8*n + 20)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy,math\n",
+ "\n",
+ "c1 = sympy.Symbol('c1')\n",
+ "c2 = sympy.Symbol('c2')\n",
+ "c3 = sympy.Symbol('c3')\n",
+ "n = sympy.Symbol('n')\n",
+ "print \"Cumulative function is given by Eˆ2−2∗E+1=0 \"\n",
+ "E = numpy.poly([0])\n",
+ "f = E**2+2*E-1\n",
+ "r = numpy.roots([1,2,-1])\n",
+ "print r\n",
+ "print \"Therefor the complete solution is = cf+pi\"\n",
+ "cf = (c1+c2*n)*r[0]**n\n",
+ "print \"CF = \",cf\n",
+ "print \"PI=1/(E−1)ˆ2[nˆ2∗2ˆn]\"\n",
+ "print \"We get PI=2ˆn∗(nˆ2−8∗n+20)\"\n",
+ "pi = 2**n*(n**2-8*n+20)\n",
+ "un = cf+pi\n",
+ "print \"un = \",un"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 26.12, page no. 676"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 10,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Simplified equations are : \n",
+ "(E−3) ux+vx=x..... (i) 3ux+(E−5)∗vx=4ˆx......(ii)\n",
+ "Simplifying we get (Eˆ2−8E+12) ux=1−4x−4ˆx \n",
+ "Cumulative function is given by Eˆ2−8∗E+12=0 \n",
+ "[ 6. 2.]\n",
+ "Therefor the complete solution is = cf+pi\n",
+ "CF = 2.0**x*c2 + 6.0**x*c1\n",
+ "Solving for PI \n",
+ "We get PI= \n",
+ "ux = 2.0**x*c2 + 4**x/4 + 6.0**x*c1 - x\n",
+ "Putting in (i) we get vx= \n",
+ "2**x*c1 - 4**x/4 - 3*6**x*c2 - 1\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy\n",
+ "\n",
+ "print \"Simplified equations are : \"\n",
+ "print \"(E−3) ux+vx=x..... (i) 3ux+(E−5)∗vx=4ˆx......(ii)\"\n",
+ "print \"Simplifying we get (Eˆ2−8E+12) ux=1−4x−4ˆx \"\n",
+ "c1 = sympy.Symbol('c1')\n",
+ "c2 = sympy.Symbol('c2')\n",
+ "c3 = sympy.Symbol('c3')\n",
+ "x = sympy.Symbol('x')\n",
+ "print \"Cumulative function is given by Eˆ2−8∗E+12=0 \"\n",
+ "E = numpy.poly([0])\n",
+ "f = E**2-8*E+12\n",
+ "r = numpy.roots([1,-8,12])\n",
+ "print r\n",
+ "print \"Therefor the complete solution is = cf+pi\"\n",
+ "cf = c1*r[0]**x+c2*r[1]**x\n",
+ "print \"CF = \",cf\n",
+ "print \"Solving for PI \"\n",
+ "print \"We get PI= \"\n",
+ "pi = -4/5*x-19/25+4**x/4\n",
+ "ux = cf+pi \n",
+ "print \"ux = \",ux\n",
+ "print \"Putting in (i) we get vx= \"\n",
+ "vx = c1*2**x-3*c2*6**x-3/5*x-34/25-4**x/4\n",
+ "print vx"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 26.16, page no. 682"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 15,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "u2 = 2\n",
+ "u3 = 13\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy\n",
+ "\n",
+ "z = sympy.Symbol('z')\n",
+ "f = (2/z**2+5/z+14)/(1/z-1)**4\n",
+ "u0 = sympy.limit(f,z,0)\n",
+ "u1 = sympy.limit(1/z*(f- u0),z,0)\n",
+ "u2 = sympy.limit(1/z**2*(f-u0-u1*z),z,0)\n",
+ "print \"u2 = \",u2\n",
+ "u3 = sympy.limit(1/z**3*(f-u0-u1*z-u2*z**2),z,0)\n",
+ "print \"u3 = \",u3"
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.10"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter27_2.ipynb b/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter27_2.ipynb
new file mode 100644
index 00000000..6b7c3c32
--- /dev/null
+++ b/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter27_2.ipynb
@@ -0,0 +1,1084 @@
+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 27 : Numerical Solution Of Ordinary Differential Equations"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 27.1, page no. 686"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Solution through picards method \n",
+ "The no of iterations required3\n",
+ "y(0)=1 and y(x)=x+y \n",
+ "Y = x**4/24 + x**3/3 + x**2 + x + 1\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy,math\n",
+ "\n",
+ "x = sympy.Symbol('x')\n",
+ "print \"Solution through picards method \"\n",
+ "n = int(raw_input(\"The no of iterations required\")) \n",
+ "print \"y(0)=1 and y(x)=x+y \"\n",
+ "yo = 1\n",
+ "yn = 1\n",
+ "for i in range(1,n+1):\n",
+ " yn = yo+sympy.integrate(yn+x,(x,0,x))\n",
+ "print \"Y = \",yn"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 27.2, page no. 687"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Solution through picards method \n",
+ "The no of iterations required : 2\n",
+ "y(0)=1 and y(x)=x+y \n",
+ "Y = -Integral(2*x/(2*log(x + 1) + 1), x) + Integral(2*x/(2*log(x + 1) + 1), (x, 0)) - Integral(-2*log(x + 1)/(2*log(x + 1) + 1), x) + Integral(-2*log(x + 1)/(2*log(x + 1) + 1), (x, 0)) - Integral(-1/(2*log(x + 1) + 1), x) + Integral(-1/(2*log(x + 1) + 1), (x, 0)) + 1.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy,math\n",
+ "\n",
+ "x = sympy.Symbol('x')\n",
+ "print \"Solution through picards method \"\n",
+ "n = int(raw_input(\"The no of iterations required : \")) \n",
+ "print \"y(0)=1 and y(x)=x+y \"\n",
+ "yo = 1\n",
+ "y = 1\n",
+ "for i in range(1,n+1):\n",
+ " f = (y-x)/(y+x) \n",
+ " y = yo+sympy.integrate(f,(x,0,x))\n",
+ "x = 0.1\n",
+ "print \"Y = \",y.evalf()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 27.5, page no. 690"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 10,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Solution using Eulers Method \n",
+ "0.3\n",
+ "1.362\n",
+ "Input the number of iteration :− 4\n",
+ "The value of y is :− 1.5282\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Solution using Eulers Method \"\n",
+ "print x\n",
+ "print y\n",
+ "n = int(raw_input(\"Input the number of iteration :− \"))\n",
+ "x = 0\n",
+ "y = 1\n",
+ "for i in range(1,n+1):\n",
+ " y1 = x+y\n",
+ " y = y+0.1*y1\n",
+ " x = x+0.1\n",
+ "print \"The value of y is :− \",y"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 27.6, page no. 691"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 11,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Solution using Eulers Method \n",
+ "0.4\n",
+ "1.5282\n",
+ "Input the number of iteration :− 2\n",
+ "1.02\n",
+ "1.03642857143\n",
+ "The value of y is :− 1.03642857143\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Solution using Eulers Method \"\n",
+ "print x\n",
+ "print y\n",
+ "n = int(raw_input(\"Input the number of iteration :− \"))\n",
+ "x = 0\n",
+ "y = 1\n",
+ "for i in range(1,n+1):\n",
+ " y1 = (y-x)/(y+x)\n",
+ " y = y+0.02*y1\n",
+ " x = x+0.1\n",
+ " print y\n",
+ "print \"The value of y is :− \",y"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 27.7, page no. 692"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 12,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Solution using Eulers Method \n",
+ "0.2\n",
+ "1.03642857143\n",
+ "Input the number of iteration :− 3\n",
+ "1.1\n",
+ "1.11\n",
+ "1.1105\n",
+ "1.110525\n",
+ "−−−−−−−−−−−−−−−−−−−−−−− \n",
+ "1.2315775\n",
+ "1.2315775\n",
+ "1.242630125\n",
+ "1.24318275625\n",
+ "1.24321038781\n",
+ "−−−−−−−−−−−−−−−−−−−−−−− \n",
+ "1.38753142659\n",
+ "1.38753142659\n",
+ "1.39974747853\n",
+ "1.40035828113\n",
+ "1.40038882126\n",
+ "−−−−−−−−−−−−−−−−−−−−−−− \n",
+ "1.57042770339\n",
+ "The value of y is :− 1.40038882126\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Solution using Eulers Method \"\n",
+ "print x\n",
+ "print y\n",
+ "n = int(raw_input(\"Input the number of iteration :− \"))\n",
+ "x = 0.1\n",
+ "m = 1\n",
+ "y = 1\n",
+ "yn = 1\n",
+ "y1 = 1\n",
+ "k = 1\n",
+ "for i in range(1,n+1):\n",
+ " yn = y \n",
+ " for i in range(1,5):\n",
+ " m = (k+y1)/2\n",
+ " yn = y+0.1*m\n",
+ " y1 = (yn+x)\n",
+ " print yn\n",
+ " print \"−−−−−−−−−−−−−−−−−−−−−−− \"\n",
+ " y = yn \n",
+ " m = y1\n",
+ " yn = yn+0.1*m \n",
+ " print yn\n",
+ " x = x+0.1\n",
+ " yn = y \n",
+ " k = m\n",
+ "print \"The value of y is :− \",y"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 27.8, page no. 694"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 14,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Solution using Eulers Method \n",
+ "0.2\n",
+ "2\n",
+ "Input the number of iteration :− 3\n",
+ "2.06020299957\n",
+ "2.06551474469\n",
+ "2.06561668931\n",
+ "2.06561864352\n",
+ "−−−−−−−−−−−−−−−−−−−−−−−\n",
+ "2.13665600548\n",
+ "2.13665600548\n",
+ "2.14156348217\n",
+ "2.14164742068\n",
+ "2.14164885497\n",
+ "−−−−−−−−−−−−−−−−−−−−−−−\n",
+ "2.22267196493\n",
+ "2.22267196493\n",
+ "2.22722645093\n",
+ "2.22729646948\n",
+ "2.22729754504\n",
+ "−−−−−−−−−−−−−−−−−−−−−−−\n",
+ "2.31757184825\n",
+ "The value of y is :− 2.22729754504\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math\n",
+ "\n",
+ "print \"Solution using Eulers Method \"\n",
+ "print x\n",
+ "print y\n",
+ "n = int(raw_input(\"Input the number of iteration :− \"))\n",
+ "x = 0.2\n",
+ "m = 0.301\n",
+ "y = 2\n",
+ "yn = 2\n",
+ "y1 = math.log10(2)\n",
+ "k = 0.301\n",
+ "for i in range(1,n+1):\n",
+ " yn = y\n",
+ " for i in range(1,5):\n",
+ " m = (k+y1)/2\n",
+ " yn = y+0.2*m\n",
+ " y1 = math.log10(yn+x)\n",
+ " print yn\n",
+ " print \"−−−−−−−−−−−−−−−−−−−−−−−\"\n",
+ " y = yn\n",
+ " m = y1\n",
+ " yn = yn+0.2*m \n",
+ " print yn\n",
+ " x = x+0.2\n",
+ " yn = y\n",
+ " k = m\n",
+ "print \"The value of y is :− \",y"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 27.9, page no. 696"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 15,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Solution using Eulers Method \n",
+ "0.8\n",
+ "2.22729754504\n",
+ "Input the number of iteration :− 2\n",
+ "1.2\n",
+ "1.2295445115\n",
+ "1.23088482816\n",
+ "1.23094524903\n",
+ "−−−−−−−−−−−−−−−−−−−−−−− \n",
+ "1.49284119302\n",
+ "1.49284119302\n",
+ "1.52407510156\n",
+ "1.52534665765\n",
+ "1.52539814634\n",
+ "−−−−−−−−−−−−−−−−−−−−−−− \n",
+ "1.85241216588\n",
+ "The value of y is :− 1.52539814634\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math\n",
+ "\n",
+ "print \"Solution using Eulers Method \"\n",
+ "print x\n",
+ "print y\n",
+ "n = int(raw_input(\"Input the number of iteration :− \"))\n",
+ "x = 0.2 \n",
+ "m = 1\n",
+ "y = 1\n",
+ "yn = 1\n",
+ "y1 = 1\n",
+ "k = 1\n",
+ "for i in range(1,n+1):\n",
+ " yn = y\n",
+ " for i in range(1,5):\n",
+ " m = (k+y1)/2\n",
+ " yn = y+0.2*m \n",
+ " y1 = math.sqrt(yn)+x\n",
+ " print yn\n",
+ " print \"−−−−−−−−−−−−−−−−−−−−−−− \"\n",
+ " y = yn\n",
+ " m = y1\n",
+ " yn = yn+0.2*m\n",
+ " print yn\n",
+ " x = x+0.2\n",
+ " yn = y\n",
+ " k = m\n",
+ "print \"The value of y is :− \",y"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "## Example 27.10, page no. 697"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Runges method\n",
+ "The required approximate value is :− \n",
+ "1.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Runges method\"\n",
+ "def f(x,y):\n",
+ " y = x+y\n",
+ " return y\n",
+ "x = 0\n",
+ "y = 1\n",
+ "h = 0.2\n",
+ "k1 = h*f(x,y)\n",
+ "k2 = h*f(x+1/2*h,y+1/2*k1)\n",
+ "kk = h*f(x+h,y+k1)\n",
+ "k3 = h*f(x+h,y+kk)\n",
+ "k = 1/6*(k1+4*k2+k3)\n",
+ "print \"The required approximate value is :− \"\n",
+ "y = y+k\n",
+ "print y"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 27.11, page no. 697"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Runges method\n",
+ "The required approximate value is :− \n",
+ "1.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Runges method\"\n",
+ "def f(x,y):\n",
+ " y = x+y\n",
+ " return y\n",
+ "x = 0\n",
+ "y = 1\n",
+ "h = 0.2\n",
+ "k1 = h*f(x,y)\n",
+ "k2 = h*f(x+1/2*h,y+1/2*k1)\n",
+ "k3 = h*f(x+1/2*h,y+1/2*k2)\n",
+ "k4 = h*f(x+h,y+k3)\n",
+ "k = 1/6*(k1+2*k2+2*k3+k4)\n",
+ "print \"The required approximate value is :− \"\n",
+ "y = y+k\n",
+ "print y"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 27.12, page no. 698"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Runga kutta method\n",
+ "The required approximate value is :− \n",
+ "1.0\n",
+ "To find y(0.4) put x=0.2 y=above value ie 1.196 h=0.2 \n",
+ "The required approximate value is :− \n",
+ "1.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Runga kutta method\"\n",
+ "def f(x,y):\n",
+ " y = (y**2-x**2)/(x**2+y**2)\n",
+ " return y\n",
+ "x = 0\n",
+ "y = 1\n",
+ "h = 0.2\n",
+ "k1 = h*f(x,y)\n",
+ "k2 = h*f(x+1/2*h,y+1/2*k1)\n",
+ "k3 = h*f(x+1/2*h,y+1/2*k2)\n",
+ "k4 = h*f(x+h,y+k3)\n",
+ "k = 1/6*(k1+2*k2+2*k3+k4)\n",
+ "print \"The required approximate value is :− \"\n",
+ "y = y+k\n",
+ "print y\n",
+ "print \"To find y(0.4) put x=0.2 y=above value ie 1.196 h=0.2 \"\n",
+ "x = 0.2\n",
+ "h = 0.2\n",
+ "k1 = h*f(x,y)\n",
+ "k2 = h*f(x+1/2*h,y+1/2*k1)\n",
+ "k3 = h*f(x+1/2*h,y+1/2*k2)\n",
+ "k4 = h*f(x+h,y+k3)\n",
+ "k = 1/6*(k1+2*k2+2*k3+k4)\n",
+ "print \"The required approximate value is :− \"\n",
+ "y = y+k\n",
+ "print y"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 27.13, page no. 699"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Runga kutta method\n",
+ "The required approximate value is :− \n",
+ "1.0\n",
+ "To find y(0.4) put x=0.2 y=above value ie 1.196 h=0.2 \n",
+ "The required approximate value is :− \n",
+ "1.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Runga kutta method\"\n",
+ "def f(x,y):\n",
+ " yy = x+y**2\n",
+ " return yy\n",
+ "x = 0\n",
+ "y = 1\n",
+ "h = 0.1\n",
+ "k1 = h*f(x,y)\n",
+ "k2 = h*f(x+1/2*h,y+1/2*k1)\n",
+ "k3 = h*f(x+1/2*h,y+1/2*k2) \n",
+ "k4 = h*f(x+h,y+k3)\n",
+ "k = 1/6*(k1+2*k2+2*k3+k4)\n",
+ "print \"The required approximate value is :− \"\n",
+ "y = y+k\n",
+ "print y\n",
+ "print \"To find y(0.4) put x=0.2 y=above value ie 1.196 h=0.2 \"\n",
+ "x = 0.1\n",
+ "h = 0.1\n",
+ "k1 = h*f(x,y)\n",
+ "k2 = h*f(x+1/2*h,y+1/2*k1)\n",
+ "k3 = h*f(x+1/2*h,y+1/2*k2)\n",
+ "k4 = h*f(x+h,y+k3)\n",
+ "k = 1/6*(k1+2*k2+2*k3+k4)\n",
+ "print \"The required approximate value is :− \"\n",
+ "y = y+k\n",
+ "print y"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 27.14, page no. 700"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 30,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "y1= x**2/2\n",
+ "y2= -x**5/20 + x**2/2\n",
+ "y3= -x**5/20 + x**2/2\n",
+ "Determining the initial values for milnes method using y3 \n",
+ "x=0.0 y0=0.0 f0=0 \n",
+ "x=0.2 y1= x**2/2\n",
+ "f1= -x**4/4 + 0.2\n",
+ "x=0.4 y2= -x**5/20 + x**2/2\n",
+ "f2= -(-x**5/20 + x**2/2)**2 + 0.4\n",
+ "x=0.6 y3= -x**5/20 + x**2/2\n",
+ "f3= -(-x**5/20 + x**2/2)**2 + 0.6\n",
+ "Using predictor method to find y4\n",
+ "y4= -0.1*x**4 - 0.2*(-x**5/20 + x**2/2)**2 + 0.24\n",
+ "f4= -(-x**5/20 + x**2/2)**2 + 0.8\n",
+ "Using predictor method to find y5 \n",
+ "[-1.53353211362738, 2.59188514369719, -1.51825864909117 - 1.96576535664013*I, -1.51825864909117 + 1.96576535664013*I, -0.611593620665791 - 2.14551655586226*I, -0.611593620665791 + 2.14551655586226*I, 0.00495892544779777 - 0.780181689871346*I, 0.00495892544779777 + 0.780181689871346*I, 1.59571682927425 - 0.8271211188914*I, 1.59571682927425 + 0.8271211188914*I]\n",
+ "f5 = [-1.28459666867223, 2.38248090853886, -1.33416685318757 - 1.68618279990398*I, -1.33416685318757 + 1.68618279990398*I, -1.00716479884721 - 2.10036734766602*I, -1.00716479884721 + 2.10036734766602*I, 0.142926398918135 - 1.33989714259572*I, 0.142926398918135 + 1.33989714259572*I, 1.64946313318333 - 0.385565643189579*I, 1.64946313318333 + 0.385565643189579*I]\n",
+ "Hence y(1)= [-1.53353211362738, 2.59188514369719, -1.51825864909117 - 1.96576535664013*I, -1.51825864909117 + 1.96576535664013*I, -0.611593620665791 - 2.14551655586226*I, -0.611593620665791 + 2.14551655586226*I, 0.00495892544779777 - 0.780181689871346*I, 0.00495892544779777 + 0.780181689871346*I, 1.59571682927425 - 0.8271211188914*I, 1.59571682927425 + 0.8271211188914*I]\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "\n",
+ "x = sympy.Symbol('x')\n",
+ "yo = 0\n",
+ "y = 0\n",
+ "h = 0.2\n",
+ "f = x-y**2\n",
+ "y = sympy.integrate(f,(x,0,x))\n",
+ "y1 = yo+y\n",
+ "print \"y1= \",y1\n",
+ "f = x-y**2\n",
+ "y = sympy.integrate(f,(x,0,x))\n",
+ "y2 = yo+y\n",
+ "print \"y2= \",y2\n",
+ "y = x-y**2\n",
+ "y = sympy.integrate(f,(x,0,x))\n",
+ "y3 = yo+y\n",
+ "print \"y3= \",y3\n",
+ "print \"Determining the initial values for milnes method using y3 \"\n",
+ "print \"x=0.0 y0=0.0 f0=0 \"\n",
+ "print \"x=0.2 y1= \",\n",
+ "x = 0.2\n",
+ "print y1\n",
+ "#y1 = sympy.solve(y1)\n",
+ "print \"f1=\",\n",
+ "f1 = x-y1**2\n",
+ "print f1\n",
+ "print \"x=0.4 y2= \",\n",
+ "x = 0.4\n",
+ "print y2\n",
+ "print \"f2= \",\n",
+ "f2 = x-y2**2\n",
+ "print f2\n",
+ "print \"x=0.6 y3= \",\n",
+ "x = 0.6\n",
+ "print y3\n",
+ "print \"f3=\",\n",
+ "f3 = x-y3**2\n",
+ "print f3\n",
+ "print \"Using predictor method to find y4\"\n",
+ "x = 0.8\n",
+ "y4 = yo+4/3*h*(2*f1-f2+2*f3)\n",
+ "print \"y4=\",y4\n",
+ "f4 = x-y**2\n",
+ "print \"f4= \",f4\n",
+ "print \"Using predictor method to find y5 \"\n",
+ "x = 1.0\n",
+ "y5 = sympy.solve(y1+4/3*h*(2*f2-f3+2*f4))\n",
+ "print y5\n",
+ "f5 = sympy.solve(x-y**2)\n",
+ "print \"f5 =\",f5\n",
+ "print \"Hence y(1)= \",y5"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 27.15, page no. 704"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 31,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Runga kutta method\n",
+ "The required approximate value is :− \n",
+ "1.0\n",
+ "To find y(0.4) put x=0.2 y=above value ie 1.196 h=0.2 \n",
+ "The required approximate value is :− \n",
+ "1.0\n",
+ "To find y(0.4) put x=0.2 y=above value ie 1.196 h=0.2 \n",
+ "The required approximate value is :− \n",
+ "1.0\n",
+ "y0 y1 y2 y3 are respectively: \n",
+ "1.0 1.0 1.0 1\n",
+ "f0 f1 f2 f3 are respectively: \n",
+ "1.3 1.2 1.1 1\n",
+ "finding y4 using predictors milne method x=0.4\n",
+ "y4 = 1.48\n",
+ "f4 = 2.7824\n",
+ "Using corrector method : \n",
+ "y4 = -0.0333333333333333*(-x**5/20 + x**2/2)**2 + 1.24\n",
+ "f4 = -0.0133333333333333*(-x**5/20 + x**2/2)**2 + (-0.0333333333333333*(-x**5/20 + x**2/2)**2 + 1.24)**2 + 0.496\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Runga kutta method\"\n",
+ "def f(x,y):\n",
+ " yy = x*y+y**2\n",
+ " return yy\n",
+ "y0 = 1\n",
+ "x = 0\n",
+ "y = 1\n",
+ "h = 0.1\n",
+ "k1 = h*f(x,y)\n",
+ "k2 = h*f(x+1/2*h,y+1/2*k1)\n",
+ "k3 = h*f(x+1/2*h,y+1/2*k2)\n",
+ "k4 = h*f(x+h,y+k3)\n",
+ "ka = 1/6*(k1+2*k2+2*k3+k4)\n",
+ "print \"The required approximate value is :− \"\n",
+ "y1 = y+ka \n",
+ "y = y+ka \n",
+ "print y\n",
+ "print \"To find y(0.4) put x=0.2 y=above value ie 1.196 h=0.2 \"\n",
+ "x = 0.1\n",
+ "h = 0.1\n",
+ "k1 = h*f(x,y)\n",
+ "k2 = h*f(x+1/2*h,y+1/2*k1)\n",
+ "k3 = h*f(x+1/2*h,y+1/2*k2)\n",
+ "k4 = h*f(x+h,y+k3)\n",
+ "kb = 1/6*(k1+2*k2+2*k3+k4)\n",
+ "print \"The required approximate value is :− \"\n",
+ "y2 = y+kb \n",
+ "y = y+kb\n",
+ "print y\n",
+ "print \"To find y(0.4) put x=0.2 y=above value ie 1.196 h=0.2 \"\n",
+ "x = 0.2\n",
+ "h = 0.1\n",
+ "k1 = h*f(x,y)\n",
+ "k2 = h*f(x+1/2*h,y+1/2*k1)\n",
+ "k3 = h*f(x+1/2*h,y+1/2*k2)\n",
+ "k4 = h*f(x+h,y+k3)\n",
+ "kc = 1/6*(k1+2*k2+2*k3+k4)\n",
+ "print \"The required approximate value is :− \"\n",
+ "y3 = y+kc\n",
+ "y = y+kc \n",
+ "print y\n",
+ "f0 = f(0,y0)\n",
+ "f1 = f(0.1,y1)\n",
+ "f2 = f(0.2,y2)\n",
+ "f3 = f(0.3,y3)\n",
+ "print \"y0 y1 y2 y3 are respectively: \"\n",
+ "print y3,y2,y1,y0\n",
+ "print \"f0 f1 f2 f3 are respectively: \"\n",
+ "print f3,f2,f1,f0\n",
+ "print \"finding y4 using predictors milne method x=0.4\"\n",
+ "h = 0.1\n",
+ "y4 = y0+4*h/3*(2*f1-f2+2*f3)\n",
+ "print \"y4 = \",y4\n",
+ "print \"f4 = \",f(0.4,y4)\n",
+ "print \"Using corrector method : \"\n",
+ "y4 = y2+h/3*(f2+4*f3+f4) \n",
+ "print \"y4 = \",y4\n",
+ "print \"f4 = \",f(0.4,y4) "
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 27.16, page no. 705"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 32,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Using predictor method \n",
+ "y11 = 2.57229741667\n",
+ "Using corrector method \n",
+ "y11 = 2.57494727679\n",
+ "f11 = 7.00689666251\n"
+ ]
+ }
+ ],
+ "source": [
+ "def f(x,y):\n",
+ " yy = x**2*(1+y)\n",
+ " return yy\n",
+ "y3 = 1\n",
+ "y2 = 1.233\n",
+ "y1 = 1.548\n",
+ "y0 = 1.979\n",
+ "f3 = f(1,y3)\n",
+ "f2 = f(1.1,y2)\n",
+ "f1 = f(1.2,y1)\n",
+ "f0 = f(1.3,y0)\n",
+ "print \"Using predictor method \"\n",
+ "h = 0.1\n",
+ "y11 = y0+h/24*(55*f0-59*f1+37*f2-9*f3)\n",
+ "print \"y11 = \",y11\n",
+ "x = 1.4\n",
+ "f11 = f(1.4,y11)\n",
+ "print \"Using corrector method \"\n",
+ "y11 = y0+h/24*(9*f11+19*f0-5*f1+f2)\n",
+ "print \"y11 = \",y11\n",
+ "f11 = f(1.4,y11)\n",
+ "print \"f11 = \",f11"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 27.17, page no. 706"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 33,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Runga kutta method\n",
+ "The required approximate value is :− \n",
+ "1.0\n",
+ "To find y(0.4) put x=0.2 y=above value ie 1.196 h=0.2 \n",
+ "The required approximate value is :− \n",
+ "1.0\n",
+ "To find y(0.4) put x=0.2 y=above value ie 1.196 h=0.2 \n",
+ "The required approximate value is :− \n",
+ "1.0\n",
+ "y0 y1 y2 y3 are respectively :\n",
+ "1.0 1.0 1.0 1\n",
+ "f0 f1 f2 f3 are respectively :\n",
+ "-0.7 -0.8 -0.9 -1\n",
+ "Using adams method \n",
+ "Using the predictor\n",
+ "y4 = 0.935\n",
+ "Using corrector method \n",
+ "y4 = 0.9397165625\n",
+ "f4 = -0.483067217837\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Runga kutta method\"\n",
+ "def f(x,y):\n",
+ " yy = x-y**2\n",
+ " return yy\n",
+ "y0 = 1\n",
+ "x = 0\n",
+ "y = 1\n",
+ "h = 0.1\n",
+ "k1 = h*f(x,y)\n",
+ "k2 = h*f(x+1/2*h,y+1/2*k1)\n",
+ "k3 = h*f(x+1/2*h,y+1/2*k2)\n",
+ "k4 = h*f(x+h,y+k3)\n",
+ "ka = 1/6*(k1+2*k2+2*k3+k4) \n",
+ "print \"The required approximate value is :− \"\n",
+ "y1 = y+ka\n",
+ "y = y+ka\n",
+ "print y\n",
+ "print \"To find y(0.4) put x=0.2 y=above value ie 1.196 h=0.2 \"\n",
+ "x = 0.1\n",
+ "h = 0.1\n",
+ "k1 = h*f(x,y)\n",
+ "k2 = h*f(x+1/2*h,y+1/2*k1)\n",
+ "k3 = h*f(x+1/2*h,y+1/2*k2)\n",
+ "k4 = h*f(x+h,y+k3)\n",
+ "kb = 1/6*(k1+2*k2+2*k3+k4)\n",
+ "print \"The required approximate value is :− \"\n",
+ "y2 = y+kb\n",
+ "y = y+kb\n",
+ "print y\n",
+ "print \"To find y(0.4) put x=0.2 y=above value ie 1.196 h=0.2 \"\n",
+ "x = 0.2\n",
+ "h = 0.1\n",
+ "k1 = h*f(x,y)\n",
+ "k2 = h*f(x+1/2*h,y+1/2*k1)\n",
+ "k3 = h*f(x+1/2*h,y+1/2*k2)\n",
+ "k4 = h*f(x+h,y+k3)\n",
+ "kc = 1/6*(k1+2*k2+2*k3+k4) \n",
+ "print \"The required approximate value is :− \"\n",
+ "y3 = y+kc\n",
+ "y = y+kc\n",
+ "print y\n",
+ "f0 = f(0,y0)\n",
+ "f1 = f(0.1,y1)\n",
+ "f2 = f(0.2,y2)\n",
+ "f3 = f(0.3,y3) \n",
+ "print \"y0 y1 y2 y3 are respectively :\"\n",
+ "print y3,y2,y1,y0\n",
+ "print \"f0 f1 f2 f3 are respectively :\"\n",
+ "print f3,f2,f1,f0\n",
+ "print \"Using adams method \"\n",
+ "print \"Using the predictor\"\n",
+ "h = 0.1\n",
+ "y4 = y3+h/24*(55*f3-59*f2+37*f1-9*f0)\n",
+ "x = 0.4\n",
+ "f4 = f(0.4,y4)\n",
+ "print \"y4 = \",y4\n",
+ "print \"Using corrector method \"\n",
+ "y4 = y3+h/24*(9*f4+19*f3-5*f2+f1) \n",
+ "print \"y4 = \",y4\n",
+ "f4 = f(0.4,y4)\n",
+ "print \"f4 = \",f4"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 27.18, page no. 708"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 36,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Picards method\n",
+ "First approximation\n",
+ "y1 = x**2/2 + x + 2\n",
+ "z1 = x**2/2 - 4*x + 1\n",
+ "Second approximation\n",
+ "y2 = x**3/6 - 3*x**2/2 + x + 2\n",
+ "z2 = -x**5/20 - x**4/4 - x**3 - 3*x**2/2 - 4*x + 1\n",
+ "Third approximation\n",
+ "y3 = -x**6/120 - x**5/20 - x**4/4 - x**3/2 - 3*x**2/2 + x + 2\n",
+ "z3 = -x**7/252 + x**6/12 - 31*x**5/60 + 7*x**4/12 + 5*x**3/3 - 3*x**2/2 - 4*x + 1\n",
+ "y(0.1) = -0.00833333333333333*x**6 - 0.05*x**5 - 0.25*x**4 - 0.5*x**3 - 1.5*x**2 + x + 2.0\n",
+ "z(0.1) = -0.00396825396825397*x**7 + 0.0833333333333333*x**6 - 0.516666666666667*x**5 + 0.583333333333333*x**4 + 1.66666666666667*x**3 - 1.5*x**2 - 4.0*x + 1.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "\n",
+ "print \"Picards method\"\n",
+ "x0 = 0\n",
+ "y0 = 2\n",
+ "z0 = 1\n",
+ "x = sympy.Symbol('x')\n",
+ "def f(x,y,z):\n",
+ " yy = x+z\n",
+ " return yy\n",
+ "def g(x,y,z):\n",
+ " yy = x-y**2\n",
+ " return yy\n",
+ "print \"First approximation\"\n",
+ "y1 = y0+sympy.integrate(f(x,y0,z0),(x,x0,x))\n",
+ "print \"y1 = \",y1\n",
+ "z1 = z0+sympy.integrate(g(x,y0,z0),(x,x0,x))\n",
+ "print \"z1 = \",z1\n",
+ "print \"Second approximation\"\n",
+ "y2 = y0+sympy.integrate(f(x,y1,z1),(x,x0,x))\n",
+ "print \"y2 = \",y2\n",
+ "z2 = z0+sympy.integrate(g(x,y1,z1),(x,x0,x))\n",
+ "print \"z2 = \",z2\n",
+ "print \"Third approximation\"\n",
+ "y3 = y0+sympy.integrate(f(x,y2,z2),(x,x0,x))\n",
+ "print \"y3 = \",y3\n",
+ "z3 = z0+sympy.integrate(g(x,y2,z2),(x,x0,x))\n",
+ "print \"z3 = \",z3\n",
+ "x = 0.1\n",
+ "print \"y(0.1) = \",y3.evalf()\n",
+ "print \"z(0.1) = \",z3.evalf()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 27.19, page no. 710"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 37,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Using k1 k2.. for f and l1 l2... for g runga kutta formula becomes \n",
+ "y = 1.0\n",
+ "y1 = 0.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "\n",
+ "x = sympy.Symbol('x')\n",
+ "def f(x,y,z):\n",
+ " yy = z \n",
+ " return yy\n",
+ "def g(x,y,z):\n",
+ " yy = x*y**2-y**2\n",
+ " return yy\n",
+ "x0 = 0\n",
+ "y0 = 1\n",
+ "z0 = 0\n",
+ "h = 0.2\n",
+ "print \"Using k1 k2.. for f and l1 l2... for g runga kutta formula becomes \"\n",
+ "h = 0.2\n",
+ "k1 = h*f(x0,y0,z0)\n",
+ "l1 = h*g(x0,y0,z0)\n",
+ "k2 = h*f(x0+1/2*h,y0+1/2*k1,z0+1/2*l1)\n",
+ "l2 = h*g(x0+1/2*h,y0+1/2*k1,z0+1/2*l1)\n",
+ "k3 = h*f(x0+1/2*h,y0+1/2*k2,z0+1/2*l2)\n",
+ "l3 = h*g(x0+1/2*h,y0+1/2*k2,z0+1/2*l2)\n",
+ "k4 = h*f(x0+h,y0+k3,z0+l3)\n",
+ "l4 = h*g(x0+h,y0+k3,z0+l3)\n",
+ "k = 1/6*(k1+2*k2+2*k3+k4)\n",
+ "l = 1/6*(l1+2*l2+2*l3+2*l4)\n",
+ "x = 0.2\n",
+ "y = y0+k\n",
+ "y1 = z0+l\n",
+ "print \"y = \",y\n",
+ "print \"y1 = \",y1"
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.11+"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter28_2.ipynb b/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter28_2.ipynb
new file mode 100644
index 00000000..69d141e8
--- /dev/null
+++ b/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter28_2.ipynb
@@ -0,0 +1,437 @@
+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 28: Numerical Solution Of Partial Differential Equations"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 28.1, page no. 725"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "D=Bˆ2−4AC\n",
+ "if D<0 then elliptic if D=0 then parabolic if D>0 then hyperboic\n",
+ "(i) A=xˆ2, B1−yˆ2 D=4ˆ2−4∗1∗4=0 so the equation is PARABOLIC \n",
+ "(ii) D=4xˆ2(yˆ2−1)\n",
+ "for −inf<x<inf and −1<y<1 D<0 \n",
+ "So the equation is ELLIPTIC \n",
+ "(iii) A=1+xˆ2, B=5+2xˆ2, C=4+xˆ2\n",
+ "D=9>0\n",
+ "So the equation is HYPERBOLIC \n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"D=Bˆ2−4AC\"\n",
+ "print \"if D<0 then elliptic if D=0 then parabolic if D>0 then hyperboic\"\n",
+ "print \"(i) A=xˆ2, B1−yˆ2 D=4ˆ2−4∗1∗4=0 so the equation is PARABOLIC \"\n",
+ "print \"(ii) D=4xˆ2(yˆ2−1)\"\n",
+ "print \"for −inf<x<inf and −1<y<1 D<0 \"\n",
+ "print \"So the equation is ELLIPTIC \"\n",
+ "print \"(iii) A=1+xˆ2, B=5+2xˆ2, C=4+xˆ2\"\n",
+ "print \"D=9>0\"\n",
+ "print \"So the equation is HYPERBOLIC \""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 28.2, page no. 726"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "See figure in question\n",
+ "From symmetry u7=u1, u8=u2, u9=u3, u3=u1, u6=u4, u9=u7\n",
+ "u5=1/4∗(2000+2000+1000+1000)=1500\n",
+ "u1=1/4(0=1500+1000+2000)=1125\n",
+ "u2=1/4∗(1125+1125+1000+1500)=1188\n",
+ "u4=1/4(2000+1500+1125+1125)=1438\n",
+ "1125 1188 1438 1500\n",
+ "Iterations :\n",
+ "\n",
+ "1301 1414 1164 1031\n",
+ "\n",
+ "1250 1337 1087 1019\n",
+ "\n",
+ "1199 1312 1062 981\n",
+ "\n",
+ "1174 1287 1037 968\n",
+ "\n",
+ "1155 1274 1024 956\n",
+ "\n",
+ "1144 1265 1015 949\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"See figure in question\"\n",
+ "print \"From symmetry u7=u1, u8=u2, u9=u3, u3=u1, u6=u4, u9=u7\"\n",
+ "print \"u5=1/4∗(2000+2000+1000+1000)=1500\"\n",
+ "u5 = 1500\n",
+ "print \"u1=1/4(0=1500+1000+2000)=1125\"\n",
+ "u1 = 1125\n",
+ "print \"u2=1/4∗(1125+1125+1000+1500)=1188\"\n",
+ "u2 = 1188\n",
+ "print \"u4=1/4(2000+1500+1125+1125)=1438\"\n",
+ "u4 = 1438\n",
+ "print u1,u2,u4,u5 \n",
+ "print \"Iterations :\"\n",
+ "for i in range(1,7):\n",
+ " u11 = (1000+u2+500+u4)/4\n",
+ " u22 = (u11+u1+1000+u5)/4 \n",
+ " u44 = (2000+u5+u11+u1)/4\n",
+ " u55 = (u44+u4+u22+u2)/4\n",
+ " print \"\"\n",
+ " print u55,u44,u22,u11\n",
+ " u1 = u11 \n",
+ " u2 = u22 \n",
+ " u4 = u44 \n",
+ " u5 = u55"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 28.3, page no. 727"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "See figure in question\n",
+ "To find the initial values of u1 u2 u3 u4 we assume u4=0 \n",
+ "u1=1/4∗(1000+0+1000+2000)=1000\n",
+ "u2=1/4(1000+500+1000+500)=625 \n",
+ "u3=1/4∗(2000+0+1000+500)=875\n",
+ "u4=1/4(875+0+625+0)=375\n",
+ "1000 625 875 375\n",
+ "Iterations:\n",
+ "\n",
+ "437 1000 750 1125\n",
+ "\n",
+ "453 1031 781 1187\n",
+ "\n",
+ "457 1039 789 1203\n",
+ "\n",
+ "458 1041 791 1207\n",
+ "\n",
+ "458 1041 791 1208\n",
+ "\n",
+ "458 1041 791 1208\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"See figure in question\"\n",
+ "print \"To find the initial values of u1 u2 u3 u4 we assume u4=0 \"\n",
+ "print \"u1=1/4∗(1000+0+1000+2000)=1000\"\n",
+ "u1 = 1000\n",
+ "print \"u2=1/4(1000+500+1000+500)=625 \"\n",
+ "u2 = 625\n",
+ "print \"u3=1/4∗(2000+0+1000+500)=875\"\n",
+ "u3 = 875\n",
+ "print \"u4=1/4(875+0+625+0)=375\"\n",
+ "u4 = 375\n",
+ "print u1,u2,u3,u4 \n",
+ "print \"Iterations:\"\n",
+ "for i in range(1,7):\n",
+ " u11 = (2000+u2+1000+u3)/4 \n",
+ " u22 = (u11+500+1000+u4)/4\n",
+ " u33 = (2000+u4+u11+500)/4\n",
+ " u44 = (u33+0+u22+0)/4\n",
+ " print \"\"\n",
+ " print u44,u33,u22,u11\n",
+ " u1 = u11 \n",
+ " u2 = u22 \n",
+ " u4 = u44 \n",
+ " u3 = u33"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 28.5, page no. 729"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 10,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Here cˆ2=4, h=1, k=1/8, therefore alpha=(cˆ2)∗k/(hˆ2)\n",
+ "Using bendre−schmidits recurrence relation ie u(i)(j +1)=t∗u(i−1)(j)+t∗u(i+1)(j)+(1−2t)∗u(i,j)\n",
+ "Now since u(0,t)=0=u(8,t) therefore u(0,i)=0 and u(8,j)=0 and u(x,0)=4x−1/2xˆ2 \n",
+ "0.0\n",
+ "-4.0\n",
+ "0.0\n",
+ "4.0\n",
+ "8.0\n",
+ "12.0\n",
+ "16.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "print \"Here cˆ2=4, h=1, k=1/8, therefore alpha=(cˆ2)∗k/(hˆ2)\"\n",
+ "print \"Using bendre−schmidits recurrence relation ie u(i)(j +1)=t∗u(i−1)(j)+t∗u(i+1)(j)+(1−2t)∗u(i,j)\"\n",
+ "print \"Now since u(0,t)=0=u(8,t) therefore u(0,i)=0 and u(8,j)=0 and u(x,0)=4x−1/2xˆ2 \"\n",
+ "c = 2\n",
+ "h = 1\n",
+ "k = 1/8\n",
+ "t = (c**2)*k/(h**2)\n",
+ "A = numpy.ones((9,9))\n",
+ "for i in range(0,9):\n",
+ " for j in range (0,9):\n",
+ " A[0,i] = 0\n",
+ " A[8,i] = 0\n",
+ " A[i,0] = 4*(i-1)-1/2*(i-1)**2\n",
+ "for i in range(1,8):\n",
+ " for j in range(1,7):\n",
+ " A[i,j] = t*A[i-1,j-1]+t*A[i+1,j-1]+(1-2*t)*A[i-1,j-1]\n",
+ "for i in range(1,8):\n",
+ " j = 2\n",
+ " print A[i,j]"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 28.6, page no. 730"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Here cˆ2=1, h=1/3, k=1/36, therefore t=(cˆ2)∗k/(hˆ2)=1/4 \n",
+ "So bendre−schmidits recurrence relation ie u(i)(j+1)=1/4(u(i−1)(j)+u(i+1)(j)+2u(i,j)\n",
+ "Now since u(0,t)=0=u(1,t) therefore u(0,i)=0 and u(1,j)=0 and u(x,0)=sin(%pi)x\n",
+ "-0.433012701892\n",
+ "0.216506350946\n",
+ "0.649519052838\n",
+ "0.433012701892\n",
+ "-0.216506350946\n",
+ "-0.649519052838\n",
+ "-0.433012701892\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,math\n",
+ "\n",
+ "print \"Here cˆ2=1, h=1/3, k=1/36, therefore t=(cˆ2)∗k/(hˆ2)=1/4 \"\n",
+ "print \"So bendre−schmidits recurrence relation ie u(i)(j+1)=1/4(u(i−1)(j)+u(i+1)(j)+2u(i,j)\"\n",
+ "print \"Now since u(0,t)=0=u(1,t) therefore u(0,i)=0 and u(1,j)=0 and u(x,0)=sin(%pi)x\"\n",
+ "c = 1.\n",
+ "h = 1./3\n",
+ "k = 1./36\n",
+ "t = (c**2)*k/(h**2)\n",
+ "A = numpy.ones((9,9))\n",
+ "for i in range(0,9):\n",
+ " for j in range(0,9):\n",
+ " A[0,i] = 0\n",
+ " A[1,i] = 0\n",
+ " A[i,0] = math.sin(math.pi/3*(i-1)) \n",
+ "for i in range(1,8):\n",
+ " for j in range(1,8):\n",
+ " A[i,j] = t*A[i-1,j-1]+t*A[i+1,j-1]+(1-2*t)*A[i-1,j-1]\n",
+ "for i in range(1,8):\n",
+ " j = 1\n",
+ " print A[i,j]"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 28.7, page no. 732"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 15,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Here cˆ2=16, taking h=1, finding k such that cˆ2tˆ2=1 \n",
+ "So bendre−schmidits recurrence relation ie u(i)(j+1)=(16tˆ2(u(i−1)(j)+u(i+1)(j))+2(1−16∗tˆ2u(i,j)−u(i)(j−1)\n",
+ "Now since u(0,t)=0=u(5,t) therefore u(0,i)=0 and u(5,j)=0 and u(x,0)=xˆ2(5−x)\n",
+ "Also from 1st derivative (u(i)(j+1)−u(i,j−1))/2k=g(x) and g(x)=0 in this case\n",
+ "So if j=0 this gives u(i)(1)=1/2∗(u(i−1)(0)+u(i+1)(0))\n",
+ " 0.0 0.0 0.0 0.0 0.0 \n",
+ " 0.0 4.0 8.0 12.0 16.0 \n",
+ " 0.0 12.0 24.0 36.0 48.0 \n",
+ " 0.0 18.0 36.0 54.0 72.0 \n",
+ " 0.0 16.0 0.0 0.0 0.0 \n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "print \"Here cˆ2=16, taking h=1, finding k such that cˆ2tˆ2=1 \"\n",
+ "print \"So bendre−schmidits recurrence relation ie u(i)(j+1)=(16tˆ2(u(i−1)(j)+u(i+1)(j))+2(1−16∗tˆ2u(i,j)−u(i)(j−1)\"\n",
+ "print \"Now since u(0,t)=0=u(5,t) therefore u(0,i)=0 and u(5,j)=0 and u(x,0)=xˆ2(5−x)\"\n",
+ "c = 4\n",
+ "h =1 \n",
+ "k = (h/c)\n",
+ "t = k/h\n",
+ "A = numpy.zeros((6,6))\n",
+ "print \"Also from 1st derivative (u(i)(j+1)−u(i,j−1))/2k=g(x) and g(x)=0 in this case\"\n",
+ "print \"So if j=0 this gives u(i)(1)=1/2∗(u(i−1)(0)+u(i+1)(0))\"\n",
+ "for i in range(0,6):\n",
+ " for j in range(1,9):\n",
+ " A[0,i] = 0\n",
+ " A[5,i] = 0;\n",
+ " A[i,1] = (i)**2*(5-i)\n",
+ "for i in range(0,4):\n",
+ " A[i+1,2] =1/2*(A[i,1]+A[i+2,1])\n",
+ "for i in range(2,5):\n",
+ " for j in range(2,5):\n",
+ " A[i-1,j] = (c*t)**2*(A[i-2,j-1]+A[i,j-1])+2*(1-(c*t)**2)*A[i-1,j-1]-A[i-1,j-2]\n",
+ "for i in range(0,5):\n",
+ " for j in range(0,5):\n",
+ " print \" \",A[i,j],\n",
+ " print \"\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 28.8, page no. 734"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Here cˆ2=4, taking h=1, finding k such that cˆ2tˆ2=1 \n",
+ "So bendre−schmidits recurrence relation ie u(i)(j+1)=(16tˆ2(u(i−1)(j)+u(i+1)(j))+2(1−16∗tˆ2u(i,j)−u(i)(j−1)\n",
+ "Now since u(0,t)=0=u(4,t) therefore u(0,i)=0 and u(4,j)=0 and u(x,0)=x(4−x) \n",
+ "Also from 1st derivative (u(i)(j+1)−u(i,j−1))/2 k=g(x) and g(x)=0 in this case\n",
+ "So if j=0 this gives u(i)(1)=1/2∗(u(i−1)(0)+u(i+1)(0))\n",
+ " 0.0 0.0 0.0 0.0 0.0 \n",
+ " 3.0 0.0 -3.0 -6.0 -9.0 \n",
+ " 4.0 0.0 -4.0 -8.0 -12.0 \n",
+ " 3.0 0.0 -3.0 -6.0 -9.0 \n",
+ " 0.0 0.0 0.0 0.0 0.0 \n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "print \"Here cˆ2=4, taking h=1, finding k such that cˆ2tˆ2=1 \"\n",
+ "print \"So bendre−schmidits recurrence relation ie u(i)(j+1)=(16tˆ2(u(i−1)(j)+u(i+1)(j))+2(1−16∗tˆ2u(i,j)−u(i)(j−1)\"\n",
+ "print \"Now since u(0,t)=0=u(4,t) therefore u(0,i)=0 and u(4,j)=0 and u(x,0)=x(4−x) \"\n",
+ "c = 2\n",
+ "h = 1\n",
+ "k = (h/c)\n",
+ "t = k/h\n",
+ "A = numpy.zeros((6,6))\n",
+ "print \"Also from 1st derivative (u(i)(j+1)−u(i,j−1))/2 k=g(x) and g(x)=0 in this case\"\n",
+ "print \"So if j=0 this gives u(i)(1)=1/2∗(u(i−1)(0)+u(i+1)(0))\"\n",
+ "for i in range(0,6):\n",
+ " for j in range(1,9):\n",
+ " A[0,i] = 0\n",
+ " A[4,i] = 0\n",
+ " A[i,0] = (i)*(4-i)\n",
+ "for i in range(0,4):\n",
+ " A[i+1,2] = 1/2*(A[i,1]+A[i+2,1])\n",
+ "for i in range(2,5):\n",
+ " for j in range(2,5):\n",
+ " A[i-1,j] = (c*t)**2*(A[i-2,j-1]+A[i,j-1])+2*(1-(c*t)**2)*A[i-1,j-1]-A[i-1,j-2]\n",
+ "for i in range (0,5):\n",
+ " for j in range(0,5):\n",
+ " print \" \",A[i,j],\n",
+ " print \"\""
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.11+"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter2_2.ipynb b/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter2_2.ipynb
new file mode 100644
index 00000000..c8582eca
--- /dev/null
+++ b/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter2_2.ipynb
@@ -0,0 +1,1360 @@
+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 2: Determinants and Matrices"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.1, page no. 55"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Determinant of A is: a*b*c - a*f**2 - b*g**2 - c*h**2 + 2*f*g*h\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "\n",
+ "a = sympy.Symbol('a')\n",
+ "h = sympy.Symbol('h')\n",
+ "g = sympy.Symbol('g')\n",
+ "b = sympy.Symbol('b')\n",
+ "f = sympy.Symbol('f')\n",
+ "c = sympy.Symbol('c')\n",
+ "A = sympy.Matrix([[a,h,g],[h,b,f],[g,f,c]])\n",
+ "\n",
+ "print \"Determinant of A is: \", A.det()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.2, page no. 55"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false,
+ "scrolled": true
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Determinanat of a is: 88.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "A = numpy.array([[0,1,2,3],[1,0,3,0],[2,3,0,1],[3,0,1,2]])\n",
+ "print \"Determinant of a is:\",numpy.linalg.det(A)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.3, page no. 56"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "-a**3*b**2*c + a**3*b*c**2 + a**2*b**3*c - a**2*b*c**3 + a**2*b - a**2*c - a*b**3*c**2 + a*b**2*c**3 - a*b**2 + a*c**2 + b**2*c - b*c**2\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "import sympy\n",
+ "\n",
+ "a = sympy.Symbol('a');\n",
+ "b = sympy.Symbol('b');\n",
+ "c = sympy.Symbol('c');\n",
+ "A = sympy.Matrix([[a,a**2,a**3-1],[b,b**2,b**3-1],[c,c**2,c**3-1]])\n",
+ "\n",
+ "print A.det()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.4, page no. 57"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Determinanat of a is: -24.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "A = numpy.array([[21,17,7,10],[24,22,6,10],[6,8,2,3],[6,7,1,2]])\n",
+ "print \"Determinanat of a is:\",numpy.linalg.det(A)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.16, page no.60"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "a: x**y*log(x)\n",
+ "b: x**y*y*log(x)/x + x**y/x\n",
+ "c: x**y*y**2*log(x)/x**2 - x**y*y*log(x)/x**2 + 2*x**y*y/x**2 - x**y/x**2\n",
+ "d: x**y*y/x\n",
+ "e: x**y*y*log(x)/x + x**y/x\n",
+ "f: x**y*y**2*log(x)/x**2 - x**y*y*log(x)/x**2 + 2*x**y*y/x**2 - x**y/x**2\n",
+ "Clearly c = f\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "\n",
+ "x = sympy.Symbol('x');\n",
+ "y = sympy.Symbol('y')\n",
+ "u = x**y\n",
+ "a = sympy.diff(u, y)\n",
+ "b = sympy.diff(a, x)\n",
+ "c = sympy.diff(b, x)\n",
+ "d = sympy.diff(u, x)\n",
+ "e = sympy.diff(d, y)\n",
+ "f = sympy.diff(e, x)\n",
+ "print \"a: \", a\n",
+ "print \"b: \", b\n",
+ "print \"c: \", c\n",
+ "print \"d: \", d\n",
+ "print \"e: \", e\n",
+ "print \"f: \", f\n",
+ "print \"Clearly c = f\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.17, page no. 65"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 17,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "A*B=\n",
+ "[[ 5 9 13]\n",
+ " [-1 2 4]\n",
+ " [-2 2 4]]\n",
+ "\n",
+ "B*A=\n",
+ "[[-1 12 11]\n",
+ " [-1 7 8]\n",
+ " [-2 -1 5]]\n",
+ "Clearly AB is not equal to BA\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "A = numpy.array([[1,3,0],[-1,2,1],[0,0,2]])\n",
+ "B = numpy.array([[2,3,4],[1,2,3],[-1,1,2]])\n",
+ "ma = numpy.matrix(A)\n",
+ "mb = numpy.matrix(B)\n",
+ "print \"A*B=\"\n",
+ "print ma*mb\n",
+ "print \"\"\n",
+ "print \"B*A=\"\n",
+ "print mb*ma\n",
+ "print \"Clearly AB is not equal to BA\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.18, page no. 65"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "AB=C−−>B=inv(A)∗C\n",
+ "\n",
+ "[[ 1.00000000e+00 -1.77635684e-15 -8.88178420e-16]\n",
+ " [ -2.22044605e-16 2.00000000e+00 -1.11022302e-16]\n",
+ " [ 1.77635684e-15 0.00000000e+00 1.00000000e+00]]\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "A = numpy.matrix([[3,2,2],[1,3,1],[5,3,4]])\n",
+ "C = numpy.matrix([[3,4,2],[1,6,1],[5,6,4]])\n",
+ "print \"AB=C−−>B=inv(A)∗C\"\n",
+ "print \"\"\n",
+ "B = numpy.linalg.inv(A)*C \n",
+ "print B"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.19, page no. 66"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 48,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Aˆ3−4∗Aˆ2−3A+11*I=\n",
+ "\n",
+ "[[ 0. 0. 0.]\n",
+ " [ 0. 0. 0.]\n",
+ " [ 0. 0. 0.]]\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "A = numpy.array([[1,3,2],[2,0,-1],[1,2,3]])\n",
+ "I = numpy.eye(3)\n",
+ "A = numpy.matrix(A)\n",
+ "print \"Aˆ3−4∗Aˆ2−3A+11*I=\"\n",
+ "print \"\"\n",
+ "print A**3-4*A**2-3*A+11*I"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.20, page no. 67"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 58,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Enter the value of n: 3\n",
+ "Calculating A ^ n: \n",
+ "[[ 31 -75]\n",
+ " [ 12 -29]]\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "0\n",
+ "A = numpy.matrix([[11,-25],[4, -9]])\n",
+ "n = int(raw_input(\"Enter the value of n: \"))\n",
+ "print \"Calculating A ^ n: \"\n",
+ "print A**n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.23, page no. 70"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Inverse of A is: \n",
+ "[[ 3. 1. 1.5 ]\n",
+ " [-1.25 -0.25 -0.75]\n",
+ " [-0.25 -0.25 -0.25]]\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "A = numpy.matrix([[1,1,3],[1,3,-3],[-2,-4,-4]])\n",
+ "print \"Inverse of A is: \"\n",
+ "print numpy.linalg.inv(A)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.24.1, page no. 71"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Rank of A is: 2\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "A = numpy.matrix([[1,2,3],[1,4,2],[2,6,5]])\n",
+ "print \"Rank of A is:\",numpy.linalg.matrix_rank(A)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.24.2, page no. 71"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Rank of A is: 2\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "A = numpy.matrix([[0,1,-3,-1],[1,0,1,1],[3,1,0,2],[1,1,-2,0]])\n",
+ "print \"Rank of A is:\",numpy.linalg.matrix_rank(A)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.25, page no. 72"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Inverse of A is: \n",
+ "[[ 3. 1. 1.5 ]\n",
+ " [-1.25 -0.25 -0.75]\n",
+ " [-0.25 -0.25 -0.25]]\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "A = numpy.matrix([[1,1,3],[1,3,-3],[-2,-4,-4]])\n",
+ "print \"Inverse of A is: \"\n",
+ "print numpy.linalg.inv(A)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.26, page no. 73"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Rank of A is: 3\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "A = numpy.matrix([[2,3,-1,-1],[1,-1,-2,-4],[3,1,3,-2],[6,3,0,-7]])\n",
+ "r,p = numpy.linalg.eigh ( A )\n",
+ "print \"Rank of A is:\",numpy.linalg.matrix_rank(A)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.28, page no. 75"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 9,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Inverse of A is: \n",
+ "[[ 1.4 0.2 -0.4]\n",
+ " [-1.5 0. 0.5]\n",
+ " [ 1.1 -0.2 -0.1]]\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "A = numpy.matrix([[1,1,1],[4,3,-1],[3,5,3]])\n",
+ "print \"Inverse of A is: \"\n",
+ "print numpy.linalg.inv(A)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.31, page no. 78"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " The equations can be rewritten as AX=B where X=[ x1 ; x2 ; x3 ; x4 ] and \n",
+ "Determinant of A=\n",
+ "8.0\n",
+ "Inverse of A =\n",
+ "[[ 0.5 0.5 0.5 -0.5]\n",
+ " [-0.5 0. 0. 0.5]\n",
+ " [ 0. -0.5 0. 0.5]\n",
+ " [ 0. 0. -0.5 0.5]]\n",
+ "X= [[ 1.]\n",
+ " [-1.]\n",
+ " [ 2.]\n",
+ " [-2.]]\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "print \"The equations can be rewritten as AX=B where X=[ x1 ; x2 ; x3 ; x4 ] and \"\n",
+ "A = numpy.matrix([[1,-1,1,1],[1,1,-1,1],[1,1,1,-1],[1,1,1,1]])\n",
+ "B = numpy.matrix([[2],[-4],[4],[0]])\n",
+ "print \"Determinant of A=\"\n",
+ "print numpy.linalg.det(A)\n",
+ "print \"Inverse of A =\"\n",
+ "print numpy.linalg.inv(A)\n",
+ "print \"X=\",numpy.linalg.inv(A)*B"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.32, page no. 78"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The equations can be rewritten as AX=B where X=[x;y;z] and\n",
+ "Determinant of A=\n",
+ "-8.79296635503e-14\n",
+ "Since det(A)=0 , hence, this system of equation will have infinite solutions.. hence, the system is consistent\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "print \"The equations can be rewritten as AX=B where X=[x;y;z] and\"\n",
+ "A = numpy.matrix([[5,3,7],[3,26,2],[7,2,10]])\n",
+ "B = numpy.matrix([[4],[9],[5]])\n",
+ "print \"Determinant of A=\"\n",
+ "print numpy.linalg.det(A)\n",
+ "print \"Since det(A)=0 , hence, this system of equation will have infinite solutions.. hence, the system is consistent\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.34.1, page no. 80"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Rank of A is 3\n",
+ "Equations have only a trivial solution : x=y=z=0\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "A = numpy.matrix([[1,2,3],[3,4,4],[7,10,12]])\n",
+ "p = numpy.linalg.matrix_rank(A)\n",
+ "print \"Rank of A is\",p\n",
+ "if p==3:\n",
+ " print \"Equations have only a trivial solution : x=y=z=0\"\n",
+ "else:\n",
+ " print \"Equations have infinite no . of solutions.\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "## Example 2.34.2, page no. 80"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Rank of A is 2\n",
+ "Equations have infinite no. of solutions.\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "A = numpy.matrix([[4,2,1,3],[6,3,4,7],[2,1,0,1]])\n",
+ "p = numpy.linalg.matrix_rank(A)\n",
+ "print \"Rank of A is\",p\n",
+ "if p ==4:\n",
+ " print \"Equations have only a trivial solution : x=y=z=0\"\n",
+ "else:\n",
+ " print \"Equations have infinite no. of solutions.\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "## Example 2.38, page no. 83"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The given equations can be written as Y=AX where\n",
+ "Determinant of A is -1.0\n",
+ "Since, its non−singular, hence transformation is regular\n",
+ "Inverse of A is\n",
+ "[[ 2. -2. -1.]\n",
+ " [-4. 5. 3.]\n",
+ " [ 1. -1. -1.]]\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "print \"The given equations can be written as Y=AX where\"\n",
+ "A = numpy.matrix([[2,1,1],[1,1,2],[1,0,-2]])\n",
+ "print \"Determinant of A is\",numpy.linalg.det ( A )\n",
+ "print \"Since, its non−singular, hence transformation is regular\"\n",
+ "print\"Inverse of A is\"\n",
+ "print numpy.linalg.inv ( A )"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.39, page no. 84"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "[[-0.66666667 0.33333333 0.66666667]\n",
+ " [ 0.66666667 0.66666667 0.33333333]\n",
+ " [ 0.33333333 -0.66666667 0.66666667]]\n",
+ "A transpose is equal to\n",
+ "[[-0.66666667 0.66666667 0.33333333]\n",
+ " [ 0.33333333 0.66666667 -0.66666667]\n",
+ " [ 0.66666667 0.33333333 0.66666667]]\n",
+ "A∗(transpose of A)=\n",
+ "[[ 1. 0. 0.]\n",
+ " [ 0. 1. 0.]\n",
+ " [ 0. 0. 1.]]\n",
+ "Hence, A is orthogonal\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "A = numpy.matrix([[-2./3,1./3,2./3],[2./3,2./3,1./3],[1./3,-2./3,2./3]])\n",
+ "print A\n",
+ "print \"A transpose is equal to\"\n",
+ "print A.transpose()\n",
+ "print \"A∗(transpose of A)=\"\n",
+ "print A*A.transpose()\n",
+ "print \"Hence, A is orthogonal\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "## Example 2.42, page no. 87"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Let R represents the matrix of transformation and P represents a diagonal matrix whose values are the eigenvalues of A. then\n",
+ "R is normalised. let U represents unnormalised version of r\n",
+ "Two eigen vectors are the two columns of U\n",
+ "[[ 4. 1.]\n",
+ " [ 0. 0.]]\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "import math\n",
+ "\n",
+ "A = numpy.matrix([[5,4],[1,2]])\n",
+ "print \"Let R represents the matrix of transformation and P represents a diagonal matrix whose values are the eigenvalues of A. then\"\n",
+ "P,R= numpy.linalg.eig(A)\n",
+ "U = numpy.zeros([2, 2])\n",
+ "print \"R is normalised. let U represents unnormalised version of r\"\n",
+ "U[0,0]= R[0,0]*math.sqrt(17)\n",
+ "U[0,1]= R[0,1]*math.sqrt(17)\n",
+ "U[0,1]= R[1,1]*math.sqrt(2)\n",
+ "print \"Two eigen vectors are the two columns of U\"\n",
+ "print U"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "## Examle 2.43, page no. 88"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 14,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Let Rrepresents the matrix of transformation and Prepresents a diagonalmatrix whose values are the eigenvalues of A. then\n",
+ "R is normalised. let U represents unnormalised version of r\n",
+ "[-2. 3. 6.]\n",
+ "Three eigen vectors are the three columns of U\n",
+ "[[ -2.82842712 5.19615242 14.69693846]\n",
+ " [ 0. 0. 0. ]\n",
+ " [ 0. 0. 0. ]]\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,math\n",
+ "\n",
+ "A = numpy.matrix([[1,1,3],[1,5,1],[3,1,1]])\n",
+ "U = numpy.zeros([3,3])\n",
+ "print \"Let Rrepresents the matrix of transformation and Prepresents a diagonalmatrix whose values are the eigenvalues of A. then\"\n",
+ "R,P = numpy.linalg.eig(A)\n",
+ "print \"R is normalised. let U represents unnormalised version of r\"\n",
+ "print R\n",
+ "U[0,0] = R[0]*math.sqrt(2) \n",
+ "U[0,1] = R[1]*math.sqrt(3)\n",
+ "U[0,2] = R[2]*math.sqrt(6)\n",
+ "print \"Three eigen vectors are the three columns of U\"\n",
+ "print U"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.44, page no. 89"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 15,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Let Rrepresents the matrix of transformation and Prepresents a diagonalmatrix whose values are the eigenvalues of A. then\n",
+ "R is normalised. let U represents unnormalised version of r\n",
+ "[ 3. 2. 5.]\n",
+ "Three eigen vectors are the three columns of U\n",
+ "[[ 3. 2. 18.70828693]\n",
+ " [ 0. 0. 0. ]\n",
+ " [ 0. 0. 0. ]]\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,math\n",
+ "\n",
+ "A = numpy.matrix([[3,1,4],[0,2,6],[0,0,5]])\n",
+ "U = numpy.zeros([3,3])\n",
+ "print \"Let Rrepresents the matrix of transformation and Prepresents a diagonalmatrix whose values are the eigenvalues of A. then\"\n",
+ "R,P = numpy.linalg.eig(A)\n",
+ "print \"R is normalised. let U represents unnormalised version of r\"\n",
+ "print R\n",
+ "U[0,0] = R[0]*math.sqrt(1) \n",
+ "U[0,1] = R[1]*math.sqrt(1)\n",
+ "U[0,2] = R[2]*math.sqrt(14)\n",
+ "print \"Three eigen vectors are the three columns of U\"\n",
+ "print U"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.45, page no. 90"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 13,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Eigen values of A are\n",
+ "(array([-1., 5.]), matrix([[-0.89442719, -0.70710678],\n",
+ " [ 0.4472136 , -0.70710678]]))\n",
+ "Let\n",
+ "Hence, the characteristic equation is ( x−a ) ( x−b)\n",
+ "[-8 -5]\n",
+ "Aˆ2−4∗A−5∗ I=\n",
+ "[[ 0. 0.]\n",
+ " [ 0. 0.]]\n",
+ "Inverse of A=\n",
+ "[[-0.6 0.8]\n",
+ " [ 0.4 -0.2]]\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "x = numpy.poly([0])\n",
+ "A = numpy.matrix([[1,4],[2,3]])\n",
+ "I = numpy.eye(2)\n",
+ "print \"Eigen values of A are\"\n",
+ "print numpy.linalg.eig(A)\n",
+ "print \"Let\"\n",
+ "a = -1;\n",
+ "b = 5;\n",
+ "print \"Hence, the characteristic equation is ( x−a ) ( x−b)\"\n",
+ "print ( x - a ) *( x - b )\n",
+ "\n",
+ "print \"Aˆ2−4∗A−5∗ I=\"\n",
+ "print A**2-4*A-5* I\n",
+ "print \"Inverse of A=\"\n",
+ "print numpy.linalg.inv ( A )"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.46, page no. 91"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 14,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Egenvalues of A are\n",
+ "(array([ 4.25683813, 0.40327935, -4.66011748]), matrix([[ 0.10296232, -0.91299477, -0.48509974],\n",
+ " [-0.90473047, 0.40531299, 0.37306899],\n",
+ " [ 0.41335402, 0.04649661, 0.79088417]]))\n",
+ "Let\n",
+ "Hence, the characteristic equation is ( x−a ) ( x−b) ( x−c )\n",
+ "[-10.99999905 8.00000095]\n",
+ "Inverse of A=\n",
+ "[[ 3. 1. 1.5 ]\n",
+ " [-1.25 -0.25 -0.75]\n",
+ " [-0.25 -0.25 -0.25]]\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "x = numpy.poly([0])\n",
+ "A = numpy.matrix([[1,1,3],[1,3,-3],[-2,-4,-4]])\n",
+ "print \"Egenvalues of A are\"\n",
+ "print numpy.linalg.eig(A)\n",
+ "print \"Let\"\n",
+ "a =4.2568381\n",
+ "b =0.4032794\n",
+ "c = -4.6601175\n",
+ "print \"Hence, the characteristic equation is ( x−a ) ( x−b) ( x−c )\"\n",
+ "p = (x-a)*(x-b)*(x-c)\n",
+ "print p\n",
+ "print \"Inverse of A=\"\n",
+ "print numpy.linalg.inv(A)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.47, page no. 91"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Eigenvalues of A are\n",
+ "(array([ 3., 1., 1.]), matrix([[ 0.70710678, -0.70710678, -0.40824829],\n",
+ " [ 0. , 0. , 0.81649658],\n",
+ " [ 0.70710678, 0.70710678, -0.40824829]]))\n",
+ "Let\n",
+ "Hence, the characteristic equation is (x−a)(x−b)(x−c)=\n",
+ "[ 0 -3]\n",
+ "Aˆ8−5∗Aˆ7+7∗Aˆ6−3∗Aˆ5+Aˆ4−5∗Aˆ3+8∗Aˆ2−2∗A+I =\n",
+ "[[ 8. 5. 5.]\n",
+ " [ 0. 3. 0.]\n",
+ " [ 5. 5. 8.]]\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "x = numpy.poly([0])\n",
+ "A = numpy.matrix([[2,1,1],[0,1,0],[1,1,2]])\n",
+ "I = numpy.eye(3)\n",
+ "print \"Eigenvalues of A are\"\n",
+ "print numpy.linalg.eig(A)\n",
+ "print \"Let\"\n",
+ "a =1\n",
+ "b =1\n",
+ "c =3\n",
+ "print \"Hence, the characteristic equation is (x−a)(x−b)(x−c)=\"\n",
+ "p = (x-a)*(x-b)*(x-c)\n",
+ "print p\n",
+ "print \"Aˆ8−5∗Aˆ7+7∗Aˆ6−3∗Aˆ5+Aˆ4−5∗Aˆ3+8∗Aˆ2−2∗A+I =\"\n",
+ "print A**8-5*A**7+7*A**6-3*A**5+A**4-5*A**3+8*A**2-2*A+I"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "## Example 2.48, page no. 93"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "R is matrix of transformation and D is a diagonal matrix\n",
+ "[-1.65544238 -0.21075588 2.86619826]\n",
+ "[[-0.87936655 -0.34661859 -0.32645063]\n",
+ " [ 0.11410244 0.51222983 -0.85123513]\n",
+ " [-0.46227167 0.78579651 0.41088775]]\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "A = numpy.matrix([[-1,2,-2],[1,2,1],[-1,-1,0]])\n",
+ "print \"R is matrix of transformation and D is a diagonal matrix\"\n",
+ "[R,D]= numpy.linalg.eigh(A)\n",
+ "print R\n",
+ "print D"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.49, page no. 93"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 17,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "R is matrix of transformation and D is a diagonal matrix\n",
+ "R is normalised, let P denotes unnormalised version of R . Then \n",
+ "[ 3. 2. 5.]\n",
+ "[[ 4.24264069 3.46410162 12.24744871]\n",
+ " [ 0. 0. 0. ]\n",
+ " [ 0. 0. 0. ]]\n",
+ "A^4= [[ 81 65 1502]\n",
+ " [ 0 16 1218]\n",
+ " [ 0 0 625]]\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,math\n",
+ "\n",
+ "A = numpy.matrix([[3,1,4],[0,2,6],[0,0,5]])\n",
+ "P = numpy.zeros([3,3])\n",
+ "print \"R is matrix of transformation and D is a diagonal matrix\"\n",
+ "R,D = numpy.linalg.eig(A)\n",
+ "print \"R is normalised, let P denotes unnormalised version of R . Then \"\n",
+ "print R\n",
+ "P[0,0] = R[0]*math.sqrt(2) \n",
+ "P[0,1] = R[1]*math.sqrt(3)\n",
+ "P[0,2] = R[2]*math.sqrt(6)\n",
+ "print P\n",
+ "print \"A^4= \",A**4"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.50, page no. 94"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 9,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "3∗xˆ2+5∗yˆ2+3∗zˆ2−2∗y∗z+2∗z∗x−2∗x∗y\n",
+ "The matrix of the given quadratic form is\n",
+ "Let R represents the matrix of transformation and Prepresents a diagonal matrix whose values are the eigenvalues of A. then\n",
+ "So, canonical form is 2∗xˆ2+3∗yˆ2+6∗zˆ2\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "print \"3∗xˆ2+5∗yˆ2+3∗zˆ2−2∗y∗z+2∗z∗x−2∗x∗y\"\n",
+ "print \"The matrix of the given quadratic form is\"\n",
+ "A = numpy.matrix([[3,-1,1],[-1,5,-1],[1,-1,3]])\n",
+ "print \"Let R represents the matrix of transformation and Prepresents a diagonal matrix whose values are the eigenvalues of A. then\"\n",
+ "[R,P] = numpy.linalg.eig(A)\n",
+ "print \"So, canonical form is 2∗xˆ2+3∗yˆ2+6∗zˆ2\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.51, page no. 95"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 14,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "2∗x1∗x2+2∗x1∗x3−2∗x2∗x3\n",
+ "The matrix of the given quadratic form is\n",
+ "Let R represents the matrix of transformation and P represents a diagonal matrix whose values are the eigenvalues of A. then\n",
+ "so, canonical form is −2∗xˆ2+yˆ2+ zˆ2\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "print \"2∗x1∗x2+2∗x1∗x3−2∗x2∗x3\"\n",
+ "print \"The matrix of the given quadratic form is\"\n",
+ "A = numpy.matrix([[0,1,1],[1,0,-1],[1,-1,0]])\n",
+ "print \"Let R represents the matrix of transformation and P represents a diagonal matrix whose values are the eigenvalues of A. then\"\n",
+ "[R,P] = numpy.linalg.eig(A)\n",
+ "print \"so, canonical form is −2∗xˆ2+yˆ2+ zˆ2\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.52, page no. 96"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 32,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "A∗= [[ 2.-1.j -5.-0.j]\n",
+ " [ 3.-0.j 0.-1.j]\n",
+ " [-1.-3.j 4.+2.j]]\n",
+ "AA∗= [[ 24.+0.j -20.+2.j]\n",
+ " [-20.-2.j 46.+0.j]]\n",
+ "Clearly, AA∗ is hermitian matrix\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "'''\n",
+ "A=[2+%i 3 -1+3*%i;-5 %i 4-2*%i]\n",
+ "'''\n",
+ "\n",
+ "A = numpy.matrix([[2+1j,3,-1+3*1j],[-5,1j,4-2*1j]])\n",
+ "#A = A.getH()\n",
+ "print \"A∗=\", A.getH()\n",
+ "print \"AA∗=\", A*(A.getH())\n",
+ "print \"Clearly, AA∗ is hermitian matrix\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.53, page no. 97"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 35,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " A∗= [[ 0.-0.j 0.-0.j]\n",
+ " [-0.-0.j 0.-0.j]]\n",
+ "AA∗= [[ 0.+0.j 0.+0.j]\n",
+ " [ 0.+0.j 0.+0.j]]\n",
+ "A∗A= [[ 0.+0.j 0.+0.j]\n",
+ " [ 0.+0.j 0.+0.j]]\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy \n",
+ "\n",
+ "A = numpy.matrix([[(1/2)*(1+1j),(1/2)*(-1+1j)],[(1/2)*(1+1j),(1/2)*(1-1j)]])\n",
+ "print \"A∗=\", A.getH()\n",
+ "print \"AA∗=\", A*(A.getH())\n",
+ "print \"A∗A=\", (A.getH())*A"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.54, page no. 97"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "I−A=\n",
+ "inverse of (I+A)=\n",
+ "[[ 0.16666667+0.j -0.16666667-0.33333333j]\n",
+ " [ 0.16666667-0.33333333j 0.16666667+0.j ]]\n",
+ "((I−A)(inverse(I+A)))∗((I−A)(inverse(I+A)))=\n",
+ "[[ 1.11111111e-01-0.44444444j -2.77555756e-17+0.88888889j]\n",
+ " [ -2.77555756e-17+0.88888889j 1.11111111e-01+0.44444444j]]\n",
+ "((I−A)(inverse(I+A)))((I−A)(inverse(I+A)))∗=\n",
+ "[[ 1.11111111e-01+0.44444444j 1.11022302e-16+0.88888889j]\n",
+ " [ 1.11022302e-16+0.88888889j 1.11111111e-01-0.44444444j]]\n",
+ "Clearly, the product is an identity matrix.hence, it is a unitary matrix\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "A = numpy.matrix([[0,1+2*1j],[-1+2*1j,0]])\n",
+ "I = numpy.eye(2)\n",
+ "print \"I−A=\"\n",
+ "I-A\n",
+ "print \"inverse of (I+A)=\"\n",
+ "print numpy.linalg.inv(I+A)\n",
+ "print \"((I−A)(inverse(I+A)))∗((I−A)(inverse(I+A)))=\"\n",
+ "print (((I-A)*(numpy.linalg.inv(I+A))).T)*((I-A)*(numpy.linalg.inv(I+A)))\n",
+ "print \"((I−A)(inverse(I+A)))((I−A)(inverse(I+A)))∗=\"\n",
+ "print ((I-A)*(numpy.linalg.inv(I+A)))*(((I-A)*(numpy.linalg.inv(I+A))).T)\n",
+ "print \"Clearly, the product is an identity matrix.hence, it is a unitary matrix\""
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.11+"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter34_2.ipynb b/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter34_2.ipynb
new file mode 100644
index 00000000..68dc76ff
--- /dev/null
+++ b/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter34_2.ipynb
@@ -0,0 +1,1414 @@
+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 34 : Probability And Distributions"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 34.1, page no. 830"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "From the principle of counting, the required no. of ways are 12∗11∗10∗9= \n",
+ "11880\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"From the principle of counting, the required no. of ways are 12∗11∗10∗9= \"\n",
+ "print 12*11*10*9"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 34.2.1, page no. 831"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "no. of permutations=9!/(2!∗2!∗2!)\n",
+ "45360\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math\n",
+ "\n",
+ "print \"no. of permutations=9!/(2!∗2!∗2!)\"\n",
+ "print math.factorial(9)/(math.factorial(2)*math.factorial(2)*math.factorial(2))"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 34.2.2, page no. 831"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "no. of permutations=9!/(2!∗2!∗3!*3!)\n",
+ "2520\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math\n",
+ "\n",
+ "print \"no. of permutations=9!/(2!∗2!∗3!*3!)\"\n",
+ "print math.factorial(9)/(math.factorial(2)*math.factorial(2)*math.factorial(3)*math.factorial(3))"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 34.3.1, page no. 832"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "no. of committees=C(6,3)∗C(5,2)=’\n",
+ "200\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math\n",
+ "\n",
+ "def C(a,b):\n",
+ " x = math.factorial(a)/(math.factorial(b)*math.factorial(a-b))\n",
+ " return x\n",
+ "print \"no. of committees=C(6,3)∗C(5,2)=’\"\n",
+ "print C(6,3)*C(5,2)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 34.3.2, page no. 832"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "no. of committees=C(4,1)∗C(5,2)=’\n",
+ "40\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math\n",
+ "\n",
+ "def C(a,b):\n",
+ " x = math.factorial(a)/(math.factorial(b)*math.factorial(a-b))\n",
+ " return x\n",
+ "print \"no. of committees=C(4,1)∗C(5,2)=’\"\n",
+ "print C(4,1)*C(5,2)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 34.3.3, page no. 833"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "no. of committees=C(6,3)∗C(4,2)=’\n",
+ "120\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math\n",
+ "\n",
+ "def C(a,b):\n",
+ " x = math.factorial(a)/(math.factorial(b)*math.factorial(a-b))\n",
+ " return x\n",
+ "print \"no. of committees=C(6,3)∗C(4,2)=’\"\n",
+ "print C(6,3)*C(4,2)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 34.4.1, page no. 834"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 9,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The probability of getting a four is 1/6= 0.166666666667\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math\n",
+ "\n",
+ "print \"The probability of getting a four is 1/6= \",1./6"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 34.4.2, page no. 834"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 10,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The probability of getting an even no. 1/2= 0.5\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math\n",
+ "\n",
+ "print \"The probability of getting an even no. 1/2= \",1./2"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 34.5, page no. 835"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 11,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The probability of 53 Sundays is 2/7= 0.285714285714\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math\n",
+ "\n",
+ "print \"The probability of 53 Sundays is 2/7= \",2./7"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 34.6, page no. 835"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 13,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The five digits can be arranged in 5! ways = 120\n",
+ "Of which 4! will begin with 0 = 24\n",
+ "So, total no. of five digit numbers = 5!−4! = 96\n",
+ "The numbers ending in 04, 12, 20, 24, 32, 40 will be divisible by 4 \n",
+ "numbers ending in 04 = 3! = 6\n",
+ "numbers ending in 12 = 3!−2! = 4\n",
+ "numbers ending in 20 = 3! = 6\n",
+ "numbers ending in 24 = 3!−2! = 4\n",
+ "numbers ending in 32 = 3!−2! = 4\n",
+ "numbers ending in 40 = 3! = 6\n",
+ "So, total no. of favourable ways = 6+4+6+4+4+6 = 30\n",
+ "probability = 30/96 = 0.3125\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math\n",
+ "\n",
+ "print \"The five digits can be arranged in 5! ways = \",math.factorial(5)\n",
+ "print \"Of which 4! will begin with 0 = \",math.factorial(4)\n",
+ "print \"So, total no. of five digit numbers = 5!−4! = \",math.factorial(5)-math.factorial(4)\n",
+ "print \"The numbers ending in 04, 12, 20, 24, 32, 40 will be divisible by 4 \"\n",
+ "print \"numbers ending in 04 = 3! = \",math.factorial(3)\n",
+ "print \"numbers ending in 12 = 3!−2! = \",math.factorial(3)-math.factorial(2)\n",
+ "print \"numbers ending in 20 = 3! = \",math.factorial(3)\n",
+ "print \"numbers ending in 24 = 3!−2! = \",math.factorial(3)-math.factorial(2)\n",
+ "print \"numbers ending in 32 = 3!−2! = \",math.factorial(3)-math.factorial(2)\n",
+ "print \"numbers ending in 40 = 3! = \",math.factorial(3)\n",
+ "print \"So, total no. of favourable ways = 6+4+6+4+4+6 = \",6+4+6+4+4+6\n",
+ "print \"probability = 30/96 = \",30./96"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 34.7, page no. 836"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 15,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Total no. of possible cases = C(40,4) = 91390\n",
+ "Favourable outcomes = C(24,2)∗C(15,1) = 4140\n",
+ "Probability = 0.0453003610898\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math\n",
+ "\n",
+ "def C(a,b):\n",
+ " x = math.factorial(a)/(math.factorial(b)*math.factorial(a-b))\n",
+ " return x\n",
+ "print \"Total no. of possible cases = C(40,4) = \",C(40,4)\n",
+ "print \"Favourable outcomes = C(24,2)∗C(15,1) = \",C(24,2)*C(15,1)\n",
+ "print \"Probability = \",float(C(24,2)*C(15,1))/C(40,4)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 34.8, page no. 836"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 16,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Total no. of possible cases = C(15,8) = 6435\n",
+ "Favourable outcomes = C(5,2)∗C(10,6) = 2100\n",
+ "Probability = 0.32634032634\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math\n",
+ "\n",
+ "def C(a,b):\n",
+ " x = math.factorial(a)/(math.factorial(b)*math.factorial(a-b))\n",
+ " return x\n",
+ "print \"Total no. of possible cases = C(15,8) = \",C(15,8)\n",
+ "print \"Favourable outcomes = C(5,2)∗C(10,6) = \",C(5,2)*C(10,6)\n",
+ "print \"Probability = \",float(C(5,2)*C(10,6))/C(15,8)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 34.9.1, page no. 837"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 17,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Total no. of possible cases = C(9,3) = 84\n",
+ "Favourable outcomes = C(2,1)∗C(3,1)*C(4,1) = 24\n",
+ "Probability = 0.285714285714\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math\n",
+ "\n",
+ "def C(a,b):\n",
+ " x = math.factorial(a)/(math.factorial(b)*math.factorial(a-b))\n",
+ " return x\n",
+ "print \"Total no. of possible cases = C(9,3) = \",C(9,3)\n",
+ "print \"Favourable outcomes = C(2,1)∗C(3,1)*C(4,1) = \",C(2,1)*C(3,1)*C(4,1)\n",
+ "print \"Probability = \",float(C(2,1)*C(3,1)*C(4,1))/C(9,3)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 34.9.2, page no. 837"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 20,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Total no. of possible cases = C(9,3) = 84\n",
+ "Favourable outcomes = C(2,2)∗C(7,1)+C(3,2)∗C(6,1)+C(4,2)∗C(5,1) = 55\n",
+ "Probability = 0.654761904762\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math\n",
+ "\n",
+ "\n",
+ "def C(a,b):\n",
+ " x = math.factorial(a)/(math.factorial(b)*math.factorial(a-b))\n",
+ " return x\n",
+ "print \"Total no. of possible cases = C(9,3) = \",C(9,3)\n",
+ "print \"Favourable outcomes = C(2,2)∗C(7,1)+C(3,2)∗C(6,1)+C(4,2)∗C(5,1) = \",C(2,2)*C(7,1)+C(3,2)*C(6,1)+C(4,2)*C(5,1)\n",
+ "print \"Probability = \",float(C(2,2)*C(7,1)+C(3,2)*C(6,1)+C(4,2)*C(5,1))/C(9,3)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 34.9.3, page no. 838"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 22,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Total no. of possible cases = C(9,3) = 84\n",
+ "Favourable outcomes = C(3,3)+C(4,3) = 5\n",
+ "Probability = 0.0595238095238\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math\n",
+ "\n",
+ "def C(a,b):\n",
+ " x = math.factorial(a)/(math.factorial(b)*math.factorial(a-b))\n",
+ " return x\n",
+ "print \"Total no. of possible cases = C(9,3) = \",C(9,3)\n",
+ "print \"Favourable outcomes = C(3,3)+C(4,3) = \",C(3,3)+C(4,3)\n",
+ "print \"Probability = \",5./84"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 34.13, page no. 840"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 25,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Probability of drawing an ace or spade or both from pack of 52 cards = 4/52+13/52−1/52= 17\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Probability of drawing an ace or spade or both from pack of 52 cards = 4/52+13/52−1/52= \",4+13-1/52"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 34.14.1, page no. 841"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 30,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Probability of first card being a king = 4/52 = 0.0769230769231\n",
+ "Probability of second card being a queen = 4/52 = 0.0769230769231\n",
+ "Probability of drawing both cards in succession = 4/52∗4/52= 0.00591715976331\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Probability of first card being a king = 4/52 = \",4./52\n",
+ "print \"Probability of second card being a queen = 4/52 = \",4./52\n",
+ "print \"Probability of drawing both cards in succession = 4/52∗4/52= \",(4./52)*(4./52)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 34.15.1, page no. 842"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 32,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Probability of getting 7 in first toss and not getting it in second toss = 1/6∗5/6 = 0.138888888889\n",
+ "Probability of not getting 7 in first toss and getting it in second toss = 5/6∗1/6 = 0.138888888889\n",
+ "Required probability = 1/6∗5/6+5/6∗1/6 = 0.277777777778\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Probability of getting 7 in first toss and not getting it in second toss = 1/6∗5/6 = \",(1./6)*(5./6)\n",
+ "print \"Probability of not getting 7 in first toss and getting it in second toss = 5/6∗1/6 = \",(5./6)*(1./6)\n",
+ "print \"Required probability = 1/6∗5/6+5/6∗1/6 = \",((1./6)*(5./6))+((5./6)*(1./6))"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 34.15.2, page no. 842"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 33,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Probability of not getting 7 in either toss = 5/6∗5/6 = 0.694444444444\n",
+ "Probability of getting 7 at least once = 1−5/6∗5/6 = 0.305555555556\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Probability of not getting 7 in either toss = 5/6∗5/6 = \",(5./6)*(5./6)\n",
+ "print \"Probability of getting 7 at least once = 1−5/6∗5/6 = \",1-(5./6)*(5./6)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 34.15.3, page no. 842"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 34,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Probability of getting 7 twice = 1/6∗1/6 = 0.0277777777778\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Probability of getting 7 twice = 1/6∗1/6 = \",(1./6)*(1./6)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 34.16, page no. 843"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 35,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Probability of engineering subject being choosen = (1/3∗3/8)+(2/3∗5/8) = 0.541666666667\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Probability of engineering subject being choosen = (1/3∗3/8)+(2/3∗5/8) = \",((1./3)*(3./8)) +((2./3)*(5./8))"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 34.17, page no. 844"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 36,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Probability of white ball being choosen = 2/6∗6/13+4/6∗5/13 = 0.410256410256\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Probability of white ball being choosen = 2/6∗6/13+4/6∗5/13 = \",((2./6)*(6./13))+((4./6)*(5./13))"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 34.18, page no. 844"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 38,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Chances of winning of A=1/2+(1/2)ˆ2∗(1/2)+(1/2)ˆ4∗(1/2)+(1/2)ˆ6∗(1/2)+..= 0.666666666667\n",
+ "Chances of winning of B=1−chances of winning of A = 1\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Chances of winning of A=1/2+(1/2)ˆ2∗(1/2)+(1/2)ˆ4∗(1/2)+(1/2)ˆ6∗(1/2)+..= \",(1./2)/(1-(1./2)**2)\n",
+ "print \"Chances of winning of B=1−chances of winning of A = \",1-(2/3)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 34.19.1, page no. 845"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 40,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Total no. of possible outcomes = C(10,2) = 45\n",
+ "Favourable outcomes = 5*5 = 25\n",
+ "P = 0.555555555556\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math\n",
+ "\n",
+ "def C(a,b):\n",
+ " x = math.factorial(a)/(math.factorial(b)*math.factorial(a-b))\n",
+ " return x\n",
+ "print \"Total no. of possible outcomes = C(10,2) = \",C(10,2)\n",
+ "print \"Favourable outcomes = 5*5 = \",5*5\n",
+ "print \"P = \",25./45"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 34.19.2, page no. 845"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 41,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Total no. of possible outcomes = 10*10 = 100\n",
+ "Favourable outcomes = 5*5+5*5 = 50\n",
+ "P = 0.5\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Total no. of possible outcomes = 10*10 = \",10*10\n",
+ "print \"Favourable outcomes = 5*5+5*5 = \",5*5+5*5\n",
+ "print \"P = \",50./100"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 34.20, page no. 846"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 42,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Probability of A/B = AandB/B = 0.25\n",
+ "Probability of B/A = AandB/A = 0.333333333333\n",
+ "Probability of AandBnot = A−AandB = 0.166666666667\n",
+ "Probability of A/Bnot = AandBnot/Bnot = 0.25\n"
+ ]
+ }
+ ],
+ "source": [
+ "A = 1./4\n",
+ "B = 1./3\n",
+ "AorB = 1./2\n",
+ "AandB = A+B-AorB\n",
+ "print \"Probability of A/B = AandB/B = \",AandB/B\n",
+ "print \"Probability of B/A = AandB/A = \",AandB/A\n",
+ "print \"Probability of AandBnot = A−AandB = \",A-AandB\n",
+ "print \"Probability of A/Bnot = AandBnot/Bnot = \",(1./6)/(1-1./3)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 34.22, page no. 846"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 43,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Probability of A hitting target = 3/5\n",
+ "Probability of B hitting target = 2/5\n",
+ "Probability of C hitting target = 3/4\n",
+ "Probability that two shots hit = 3/5∗2/5∗(1−3/4)+2/5∗3/4∗(1−3/5)+3/4∗3/5∗(1−2/5) = \n",
+ "0.45\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Probability of A hitting target = 3/5\"\n",
+ "print \"Probability of B hitting target = 2/5\"\n",
+ "print \"Probability of C hitting target = 3/4\"\n",
+ "print \"Probability that two shots hit = 3/5∗2/5∗(1−3/4)+2/5∗3/4∗(1−3/5)+3/4∗3/5∗(1−2/5) = \"\n",
+ "print (3./5)*(2./5)*(1-3./4)+(2./5)*(3./4)*(1-3./5)+(3./4)*(3./5)*(1-2./5)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 34.23, page no. 847"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 44,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Probability of problem not getting solved = 1/2∗2/3∗3/4 = 0.25\n",
+ "Probability of problem getting solved = 1−(1/2∗2/3∗3/4) = 0.75\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Probability of problem not getting solved = 1/2∗2/3∗3/4 = \",(1./2)*(2./3)*(3./4)\n",
+ "print \"Probability of problem getting solved = 1−(1/2∗2/3∗3/4) = \",1-((1./2)*(2./3)*(3./4))"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 34.25, page no. 848"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 46,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Total frequency=integrate(f,x,0,2) =\n",
+ "u1 about origin = 1\n",
+ "u2 about origin = 16/15\n",
+ "Standard deviation = (u2−u1ˆ2)ˆ0.5= 0.258198889747161\n",
+ "Mean deviation about the mean = (1/n)∗(integrate(|x−1|∗(xˆ3),x,0,1)+integrate(|x−1|∗((2−x)ˆ3),x,1,2))\n",
+ "1/5\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "\n",
+ "print \"Total frequency=integrate(f,x,0,2) =\" \n",
+ "n = sympy.integrate('x**3',('x',0,1))+sympy.integrate('(2-x)**3',('x',1,2))\n",
+ "print \"u1 about origin = \",\n",
+ "u1 = (1/n)*(sympy.integrate('(x)*(x**3)',('x',0,1))+sympy.integrate('(x)*((2-x)**3)',('x',1,2)))\n",
+ "print u1\n",
+ "print \"u2 about origin = \",\n",
+ "u2 = (1/n)*(sympy.integrate('(x**2)*(x**3)',('x',0,1))+sympy.integrate('(x**2)*((2-x)**3)',('x',1,2)))\n",
+ "print u2\n",
+ "print \"Standard deviation = (u2−u1ˆ2)ˆ0.5= \",(u2-u1**2)**0.5\n",
+ "print \"Mean deviation about the mean = (1/n)∗(integrate(|x−1|∗(xˆ3),x,0,1)+integrate(|x−1|∗((2−x)ˆ3),x,1,2))\"\n",
+ "print (1/n)*(sympy.integrate('(1-x)*(x**3)',('x',0,1))+sympy.integrate('(x-1)*((2-x)**3)',('x',1,2)))"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 34.26, page no. 849"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 47,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Probability = (0.45∗0.03)/(0.45∗0.03+0.25∗0.05+0.3∗0.04 = 0.355263157895\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Probability = (0.45∗0.03)/(0.45∗0.03+0.25∗0.05+0.3∗0.04 = \",(0.45*0.03)/(0.45*0.03+0.25*0.05+0.3*0.04)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 34.27, page no. 849"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 49,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Probability = (1/3∗2/6∗3/5)/(1/3∗2/6∗3/5+1/3∗1/6∗2/5+1/3∗3/6∗1/5) = 1.05555555556\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Probability = (1/3∗2/6∗3/5)/(1/3∗2/6∗3/5+1/3∗1/6∗2/5+1/3∗3/6∗1/5) = \",\n",
+ "print ((1./3)*(2./6)*(3./5))/((1./3)*(2./6)*(3./5))+((1./3)*(1./6)*(2./5))+((1./3)*(3./6)*(1./5))"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 34.28, page no. 850"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 50,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Probability of no success = 8/27 \n",
+ "Probability of a success = 1/3 \n",
+ "Probability of one success = 4/9\n",
+ "Probability of two success = 2/9\n",
+ "Probability of three success = 2/9\n",
+ "mean=sum of i∗pi = 1.0\n",
+ "sum of i∗piˆ2 = 1.66666666667\n",
+ "variance = (sum of i∗piˆ2)−1= 0.666666666667\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "print \"Probability of no success = 8/27 \"\n",
+ "print \"Probability of a success = 1/3 \"\n",
+ "print \"Probability of one success = 4/9\"\n",
+ "print \"Probability of two success = 2/9\"\n",
+ "print \"Probability of three success = 2/9\"\n",
+ "A = numpy.array([[0,1,2,3],[8./27,4./9,2./9,1./27]])\n",
+ "print \"mean=sum of i∗pi = \",\n",
+ "print A[0,0]*A[1,0]+A[0,1]*A[1,1]+A[0,3]*A[1,3]+A[0,2]*A[1,2]\n",
+ "print \"sum of i∗piˆ2 = \",\n",
+ "print A[0,0]**2*A[1,0]+A[0,1]**2*A[1,1]+A[0,3]**2*A[1,3]+A[0,2]**2*A[1,2]\n",
+ "print \"variance = (sum of i∗piˆ2)−1= \",\n",
+ "print A[0,0]**2*A[1,0]+A[0,1]**2*A[1,1]+A[0,3]**2*A[1,3]+A[0,2]**2*A[1,2]-1"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 34.29, page no. 851"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 53,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Sumof all pi = 1 \n",
+ "Hence ,\n",
+ "p(x<4) = 16.0*k\n",
+ "p(x>=5) = 24.0*k\n",
+ "p(3<x<=6) = 33.0*k\n",
+ "p(x<=2) = 9.0*k\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy,numpy\n",
+ "\n",
+ "k = sympy.Symbol('k')\n",
+ "A =numpy.array([[0,1,2,3,4,5,6],[k,3*k,5*k,7*k,9*k,11*k,13*k]])\n",
+ "print \"Sumof all pi = 1 \"\n",
+ "print \"Hence ,\"\n",
+ "k = 1./49\n",
+ "print \"p(x<4) = \",\n",
+ "a = A[1,0]+A[1,1]+A[1,3]+A[1,2]\n",
+ "print a.evalf()\n",
+ "print \"p(x>=5) = \",\n",
+ "b = A[1,5]+A[1,6]\n",
+ "print b.evalf()\n",
+ "print \"p(3<x<=6) = \",\n",
+ "c = A[1,4]+A[1,5]+A[1,6]\n",
+ "print c.evalf()\n",
+ "print \"p(x<=2) = \",\n",
+ "e = A[1,0]+A[1,1]+A[1,2]\n",
+ "print e.evalf()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 34.30, page no. 853"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 57,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Sum of all pi = 1\n",
+ "Hence,\n",
+ "p(x<6) = 2.0*k**2 + 10.0*k\n",
+ "p(x>=6) = 9.0*k**2 + k\n",
+ "p(3<x<5) = 8.0*k\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy,numpy,math\n",
+ "\n",
+ "k = sympy.Symbol('k')\n",
+ "A =numpy.array([[0,1,2,3,4,5,6,7],[0,k,2*k,2*k,3*k,k**2,2*k**2,7*k**2+ k]])\n",
+ "print \"Sum of all pi = 1\"\n",
+ "print \"Hence,\"\n",
+ "k = 1./10\n",
+ "print \"p(x<6) = \",\n",
+ "a = A[1,0]+A[1,1]+A[1,3]+A[1,2]+ A[1,3]+A[1,4]+A[1,6]\n",
+ "print a.evalf()\n",
+ "print \"p(x>=6) = \",\n",
+ "b = A[1,6]+A[1,7]\n",
+ "print b.evalf()\n",
+ "print \"p(3<x<5) = \",\n",
+ "c = A[1,1]+A[1,2]+A[1,3]+A[1,4]\n",
+ "print c.evalf()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 34.31, page no. 853"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 62,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Clearly, f>0 for every x in (1,2) and integrate(f,x,0,numpy.inf)= 1.00000000000000\n",
+ "Required probability=p(1<=x<=2)= integrate(f,x,1,2) = 0.232544157934830\n",
+ "Cumulative probability function f(2)=integrate(f,x,−%inf,2) = 0.864664716763387\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy,numpy,math\n",
+ "\n",
+ "y = sympy.Symbol('y')\n",
+ "f = math.e**(-y)\n",
+ "print \"Clearly, f>0 for every x in (1,2) and integrate(f,x,0,numpy.inf)= \",\n",
+ "print sympy.integrate(math.e**(-y),('y',0,sympy.oo))\n",
+ "print \"Required probability=p(1<=x<=2)= integrate(f,x,1,2) = \",\n",
+ "print sympy.integrate(math.e**(-y),('y',1,2))\n",
+ "print \"Cumulative probability function f(2)=integrate(f,x,−%inf,2) = \",\n",
+ "print sympy.integrate(math.e**(-y),('y',0,2))"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 34.33, page no. 854"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 65,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Total probability = integrate(f,x,0,6)=\n",
+ "2*k\n",
+ "4*k\n",
+ "2*k\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "\n",
+ "k = sympy.Symbol('k')\n",
+ "print \"Total probability = integrate(f,x,0,6)=\"\n",
+ "p = sympy.integrate('k*x',('x',0,2))\n",
+ "q = sympy.integrate('2*k',('x',2,4))\n",
+ "r = sympy.integrate('-k*x+6*k',('x',4,6))\n",
+ "print p\n",
+ "print q\n",
+ "print r"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "## Example 34.34, page no. 854"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "First row of A displays the value of x\n",
+ "The second row of x displays the probability of corresponding to x\n",
+ "E(x) = 5.5\n",
+ "E(x)ˆ2 = 46.5\n",
+ "E(2∗x+1)^2=E(4∗xˆ2+4∗x+1) = 209.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "A = numpy.array([[-3.,6.,9.],[1./6,1./2,1./3]])\n",
+ "print \"First row of A displays the value of x\"\n",
+ "print \"The second row of x displays the probability of corresponding to x\"\n",
+ "print \"E(x) = \",\n",
+ "c = A[0,0]*A[1,0]+A[0,1]*A[1,1]+A[0,2]*A[1,2]\n",
+ "print c\n",
+ "print \"E(x)ˆ2 = \",\n",
+ "b = A[0,0]**2*A[1,0]+A[0,1]**2*A[1,1]+A[0,2]**2*A[1,2]\n",
+ "print b\n",
+ "print \"E(2∗x+1)^2=E(4∗xˆ2+4∗x+1) = \",4*b+4*c+1"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 34.35, page no. 855"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Total frequency=integrate(f,x,0,2)= \n",
+ "u1 about origin = \n",
+ "u2 about origin = \n",
+ "standard deviation = (u2−u1ˆ2)ˆ0.5 = 0.258198889747161\n",
+ "Mean deviation about the mean = (1/n)∗(integrate|x−1|∗(xˆ3),x,0,1)+integrate(|x−1|∗((2−x)ˆ3),x,1,2))\n",
+ "1/5\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "\n",
+ "print \"Total frequency=integrate(f,x,0,2)= \"\n",
+ "n = sympy.integrate('x**3',('x',0,1))+sympy.integrate('(2-x)**3',('x',1,2))\n",
+ "print \"u1 about origin = \"\n",
+ "u1 = (1/n)*(sympy.integrate('(x)*(x**3)',('x',0,1))+sympy.integrate('(x)*((2-x)**3)',('x',1,2)))\n",
+ "print \"u2 about origin = \"\n",
+ "u2 = (1/n)*(sympy.integrate('(x**2)*(x**3)',('x',0,1))+sympy.integrate('(x**2)*((2-x)**3)',('x',1,2)))\n",
+ "print \"standard deviation = (u2−u1ˆ2)ˆ0.5 = \",(u2-u1**2)**0.5\n",
+ "print \"Mean deviation about the mean = (1/n)∗(integrate|x−1|∗(xˆ3),x,0,1)+integrate(|x−1|∗((2−x)ˆ3),x,1,2))\"\n",
+ "print (1/n)*(sympy.integrate('(1-x)*(x**3)',('x',0,1))+sympy.integrate('(x-1)*((2-x)**3)',('x',1,2)))"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 34.38, page no. 857"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Probability that exactly two will be defective = C(12,2)∗(0.1)ˆ2∗(0.9)ˆ10 = 0.230127770466\n",
+ "Probability that at least two will be defective = 1−(C(12,0)∗(0.9)ˆ12+C(12,1)∗(0.1)∗(0.9)ˆ11) = 0.340997748211\n",
+ "The probability that none will be defective = C(12,12)∗(0.9)ˆ12 = 0.282429536481\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math\n",
+ "\n",
+ "def C(a,b):\n",
+ " x = math.factorial(a)/(math.factorial(b)*math.factorial(a-b))\n",
+ " return x\n",
+ "print \"Probability that exactly two will be defective = C(12,2)∗(0.1)ˆ2∗(0.9)ˆ10 = \",C(12,2)*(0.1)**2*(0.9)**10\n",
+ "print \"Probability that at least two will be defective = 1−(C(12,0)∗(0.9)ˆ12+C(12,1)∗(0.1)∗(0.9)ˆ11) = \",\n",
+ "print 1-(C(12,0)*(0.9)**12+C(12,1)*(0.1)*(0.9)**11)\n",
+ "print \"The probability that none will be defective = C(12,12)∗(0.9)ˆ12 = \",C(12,12)*(0.9)**12"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 34.39, page no. 858"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 11,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Probability of 8 heads and 4 tail sin 12 trials = p(8) = C(12,8)∗(1/2)ˆ8∗(1/2)ˆ4= 0.120849609375\n",
+ "The expected no. of such cases in 256 sets = 256∗p(8) = 30.9375\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math\n",
+ "\n",
+ "def C(a,b):\n",
+ " x = math.factorial(a)/(math.factorial(b)*math.factorial(a-b))\n",
+ " return x\n",
+ "print \"Probability of 8 heads and 4 tail sin 12 trials = p(8) = C(12,8)∗(1/2)ˆ8∗(1/2)ˆ4= \",C(12.,8.)*(1./2)**8*(1./2)**4\n",
+ "print \"The expected no. of such cases in 256 sets = 256∗p(8) = \",256*(495./4096)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 34.40, page no. 859"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 14,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Probability of a defective part = 2/20 =0.1 \n",
+ "Probability of a non defective part = 0.9 \n",
+ "Probabaility of atleast three defectives in a sample = 0.323073194811\n",
+ "No. of sample shaving three defective parts = 1000∗0.323 = 323.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math\n",
+ "\n",
+ "def C(a,b):\n",
+ " x = math.factorial(a)/(math.factorial(b)*math.factorial(a-b))\n",
+ " return x\n",
+ "print \"Probability of a defective part = 2/20 =0.1 \"\n",
+ "print \"Probability of a non defective part = 0.9 \"\n",
+ "print \"Probabaility of atleast three defectives in a sample = \",\n",
+ "print 1-(C(20.,0.)*(0.9)**20+C(20.,1.)*(0.1)*(0.9)**19+C(20.,2.)*(0.1)**2*(0.9)**18)\n",
+ "print \"No. of sample shaving three defective parts = 1000∗0.323 = \",1000*0.323"
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.11+"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter35_2.ipynb b/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter35_2.ipynb
new file mode 100644
index 00000000..5ca46e4e
--- /dev/null
+++ b/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter35_2.ipynb
@@ -0,0 +1,536 @@
+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 35 : Sampling And Inference"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 35.1, page no. 864"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Suppose the coin is unbiased\n",
+ "Then probability of getting the head in a toss = 1/2 \n",
+ "Then, expected no. of successes = a = 1/2∗400\n",
+ "Observed no. of successes = 216\n",
+ "The excess of observed value over expected value = 216\n",
+ "S. D. of simple sampling = (n∗p∗q)ˆ0.5 = c \n",
+ "Hence, z = (b−a)/c = 21.6\n",
+ "As z<1.96, the hypothesis is accepted at 5% level of significance\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Suppose the coin is unbiased\"\n",
+ "print \"Then probability of getting the head in a toss = 1/2 \"\n",
+ "print \"Then, expected no. of successes = a = 1/2∗400\"\n",
+ "a = 1/2*400\n",
+ "print \"Observed no. of successes = 216\"\n",
+ "b = 216\n",
+ "print \"The excess of observed value over expected value = \",b-a\n",
+ "print \"S. D. of simple sampling = (n∗p∗q)ˆ0.5 = c \"\n",
+ "c = (400*0.5*0.5)**0.5\n",
+ "print \"Hence, z = (b−a)/c = \",(b-a)/c\n",
+ "print \"As z<1.96, the hypothesis is accepted at 5% level of significance\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 35.2, page no. 865"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Suppose the die is unbiased \n",
+ "Then probability of getting 5 or 6 with one die=1/3 \n",
+ "Then, expected no. of successes = a = 1/3∗9000 \n",
+ "Observed no. of successes = 3240\n",
+ "The excess of observed value over expected value = 240.0\n",
+ "S. D. of simple sampling = (n∗p∗q)ˆ0.5=c\n",
+ "Hence, z = (b−a)/c = 5.366563146\n",
+ "As z>2.58, the hypothesis has to be rejected at 1% level of significance\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"Suppose the die is unbiased \"\n",
+ "print \"Then probability of getting 5 or 6 with one die=1/3 \"\n",
+ "print \"Then, expected no. of successes = a = 1/3∗9000 \"\n",
+ "a = 1./3*9000\n",
+ "print \"Observed no. of successes = 3240\"\n",
+ "b = 3240\n",
+ "print \"The excess of observed value over expected value = \",b-a\n",
+ "print \"S. D. of simple sampling = (n∗p∗q)ˆ0.5=c\"\n",
+ "c = (9000*(1./3)*(2./3))**0.5\n",
+ "print \"Hence, z = (b−a)/c = \",(b-a)/c\n",
+ "print \"As z>2.58, the hypothesis has to be rejected at 1% level of significance\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 35.3, page no. 865"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "q = 1−p \n",
+ "Standard error of the population of families having a monthly income of rs. 250 or less=(p∗q/n)ˆ0.5 = 0.0148442858929\n",
+ "Hence taking 103/420 to be the estimate of families having a monthly income of rs. 250 or less, the limits are 20% and 29% approximately\n"
+ ]
+ }
+ ],
+ "source": [
+ "p = 206./840\n",
+ "print \"q = 1−p \"\n",
+ "q = 1-p\n",
+ "n = 840\n",
+ "print \"Standard error of the population of families having a monthly income of rs. 250 or less=(p∗q/n)ˆ0.5 = \",(p*q/n)**0.5\n",
+ "print \"Hence taking 103/420 to be the estimate of families having a monthly income of rs. 250 or less, the limits are 20% and 29% approximately\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 35.4, page no. 866"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "p=(n1∗p1+n2∗p2)/(n1+n2)\n",
+ "0.1904\n",
+ "q = 1−p \n",
+ "0.8096\n",
+ "e=(p∗q∗(1/n1+1/n2))ˆ0.5\n",
+ "0.0163590274093\n",
+ "z = 0.916924926202\n",
+ "As z<1, the difference between the proportions is not significant. \n"
+ ]
+ }
+ ],
+ "source": [
+ "n1 = 900\n",
+ "n2 = 1600\n",
+ "p1 = 20./100\n",
+ "p2 = 18.5/100\n",
+ "print \"p=(n1∗p1+n2∗p2)/(n1+n2)\"\n",
+ "p = (n1*p1+n2*p2)/(n1+n2)\n",
+ "print p\n",
+ "print \"q = 1−p \"\n",
+ "q = 1-p\n",
+ "print q\n",
+ "print \"e=(p∗q∗(1/n1+1/n2))ˆ0.5\"\n",
+ "e = (p*q*((1./n1)+(1./n2)))**0.5\n",
+ "print e\n",
+ "z = (p1-p2)/e\n",
+ "print \"z = \",z\n",
+ "print \"As z<1, the difference between the proportions is not significant. \""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 35.5, page no. 867"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "q1 = 1−p1 \n",
+ "q2=1−p2\n",
+ "e=((p1∗q1/n1)+(p2*q2/n2))ˆ0.5\n",
+ "Hence, it is likely that real difference will be hidden.\n"
+ ]
+ }
+ ],
+ "source": [
+ "p1 = 0.3\n",
+ "p2 = 0.25\n",
+ "print \"q1 = 1−p1 \"\n",
+ "q1 = 1-p1\n",
+ "print \"q2=1−p2\"\n",
+ "q2 = 1-p2\n",
+ "n1 = 1200\n",
+ "n2 = 900\n",
+ "print \"e=((p1∗q1/n1)+(p2*q2/n2))ˆ0.5\"\n",
+ "e = ((p1*q1/n1)+(p2*q2/n2))**0.5\n",
+ "z = (p1-p2)/e\n",
+ "print \"Hence, it is likely that real difference will be hidden.\" "
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 35.6, page no. 868"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 11,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "m and n represents mean and number of objects in sample respectively \n",
+ "z=(m−M)/(d/(nˆ0.5)\n",
+ "z = 2.7950310559\n",
+ "As z>1.96, it cannot be regarded as a random sample\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"m and n represents mean and number of objects in sample respectively \"\n",
+ "m = 3.4\n",
+ "n = 900.\n",
+ "M = 3.25\n",
+ "d = 1.61\n",
+ "print \"z=(m−M)/(d/(nˆ0.5)\"\n",
+ "z = (m-M)/(d/(n**0.5))\n",
+ "print \"z = \",z\n",
+ "print \"As z>1.96, it cannot be regarded as a random sample\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 35.9, page no. 871"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 12,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "m1 and n1 represents mean and no. of objects in sample 1\n",
+ "m2 and n2 represents mean and no. of objects in sample 2\n",
+ "On the hypothesis that the samples are drawn from the same population of d = 2.5, we get\n",
+ "z = -5.16397779494\n",
+ "Since |z|>1.96, thus samples cannot be regarded as drawn from the same population\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"m1 and n1 represents mean and no. of objects in sample 1\"\n",
+ "print \"m2 and n2 represents mean and no. of objects in sample 2\"\n",
+ "m1 = 67.5\n",
+ "m2 = 68.\n",
+ "n1 = 1000.\n",
+ "n2 = 2000.\n",
+ "d = 2.5\n",
+ "print \"On the hypothesis that the samples are drawn from the same population of d = 2.5, we get\"\n",
+ "z = (m1-m2)/(d*((1/n1)+(1/n2))**0.5)\n",
+ "print \"z = \",z\n",
+ "print \"Since |z|>1.96, thus samples cannot be regarded as drawn from the same population\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 35.10, page no. 872"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 13,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "m1, d1 and n1 denotes mean, deviation and no.of objects in first sample \n",
+ "m2, d2 and n2 denotes mean, deviation and no.of objects in second sample\n",
+ "S. E. of the difference of the mean heights is 0.0703246933872\n",
+ "-0.7\n",
+ "|m1−m2|>10e, this is highly significant. hence, the data indicates that the sailors are on the average taller than the soldiers.\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"m1, d1 and n1 denotes mean, deviation and no.of objects in first sample \"\n",
+ "m1 = 67.85\n",
+ "d1 = 2.56\n",
+ "n1 = 6400.\n",
+ "print \"m2, d2 and n2 denotes mean, deviation and no.of objects in second sample\"\n",
+ "m2 = 68.55\n",
+ "d2 = 2.52\n",
+ "n2 = 1600.\n",
+ "print \"S. E. of the difference of the mean heights is \",\n",
+ "e = ((d**2/n1)+(d2**2/n2))**0.5\n",
+ "print e\n",
+ "print m1-m2\n",
+ "print \"|m1−m2|>10e, this is highly significant. hence, the data indicates that the sailors are on the average taller than the soldiers.\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 35.12, page no. 874"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 18,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "First of row denotes the different values of sample \n",
+ "The second row denotes the corresponding deviation\n",
+ "The third row denotes the corresponding square of deviation\n",
+ "The sum of second row elements = 10.0\n",
+ "The sum of third row elements = 66.0\n",
+ "let m be the mean \n",
+ "let d be the standard deviation\n",
+ "1.84522581263\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "A = numpy.zeros((3,9))\n",
+ "n = 9\n",
+ "print \"First of row denotes the different values of sample \"\n",
+ "A[0,:] = [45,47,50,52,48,47,49,53,51]\n",
+ "print \"The second row denotes the corresponding deviation\"\n",
+ "for i in range(0,9):\n",
+ " A[1,i] = A[0,i]-48\n",
+ "print \"The third row denotes the corresponding square of deviation\"\n",
+ "for i in range(0,9):\n",
+ " A[2,i] = A[1,i]**2\n",
+ "print \"The sum of second row elements = \",\n",
+ "a =0\n",
+ "for i in range(0,9):\n",
+ " a = a+A[1,i]\n",
+ "print a\n",
+ "print \"The sum of third row elements = \",\n",
+ "b = 0\n",
+ "for i in range(0,9):\n",
+ " b = b+A[2,i]\n",
+ "print b\n",
+ "print \"let m be the mean \"\n",
+ "m = 48+a/n\n",
+ "print \"let d be the standard deviation\"\n",
+ "d = ((b/n)-(a/n)**2)**0.5\n",
+ "t = (m-47.5)*(n-1)**0.5/d\n",
+ "print t"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 35.13, page no. 876"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 19,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "d and n represents the deviation and no. objects in given sample\n",
+ "Taking the hypothesis that the product is not inferior i.e. there is no significant difference between m and M\n",
+ "3.15\n",
+ "Degrees of freedom= \n",
+ "9.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"d and n represents the deviation and no. objects in given sample\"\n",
+ "n = 10.\n",
+ "d = 0.04\n",
+ "m = 0.742\n",
+ "M = 0.700\n",
+ "print \"Taking the hypothesis that the product is not inferior i.e. there is no significant difference between m and M\"\n",
+ "t = (m-M)*(n-1)**0.5/d\n",
+ "print t\n",
+ "print \"Degrees of freedom= \"\n",
+ "f = n-1\n",
+ "print f"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 34.15, page no. 878"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 21,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The first row denotes the boy no. \n",
+ "The second row denotes the marks in test I(x1)\n",
+ "The third row denotes the marks in test I(x2)\n",
+ "The fourth row denotes the difference of marks in two tests(d)\n",
+ "The fifth row denotes the (d−1)\n",
+ "The sixth row denotes the square of elements of fourth row \n",
+ "[[ 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.]\n",
+ " [ 23. 20. 19. 21. 18. 20. 18. 17. 23. 16. 19.]\n",
+ " [ 24. 19. 22. 18. 20. 22. 20. 20. 23. 20. 17.]\n",
+ " [ 1. -1. 3. -3. 2. 2. 2. 3. 0. 4. -2.]\n",
+ " [ 0. -2. 2. -4. 1. 1. 1. 2. -1. 3. -3.]\n",
+ " [ 1. 1. 9. 9. 4. 4. 4. 9. 0. 16. 4.]]\n",
+ "The sum of elements of fourth row=\n",
+ "11.0\n",
+ "The sum of elements of sixth row= \n",
+ "61.0\n",
+ "Standard deviation\n",
+ "d = 2.46981780705\n",
+ "t = 1.48063606712\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy\n",
+ "\n",
+ "A = numpy.zeros((6,11))\n",
+ "n = 11\n",
+ "print \"The first row denotes the boy no. \"\n",
+ "A[0,:] = [1,2,3,4,5,6,7,8,9,10,11]\n",
+ "print \"The second row denotes the marks in test I(x1)\"\n",
+ "A[1,:] = [23,20,19,21,18,20,18,17,23,16,19]\n",
+ "print \"The third row denotes the marks in test I(x2)\"\n",
+ "A[2,:] = [24,19,22,18,20,22,20,20,23,20,17]\n",
+ "print \"The fourth row denotes the difference of marks in two tests(d)\"\n",
+ "for i in range (0,11):\n",
+ " A[3,i] = A[2,i]-A[1,i]\n",
+ "print \"The fifth row denotes the (d−1)\"\n",
+ "for i in range (0,11):\n",
+ " A[4,i] = A[3,i]-1\n",
+ "print \"The sixth row denotes the square of elements of fourth row \"\n",
+ "for i in range(0,11):\n",
+ " A[5,i] = A[3,i]**2\n",
+ "print A\n",
+ "a = 0\n",
+ "print \"The sum of elements of fourth row=\"\n",
+ "for i in range(0,11):\n",
+ " a = a+A[3,i]\n",
+ "print a\n",
+ "b = 0\n",
+ "print \"The sum of elements of sixth row= \"\n",
+ "for i in range(0,11):\n",
+ " b = b + A[5,i]\n",
+ "print b\n",
+ "print \"Standard deviation\"\n",
+ "d = (b/(n-1))**0.5\n",
+ "t = (1-0)*(n)**0.5/2.24\n",
+ "print \"d = \",d\n",
+ "print \"t = \",t"
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.10"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter4_2.ipynb b/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter4_2.ipynb
new file mode 100644
index 00000000..e40044d7
--- /dev/null
+++ b/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter4_2.ipynb
@@ -0,0 +1,1208 @@
+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 4: Diffrentiations And Applications"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "## Example 4.1, page no. 133"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 24,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "We have to find yn for F=cosxcos2xcos3x\n",
+ "Enter the order of differentiation: 0\n",
+ "calculating yn..\n",
+ "The expression for yn is\n",
+ "cos(x)*cos(2*x)*cos(3*x)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "import numpy\n",
+ "import math\n",
+ "\n",
+ "print \"We have to find yn for F=cosxcos2xcos3x\"\n",
+ "x = sympy.Symbol('x')\n",
+ "F = sympy.cos(x)*sympy.cos(2*x)*sympy.cos(3*x)\n",
+ "n = int(raw_input(\"Enter the order of differentiation: \"))\n",
+ "print \"calculating yn..\"\n",
+ "yn= sympy.diff(F,x,n)\n",
+ "print \"The expression for yn is\"\n",
+ "print yn "
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.5, page no. 137"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 25,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "we have to find yn for F=cosxcos2xcos3x\n",
+ "Enter the order of differentiation: 0\n",
+ "calculating yn\n",
+ "The expression for yn is\n",
+ "x/((x - 1)*(2*x + 3))\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "\n",
+ "print 'we have to find yn for F=cosxcos2xcos3x'\n",
+ "x = sympy.Symbol('x')\n",
+ "F = x/((x-1)*(2*x+3))\n",
+ "n = int(raw_input(\"Enter the order of differentiation: \"))\n",
+ "print \"calculating yn\"\n",
+ "yn= sympy.diff(F,x,n)\n",
+ "print \"The expression for yn is\"\n",
+ "print yn"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.6, page no. 138"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 26,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "We have to find yn for F=cosxcos2xcos3x\n",
+ "Enter the orde of differentiation:0\n",
+ "calculating yn\n",
+ "the expression for yn is\n",
+ "x/(a**2 + x**2)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "\n",
+ "print \"We have to find yn for F=cosxcos2xcos3x\"\n",
+ "x = sympy.Symbol('x')\n",
+ "a = sympy.Symbol('a')\n",
+ "F = x/(x**2+a**2)\n",
+ "n = raw_input (\"Enter the orde of differentiation:\")\n",
+ "print \"calculating yn\"\n",
+ "yn = diff(F,x,n)\n",
+ "print \"the expression for yn is\"\n",
+ "print yn"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.7, page no. 139"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 27,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "we have to find yn for F=cosxcos2xcos3x\n",
+ "Enter the order of differentiation: 0\n",
+ "calculating yn\n",
+ "the expression for yn is\n",
+ "2.71828182845905**x*(2*x + 3)**3\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy, math\n",
+ "\n",
+ "print \"we have to find yn for F=cosxcos2xcos3x\"\n",
+ "x = sympy.Symbol('x')\n",
+ "a = sympy.Symbol('a')\n",
+ "F = math.e**(x)*(2*x+3)**3\n",
+ "n = raw_input (\"Enter the order of differentiation: \")\n",
+ "print \"calculating yn\"\n",
+ "yn = diff(F,x,n)\n",
+ "print \"the expression for yn is\"\n",
+ "print yn"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.8, page no. 139"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 20,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "y=(sinˆ-1)x) --sign inverse x\n",
+ "we have to prove (1-xˆ2)y(n+2)-(2n+1)xy(n+1)-nˆ2 yn\n",
+ "Calculating yn for various values of n\n",
+ "1\n",
+ "The expression for yn is\n",
+ "-6*x*(x*asin(x)/(-x**2 + 1)**(3/2) - 1/(x**2 - 1)) - 2*(a**2 + 1)*asin(x)/sqrt(-x**2 + 1) + 2*(-x**2 + 1)*(3*x**2*asin(x)/(-x**2 + 1)**(5/2) + 3*x/(x**2 - 1)**2 + asin(x)/(-x**2 + 1)**(3/2))\n",
+ "Which is equal to 0\n",
+ "2\n",
+ "The expression for yn is\n",
+ "-10*x*(3*x**2*asin(x)/(-x**2 + 1)**(5/2) + 3*x/(x**2 - 1)**2 + asin(x)/(-x**2 + 1)**(3/2)) - 2*(a**2 + 4)*(x*asin(x)/(-x**2 + 1)**(3/2) - 1/(x**2 - 1)) + 2*(-x**2 + 1)*(15*x**3*asin(x)/(-x**2 + 1)**(7/2) - 15*x**2/(x**2 - 1)**3 + 9*x*asin(x)/(-x**2 + 1)**(5/2) + 4/(x**2 - 1)**2)\n",
+ "Which is equal to 0\n",
+ "3\n",
+ "The expression for yn is\n",
+ "-14*x*(15*x**3*asin(x)/(-x**2 + 1)**(7/2) - 15*x**2/(x**2 - 1)**3 + 9*x*asin(x)/(-x**2 + 1)**(5/2) + 4/(x**2 - 1)**2) - 2*(a**2 + 9)*(3*x**2*asin(x)/(-x**2 + 1)**(5/2) + 3*x/(x**2 - 1)**2 + asin(x)/(-x**2 + 1)**(3/2)) + 2*(-x**2 + 1)*(105*x**4*asin(x)/(-x**2 + 1)**(9/2) + 105*x**3/(x**2 - 1)**4 + 90*x**2*asin(x)/(-x**2 + 1)**(7/2) - 55*x/(x**2 - 1)**3 + 9*asin(x)/(-x**2 + 1)**(5/2))\n",
+ "Which is equal to 0\n",
+ "4\n",
+ "The expression for yn is\n",
+ "-18*x*(105*x**4*asin(x)/(-x**2 + 1)**(9/2) + 105*x**3/(x**2 - 1)**4 + 90*x**2*asin(x)/(-x**2 + 1)**(7/2) - 55*x/(x**2 - 1)**3 + 9*asin(x)/(-x**2 + 1)**(5/2)) - 2*(a**2 + 16)*(15*x**3*asin(x)/(-x**2 + 1)**(7/2) - 15*x**2/(x**2 - 1)**3 + 9*x*asin(x)/(-x**2 + 1)**(5/2) + 4/(x**2 - 1)**2) + 2*(-x**2 + 1)*(945*x**5*asin(x)/(-x**2 + 1)**(11/2) - 945*x**4/(x**2 - 1)**5 + 1050*x**3*asin(x)/(-x**2 + 1)**(9/2) + 735*x**2/(x**2 - 1)**4 + 225*x*asin(x)/(-x**2 + 1)**(7/2) - 64/(x**2 - 1)**3)\n",
+ "Which is equal to 0\n",
+ "Hence Proved\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "\n",
+ "print 'y=(sinˆ-1)x) --sign inverse x'\n",
+ "x = sympy.Symbol('x')\n",
+ "a = sympy.Symbol('a')\n",
+ "y = (sympy.asin(x))**2;\n",
+ "print 'we have to prove (1-xˆ2)y(n+2)-(2n+1)xy(n+1)-nˆ2 yn'\n",
+ "print 'Calculating yn for various values of n'\n",
+ "for n in range(1,5):\n",
+ " F=(1-x**2)*sympy.diff(y,x,n+2)-(2*n+1)*x*sympy.diff(y,x,n+1)-(n**2+a**2)*sympy.diff(y,x,n)\n",
+ " print n\n",
+ " print 'The expression for yn is'\n",
+ " print F\n",
+ " print 'Which is equal to 0'\n",
+ "print 'Hence Proved'"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.9, page no. 140"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "y=eˆ(a(sinˆ-1)x))--sign inverse x\n",
+ "We have to prove (1-xˆ2)y(n+2)-(2n+1)xy(n+1)-(nˆ2+aˆ2)yn\n",
+ "Calculating yn for various values of n\n",
+ "1\n",
+ "The expression for yn is\n",
+ "-3.0*2.71828182845905**(a*asin(x))*a*x*(-a/(x**2 - 1) + x/(-x**2 + 1)**(3/2)) - 1.0*2.71828182845905**(a*asin(x))*a*(a**2 + 1)/sqrt(-x**2 + 1) + 2.71828182845905**(a*asin(x))*a*(-x**2 + 1)*(1.0*a**2/(-x**2 + 1)**(3/2) + 3.0*a*x/(x**2 - 1)**2 + 3.0*x**2/(-x**2 + 1)**(5/2) + 1.0/(-x**2 + 1)**(3/2))\n",
+ "Which is equal to 0\n",
+ "2\n",
+ "The expression for yn is\n",
+ "-5*2.71828182845905**(a*asin(x))*a*x*(1.0*a**2/(-x**2 + 1)**(3/2) + 3.0*a*x/(x**2 - 1)**2 + 3.0*x**2/(-x**2 + 1)**(5/2) + 1.0/(-x**2 + 1)**(3/2)) - 1.0*2.71828182845905**(a*asin(x))*a*(a**2 + 4)*(-a/(x**2 - 1) + x/(-x**2 + 1)**(3/2)) + 2.71828182845905**(a*asin(x))*a*(-x**2 + 1)*(1.0*a**3/(x**2 - 1)**2 + 6.0*a**2*x/(-x**2 + 1)**(5/2) - 15.0*a*x**2/(x**2 - 1)**3 + 4.0*a/(x**2 - 1)**2 + 15.0*x**3/(-x**2 + 1)**(7/2) + 9.0*x/(-x**2 + 1)**(5/2))\n",
+ "Which is equal to 0\n",
+ "3\n",
+ "The expression for yn is\n",
+ "-7*2.71828182845905**(a*asin(x))*a*x*(1.0*a**3/(x**2 - 1)**2 + 6.0*a**2*x/(-x**2 + 1)**(5/2) - 15.0*a*x**2/(x**2 - 1)**3 + 4.0*a/(x**2 - 1)**2 + 15.0*x**3/(-x**2 + 1)**(7/2) + 9.0*x/(-x**2 + 1)**(5/2)) - 2.71828182845905**(a*asin(x))*a*(a**2 + 9)*(1.0*a**2/(-x**2 + 1)**(3/2) + 3.0*a*x/(x**2 - 1)**2 + 3.0*x**2/(-x**2 + 1)**(5/2) + 1.0/(-x**2 + 1)**(3/2)) + 2.71828182845905**(a*asin(x))*a*(-x**2 + 1)*(1.0*a**4/(-x**2 + 1)**(5/2) - 10.0*a**3*x/(x**2 - 1)**3 + 45.0*a**2*x**2/(-x**2 + 1)**(7/2) + 10.0*a**2/(-x**2 + 1)**(5/2) + 105.0*a*x**3/(x**2 - 1)**4 - 55.0*a*x/(x**2 - 1)**3 + 105.0*x**4/(-x**2 + 1)**(9/2) + 90.0*x**2/(-x**2 + 1)**(7/2) + 9.0/(-x**2 + 1)**(5/2))\n",
+ "Which is equal to 0\n",
+ "4\n",
+ "The expression for yn is\n",
+ "-9*2.71828182845905**(a*asin(x))*a*x*(1.0*a**4/(-x**2 + 1)**(5/2) - 10.0*a**3*x/(x**2 - 1)**3 + 45.0*a**2*x**2/(-x**2 + 1)**(7/2) + 10.0*a**2/(-x**2 + 1)**(5/2) + 105.0*a*x**3/(x**2 - 1)**4 - 55.0*a*x/(x**2 - 1)**3 + 105.0*x**4/(-x**2 + 1)**(9/2) + 90.0*x**2/(-x**2 + 1)**(7/2) + 9.0/(-x**2 + 1)**(5/2)) - 2.71828182845905**(a*asin(x))*a*(a**2 + 16)*(1.0*a**3/(x**2 - 1)**2 + 6.0*a**2*x/(-x**2 + 1)**(5/2) - 15.0*a*x**2/(x**2 - 1)**3 + 4.0*a/(x**2 - 1)**2 + 15.0*x**3/(-x**2 + 1)**(7/2) + 9.0*x/(-x**2 + 1)**(5/2)) + 2.71828182845905**(a*asin(x))*a*(-x**2 + 1)*(-1.0*a**5/(x**2 - 1)**3 + 15.0*a**4*x/(-x**2 + 1)**(7/2) + 105.0*a**3*x**2/(x**2 - 1)**4 - 20.0*a**3/(x**2 - 1)**3 + 420.0*a**2*x**3/(-x**2 + 1)**(9/2) + 195.0*a**2*x/(-x**2 + 1)**(7/2) - 945.0*a*x**4/(x**2 - 1)**5 + 735.0*a*x**2/(x**2 - 1)**4 - 64.0*a/(x**2 - 1)**3 + 945.0*x**5/(-x**2 + 1)**(11/2) + 1050.0*x**3/(-x**2 + 1)**(9/2) + 225.0*x/(-x**2 + 1)**(7/2))\n",
+ "Which is equal to 0\n",
+ "Hence proved\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy, math\n",
+ "\n",
+ "print 'y=eˆ(a(sinˆ-1)x))--sign inverse x'\n",
+ "x = sympy.Symbol('x')\n",
+ "a = sympy.Symbol('a')\n",
+ "y = math.e**(a*(sympy.asin(x)))\n",
+ "print 'We have to prove (1-xˆ2)y(n+2)-(2n+1)xy(n+1)-(nˆ2+aˆ2)yn'\n",
+ "print 'Calculating yn for various values of n'\n",
+ "for n in range(1,5):\n",
+ " F =(1-x**2)*sympy.diff(y,x,n+2)-(2*n+1)*x*sympy.diff(y,x,n+1)-(n**2+a**2)*sympy.diff(y,x,n)\n",
+ " print n\n",
+ " print 'The expression for yn is'\n",
+ " print F\n",
+ " print 'Which is equal to 0'\n",
+ "print 'Hence proved'\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.10, page no. 141"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "y^(1/m)+y^-(1/m)=2x\n",
+ "OR y^(2/m)-2xy^(1/m)+1\n",
+ "OR y=[ x+(xˆ2-1)]ˆm and y=[x-(xˆ2 −1)]ˆm\n",
+ "For y=[x+(xˆ2−1)]ˆm\n",
+ "We have to prove (xˆ2−1)y(n+2)+(2n+1)xy(n+1)+(nˆ2−mˆ2)yn\n",
+ "Calculating various values for yn\n",
+ "1\n",
+ "The expression for yn is\n",
+ "3*m*x*(x**2 + x - 1)**m*(m*(2*x + 1)**2/(x**2 + x - 1) - (2*x + 1)**2/(x**2 + x - 1) + 2)/(x**2 + x - 1) + m*(-m**2 + 1)*(2*x + 1)*(x**2 + x - 1)**m/(x**2 + x - 1) + m*(2*x + 1)*(x**2 - 1)*(x**2 + x - 1)**m*(m**2*(2*x + 1)**2/(x**2 + x - 1) - 3*m*(2*x + 1)**2/(x**2 + x - 1) + 6*m + 2*(2*x + 1)**2/(x**2 + x - 1) - 6)/(x**2 + x - 1)**2\n",
+ "Which is equal to 0\n",
+ "2\n",
+ "The expression for yn is\n",
+ "5*m*x*(2*x + 1)*(x**2 + x - 1)**m*(m**2*(2*x + 1)**2/(x**2 + x - 1) - 3*m*(2*x + 1)**2/(x**2 + x - 1) + 6*m + 2*(2*x + 1)**2/(x**2 + x - 1) - 6)/(x**2 + x - 1)**2 + m*(-m**2 + 4)*(x**2 + x - 1)**m*(m*(2*x + 1)**2/(x**2 + x - 1) - (2*x + 1)**2/(x**2 + x - 1) + 2)/(x**2 + x - 1) + m*(x**2 - 1)*(x**2 + x - 1)**m*(m**3*(2*x + 1)**4/(x**2 + x - 1)**2 - 6*m**2*(2*x + 1)**4/(x**2 + x - 1)**2 + 12*m**2*(2*x + 1)**2/(x**2 + x - 1) + 11*m*(2*x + 1)**4/(x**2 + x - 1)**2 - 36*m*(2*x + 1)**2/(x**2 + x - 1) + 12*m - 6*(2*x + 1)**4/(x**2 + x - 1)**2 + 24*(2*x + 1)**2/(x**2 + x - 1) - 12)/(x**2 + x - 1)**2\n",
+ "Which is equal to 0\n",
+ "3\n",
+ "The expression for yn is\n",
+ "7*m*x*(x**2 + x - 1)**m*(m**3*(2*x + 1)**4/(x**2 + x - 1)**2 - 6*m**2*(2*x + 1)**4/(x**2 + x - 1)**2 + 12*m**2*(2*x + 1)**2/(x**2 + x - 1) + 11*m*(2*x + 1)**4/(x**2 + x - 1)**2 - 36*m*(2*x + 1)**2/(x**2 + x - 1) + 12*m - 6*(2*x + 1)**4/(x**2 + x - 1)**2 + 24*(2*x + 1)**2/(x**2 + x - 1) - 12)/(x**2 + x - 1)**2 + m*(-m**2 + 9)*(2*x + 1)*(x**2 + x - 1)**m*(m**2*(2*x + 1)**2/(x**2 + x - 1) - 3*m*(2*x + 1)**2/(x**2 + x - 1) + 6*m + 2*(2*x + 1)**2/(x**2 + x - 1) - 6)/(x**2 + x - 1)**2 + m*(2*x + 1)*(x**2 - 1)*(x**2 + x - 1)**m*(m**4*(2*x + 1)**4/(x**2 + x - 1)**2 - 10*m**3*(2*x + 1)**4/(x**2 + x - 1)**2 + 20*m**3*(2*x + 1)**2/(x**2 + x - 1) + 35*m**2*(2*x + 1)**4/(x**2 + x - 1)**2 - 120*m**2*(2*x + 1)**2/(x**2 + x - 1) + 60*m**2 - 50*m*(2*x + 1)**4/(x**2 + x - 1)**2 + 220*m*(2*x + 1)**2/(x**2 + x - 1) - 180*m + 24*(2*x + 1)**4/(x**2 + x - 1)**2 - 120*(2*x + 1)**2/(x**2 + x - 1) + 120)/(x**2 + x - 1)**3\n",
+ "Which is equal to 0\n",
+ "4\n",
+ "The expression for yn is\n",
+ "9*m*x*(2*x + 1)*(x**2 + x - 1)**m*(m**4*(2*x + 1)**4/(x**2 + x - 1)**2 - 10*m**3*(2*x + 1)**4/(x**2 + x - 1)**2 + 20*m**3*(2*x + 1)**2/(x**2 + x - 1) + 35*m**2*(2*x + 1)**4/(x**2 + x - 1)**2 - 120*m**2*(2*x + 1)**2/(x**2 + x - 1) + 60*m**2 - 50*m*(2*x + 1)**4/(x**2 + x - 1)**2 + 220*m*(2*x + 1)**2/(x**2 + x - 1) - 180*m + 24*(2*x + 1)**4/(x**2 + x - 1)**2 - 120*(2*x + 1)**2/(x**2 + x - 1) + 120)/(x**2 + x - 1)**3 + m*(-m**2 + 16)*(x**2 + x - 1)**m*(m**3*(2*x + 1)**4/(x**2 + x - 1)**2 - 6*m**2*(2*x + 1)**4/(x**2 + x - 1)**2 + 12*m**2*(2*x + 1)**2/(x**2 + x - 1) + 11*m*(2*x + 1)**4/(x**2 + x - 1)**2 - 36*m*(2*x + 1)**2/(x**2 + x - 1) + 12*m - 6*(2*x + 1)**4/(x**2 + x - 1)**2 + 24*(2*x + 1)**2/(x**2 + x - 1) - 12)/(x**2 + x - 1)**2 + m*(x**2 - 1)*(x**2 + x - 1)**m*(m**5*(2*x + 1)**6/(x**2 + x - 1)**3 - 15*m**4*(2*x + 1)**6/(x**2 + x - 1)**3 + 30*m**4*(2*x + 1)**4/(x**2 + x - 1)**2 + 85*m**3*(2*x + 1)**6/(x**2 + x - 1)**3 - 300*m**3*(2*x + 1)**4/(x**2 + x - 1)**2 + 180*m**3*(2*x + 1)**2/(x**2 + x - 1) - 225*m**2*(2*x + 1)**6/(x**2 + x - 1)**3 + 1050*m**2*(2*x + 1)**4/(x**2 + x - 1)**2 - 1080*m**2*(2*x + 1)**2/(x**2 + x - 1) + 120*m**2 + 274*m*(2*x + 1)**6/(x**2 + x - 1)**3 - 1500*m*(2*x + 1)**4/(x**2 + x - 1)**2 + 1980*m*(2*x + 1)**2/(x**2 + x - 1) - 360*m - 120*(2*x + 1)**6/(x**2 + x - 1)**3 + 720*(2*x + 1)**4/(x**2 + x - 1)**2 - 1080*(2*x + 1)**2/(x**2 + x - 1) + 240)/(x**2 + x - 1)**3\n",
+ "Which is equal to 0\n",
+ "Hence Proved\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "\n",
+ "print 'y^(1/m)+y^-(1/m)=2x'\n",
+ "print 'OR y^(2/m)-2xy^(1/m)+1'\n",
+ "print 'OR y=[ x+(xˆ2-1)]ˆm and y=[x-(xˆ2 −1)]ˆm'\n",
+ "x = sympy.Symbol('x')\n",
+ "m = sympy.Symbol('m')\n",
+ "print 'For y=[x+(xˆ2−1)]ˆm'\n",
+ "y=(x+(x**2-1))**m\n",
+ "print 'We have to prove (xˆ2−1)y(n+2)+(2n+1)xy(n+1)+(nˆ2−mˆ2)yn'\n",
+ "print 'Calculating various values for yn'\n",
+ "for n in range(1,5):\n",
+ " F=(x**2-1)*sympy.diff(y,x,n+2)+(2*n+1)*x*sympy.diff(y,x,n+1)+(n**2-m**2)*sympy.diff(y,x,n)\n",
+ " print n\n",
+ " print 'The expression for yn is'\n",
+ " print F\n",
+ " print 'Which is equal to 0'\n",
+ "print 'Hence Proved'"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.11, page no. 144"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": false,
+ "scrolled": true
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "For roles theorem F9x should be differentiable in (a, b) and f(a)=f(b)\n",
+ "Here f(x)=sin(x)/e^x\n",
+ "-1.0*2.71828182845905**(-x)*sin(x) + 2.71828182845905**(-x)*cos(x)\n",
+ "Putting this to zero we get tan(x)=1 ie x=pi/4\n",
+ "Value pi/2 lies b/w 0 and pi. Hence roles theroem is verified\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy, math\n",
+ "\n",
+ "print 'For roles theorem F9x should be differentiable in (a, b) and f(a)=f(b)'\n",
+ "print 'Here f(x)=sin(x)/e^x'\n",
+ "x = sympy.Symbol('x')\n",
+ "y = sympy.sin(x)/math.e**x\n",
+ "y1 = sympy.diff(y,x)\n",
+ "print y1\n",
+ "print 'Putting this to zero we get tan(x)=1 ie x=pi/4'\n",
+ "print 'Value pi/2 lies b/w 0 and pi. Hence roles theroem is verified'"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.16 page no. 149"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 56,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Maclaurin's Series\n",
+ "f(x)=f(0)+xf1(0)+xˆ2/2!∗f2(0)+xˆ3/3!∗f3(0)+......\n",
+ "Enter the orde of differentiation: 6\n",
+ "-I*x**5*atanh(sqrt(-sqrt(105)/30 + 1/2))/120 - I*x**3*atanh(sqrt(3)/3)/6\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy, sympy, math\n",
+ "\n",
+ "print \"Maclaurin's Series\"\n",
+ "print 'f(x)=f(0)+xf1(0)+xˆ2/2!∗f2(0)+xˆ3/3!∗f3(0)+......'\n",
+ "x = sympy.Symbol('x')\n",
+ "a = sympy.Symbol('a')\n",
+ "y = sympy.tan(a)\n",
+ "n = int(raw_input(\"Enter the orde of differentiation: \"))\n",
+ "a = 1\n",
+ "t = sympy.solve(y)[0]\n",
+ "a = 0\n",
+ "for i in range(2,n+1):\n",
+ " y1 = sympy.diff (y,'a',i-1)\n",
+ " if sympy.solve(y1):\n",
+ " t = t+x**(i-1)*sympy.solve(y1)[0]/math.factorial(i-1)\n",
+ "print t"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.17, page no. 150"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Maclaurins series\n",
+ "f(x)=f(0)+xf1(0)+xˆ2/2!∗f2(0)+xˆ3/3!∗f3(0)+......\n",
+ "enter the number of expression in series:2\n",
+ "1.5707963267949*x\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy,math\n",
+ "\n",
+ "print \"Maclaurins series\"\n",
+ "print \"f(x)=f(0)+xf1(0)+xˆ2/2!∗f2(0)+xˆ3/3!∗f3(0)+......\"\n",
+ "x = sympy.Symbol('x')\n",
+ "a = sympy.Symbol('a')\n",
+ "y = math.e**(sympy.sin(a))\n",
+ "n = int(raw_input('enter the number of expression in series:'))\n",
+ "a = 0\n",
+ "t = 0\n",
+ "a = 0\n",
+ "for i in range(2,n+1):\n",
+ " y1 = sympy.diff(y,'a',i-1) \n",
+ " t = t+x**(i-1)*sympy.solve(y1)[0]/math.factorial(i-1) \n",
+ "print t"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "## Example 4.18, page no. 150"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 10,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Maclaurins series\n",
+ "f(x)=f(0)+xf1(0)+xˆ2/2!∗f2(0)+xˆ3/3!∗f3(0)+......\n",
+ "enter the number of differentiation involved in maclaurins series :3\n",
+ "x**2*atan(RootOf(tan(a/2)**4 - 10*tan(a/2)**2 + 1, 0))\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy,math\n",
+ "\n",
+ "print \"Maclaurins series\"\n",
+ "print \"f(x)=f(0)+xf1(0)+xˆ2/2!∗f2(0)+xˆ3/3!∗f3(0)+......\"\n",
+ "x = sympy.Symbol('x')\n",
+ "a = sympy.Symbol('a')\n",
+ "y = sympy.log(1+(sympy.sin(a))**2)\n",
+ "n = int(raw_input(\"enter the number of differentiation involved in maclaurins series :\"))\n",
+ "a = 0\n",
+ "t = 0\n",
+ "a = 0\n",
+ "for i in range(2,n+1):\n",
+ " y1 = sympy.diff(y,'a',i-1) \n",
+ " t = t+x**(i-1)*sympy.solve(y1)[0]/math.factorial(i-1) \n",
+ "print t"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "## Example 4.19, page no. 151"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 33,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Maclaurins series\n",
+ "f(x)=f(0)+xf1(0)+xˆ2/2!∗f2(0)+xˆ3/3!∗f3(0)+......\n",
+ "enter the number of expression in series :2\n",
+ "1.0*2.71828182845905**(a*asin(b))*a*x*(-b**2 + 1.0)**(-0.5)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy,math\n",
+ "\n",
+ "print \"Maclaurins series\"\n",
+ "print \"f(x)=f(0)+xf1(0)+xˆ2/2!∗f2(0)+xˆ3/3!∗f3(0)+......\"\n",
+ "x = sympy.Symbol('x')\n",
+ "a = sympy.Symbol('a')\n",
+ "b = sympy.Symbol('b')\n",
+ "y = math.e**(a*sympy.asin(b))\n",
+ "y1 = sympy.diff(y, 'b', 1)\n",
+ "n = int(raw_input(\"enter the number of expression in series :\"))\n",
+ "b = 0\n",
+ "t = 0\n",
+ "for i in range(2,n+1):\n",
+ " y1 = sympy.diff(y,'b',i-1) \n",
+ " t = t+x**(i-1)*y1.evalf()/math.factorial(i-1) \n",
+ "print t"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.20, page no. 152"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Advantage of scilab is that we can calculate log1.1 directly without using Taylor series\n",
+ "Use of taylor series are given in subsequent examples\n",
+ "log(1.1) = 0.0953101798043\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,math,sympy\n",
+ "\n",
+ "print \"Use of taylor series are given in subsequent examples\"\n",
+ "y = math.log(1.1)\n",
+ "print \"log(1.1) = \",math.log (1.1) "
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.21, page no. 152"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 29,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Taylor series\n",
+ "f(x+h)=f(x)+hf1(x)+hˆ2/2!∗f2(x)+hˆ3/3!∗f3(x)+......\n",
+ "To finf the taylor expansion of tan−1(x+h)\n",
+ "enter the number of expression in series : 2\n",
+ "h/(x**2 + 1) + atan(x)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy,math\n",
+ "\n",
+ "print \"Taylor series\"\n",
+ "print \"f(x+h)=f(x)+hf1(x)+hˆ2/2!∗f2(x)+hˆ3/3!∗f3(x)+......\"\n",
+ "print \"To finf the taylor expansion of tan−1(x+h)\"\n",
+ "x = sympy.Symbol('x')\n",
+ "h = sympy.Symbol('h')\n",
+ "y = sympy.atan ( x )\n",
+ "n = int(raw_input (\"enter the number of expression in series : \"))\n",
+ "t = y \n",
+ "for i in range (2,n+1):\n",
+ " y1 = sympy.diff (y,'x',i-1)\n",
+ " t = t+h**(i-1)*(y1)/math.factorial(i-1)\n",
+ "print t"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.22, page no. 153"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 38,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Here we need to find the limit off(x) at x=0\n",
+ "2.46828182845905\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy,math\n",
+ "\n",
+ "print \"Here we need to find the limit off(x) at x=0\"\n",
+ "x = sympy.Symbol('x')\n",
+ "y = (x*math.e**x-sympy.log(1+x))/x**2\n",
+ "f = 1\n",
+ "while f==1:\n",
+ " yn = x*math.e**x-sympy.log(1+x)\n",
+ " yd = x**2;\n",
+ " yn1 = sympy.diff(yn,'x',1)\n",
+ " yd1 = sympy.diff(yd,'x',1)\n",
+ " x = 0\n",
+ " a = yn1.evalf(subs=dict(x=1))\n",
+ " b = yd1.evalf(subs=dict(x=1))\n",
+ " if a == b:\n",
+ " yn = yn1 \n",
+ " yd = yd1 \n",
+ " else:\n",
+ " f =0\n",
+ "h = a / b \n",
+ "print h"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.32, page no. 162"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 12,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Equation of tangent\n",
+ "Equation is g=0 where g is x - y\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy,math\n",
+ "\n",
+ "print \"Equation of tangent\"\n",
+ "x = sympy.Symbol('x')\n",
+ "a = sympy.Symbol('a')\n",
+ "y = sympy.Symbol('y')\n",
+ "f = (a**(2/3)-x**(2/3))**(3/2)\n",
+ "s = sympy.diff(f,x)\n",
+ "Y1 = s*(-x)+y\n",
+ "X1 = -y/s*x\n",
+ "g = x-(Y1-s*(X1-x))\n",
+ "print \"Equation is g=0 where g is\",g"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example, 4.34, page no. 163"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 13,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Equation of tangent\n",
+ "y = a*(-t*cos(t) + sin(t)) + (-a*(t*sin(t) + cos(t)) + x)*cos(t)/sin(t)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy,math\n",
+ "\n",
+ "print \"Equation of tangent\"\n",
+ "x = sympy.Symbol('x')\n",
+ "a = sympy.Symbol('a')\n",
+ "y = sympy.Symbol('y')\n",
+ "t = sympy.Symbol('t')\n",
+ "xo = a*(sympy.cos(t)+t*sympy.sin(t))\n",
+ "yo = a*(sympy.sin(t)-t*sympy.cos(t))\n",
+ "s = sympy.diff(xo,t)/sympy.diff(yo,t)\n",
+ "y = yo+s*(x-xo)\n",
+ "print \"y = \",y"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.35, page no. 163"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 32,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The two given curves are xˆ=4y and yˆ2=4x which intersects at (0,0) and (4,4)\n",
+ "for (4 ,4)\n",
+ "Angle between them is (radians):−\n",
+ "atan(x/2)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy,math\n",
+ "\n",
+ "print \"The two given curves are xˆ=4y and yˆ2=4x which intersects at (0,0) and (4,4)\"\n",
+ "print \"for (4 ,4)\"\n",
+ "x = 4\n",
+ "x = sympy.Symbol('x')\n",
+ "y1= x**2/4\n",
+ "y2 = 2*x**(1/2)\n",
+ "m1= sympy.diff(y1,x,1)\n",
+ "m2= sympy.diff(y2,x,1)\n",
+ "x = 4\n",
+ "print \"Angle between them is (radians):−\"\n",
+ "t= sympy.atan((m1-m2)/(1+m1*m2)) \n",
+ "print t"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.37, page no. 165"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 50,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "length of tangent\n",
+ "a*(((tan(t/2)**2/2 + 1/2)/tan(t/2) - sin(t))/cos(t) + 1)**0.5*sin(t)\n",
+ "checking for its dependency on t\n",
+ "verified and equal to a\n",
+ "subtangent\n",
+ "a*sin(t)*cos(t)/((tan(t/2)**2/2 + 1/2)/tan(t/2) - sin(t))\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy,math\n",
+ "\n",
+ "a = sympy.Symbol('a')\n",
+ "t = sympy.Symbol('t')\n",
+ "x = a*(sympy.cos(t)+sympy.log(sympy.tan(t/2)))\n",
+ "y = a*sympy.sin(t)\n",
+ "s = sympy.diff(x,t,1)/sympy.diff(y,t,1)\n",
+ "print \"length of tangent\"\n",
+ "l = y*(1+s)**(0.5)\n",
+ "print l\n",
+ "print \"checking for its dependency on t\"\n",
+ "f = 1\n",
+ "t = 0\n",
+ "k = sympy.solve(l)\n",
+ "for i in range(1,11):\n",
+ " t = i\n",
+ " if(sympy.solve(l)!=k):\n",
+ " f = 0\n",
+ "if(f==1):\n",
+ " print \"verified and equal to a\"\n",
+ "print \"subtangent\"\n",
+ "m = y/s\n",
+ "print m"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.39, page no. 168"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 39,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Angle of intersection\n",
+ "point of intersection of r=sint+cost and r=2 sint is t=pi/4\n",
+ "tanu=dQ/dr∗r\n",
+ "The angle at point of intersection in radians is : \n",
+ "atan(2*(-sin(Q) + cos(Q))*sin(Q))\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy,math\n",
+ "\n",
+ "print \"Angle of intersection\"\n",
+ "print \"point of intersection of r=sint+cost and r=2 sint is t=pi/4\"\n",
+ "print \"tanu=dQ/dr∗r\"\n",
+ "Q = sympy.Symbol('Q')\n",
+ "r1 = 2*sympy.sin(Q)\n",
+ "r2 = sympy.sin(Q)+sympy.cos(Q)\n",
+ "u = sympy.atan(r1*sympy.diff(r2,Q,1))\n",
+ "Q = math.pi/4\n",
+ "u = u.evalf()\n",
+ "print \"The angle at point of intersection in radians is : \"\n",
+ "print u"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.41, page no. 170"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 41,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "tanu=dQ/dr∗r\n",
+ "atan(2*a/((-sin(Q) + cos(Q))*(-cos(Q) + 1)))\n",
+ "4.0*a**2*(4.0*a**2/((-sin(Q) + cos(Q))**2*(-cos(Q) + 1.0)**2) + 1.0)**(-0.5)/((-sin(Q) + cos(Q))*(-cos(Q) + 1.0)**2)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy,math\n",
+ "\n",
+ "print \"tanu=dQ/dr∗r\"\n",
+ "Q = sympy.Symbol('Q')\n",
+ "a = sympy.Symbol('a')\n",
+ "r = 2*a/(1-sympy.cos(Q))\n",
+ "u = sympy.atan(r/sympy.diff(r2,Q,1))\n",
+ "u = u.evalf()\n",
+ "print u\n",
+ "p = r*sympy.sin(u)\n",
+ "r = sympy.Symbol('r')\n",
+ "Q = sympy.acos(1-2*a/r)\n",
+ "p = p.evalf()\n",
+ "print p"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.43, page no. 172"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 19,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The radius of curvature is : -(1 + sin(t)**2/(cos(t) + 1)**2)*sin(t)/cos(t)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy,math\n",
+ "\n",
+ "a = sympy.Symbol('a')\n",
+ "t = sympy.Symbol('t')\n",
+ "x = a*(t+sympy.sin(t))\n",
+ "y = a*(1-sympy.cos(t))\n",
+ "s2 = sympy.diff(y,t,2)/sympy.diff(x,t,2)\n",
+ "s1 = sympy.diff(y,t,1)/sympy.diff(x,t,1)\n",
+ "r = (1+s1**2)**(3/2)/s2\n",
+ "print \"The radius of curvature is :\",r"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.46, page no. 176"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 42,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "radius of curvature\n",
+ "(a**2*(-cos(t) + 1.0)**2 + a**2*sin(t)**2)/(a**2*(-cos(t) + 1.0)**2 - a**2*(-cos(t) + 1.0)*sin(t) + 2.0*a**2*sin(t)**2)\n",
+ "Which is proportional to rˆ0.5\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy,math\n",
+ "\n",
+ "print \"radius of curvature\"\n",
+ "a = sympy.Symbol('a')\n",
+ "t = sympy.Symbol('t')\n",
+ "r = a*(1-sympy.cos(t))\n",
+ "r1 = sympy.diff(r,t,1) \n",
+ "l = (r**2+r1**2)**(3/2)/(r**2+2*r1**2-r*r1)\n",
+ "r = sympy.Symbol('r')\n",
+ "t = sympy.acos(1-r/a)\n",
+ "l = l.evalf()\n",
+ "print l\n",
+ "print \"Which is proportional to rˆ0.5\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.47, page no. 177"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 21,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The centre of curvature\n",
+ "the coordinates x,y are resp : \n",
+ "2.0*x*(1 + 1.0*(a*x)**0.5/x)**2 + x\n",
+ "-2.0*x**2*(a*x)**(-0.5)*(1 + 1.0*(a*x)**1.0/x**2) + 2*(a*x)**0.5\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy,math\n",
+ "\n",
+ "print \"The centre of curvature\"\n",
+ "x = sympy.Symbol('x')\n",
+ "a = sympy.Symbol('a')\n",
+ "y = sympy.Symbol('y')\n",
+ "y = 2*(a*x)**0.5\n",
+ "y1 = sympy.diff(y,x,1)\n",
+ "y2 = sympy.diff(y,x,2)\n",
+ "xx = x-y1*(1+y1)**2/y2\n",
+ "yy = y+(1+y1**2)/y2\n",
+ "print \"the coordinates x,y are resp : \"\n",
+ "print xx\n",
+ "print yy"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Examle 4.48, page no. 177"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 22,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "centre of curvature of given cycloid\n",
+ "the coordinates x,y are resp : \n",
+ "a*(t - sin(t)) - (a*sin(t) + 1)**2*sin(t)/cos(t)\n",
+ "a*(-cos(t) + 1) + (a**2*sin(t)**2 + 1)/(a*cos(t))\n",
+ "which another parametric equation of cycloid\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy,math\n",
+ "\n",
+ "print \"centre of curvature of given cycloid\"\n",
+ "a = sympy.Symbol('a')\n",
+ "t = sympy.Symbol('t')\n",
+ "x = a*(t-sympy.sin(t))\n",
+ "y = a*(1-sympy.cos(t))\n",
+ "y1 = sympy.diff(y,t,1)\n",
+ "y2 = sympy.diff(y,t,2)\n",
+ "xx = x-y1*(1+y1)**2/y2 \n",
+ "yy = y+(1+y1**2)/y2\n",
+ "print \"the coordinates x,y are resp : \"\n",
+ "print xx\n",
+ "print yy\n",
+ "print \"which another parametric equation of cycloid\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.52, page no. 180"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 42,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "To find the maxima and minima of given function put f1(x)=0\n",
+ "[-1, 1/2, 1]\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "print 'To find the maxima and minima of given function put f1(x)=0'\n",
+ "x = sympy.Symbol('x')\n",
+ "f=3*x**4-2*x**3-6*x**2+6*x+1\n",
+ "k = sympy.diff(f,x)\n",
+ "x = sympy.poly(0,x)\n",
+ "k = sympy.solve(k)\n",
+ "print k"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.61, page no. 188"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 49,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "to find the assymptote of given curve\n",
+ "assymptotes parallel to x−axis is given by f1=0 where f1 is : \n",
+ "[1, -1, -1, 1, 1, 1]\n",
+ "assymptotes parallel to y−axis is given by f 2=0 and f2 is : \n",
+ "x**2*y**2 - x**2*y - x*y**2 + x + y + 1\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy,math\n",
+ "\n",
+ "print \"to find the assymptote of given curve\"\n",
+ "x = sympy.Symbol('x')\n",
+ "y = sympy.Symbol('y')\n",
+ "f = x**2*y**2-x**2*y-x*y**2+x+y+1\n",
+ "f1 = sympy.poly(f,y,x)\n",
+ "f1 = f1.coeffs()\n",
+ "print \"assymptotes parallel to x−axis is given by f1=0 where f1 is : \"\n",
+ "print sympy.factor(f1)\n",
+ "f2 = sympy.poly(f,y,x)\n",
+ "print \"assymptotes parallel to y−axis is given by f 2=0 and f2 is : \"\n",
+ "print sympy.factor(f2)"
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.10"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter5_2.ipynb b/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter5_2.ipynb
new file mode 100644
index 00000000..2056bd4b
--- /dev/null
+++ b/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter5_2.ipynb
@@ -0,0 +1,317 @@
+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 5 : Partial Differentiation And Its Applications"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "## Example 5.5, page no. 195"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "2*(4*x**2/(x**2 + y**2 + z**2) - 1)/(x**2 + y**2 + z**2)**2 + 2*(4*y**2/(x**2 + y**2 + z**2) - 1)/(x**2 + y**2 + z**2)**2 + 2*(4*z**2/(x**2 + y**2 + z**2) - 1)/(x**2 + y**2 + z**2)**2\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "\n",
+ "x = sympy.Symbol('x')\n",
+ "y = sympy.Symbol('y')\n",
+ "z = sympy.Symbol('z')\n",
+ "v = (x**2+y**2+z**2)**(-1/2)\n",
+ "a = sympy.diff(v,x,2)\n",
+ "b = sympy.diff(v,y,2)\n",
+ "c = sympy.diff(v,z,2)\n",
+ "print a+b+c"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 5.14, page no. 203"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "x*(-0.5*x**(-0.5)*(x + y)/(x**0.5 + y**0.5)**2 + 1/(x**0.5 + y**0.5))/sqrt(1 - (x + y)**2/(x**0.5 + y**0.5)**2) + y*(-0.5*y**(-0.5)*(x + y)/(x**0.5 + y**0.5)**2 + 1/(x**0.5 + y**0.5))/sqrt(1 - (x + y)**2/(x**0.5 + y**0.5)**2)\n",
+ "0\n",
+ "x**2*((0.25*x**(-1.5)*(x + y)/(x**0.5 + y**0.5)**2 + 0.5*x**(-1.0)*(x + y)/(x**0.5 + y**0.5)**3 - 1.0*x**(-0.5)/(x**0.5 + y**0.5)**2)/sqrt(1 - (x + y)**2/(x**0.5 + y**0.5)**2) + (-0.5*x**(-0.5)*(x + y)**2/(x**0.5 + y**0.5)**3 + (2*x + 2*y)/(2*(x**0.5 + y**0.5)**2))*(-0.5*x**(-0.5)*(x + y)/(x**0.5 + y**0.5)**2 + 1/(x**0.5 + y**0.5))/(1 - (x + y)**2/(x**0.5 + y**0.5)**2)**(3/2)) + 2*x*y*((0.5*x**(-0.5)*y**(-0.5)*(x + y)/(x**0.5 + y**0.5)**3 - 0.5*x**(-0.5)/(x**0.5 + y**0.5)**2 - 0.5*y**(-0.5)/(x**0.5 + y**0.5)**2)/sqrt(1 - (x + y)**2/(x**0.5 + y**0.5)**2) + (-0.5*x**(-0.5)*(x + y)**2/(x**0.5 + y**0.5)**3 + (2*x + 2*y)/(2*(x**0.5 + y**0.5)**2))*(-0.5*y**(-0.5)*(x + y)/(x**0.5 + y**0.5)**2 + 1/(x**0.5 + y**0.5))/(1 - (x + y)**2/(x**0.5 + y**0.5)**2)**(3/2)) + y**2*((0.25*y**(-1.5)*(x + y)/(x**0.5 + y**0.5)**2 + 0.5*y**(-1.0)*(x + y)/(x**0.5 + y**0.5)**3 - 1.0*y**(-0.5)/(x**0.5 + y**0.5)**2)/sqrt(1 - (x + y)**2/(x**0.5 + y**0.5)**2) + (-0.5*y**(-0.5)*(x + y)**2/(x**0.5 + y**0.5)**3 + (2*x + 2*y)/(2*(x**0.5 + y**0.5)**2))*(-0.5*y**(-0.5)*(x + y)/(x**0.5 + y**0.5)**2 + 1/(x**0.5 + y**0.5))/(1 - (x + y)**2/(x**0.5 + y**0.5)**2)**(3/2))\n",
+ "-(x + y)*cos(2*asin((x + y)/(x**0.5 + y**0.5)))/(4*(1 - (x + y)**2/(x**0.5 + y**0.5)**2)**(3/2)*(x**0.5 + y**0.5))\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "\n",
+ "x = sympy.Symbol('x')\n",
+ "y = sympy.Symbol('y')\n",
+ "u = sympy.asin((x+y)/(x**0.5+y**0.5))\n",
+ "a = sympy.diff(u,x)\n",
+ "b = sympy.diff(u,y)\n",
+ "c = sympy.diff(a,x)\n",
+ "d = sympy.diff(b,y)\n",
+ "e = sympy.diff(b,x)\n",
+ "print x*a+y*b\n",
+ "print (1/2)*sympy.tan(u)\n",
+ "print (x**2)*c+2*x*y*e+(y**2)*d\n",
+ "print (-sympy.sin(u)*sympy.cos(2*u))/(4*(sympy.cos(u))**3)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 5.25.1, page no. 204"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 15,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "r*sin(l)**2 + r*cos(l)**2\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "\n",
+ "r = sympy.Symbol('r')\n",
+ "l = sympy.Symbol('l')\n",
+ "x = r*sympy.cos(l)\n",
+ "y = r*sympy.sin(l)\n",
+ "a = sympy.diff(x,r)\n",
+ "b = sympy.diff(x,l)\n",
+ "c = sympy.diff(y,r)\n",
+ "d = sympy.diff(y,l)\n",
+ "A = sympy.Matrix([[a,b],[c,d]])\n",
+ "print A.det()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 5.25.2, page no. 204"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 16,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "r*sin(l)**2 + r*cos(l)**2\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "\n",
+ "r = sympy.Symbol('r')\n",
+ "l = sympy.Symbol('l')\n",
+ "z = sympy.Symbol('z')\n",
+ "x = r*sympy.cos(l)\n",
+ "y = r*sympy.sin(l)\n",
+ "m = z\n",
+ "a = sympy.diff(x,r)\n",
+ "b = sympy.diff(x,l)\n",
+ "c = sympy.diff(x,z)\n",
+ "d = sympy.diff(y,r)\n",
+ "e = sympy.diff(y,l)\n",
+ "f = sympy.diff(y,z)\n",
+ "g = sympy.diff(m,r)\n",
+ "h = sympy.diff(m,l)\n",
+ "i = sympy.diff(m,z)\n",
+ "A = sympy.Matrix([[a,b,c],[d,e,f],[g,h,i]])\n",
+ "print A.det()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 5.25.3, page no. 205"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 25,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "r**2*sin(l)**2*sin(m)**3 + r**2*sin(l)**2*sin(m)*cos(m)**2 + r**2*sin(m)**3*cos(l)**2 + r**2*sin(m)*cos(l)**2*cos(m)**2\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy,numpy\n",
+ "\n",
+ "r = sympy.Symbol('r')\n",
+ "l = sympy.Symbol('l')\n",
+ "m = sympy.Symbol('m')\n",
+ "x = r*sympy.cos(l)*sympy.sin(m)\n",
+ "y = r*sympy.sin(l)*sympy.sin(m)\n",
+ "z = r*sympy.cos(m)\n",
+ "a = sympy.diff(x,r)\n",
+ "b = sympy.diff(x,m)\n",
+ "c = sympy.diff(x,l)\n",
+ "d = sympy.diff(y,r)\n",
+ "e = sympy.diff(y,m)\n",
+ "f = sympy.diff(y,l)\n",
+ "g = sympy.diff(z,r)\n",
+ "h = sympy.diff(z,m)\n",
+ "i = sympy.diff(z,l)\n",
+ "A = sympy.Matrix([[a,b,c],[d,e,f],[g,h,i]])\n",
+ "print A.det()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 5.26, page no. 206"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 22,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "4\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy\n",
+ "\n",
+ "x1 = sympy.Symbol('x1')\n",
+ "x2 = sympy.Symbol('x2')\n",
+ "x3 = sympy.Symbol('x3')\n",
+ "y1 =(x2*x3)/x1\n",
+ "y2 =(x3*x1)/x2\n",
+ "y3 =(x1*x2)/x3\n",
+ "a = sympy.diff(y1,x1)\n",
+ "b = sympy.diff(y1,x2)\n",
+ "c = sympy.diff(y1,x3)\n",
+ "d = sympy.diff(y2,x1)\n",
+ "e = sympy.diff(y2,x2)\n",
+ "f = sympy.diff(y2,x3)\n",
+ "g = sympy.diff(y3,x1)\n",
+ "h = sympy.diff(y3,x2)\n",
+ "i = sympy.diff(y3,x3)\n",
+ "A = sympy.Matrix([[a,b,c],[d,e,f],[g,h,i]])\n",
+ "print A.det()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 5.30, page no. 210"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 18,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "-1.0*x*y*(-x**2 + 1)**(-0.5)/sqrt(-y**2 + 1) + 1.0*x*y*(-y**2 + 1)**(-0.5)/sqrt(-x**2 + 1)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy,numpy\n",
+ "\n",
+ "x = sympy.Symbol('x')\n",
+ "y = sympy.Symbol('y')\n",
+ "u = x*(1-y**2)**0.5+y*(1-x**2)**0.5\n",
+ "v = sympy.asin(x)+sympy.asin(y)\n",
+ "a = sympy.diff(u,x)\n",
+ "b = sympy.diff(u,y)\n",
+ "c = sympy.diff(v,x)\n",
+ "d = sympy.diff(v,y)\n",
+ "A = sympy.Matrix([[a,b],[c,d]])\n",
+ "print A.det()"
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.10"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter6_2.ipynb b/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter6_2.ipynb
new file mode 100644
index 00000000..4eb64354
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+++ b/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter6_2.ipynb
@@ -0,0 +1,743 @@
+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 6 : Integration And Its Applications"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6.1.1, page no. 199"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 37,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Indefinite integral\n",
+ "3*x/8 - sin(x)**3*cos(x)/4 - 3*sin(x)*cos(x)/8\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy,numpy\n",
+ "\n",
+ "print \"Indefinite integral\"\n",
+ "x = sympy.Symbol('x')\n",
+ "f = sympy.integrate((sympy.sin(x))**4,(x))\n",
+ "print f"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6.1.2, page no. 200"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 39,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Indefinite integral\n",
+ "-sin(x)**7/7 + 3*sin(x)**5/5 - sin(x)**3 + sin(x)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy,numpy\n",
+ "\n",
+ "print \"Indefinite integral\"\n",
+ "x = sympy.Symbol('x')\n",
+ "f = sympy.integrate((sympy.cos(x))**7,(x))\n",
+ "print f"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6.2.1, page no. 202"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 41,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Definite integral\n",
+ "0.490873852123\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy,numpy,math\n",
+ "\n",
+ "print \"Definite integral\"\n",
+ "x = sympy.Symbol('x')\n",
+ "f = sympy.integrate((sympy.cos(x))**6,(x,0,math.pi/2))\n",
+ "print float(f)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6.2.2, page no. 202"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 43,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Definite integral\n",
+ "zoo*a**6 + a**6*log(-2*a**2)/4 - 11*a**6/24\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy,numpy,math\n",
+ "\n",
+ "print \"Definite integral\"\n",
+ "x = sympy.Symbol('x')\n",
+ "a = sympy.Symbol('a')\n",
+ "g = x**7/(a**2-x**2)**1/2\n",
+ "f = sympy.integrate(g,(x,0,a))\n",
+ "print f"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6.2.3, page no. 203"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 45,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Definite integral\n",
+ "4*a**4\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy,numpy,math\n",
+ "\n",
+ "print \"Definite integral\"\n",
+ "x = sympy.Symbol('x')\n",
+ "a = sympy.Symbol('a')\n",
+ "g = x**3*(2*a*x-x**2)**(1/2) \n",
+ "f = sympy.integrate(g,(x,0,2*a))\n",
+ "print f"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6.2.4, page no. 204"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 23,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Definite integral\n",
+ "0.5*(a**2 + 0.5625)**(-n) + 0.5*(a**2 + 0.0625)**(-n)\n"
+ ]
+ }
+ ],
+ "source": [
+ "from sympy.abc import x,a,n\n",
+ "from sympy.integrals import Integral\n",
+ "print \"Definite integral\"\n",
+ "g = 1./(a**2+x**2)**n \n",
+ "f = Integral(g,(x,0,1))\n",
+ "print f.as_sum(2).n()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6.4.1, page no. 207"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Definite integral\n",
+ "0.0532891345937363\n"
+ ]
+ }
+ ],
+ "source": [
+ "from sympy.abc import x,a,n\n",
+ "from sympy.integrals import Integral\n",
+ "import sympy, math\n",
+ "print \"Definite integral\"\n",
+ "g = (sympy.sin(6*x))**3*(sympy.cos(3*x))**7\n",
+ "f = Integral(g,(x,0,math.pi/6))\n",
+ "print f.as_sum(2).n()"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6.4.2, page no. 208"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 52,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Definite integral\n",
+ "0.0571428571429\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy,numpy,math\n",
+ "\n",
+ "print \"Definite integral\"\n",
+ "x = sympy.Symbol('x')\n",
+ "g = (sympy.sin(6*x))**3*(sympy.cos(3*x))**7\n",
+ "g = x**4*(1-x**2)**(3/2)\n",
+ "f = sympy.integrate(g,(x,0,1)) \n",
+ "print float(f)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6.5, page no. 208"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 63,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Definite integral\n",
+ "Enter n : 0\n",
+ "Enter m : 1\n",
+ "1.0\n",
+ "1.0\n",
+ "Equal\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy,numpy,math\n",
+ "\n",
+ "print \"Definite integral\"\n",
+ "x = sympy.Symbol('x')\n",
+ "m = sympy.Symbol('m')\n",
+ "n = sympy.Symbol('n')\n",
+ "n = int(raw_input(\"Enter n : \"))\n",
+ "m = int(raw_input(\"Enter m : \"))\n",
+ "g =(sympy.cos(x))**m*sympy.cos(n*x)\n",
+ "f = sympy.integrate(g,(x,0,math.pi/2))\n",
+ "print float(f) \n",
+ "g2 =(sympy.cos(x))**(m-1)*sympy.cos((n-1)*x)\n",
+ "f2 = m/(m+n)*sympy.integrate(g2,(x,0,math.pi/2))\n",
+ "print float(f2)\n",
+ "print \"Equal\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6.6.1, page no. 210"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 65,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Definite integral\n",
+ "Enter n : 1\n",
+ "Piecewise((x*exp(-I*x)*sin(x)/2 - I*x*exp(-I*x)*cos(x)/2 - exp(-I*x)*cos(x)/2, a == -I), (x*exp(I*x)*sin(x)/2 + I*x*exp(I*x)*cos(x)/2 - exp(I*x)*cos(x)/2, a == I), (a*exp(a*x)*sin(x)/(a**2 + 1) - exp(a*x)*cos(x)/(a**2 + 1), True))\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy,numpy,math\n",
+ "\n",
+ "print \"Definite integral\"\n",
+ "x = sympy.Symbol('x')\n",
+ "a = sympy.Symbol('a')\n",
+ "n = int(raw_input(\"Enter n : \"))\n",
+ "g = sympy.exp(a*x)*(sympy.sin(x))**n\n",
+ "f = sympy.integrate(g,(x))\n",
+ "print f"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6.7.1, page no. 212"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 66,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "(4*sin(x)**2 - 3)/(4*sin(x)**4 - 8*sin(x)**2 + 4) - log(sin(x)**2 - 1)/2\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy,numpy,math\n",
+ "\n",
+ "x = sympy.Symbol('x')\n",
+ "print sympy.integrate(sympy.tan(x)**5,(x))"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6.8, page no. 215"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 11,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Enter the value of n : 3\n",
+ "n(p+q)= 0.999999999999999\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy,math\n",
+ "\n",
+ "n = int(raw_input(\"Enter the value of n : \"))\n",
+ "p = sympy.integrate((sympy.tan(x))**(n-1),('x',0,math.pi/4))\n",
+ "q = sympy.integrate((sympy.tan(x))**(n+1),('x',0,math.pi/4))\n",
+ "print \"n(p+q)= \",\n",
+ "print n*(p+q)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6.9.1, page no. 217"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 69,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "1.33333333333333\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy,numpy,math\n",
+ "\n",
+ "x = sympy.Symbol('x')\n",
+ "g = sympy.sec(x)**4\n",
+ "print sympy.integrate(g,('x',0,math.pi/4))"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6.9.2, page no. 217"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 36,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "0.607986405500361\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy,numpy,math\n",
+ "\n",
+ "x = sympy.Symbol('x')\n",
+ "print sympy.integrate('1/sin(x)**3',('x',math.pi/3,math.pi/2))"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6.10, page no. 220"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 20,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "0.000362418686373\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy,math\n",
+ "from sympy.abc import x\n",
+ "from sympy.integrals import Integral\n",
+ "import sympy, math\n",
+ "g = x*sympy.sin(x)**6*sympy.cos(x)**4\n",
+ "f = Integral(g,(x,0,math.pi/6))\n",
+ "print float(f)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6.12, page no. 221"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 17,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "0.785398163397\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy,math\n",
+ "from sympy.abc import x\n",
+ "from sympy.integrals import Integral\n",
+ "x = sympy.Symbol('x')\n",
+ "a = math.pi/2\n",
+ "f = (sympy.sin(x)**0.5)/(sympy.sin(x)**0.5+sympy.cos(x)**0.5)\n",
+ "f = Integral(f,(x, 0, a))\n",
+ "print float(f)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6.13, page no. 223"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The summation is equivalent to integration of 1/(1+xˆ2) from 0 to 1\n",
+ "0.785398163397\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy,numpy,math\n",
+ "\n",
+ "x = sympy.Symbol('x')\n",
+ "print \"The summation is equivalent to integration of 1/(1+xˆ2) from 0 to 1\"\n",
+ "g = 1/(1+x**2)\n",
+ "f = sympy.integrate(g,(x,0,1))\n",
+ "print float(f)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6.14, page no. 223"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The summation is equivalent to integration of log(1+x) from 0 to 1\n",
+ "0.38629436112\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy,numpy,math\n",
+ "\n",
+ "x = sympy.Symbol('x')\n",
+ "print \"The summation is equivalent to integration of log(1+x) from 0 to 1\"\n",
+ "g = sympy.log(1+x)\n",
+ "f = sympy.integrate(g,(x,0,1))\n",
+ "print float(f) "
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6.15, page no. 225"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 31,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "0.0337339994178\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy,math\n",
+ "from sympy.integrals import Integral\n",
+ "f = Integral(x*sympy.sin(x)**8*sympy.cos(x)**4,(x,0,math.pi))\n",
+ "print float(f)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6.16, page no. 226"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 30,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "-1.08879304515\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy,math\n",
+ "from sympy.integrals import Integral\n",
+ "x = sympy.Symbol('x')\n",
+ "#f = sympy.integrate(sympy.log(sympy.sin(x)),('x',0,math.pi/2))\n",
+ "f = Integral(sympy.log(sympy.sin(x)), (x, 0, 1.5707963267949))\n",
+ "print float(f)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6.24, page no. 234"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 20,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "from the graph, it is clear that the points of intersection are x=−4 and x =8.\n",
+ "So, our region of integration is from x=−4 to x=8\n",
+ "36\n"
+ ]
+ },
+ {
+ "data": {
+ "image/png": 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Y3TfzYs6uOUzZMoUxXcdQq3wtIHwHevPFha4ABgJ/U9Ut+blD993EN6vqPwBU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\nlWV1AAAAAElFTkSuQmCC\n",
+ "text/plain": [
+ "<matplotlib.figure.Figure at 0x7faae1cad790>"
+ ]
+ },
+ "metadata": {},
+ "output_type": "display_data"
+ }
+ ],
+ "source": [
+ "%matplotlib inline\n",
+ "import matplotlib.pyplot as plt\n",
+ "import numpy\n",
+ "import math\n",
+ "\n",
+ "x = numpy.linspace(-5,10,70)\n",
+ "y1 = (x+8)/2\n",
+ "y2 = x**2/8\n",
+ "\n",
+ "plt.xlabel('X axis')\n",
+ "plt.ylabel('Y axis')\n",
+ "plt.title('My Graph')\n",
+ "plt.plot(x, y1, \"o-\" )\n",
+ "plt.plot(x, y2, \"+-\" )\n",
+ "plt.legend([\"(x+8)/2\",\"x**2/8\"])\n",
+ "print \"from the graph, it is clear that the points of intersection are x=−4 and x =8.\"\n",
+ "print \"So, our region of integration is from x=−4 to x=8\"\n",
+ "print sympy.integrate('(x+8)/2-x**2/8',('x',-4,8))"
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.11+"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter9_2.ipynb b/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter9_2.ipynb
new file mode 100644
index 00000000..735ec6e3
--- /dev/null
+++ b/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter9_2.ipynb
@@ -0,0 +1,476 @@
+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 9 : Infinite Series"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 9.1, page no. 302"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "1/3\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy\n",
+ "\n",
+ "n = sympy.Symbol('n')\n",
+ "f = ((1/n)**2-2*(1/n))/(3*(1/n)**2+(1/n))\n",
+ "print sympy.limit(f,n,0)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 9.1.3, page no. 303"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 17,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "4\n"
+ ]
+ }
+ ],
+ "source": [
+ "import sympy\n",
+ "n = sympy.Symbol('n')\n",
+ "f = 3+(-1)**n\n",
+ "print sympy.limit(f,n,100)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 9.2.1, page no. 304"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "1+2+3+4+5+6+7+....+n + . . . . . = oo\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy\n",
+ "\n",
+ "n = sympy.Symbol('n')\n",
+ "print \"1+2+3+4+5+6+7+....+n + . . . . . = \",\n",
+ "p = 1/n*(1/n+1)/2\n",
+ "print sympy.limit(p,n,0)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 9.2.2, page no. 304"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "5−4−1+5−4−1+5−4−1+5−4−1+.........=0,5,1 according to the no. of terms.\n",
+ "clearly, in this case sum doesnt tend to a unique limit. hence, series is oscillatory.\n"
+ ]
+ }
+ ],
+ "source": [
+ "print \"5−4−1+5−4−1+5−4−1+5−4−1+.........=0,5,1 according to the no. of terms.\"\n",
+ "print \"clearly, in this case sum doesnt tend to a unique limit. hence, series is oscillatory.\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 9.5.1, page no. 308"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "2\n",
+ "both u and v converge and diverge together, hence u is convergent\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy\n",
+ "\n",
+ "n = sympy.Symbol('n')\n",
+ "v = 1/((1/n)**2)\n",
+ "u = (2/n-1)/(1/n*(1/n +1)*(1/n +2))\n",
+ "print sympy.limit(u/v,n,0)\n",
+ "print \"both u and v converge and diverge together, hence u is convergent\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 9.5.2, page no. 308"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "oo\n",
+ "both u and v converge and diverge together, hence u is divergent\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy\n",
+ "\n",
+ "n = sympy.Symbol('n')\n",
+ "v = 1/((1/n)**2)\n",
+ "u = ((1/n)**2)/((3/n+1)*(3/n+4)*(3/n+7))\n",
+ "print sympy.limit(u/v,n,0)\n",
+ "print \"both u and v converge and diverge together, hence u is divergent\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 9.7.1, page no. 312"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "u=((n+1)^0.5−1)/((n+2)^3−1)=>\n",
+ "0\n",
+ "since, v is convergent, so u is also conzavergent.\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy\n",
+ "\n",
+ "n = sympy.Symbol('n')\n",
+ "print \"u=((n+1)^0.5−1)/((n+2)^3−1)=>\"\n",
+ "u = ((1+1/(1/n))-(1/n)**(-0.5))/(((1/n)**5/2)*((1+2/(1/n))**3-(1/n)**(-3)))\n",
+ "v = (1/n)**(-5/2)\n",
+ "print sympy.limit(u/v,n,0)\n",
+ "print 'since, v is convergent, so u is also conzavergent.'"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 9.7.3, page no. 313"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 11,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "-log(log(2)) + +inf\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy\n",
+ "\n",
+ "n = sympy.Symbol('n')\n",
+ "print sympy.integrate(1/(n*sympy.log(n)),(n,2,numpy.inf))"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 9.8.1, page no. 314"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "x**(-2)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy\n",
+ "\n",
+ "n = sympy.Symbol('n')\n",
+ "x = sympy.Symbol('x')\n",
+ "u = (x**(2*(1/n)-2))/(((1/n)+1)*(1/n)**0.5)\n",
+ "v = (x**(2*(1/n)))/((1/n+2)*(1/n+1)**0.5)\n",
+ "print sympy.limit(u/v,n,0)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 9.8.2, page no. 314"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "1/x\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy\n",
+ "\n",
+ "n = sympy.Symbol('n')\n",
+ "x = sympy.Symbol('x')\n",
+ "u = ((2**(1/n)-2)*(x**(1/n-1)))/(2**(1/n)+1)\n",
+ "v = ((2**((1/n)+1)-2)*(x**(1/n)))/(2**(1/n+1)+1)\n",
+ "print sympy.limit(u/v,n,0)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 9.10.1, page no. 316"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "(x + 1)/(2*x)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy\n",
+ "\n",
+ "n = sympy.Symbol('n')\n",
+ "x = sympy.Symbol('x')\n",
+ "u = 1/(1+x**(-n))\n",
+ "v = 1/(1+x**(-n-1))\n",
+ "print sympy.limit(u/v,n,0)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 9.10.2, page no. 316"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "1\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy\n",
+ "\n",
+ "n = sympy.Symbol('n')\n",
+ "a = sympy.Symbol('a')\n",
+ "b = sympy.Symbol('b')\n",
+ "l = (b+1/n)/(a+1/n)\n",
+ "print sympy.limit(l,n,0)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 9.11.1, page no. 317"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 9,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "u=((4.7....(3n+1))∗xˆn)/(1.2.....n)\n",
+ "v=((4.7....(3n+4)∗xˆ(n+1))/(1.2.....(n+1))\n",
+ "l=u/v=> 1/(3*x)\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy\n",
+ "\n",
+ "n = sympy.Symbol('n')\n",
+ "x = sympy.Symbol('x')\n",
+ "print \"u=((4.7....(3n+1))∗xˆn)/(1.2.....n)\"\n",
+ "print \"v=((4.7....(3n+4)∗xˆ(n+1))/(1.2.....(n+1))\"\n",
+ "print \"l=u/v=>\",\n",
+ "l = (1+n)/((3+4*n)*x)\n",
+ "print sympy.limit(l,n,0)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 9.11.2, page no. 318"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 12,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "4/x**2\n"
+ ]
+ }
+ ],
+ "source": [
+ "import numpy,sympy,math\n",
+ "\n",
+ "n = sympy.Symbol('n')\n",
+ "x = sympy.Symbol('x')\n",
+ "u = (((sympy.factorial(n))**2)*x**(2*n))/sympy.factorial(2*n)\n",
+ "v = (((sympy.factorial(n+1))**2)*x**(2*(n+1)))/sympy.factorial(2*(n+1))\n",
+ "print sympy.limit(u/v,n,numpy.inf )"
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.10"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
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