diff options
.png?id=c00ea066e3f752b36e9afb0ced7084b71814c89e){kind=link}
.png?id=c00ea066e3f752b36e9afb0ced7084b71814c89e){kind=link}
.png?id=c00ea066e3f752b36e9afb0ced7084b71814c89e){kind=link}
25 files changed, 12762 insertions, 0 deletions
diff --git a/Chemical_Engineering_Thermodynamics___by_S._Sundaram/README.txt b/Chemical_Engineering_Thermodynamics___by_S._Sundaram/README.txt new file mode 100644 index 00000000..49744716 --- /dev/null +++ b/Chemical_Engineering_Thermodynamics___by_S._Sundaram/README.txt @@ -0,0 +1,10 @@ +Contributed By: Suraj Ahir +Course: btech +College/Institute/Organization: Freelancer +Department/Designation: IT +Book Title: Chemical Engineering Thermodynamics +Author: S. Sundaram +Publisher: R. N. Ahuja Book Company, New Delhi +Year of publication: 1998 +Isbn: 81-7619-001-2 +Edition: 1
\ No newline at end of file diff --git a/Electronic_Measurements_and_Instrumentation_by_Er.R.K.Rajput/README.txt b/Electronic_Measurements_and_Instrumentation_by_Er.R.K.Rajput/README.txt new file mode 100644 index 00000000..58a7e483 --- /dev/null +++ b/Electronic_Measurements_and_Instrumentation_by_Er.R.K.Rajput/README.txt @@ -0,0 +1,10 @@ +Contributed By: Gundla Keerthi vani +Course: be +College/Institute/Organization: Panzer technologies +Department/Designation: Electronics and communication +Book Title: Electronic Measurements and Instrumentation +Author: Er.R.K.Rajput +Publisher: S.Chand +Year of publication: 2015 +Isbn: 81-219-2917-2 +Edition: Third edition
\ No newline at end of file diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/Chapter9_5.ipynb b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/Chapter9_5.ipynb new file mode 100644 index 00000000..5520ce18 --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/Chapter9_5.ipynb @@ -0,0 +1,419 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:36c7c29426c0c7be14ce2ec6430ec42fa1b0070a30e5ffbe18ec0fb6fd9b8d34"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter09:Numerical Solution of Partial Differential Equations"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9.1:pg-350"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#standard five point formula\n",
+ "#example 9.1\n",
+ "#page 350\n",
+ "\n",
+ "u2=5.0;u3=1.0;\n",
+ "for i in range(0,3):\n",
+ " u1=(u2+u3+6.0)/4.0\n",
+ " u2=(u1/2.0)+(5.0/2.0)\n",
+ " u3=(u1/2.0)+(1.0/2.0)\n",
+ " print\" the values are u1=%d\\t u2=%d\\t u3=%d\\t\\n\\n\" %(u1,u2,u3)\n",
+ " \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " the values are u1=3\t u2=4\t u3=2\t\n",
+ "\n",
+ "\n",
+ " the values are u1=3\t u2=4\t u3=2\t\n",
+ "\n",
+ "\n",
+ " the values are u1=3\t u2=4\t u3=2\t\n",
+ "\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9.2:pg-351"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#solution of laplace equation by jacobi method,gauss-seidel method and SOR method\n",
+ "#example 9.2\n",
+ "#page 351\n",
+ "u1=0.25\n",
+ "u2=0.25\n",
+ "u3=0.5\n",
+ "u4=0.5 #initial values\n",
+ "print \"jacobis iteration process\\n\\n\"\n",
+ "print\"u1\\t u2\\t u3\\t u4\\t \\n\\n\"\n",
+ "print \"%f\\t %f\\t %f\\t %f\\t \\n\" %(u1,u2,u3,u4)\n",
+ "for i in range(0,7):\n",
+ " u11=(0+u2+0+u4)/4\n",
+ " u22=(u1+0+0+u3)/4\n",
+ " u33=(1+u2+0+u4)/4\n",
+ " u44=(1+0+u3+u1)/4\n",
+ " u1=u11\n",
+ " u2=u22\n",
+ " u3=u33\n",
+ " u4=u44\n",
+ " print \"%f\\t %f\\t %f\\t %f\\t \\n\" %(u11,u22,u33,u44) \n",
+ "print \" gauss seidel process\\n\\n\"\n",
+ "u1=0.25\n",
+ "u2=0.3125\n",
+ "u3=0.5625\n",
+ "u4=0.46875 #initial values\n",
+ "print \"u1\\t u2\\t u3\\t u4\\t \\n\\n\"\n",
+ "print \"%f\\t %f\\t %f\\t %f\\t \\n\" %(u1,u2,u3,u4)\n",
+ "for i in range(0,4):\n",
+ "\n",
+ " u1=(0.0+u2+0.0+u4)/4.0\n",
+ " u2=(u1+0.0+0.0+u3)/4.0\n",
+ " u3=(1.0+u2+0.0+u4)/4.0\n",
+ " u4=(1.0+0.0+u3+u1)/4.0\n",
+ " print \"%f\\t %f\\t %f\\t %f\\t \\n\" %(u1,u2,u3,u4) \n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "jacobis iteration process\n",
+ "\n",
+ "\n",
+ "u1\t u2\t u3\t u4\t \n",
+ "\n",
+ "\n",
+ "0.250000\t 0.250000\t 0.500000\t 0.500000\t \n",
+ "\n",
+ "0.187500\t 0.187500\t 0.437500\t 0.437500\t \n",
+ "\n",
+ "0.156250\t 0.156250\t 0.406250\t 0.406250\t \n",
+ "\n",
+ "0.140625\t 0.140625\t 0.390625\t 0.390625\t \n",
+ "\n",
+ "0.132812\t 0.132812\t 0.382812\t 0.382812\t \n",
+ "\n",
+ "0.128906\t 0.128906\t 0.378906\t 0.378906\t \n",
+ "\n",
+ "0.126953\t 0.126953\t 0.376953\t 0.376953\t \n",
+ "\n",
+ "0.125977\t 0.125977\t 0.375977\t 0.375977\t \n",
+ "\n",
+ " gauss seidel process\n",
+ "\n",
+ "\n",
+ "u1\t u2\t u3\t u4\t \n",
+ "\n",
+ "\n",
+ "0.250000\t 0.312500\t 0.562500\t 0.468750\t \n",
+ "\n",
+ "0.195312\t 0.189453\t 0.414551\t 0.402466\t \n",
+ "\n",
+ "0.147980\t 0.140633\t 0.385775\t 0.383439\t \n",
+ "\n",
+ "0.131018\t 0.129198\t 0.378159\t 0.377294\t \n",
+ "\n",
+ "0.126623\t 0.126196\t 0.375872\t 0.375624\t \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9.4:pg-354"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#poisson equation\n",
+ "#exaample 9.4\n",
+ "#page 354\n",
+ "u2=0.0;u4=0.0;\n",
+ "print \" u1\\t u2\\t u3\\t u4\\t\\n\\n\"\n",
+ "for i in range(0,6):\n",
+ " u1=(u2/2.0)+30.0\n",
+ " u2=(u1+u4+150.0)/4.0\n",
+ " u4=(u2/2.0)+45.0\n",
+ " print \"%0.2f\\t %0.2f\\t %0.2f\\t %0.2f\\n\" %(u1,u2,u2,u4)\n",
+ "print \" from last two iterates we conclude u1=67 u2=75 u3=75 u4=83\\n\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " u1\t u2\t u3\t u4\t\n",
+ "\n",
+ "\n",
+ "30.00\t 45.00\t 45.00\t 67.50\n",
+ "\n",
+ "52.50\t 67.50\t 67.50\t 78.75\n",
+ "\n",
+ "63.75\t 73.12\t 73.12\t 81.56\n",
+ "\n",
+ "66.56\t 74.53\t 74.53\t 82.27\n",
+ "\n",
+ "67.27\t 74.88\t 74.88\t 82.44\n",
+ "\n",
+ "67.44\t 74.97\t 74.97\t 82.49\n",
+ "\n",
+ " from last two iterates we conclude u1=67 u2=75 u3=75 u4=83\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 59
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9.6:pg-362"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#bender-schmidt formula\n",
+ "#example 9.6\n",
+ "#page 362\n",
+ "def f(x):\n",
+ " return (4*x)-(x*x)\n",
+ "#u=[f(0),f(1),f(2),f(3),f(4)]\n",
+ "u1=f(0);u2=f(1);u3=f(2);u4=f(3);u5=f(4);\n",
+ "u11=(u1+u3)/2\n",
+ "u12=(u2+u4)/2\n",
+ "u13=(u3+u5)/2\n",
+ "print \"u11=%0.2f\\t u12=%0.2f\\t u13=%0.2f\\t \\n\" %(u11,u12,u13)\n",
+ "u21=(u1+u12)/2.0\n",
+ "u22=(u11+u13)/2.0\n",
+ "u23=(u12+0)/2.0\n",
+ "print \"u21=%0.2f\\t u22=%0.2f\\t u23=%0.2f\\t \\n\" %(u21,u22,u23)\n",
+ "u31=(u1+u22)/2.0\n",
+ "u32=(u21+u23)/2.0\n",
+ "u33=(u22+u1)/2.0\n",
+ "print \"u31=%0.2f\\t u32=%0.2f\\t u33=%0.2f\\t \\n\" % (u31,u32,u33)\n",
+ "u41=(u1+u32)/2.0\n",
+ "u42=(u31+u33)/2.0\n",
+ "u43=(u32+u1)/2.0\n",
+ "print \"u41=%0.2f\\t u42=%0.2f\\t u43=%0.2f\\t \\n\" % (u41,u42,u43)\n",
+ "u51=(u1+u42)/2.0\n",
+ "u52=(u41+u43)/2.0\n",
+ "u53=(u42+u1)/2.0\n",
+ "print \"u51=%0.2f\\t u52=%0.2f\\t u53=%0.2f\\t \\n\" % (u51,u52,u53)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "u11=2.00\t u12=3.00\t u13=2.00\t \n",
+ "\n",
+ "u21=1.50\t u22=2.00\t u23=1.50\t \n",
+ "\n",
+ "u31=1.00\t u32=1.50\t u33=1.00\t \n",
+ "\n",
+ "u41=0.75\t u42=1.00\t u43=0.75\t \n",
+ "\n",
+ "u51=0.50\t u52=0.75\t u53=0.50\t \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 77
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9.7:pg-363"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "#bender-schimdt's formula and crank-nicolson formula\n",
+ "#example 9.7\n",
+ "#page 363\n",
+ "#bender -schimdt's formula\n",
+ "import math\n",
+ "from numpy import matrix\n",
+ "z=math.pi\n",
+ "def f(x,t):\n",
+ " return math.exp(z*z*t*-1)*sin(z*x)\n",
+ "#u=[f(0,0),f(0.2,0),f(0.4,0),f(0.6,0),f(0.8,0),f(1,0)]\n",
+ "u1=f(0,0)\n",
+ "u2=f(0.2,0)\n",
+ "u3=f(0.4,0)\n",
+ "u4=f(0.6,0)\n",
+ "u5=f(0.8,0)\n",
+ "u6=f(1.0,0)\n",
+ "u11=u3/2\n",
+ "u12=(u2+u4)/2\n",
+ "u13=u12\n",
+ "u14=u11\n",
+ "print \"u11=%f\\t u12=%f\\t u13=%f\\t u14=%f\\n\\n\" % (u11,u12,u13,u14)\n",
+ "u21=u12/2\n",
+ "u22=(u12+u14)/2\n",
+ "u23=u22\n",
+ "u24=u21\n",
+ "print \"u21=%f\\t u22=%f\\t u23=%f\\t u24=%f\\n\\n\" % (u21,u22,u23,u24)\n",
+ "print \"the error in the solution is: %f\\n\\n\" % (math.fabs(u22-f(0.6,0.04)))\n",
+ "#crank-nicolson formula\n",
+ "#by putting i=1,2,3,4 we obtain four equation\n",
+ "A=matrix([[4, -1, 0, 0] ,[-1, 4, -1, 0],[0, -1, 4, -1],[0, 0, -1, 4]])\n",
+ "C=matrix([[0.9510],[1.5388],[1.5388],[0.9510]])\n",
+ "X=A.I*C\n",
+ "print \"u00=%f\\t u10=%f\\t u20=%f\\t u30=%f\\t\\n\\n\" %(X[0][0],X[1][0],X[2][0],X[3][0])\n",
+ "print \"the error in the solution is: %f\\n\\n\" %(abs(X[1][0]-f(0.6,0.04)))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "u11=0.475528\t u12=0.769421\t u13=0.769421\t u14=0.475528\n",
+ "\n",
+ "\n",
+ "u21=0.384710\t u22=0.622475\t u23=0.622475\t u24=0.384710\n",
+ "\n",
+ "\n",
+ "the error in the solution is: 0.018372\n",
+ "\n",
+ "\n",
+ "u00=0.399255\t u10=0.646018\t u20=0.646018\t u30=0.399255\t\n",
+ "\n",
+ "\n",
+ "the error in the solution is: 0.005172\n",
+ "\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9.8:pg-364"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "#heat equation using crank-nicolson method\n",
+ "#example 9.8\n",
+ "#page 364\n",
+ "from numpy import matrix\n",
+ "import math\n",
+ "z=0.01878\n",
+ "#h=1/2 l=1/8,i=1\n",
+ "u01=0.0\n",
+ "u21=1.0/8.0\n",
+ "u11=(u21+u01)/6.0\n",
+ "print \" u11=%f\\n\\n\" % (u11)\n",
+ "print \"error is %f\\n\\n\" % (math.fabs(u11-z))\n",
+ "#h=1/4,l=1/8,i=1,2,3\n",
+ "A=matrix([[-3.0 ,-1.0 ,0.0],[1.0,-3.0,1.0],[0.0,1.0,-3.0]])\n",
+ "C=matrix([[0.0],[0.0],[-0.125]])\n",
+ "#here we found inverese of A then we multipy it with C\n",
+ "X=A.I*C\n",
+ "print \"u12=%f\\n\\n\" % (X[1][0])\n",
+ "print \"error is %f\\n\\n\" %(math.fabs(X[1][0]-z))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " u11=0.020833\n",
+ "\n",
+ "\n",
+ "error is 0.002053\n",
+ "\n",
+ "\n",
+ "u12=0.013889\n",
+ "\n"
+ ]
+ },
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "error is 0.004891\n",
+ "\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter1_5.ipynb b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter1_5.ipynb new file mode 100644 index 00000000..5fb2efd4 --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter1_5.ipynb @@ -0,0 +1,625 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:06d397f5a6c9d8cabedb251742216065689d5c177feee6d0eb9025d9c5dbce26"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter01:Errors in Numerical Calculations"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.1:pg-7"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 1.1\n",
+ "#rounding off\n",
+ "#page 7\n",
+ "a1=1.6583\n",
+ "a2=30.0567\n",
+ "a3=0.859378\n",
+ "a4=3.14159\n",
+ "print \"\\nthe numbers after rounding to 4 significant figures are given below\\n\"\n",
+ "print \" %f %.4g\\n'\" %(a1,a1)\n",
+ "print \" %f %.4g\\n\" %(a2,a2)\n",
+ "print \" %f %.4g\\n\" %(a3,a3)\n",
+ "print \" %f %.4g\\n\" %(a4,a4)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "the numbers after rounding to 4 significant figures are given below\n",
+ "\n",
+ " 1.658300 1.658\n",
+ "'\n",
+ " 30.056700 30.06\n",
+ "\n",
+ " 0.859378 0.8594\n",
+ "\n",
+ " 3.141590 3.142\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.2:pg-9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 1.2\n",
+ "#percentage accuracy\n",
+ "#page 9\n",
+ "import math\n",
+ "x=0.51 # the number given\n",
+ "n=2 #correcting upto 2 decimal places\n",
+ "d=math.pow(10,-n)\n",
+ "d=d/2.0\n",
+ "p=(d/x)*100 #percentage accuracy\n",
+ "print \"the percentage accuracy of %f after correcting to two decimal places is %f\" %(x,p)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the percentage accuracy of 0.510000 after correcting to two decimal places is 0.980392\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.3:pg-9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 1.3\n",
+ "#absolute and relative errors\n",
+ "#page 9\n",
+ "X=3.1428571 #approximate value of pi\n",
+ "T_X=3.1415926 # true value of pi\n",
+ "A_E=T_X-X #absolute error\n",
+ "R_E=A_E/T_X #relative error\n",
+ "print \"Absolute Error = %0.7f \\n Relative Error = %0.7f\" %(A_E,R_E)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Absolute Error = -0.0012645 \n",
+ " Relative Error = -0.0004025\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.4:pg-10"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 1.4\n",
+ "#best approximation\n",
+ "#page 10\n",
+ "X=1/3 #the actual number\n",
+ "X1=0.30\n",
+ "X2=0.33\n",
+ "X3=0.34\n",
+ "E1=abs(X-X1)\n",
+ "E2=abs(X-X2)\n",
+ "E3=abs(X-X3)\n",
+ "if E1<E2:\n",
+ " if E1<E3:\n",
+ " B_A=X1\n",
+ "elif E2<E1:\n",
+ " if E2<E3:\n",
+ " B_A=X2\n",
+ "elif E3<E2:\n",
+ " if E3<E1:\n",
+ " B_A=X3\n",
+ "print \"the best approximation of 1/3 is %f\" %(B_A)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the best approximation of 1/3 is 0.300000\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.5:pg-10"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#relative error\n",
+ "#example 1.5\n",
+ "#page 10\n",
+ "import math\n",
+ "n=8.6 # the corrected number\n",
+ "N=1 #the no is rounded to one decimal places\n",
+ "E_A=math.pow(10,-N)/2\n",
+ "E_R=E_A/n\n",
+ "print \"the relative error of the number is:%0.4f\" %(E_R)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the relative error of the number is:0.0058\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.6:pg-10"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 1.6\n",
+ "#absolute error and relative error\n",
+ "#page 10\n",
+ "import math\n",
+ "s=math.sqrt(3)+math.sqrt(5)+math.sqrt(7) #the sum square root of 3,5,7\n",
+ "n=4\n",
+ "Ea=3*(math.pow(10,-n)/2) #absolute error\n",
+ "R_E=Ea/s\n",
+ "print \"the sum of square roots is %0.4g \\n\" %(s)\n",
+ "print \"the absolute error is %f \\n\" %(Ea)\n",
+ "print \"the relative error is %f\" %(R_E)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the sum of square roots is 6.614 \n",
+ "\n",
+ "the absolute error is 0.000150 \n",
+ "\n",
+ "the relative error is 0.000023\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.7:pg-10"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#absolute error\n",
+ "#example 1.7\n",
+ "#page 10\n",
+ "n=[0.1532, 15.45, 0.0000354, 305.1, 8.12, 143.3, 0.0212, 0.643, 0.1734] #original numbers\n",
+ "#rounding all numbers to 2 decimal places\n",
+ "n=[305.1, 143.3, 0.15,15.45, 0.00, 8.12, 0.02, 0.64, 0.17] \n",
+ "sum=0;\n",
+ "#l=length(n);\n",
+ "for i in range(len(n)):\n",
+ " sum=sum+n[i];\n",
+ "\n",
+ "E_A=2*math.pow(10,-1)/2+7*math.pow(10,-2)/2\n",
+ "print \"the absolute error is:%0.2f\" %(E_A)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the absolute error is:0.14\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.8:pg-11"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#difference in 3 significant figures\n",
+ "#example 1.8\n",
+ "#page 11\n",
+ "X1=math.sqrt(6.37)\n",
+ "X2=math.sqrt(6.36)\n",
+ "d=X1-X2 #difference between two numbers\n",
+ "print \"the difference corrected to 3 significant figures is %0.3g\" %(d)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the difference corrected to 3 significant figures is 0.00198\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.9:pg-12"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#relative error\n",
+ "#example 1.10\n",
+ "#page 12\n",
+ "a=6.54\n",
+ "b=48.64\n",
+ "c=13.5\n",
+ "da=0.01\n",
+ "db=0.02\n",
+ "dc=0.03\n",
+ "s=math.pow(a,2)*math.sqrt(b)/math.pow(c,3)\n",
+ "#disp(s,'s=')\n",
+ "print \"s=%f\" %(s)\n",
+ "r_err=2*(da/a)+(db/b)/2+3*(dc/c);\n",
+ "print \"the relative error is :%f\" %(r_err)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "s=0.121241\n",
+ "the relative error is :0.009930\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.11:pg-13"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#relative error\n",
+ "#example 1.11\n",
+ "#page 13\n",
+ "import math\n",
+ "x=1\n",
+ "y=1\n",
+ "z=1\n",
+ "u=(5*x*math.pow(y,3))/math.pow(z,3)\n",
+ "dx=0.001\n",
+ "dy=0.001\n",
+ "dz=0.001\n",
+ "max=((5*math.pow(y,2))/math.pow(z,3))*dx+((10*x*y)/math.pow(z,3))*dy+((15*x*math.pow(y,2))/math.pow(z,4))*dz\n",
+ "e=max/u\n",
+ "print \" the relative error is :%f\" %(e)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " the relative error is :0.006000\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.12:pg-12"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#taylor series\n",
+ "#example 1.12\n",
+ "#page 12\n",
+ "import math\n",
+ "def f(x):\n",
+ " return math.pow(x,3)+5*x-10\n",
+ "def f1(x):\n",
+ " return 3*math.pow(x,2)-6*x+5\n",
+ "def f2(x):\n",
+ " return 6*x-6\n",
+ "def f3(x):\n",
+ " return 6\n",
+ "D=[0,f(0), f1(0), f2(0), f3(0)]\n",
+ "S1=0;\n",
+ "h=1;\n",
+ "for i in range(1,5):\n",
+ " S1=S1+math.pow(h,i-1)*D[i]/math.factorial(i-1)\n",
+ " \n",
+ "print \"the third order taylors series approximation of f(1) is :%d\" %(S1)\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the third order taylors series approximation of f(1) is :-7\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.13:pg-16"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#taylor series\n",
+ "#example 1.13\n",
+ "#page 16\n",
+ "import math\n",
+ "def f(x):\n",
+ " return math.sin(x)\n",
+ "def f1(x):\n",
+ " return math.cos(x)\n",
+ "def f2(x):\n",
+ " return -1*math.sin(x)\n",
+ "def f3(x):\n",
+ " return -1*math.cos(x)\n",
+ "def f4(x):\n",
+ " return math.sin(x)\n",
+ "def f5(x):\n",
+ " return math.cos(x)\n",
+ "def f6(x):\n",
+ " return -1*math.sin(x)\n",
+ "def f7(x):\n",
+ " return -1*math.cos(x)\n",
+ "D=[0,f(math.pi/6), f1(math.pi/6), f2(math.pi/6), f3(math.pi/6), f4(math.pi/6), f5(math.pi/6), f6(math.pi/6), f7(math.pi/6)]\n",
+ "S1=0\n",
+ "h=math.pi/6\n",
+ "print \"order of approximation computed value of sin(pi/3) absolute eror\\n\\n\"\n",
+ "for j in range(1,10):\n",
+ " for i in range(1,j):\n",
+ " S1=S1+math.pow(h,i-1)*D[i]/math.factorial(i-1) \n",
+ " print \"%d %0.9f %0.9f\\n\" %(j,S1,abs(math.sin(math.pi/3)-S1))\n",
+ " S1=0\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "order of approximation computed value of sin(pi/3) absolute eror\n",
+ "\n",
+ "\n",
+ "1 0.000000000 0.866025404\n",
+ "\n",
+ "2 0.500000000 0.366025404\n",
+ "\n",
+ "3 0.953449841 0.087424437\n",
+ "\n",
+ "4 0.884910922 0.018885518\n",
+ "\n",
+ "5 0.864191614 0.001833790\n",
+ "\n",
+ "6 0.865757475 0.000267929\n",
+ "\n",
+ "7 0.866041490 0.000016087\n",
+ "\n",
+ "8 0.866027181 0.000001777\n",
+ "\n",
+ "9 0.866025327 0.000000077\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.14:pg-18"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#maclaurins expansion\n",
+ "#example 1.14\n",
+ "#page 18\n",
+ "n=8 #correct to 8 decimal places\n",
+ "x=1\n",
+ "for i in range(1,50):\n",
+ " if x/math.factorial(i)<math.pow(10,-8)/2:\n",
+ " c=i\n",
+ " break \n",
+ "print \"no. of terms needed to correct to 8 decimal places is : %d \" %(c)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "no. of terms needed to correct to 8 decimal places is : 2 \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.15:pg-18"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#series apprixamation\n",
+ "#example 1.15\n",
+ "#page 18\n",
+ "import math\n",
+ "x=.09090909 # 1/11 =.09090909\n",
+ "S1=0\n",
+ "for i in range(1,5,2):\n",
+ " S1=S1+math.pow(x,i)/i\n",
+ "print \"value of log(1.2) is : %0.8f\\n\\n\" %(2*S1)\n",
+ "c=0\n",
+ "for i in range(1,50):\n",
+ " if math.pow(.09090909,i)/i<2*math.pow(10,-7):\n",
+ " c=i\n",
+ " break\n",
+ "print \"min no of terms needed to get value wuth same accuracy is :%d\" %(c)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "value of log(1.2) is : 0.18231906\n",
+ "\n",
+ "\n",
+ "min no of terms needed to get value wuth same accuracy is :6\n"
+ ]
+ }
+ ],
+ "prompt_number": 74
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter2_5.ipynb b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter2_5.ipynb new file mode 100644 index 00000000..b05822b8 --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter2_5.ipynb @@ -0,0 +1,2186 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:c15048f3e5cf285ff659045561f964427ecffd1f27edf5e1ca66863c0f72f448"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter02:Solution of Algebric and Transcendental Equations"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.1:pg-24"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.1\n",
+ "#bisection method\n",
+ "#page 24\n",
+ "import math\n",
+ "def f(x):\n",
+ " return math.pow(x,3)-x-1\n",
+ "x1=1\n",
+ "x2=2 #f(1) is negative and f(2) is positive\n",
+ "d=0.0001 #for accuracy of root\n",
+ "c=1\n",
+ "print \"Succesive approximations \\t x1\\t \\tx2\\t \\tm\\t \\tf(m)\\n\"\n",
+ "while abs(x1-x2)>d:\n",
+ " \n",
+ " m=(x1+x2)/2.0\n",
+ " print \" \\t%f\\t%f\\t%f\\t%f\\n\" %(x1,x2,m,f(m))\n",
+ " if f(m)*f(x1)>0.0:\n",
+ " x1=m\n",
+ " else:\n",
+ " x2=m \n",
+ " c=c+1 # to count number of iterations \n",
+ "print \"the solution of equation after %i iteration is %g\" %(c,m)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Succesive approximations \t x1\t \tx2\t \tm\t \tf(m)\n",
+ "\n",
+ " \t1.000000\t2.000000\t1.500000\t0.875000\n",
+ "\n",
+ " \t1.000000\t1.500000\t1.250000\t-0.296875\n",
+ "\n",
+ " \t1.250000\t1.500000\t1.375000\t0.224609\n",
+ "\n",
+ " \t1.250000\t1.375000\t1.312500\t-0.051514\n",
+ "\n",
+ " \t1.312500\t1.375000\t1.343750\t0.082611\n",
+ "\n",
+ " \t1.312500\t1.343750\t1.328125\t0.014576\n",
+ "\n",
+ " \t1.312500\t1.328125\t1.320312\t-0.018711\n",
+ "\n",
+ " \t1.320312\t1.328125\t1.324219\t-0.002128\n",
+ "\n",
+ " \t1.324219\t1.328125\t1.326172\t0.006209\n",
+ "\n",
+ " \t1.324219\t1.326172\t1.325195\t0.002037\n",
+ "\n",
+ " \t1.324219\t1.325195\t1.324707\t-0.000047\n",
+ "\n",
+ " \t1.324707\t1.325195\t1.324951\t0.000995\n",
+ "\n",
+ " \t1.324707\t1.324951\t1.324829\t0.000474\n",
+ "\n",
+ " \t1.324707\t1.324829\t1.324768\t0.000214\n",
+ "\n",
+ "the solution of equation after 15 iteration is 1.32477'\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.2:pg-25"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.2\n",
+ "#bisection method\n",
+ "#page 25\n",
+ "import math\n",
+ "def f(x):\n",
+ " return math.pow(x,3)-2*x-5\n",
+ "x1=2 \n",
+ "x2=3 #f(2) is negative and f(3) is positive\n",
+ "d=0.0001 #for accuracy of root\n",
+ "c=1\n",
+ "print \"Succesive approximations \\t x1\\t \\tx2\\t \\tm\\t \\tf(m)\\n\"\n",
+ "while abs(x1-x2)>d:\n",
+ " m=(x1+x2)/2.0\n",
+ " print \" \\t%f\\t%f\\t%f\\t%f\\n\" %(x1,x2,m,f(m))\n",
+ " if f(m)*f(x1)>0:\n",
+ " x1=m\n",
+ " else:\n",
+ " x2=m \n",
+ " c=c+1;# to count number of iterations \n",
+ "print \"the solution of equation after %i iteration is %0.4g\" %(c,m)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Succesive approximations \t x1\t \tx2\t \tm\t \tf(m)\n",
+ "\n",
+ " \t2.000000\t3.000000\t2.500000\t5.625000\n",
+ "\n",
+ " \t2.000000\t2.500000\t2.250000\t1.890625\n",
+ "\n",
+ " \t2.000000\t2.250000\t2.125000\t0.345703\n",
+ "\n",
+ " \t2.000000\t2.125000\t2.062500\t-0.351318\n",
+ "\n",
+ " \t2.062500\t2.125000\t2.093750\t-0.008942\n",
+ "\n",
+ " \t2.093750\t2.125000\t2.109375\t0.166836\n",
+ "\n",
+ " \t2.093750\t2.109375\t2.101562\t0.078562\n",
+ "\n",
+ " \t2.093750\t2.101562\t2.097656\t0.034714\n",
+ "\n",
+ " \t2.093750\t2.097656\t2.095703\t0.012862\n",
+ "\n",
+ " \t2.093750\t2.095703\t2.094727\t0.001954\n",
+ "\n",
+ " \t2.093750\t2.094727\t2.094238\t-0.003495\n",
+ "\n",
+ " \t2.094238\t2.094727\t2.094482\t-0.000771\n",
+ "\n",
+ " \t2.094482\t2.094727\t2.094604\t0.000592\n",
+ "\n",
+ " \t2.094482\t2.094604\t2.094543\t-0.000090\n",
+ "\n",
+ "the solution of equation after 15 iteration is 2.095\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.3:pg-26"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.3\n",
+ "#bisection method\n",
+ "#page 26\n",
+ "import math\n",
+ "def f(x):\n",
+ " return math.pow(x,3)+math.pow(x,2)+x+7\n",
+ "x1=-3\n",
+ "x2=-2 #f(-3) is negative and f(-2) is positive\n",
+ "d=0.0001 #for accuracy of root\n",
+ "c=1\n",
+ "print \"Succesive approximations \\t x1\\t \\tx2\\t \\tm\\t \\tf(m)\\n\"\n",
+ "while abs(x1-x2)>d:\n",
+ " m=(x1+x2)/2.0\n",
+ " print \" \\t%f\\t%f\\t%f\\t%f\\n\" %(x1,x2,m,f(m))\n",
+ " if f(m)*f(x1)>0:\n",
+ " x1=m\n",
+ " else:\n",
+ " x2=m \n",
+ " c=c+1 # to count number of iterations \n",
+ "print \"the solution of equation after %i iteration is %0.4g\" %(c,m)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Succesive approximations \t x1\t \tx2\t \tm\t \tf(m)\n",
+ "\n",
+ " \t-3.000000\t-2.000000\t-2.500000\t-4.875000\n",
+ "\n",
+ " \t-2.500000\t-2.000000\t-2.250000\t-1.578125\n",
+ "\n",
+ " \t-2.250000\t-2.000000\t-2.125000\t-0.205078\n",
+ "\n",
+ " \t-2.125000\t-2.000000\t-2.062500\t0.417725\n",
+ "\n",
+ " \t-2.125000\t-2.062500\t-2.093750\t0.111481\n",
+ "\n",
+ " \t-2.125000\t-2.093750\t-2.109375\t-0.045498\n",
+ "\n",
+ " \t-2.109375\t-2.093750\t-2.101562\t0.033315\n",
+ "\n",
+ " \t-2.109375\t-2.101562\t-2.105469\t-0.006010\n",
+ "\n",
+ " \t-2.105469\t-2.101562\t-2.103516\t0.013673\n",
+ "\n",
+ " \t-2.105469\t-2.103516\t-2.104492\t0.003836\n",
+ "\n",
+ " \t-2.105469\t-2.104492\t-2.104980\t-0.001086\n",
+ "\n",
+ " \t-2.104980\t-2.104492\t-2.104736\t0.001376\n",
+ "\n",
+ " \t-2.104980\t-2.104736\t-2.104858\t0.000145\n",
+ "\n",
+ " \t-2.104980\t-2.104858\t-2.104919\t-0.000470\n",
+ "\n",
+ "the solution of equation after 15 iteration is -2.105\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.4:pg-26"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.4\n",
+ "#bisection method\n",
+ "#page 26\n",
+ "import math\n",
+ "def f(x):\n",
+ " return x*math.exp(x)-1\n",
+ "x1=0 \n",
+ "x2=1 #f(0) is negative and f(1) is positive\n",
+ "d=0.0005 #maximun tolerance value\n",
+ "c=1\n",
+ "print \"Succesive approximations \\t x1\\t \\tx2\\t \\tm\\t \\ttol\\t \\tf(m)\\n\"\n",
+ "while abs((x2-x1)/x2)>d:\n",
+ " m=(x1+x2)/2.0 #tolerance value for each iteration\n",
+ " tol=((x2-x1)/x2)*100\n",
+ " print \" \\t%f\\t%f\\t%f\\t%f\\t%f\\n\" %(x1,x2,m,tol,f(m))\n",
+ " if f(m)*f(x1)>0:\n",
+ " x1=m\n",
+ " else:\n",
+ " x2=m \n",
+ " c=c+1 # to count number of iterations \n",
+ "print \"the solution of equation after %i iteration is %0.4g\" %(c,m)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Succesive approximations \t x1\t \tx2\t \tm\t \ttol\t \tf(m)\n",
+ "\n",
+ " \t0.000000\t1.000000\t0.500000\t100.000000\t-0.175639\n",
+ "\n",
+ " \t0.500000\t1.000000\t0.750000\t50.000000\t0.587750\n",
+ "\n",
+ " \t0.500000\t0.750000\t0.625000\t33.333333\t0.167654\n",
+ "\n",
+ " \t0.500000\t0.625000\t0.562500\t20.000000\t-0.012782\n",
+ "\n",
+ " \t0.562500\t0.625000\t0.593750\t10.000000\t0.075142\n",
+ "\n",
+ " \t0.562500\t0.593750\t0.578125\t5.263158\t0.030619\n",
+ "\n",
+ " \t0.562500\t0.578125\t0.570312\t2.702703\t0.008780\n",
+ "\n",
+ " \t0.562500\t0.570312\t0.566406\t1.369863\t-0.002035\n",
+ "\n",
+ " \t0.566406\t0.570312\t0.568359\t0.684932\t0.003364\n",
+ "\n",
+ " \t0.566406\t0.568359\t0.567383\t0.343643\t0.000662\n",
+ "\n",
+ " \t0.566406\t0.567383\t0.566895\t0.172117\t-0.000687\n",
+ "\n",
+ " \t0.566895\t0.567383\t0.567139\t0.086059\t-0.000013\n",
+ "\n",
+ "the solution of equation after 13 iteration is 0.5671\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.5:pg-27"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.5\n",
+ "#bisection method\n",
+ "#page 27\n",
+ "import math\n",
+ "def f(x):\n",
+ " return 4*math.exp(-x)*math.sin(x)-1\n",
+ "x1=0 \n",
+ "x2=0.5 #f(0) is negative and f(1) is positive\n",
+ "d=0.0001 #for accuracy of root\n",
+ "c=1 \n",
+ "print \"Succesive approximations \\t x1\\t \\tx2\\t \\tm\\t \\t \\tf(m)\\n\"\n",
+ "while abs(x2-x1)>d:\n",
+ " m=(x1+x2)/2.0\n",
+ " print \" \\t%f\\t%f\\t%f\\t%f\\n\" %(x1,x2,m,f(m))\n",
+ " if f(m)*f(x1)>0:\n",
+ " x1=m\n",
+ " else:\n",
+ " x2=m \n",
+ " c=c+1 # to count number of iterations \n",
+ "print \"the solution of equation after %i iteration is %0.3g\" %(c,m)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Succesive approximations \t x1\t \tx2\t \tm\t \t \tf(m)\n",
+ "\n",
+ " \t0.000000\t0.500000\t0.250000\t-0.229286\n",
+ "\n",
+ " \t0.250000\t0.500000\t0.375000\t0.006941\n",
+ "\n",
+ " \t0.250000\t0.375000\t0.312500\t-0.100293\n",
+ "\n",
+ " \t0.312500\t0.375000\t0.343750\t-0.044068\n",
+ "\n",
+ " \t0.343750\t0.375000\t0.359375\t-0.017925\n",
+ "\n",
+ " \t0.359375\t0.375000\t0.367188\t-0.005334\n",
+ "\n",
+ " \t0.367188\t0.375000\t0.371094\t0.000842\n",
+ "\n",
+ " \t0.367188\t0.371094\t0.369141\t-0.002236\n",
+ "\n",
+ " \t0.369141\t0.371094\t0.370117\t-0.000694\n",
+ "\n",
+ " \t0.370117\t0.371094\t0.370605\t0.000075\n",
+ "\n",
+ " \t0.370117\t0.370605\t0.370361\t-0.000310\n",
+ "\n",
+ " \t0.370361\t0.370605\t0.370483\t-0.000118\n",
+ "\n",
+ " \t0.370483\t0.370605\t0.370544\t-0.000022\n",
+ "\n",
+ "the solution of equation after 14 iteration is 0.371\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.6:pg-28"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.6\n",
+ "#false position method\n",
+ "#page 28\n",
+ "import math\n",
+ "def f(x):\n",
+ " return x**3-2*x-5\n",
+ "a=2.0\n",
+ "b=3.0 #f(2) is negative and f(3)is positive\n",
+ "d=0.00001\n",
+ "print \"succesive iterations \\ta\\t b\\t f(a)\\t f(b)\\t\\ x1\\n\"\n",
+ "for i in range(1,25):\n",
+ " x1=b*f(a)/(f(a)-f(b))+a*f(b)/(f(b)-f(a))\n",
+ " if(f(a)*f(x1))>0:\n",
+ " b=x1\n",
+ " else:\n",
+ " a=x1\n",
+ " if abs(f(x1))<d:\n",
+ " break\n",
+ " print \" \\t%f %f %f %f %f\\n\" %(a,b,f(a),f(b),x1)\n",
+ "print \"the root of the equation is %f\" %(x1)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "succesive iterations \ta\t b\t f(a)\t f(b)\t\\ x1\n",
+ "\n",
+ " \t2.000000 2.058824 -1.000000 -0.390800 2.058824\n",
+ "\n",
+ " \t2.096559 2.058824 0.022428 -0.390800 2.096559\n",
+ "\n",
+ " \t2.094511 2.058824 -0.000457 -0.390800 2.094511\n",
+ "\n",
+ "the root of the equation is 2.094552\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.7:pg-29"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.7\n",
+ "#false position method\n",
+ "#page 29\n",
+ "def f(x):\n",
+ " return x**2.2-69\n",
+ "a=5.0\n",
+ "b=6.0 #f(5) is negative and f(6)is positive\n",
+ "d=0.00001\n",
+ "print \"succesive iterations \\ta\\t b\\t f(a)\\t f(b)\\t\\ x1\\n\"\n",
+ "for i in range(1,25):\n",
+ " x1=b*f(a)/(f(a)-f(b))+a*f(b)/(f(b)-f(a));\n",
+ " if(f(a)*f(x1))>0:\n",
+ " b=x1\n",
+ " else:\n",
+ " a=x1\n",
+ " if abs(f(x1))<d:\n",
+ " break\n",
+ " print \" \\t%f %f %f %f %f\\n\" %(a,b,f(a),f(b),x1)\n",
+ "print \"the root of the equation is %f\" %(x1)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "succesive iterations \ta\t b\t f(a)\t f(b)\t\\ x1\n",
+ "\n",
+ " \t7.027228 6.000000 3.933141 -17.485113 7.027228\n",
+ "\n",
+ " \t6.838593 6.000000 -0.304723 -17.485113 6.838593\n",
+ "\n",
+ " \t6.853467 6.000000 0.024411 -17.485113 6.853467\n",
+ "\n",
+ " \t6.852277 6.000000 -0.001950 -17.485113 6.852277\n",
+ "\n",
+ " \t6.852372 6.000000 0.000156 -17.485113 6.852372\n",
+ "\n",
+ " \t6.852365 6.000000 -0.000012 -17.485113 6.852365\n",
+ "\n",
+ "the root of the equation is 6.852365\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.8:pg-29"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.8\n",
+ "#false position method\n",
+ "#page 29\n",
+ "import math\n",
+ "def f(x):\n",
+ " return 2*x-log10(x)-7\n",
+ "a=3.0\n",
+ "b=4.0 #f(3) is negative and f(4)is positive\n",
+ "d=0.00001\n",
+ "print \"succesive iterations \\ta\\t b\\t f(a)\\t f(b)\\t\\ x1\\n\"\n",
+ "for i in range(1,25):\n",
+ " x1=b*f(a)/(f(a)-f(b))+a*f(b)/(f(b)-f(a))\n",
+ " if(f(a)*f(x1))>0:\n",
+ " b=x1\n",
+ " else:\n",
+ " a=x1\n",
+ " if abs(f(x1))<d:\n",
+ " break\n",
+ " print \" \\t%f %f %f %f %f\\n\" %(a,b,f(a),f(b),x1)\n",
+ "print \"the root of the equation is %0.4g\" %(x1)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "succesive iterations \ta\t b\t f(a)\t f(b)\t\\ x1\n",
+ "\n",
+ " \t3.000000 3.787772 -1.477121 -0.002839 3.787772\n",
+ "\n",
+ " \t3.789289 3.787772 0.000021 -0.002839 3.789289\n",
+ "\n",
+ "the root of the equation is 3.789\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.9:pg-30"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.9\n",
+ "#false position method\n",
+ "#page 30\n",
+ "import math\n",
+ "def f(x):\n",
+ " return 4*math.exp(-x)*math.sin(x)-1\n",
+ "a=0.0\n",
+ "b=0.5 #f(0) is negative and f(0.5)is positive\n",
+ "d=0.00001\n",
+ "print \"succesive iterations \\ta\\t b\\t f(a)\\t f(b)\\t\\ x1\\n\"\n",
+ "for i in range(1,25):\n",
+ " x1=b*f(a)/(f(a)-f(b))+a*f(b)/(f(b)-f(a))\n",
+ " if(f(a)*f(x1))>0:\n",
+ " b=x1\n",
+ " else:\n",
+ " a=x1\n",
+ " if abs(f(x1))<d:\n",
+ " break\n",
+ " print \" \\t%f %f %f %f %f\\n\" %(a,b,f(a),f(b),x1)\n",
+ "print \"the root of the equation is %f\" %(x1)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "succesive iterations \ta\t b\t f(a)\t f(b)\t\\ x1\n",
+ "\n",
+ " \t0.429869 0.500000 0.084545 0.163145 0.429869\n",
+ "\n",
+ " \t0.354433 0.500000 -0.026054 0.163145 0.354433\n",
+ "\n",
+ " \t0.374479 0.500000 0.006132 0.163145 0.374479\n",
+ "\n",
+ " \t0.369577 0.500000 -0.001547 0.163145 0.369577\n",
+ "\n",
+ " \t0.370802 0.500000 0.000384 0.163145 0.370802\n",
+ "\n",
+ " \t0.370497 0.500000 -0.000096 0.163145 0.370497\n",
+ "\n",
+ " \t0.370573 0.500000 0.000024 0.163145 0.370573\n",
+ "\n",
+ "the root of the equation is 0.370554\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.10:pg-33"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.10\n",
+ "#iteration method\n",
+ "#page 33\n",
+ "import math\n",
+ "def f(x):\n",
+ " return 1/(math.sqrt(x+1))\n",
+ "x1=0.75\n",
+ "x2=0.0\n",
+ "n=1\n",
+ "d=0.0001 #accuracy opto 10^-4\n",
+ "c=0 #to count no of iterations \n",
+ "print \"successive iterations \\t\\x01\\tf(x1)\\n\"\n",
+ "while abs(x1-x2)>d:\n",
+ " print \" \\t%f %f\\n\" %(x1,f(x1))\n",
+ " x2=x1\n",
+ " x1=f(x1)\n",
+ " c=c+1\n",
+ "print \" the root of the eqaution after %i iteration is %0.4g\" %(c,x1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \t\u0001\tf(x1)\n",
+ "\n",
+ " \t0.750000 0.755929\n",
+ "\n",
+ " \t0.755929 0.754652\n",
+ "\n",
+ " \t0.754652 0.754926\n",
+ "\n",
+ " \t0.754926 0.754867\n",
+ "\n",
+ " the root of the eqaution after 4 iteration is 0.7549\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.11:pg-34"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.11\n",
+ "#iteration method\n",
+ "#page34\n",
+ "import math\n",
+ "def f(x):\n",
+ " return cos(x)/2.0+3.0/2.0\n",
+ "x1=1.5 # as roots lies between 3/2 and pi/2\n",
+ "x2=0\n",
+ "d=0.0001 # accuracy opto 10^-4\n",
+ "c=0 # to count no of iterations \n",
+ "print \"successive iterations \\t\\x01\\tf(x1)\\n\"\n",
+ "while abs(x2-x1)>d:\n",
+ " \n",
+ " print \" \\t%f %f\\n\" %(x1,f(x1))\n",
+ " x2=x1\n",
+ " x1=f(x1)\n",
+ " c=c+1\n",
+ "print \" the root of the eqaution after %i iteration is %0.4g\" %(c,x1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \t\u0001\tf(x1)\n",
+ "\n",
+ " \t1.500000 1.535369\n",
+ "\n",
+ " \t1.535369 1.517710\n",
+ "\n",
+ " \t1.517710 1.526531\n",
+ "\n",
+ " \t1.526531 1.522126\n",
+ "\n",
+ " \t1.522126 1.524326\n",
+ "\n",
+ " \t1.524326 1.523227\n",
+ "\n",
+ " \t1.523227 1.523776\n",
+ "\n",
+ " \t1.523776 1.523502\n",
+ "\n",
+ " \t1.523502 1.523639\n",
+ "\n",
+ " \t1.523639 1.523570\n",
+ "\n",
+ " the root of the eqaution after 10 iteration is 1.524\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.12:pg-35"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.12\n",
+ "#iteration method\n",
+ "#page 35\n",
+ "import math\n",
+ "def f(x):\n",
+ " return math.exp(-x)\n",
+ "x1=1.5 # as roots lies between 0 and 1\n",
+ "x2=0\n",
+ "d=0.0001 # accuracy opto 10^-4\n",
+ "c=0 # to count no of iterations \n",
+ "print \"successive iterations \\t x1 \\t f(x1)\\n\"\n",
+ "while abs(x2-x1)>d:\n",
+ " \n",
+ " print \" \\t%f %f\\n\" %(x1,f(x1))\n",
+ " x2=x1\n",
+ " x1=f(x1)\n",
+ " c=c+1\n",
+ "print \" the root of the eqaution after %i iteration is %0.4g\" %(c,x1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \t x1 \t f(x1)\n",
+ "\n",
+ " \t1.500000 0.223130\n",
+ "\n",
+ " \t0.223130 0.800011\n",
+ "\n",
+ " \t0.800011 0.449324\n",
+ "\n",
+ " \t0.449324 0.638059\n",
+ "\n",
+ " \t0.638059 0.528317\n",
+ "\n",
+ " \t0.528317 0.589597\n",
+ "\n",
+ " \t0.589597 0.554551\n",
+ "\n",
+ " \t0.554551 0.574330\n",
+ "\n",
+ " \t0.574330 0.563082\n",
+ "\n",
+ " \t0.563082 0.569451\n",
+ "\n",
+ " \t0.569451 0.565836\n",
+ "\n",
+ " \t0.565836 0.567885\n",
+ "\n",
+ " \t0.567885 0.566723\n",
+ "\n",
+ " \t0.566723 0.567382\n",
+ "\n",
+ " \t0.567382 0.567008\n",
+ "\n",
+ " \t0.567008 0.567220\n",
+ "\n",
+ " \t0.567220 0.567100\n",
+ "\n",
+ " \t0.567100 0.567168\n",
+ "\n",
+ " the root of the eqaution after 18 iteration is 0.5672\n"
+ ]
+ }
+ ],
+ "prompt_number": 34
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.13:pg-35"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.13\n",
+ "#iteration method\n",
+ "#page 35\n",
+ "import math\n",
+ "def f(x):\n",
+ " return 1+math.sin(x)/10\n",
+ "x1=1.0 # as roots lies between 1 and pi evident from graph\n",
+ "x2=0\n",
+ "d=0.0001 # accuracy opto 10^-4\n",
+ "c=0 # to count no of iterations \n",
+ "print \"successive iterations \\t x1 \\t f(x1)\\n\"\n",
+ "while abs(x2-x1)>d:\n",
+ " print \" \\t%f %f\\n\" %(x1,f(x1))\n",
+ " x2=x1\n",
+ " x1=f(x1)\n",
+ " c=c+1\n",
+ "print \" the root of the eqaution after %i iteration is %0.4g\" %(c,x1)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \t x1 \t f(x1)\n",
+ "\n",
+ " \t1.000000 1.084147\n",
+ "\n",
+ " \t1.084147 1.088390\n",
+ "\n",
+ " \t1.088390 1.088588\n",
+ "\n",
+ " \t1.088588 1.088597\n",
+ "\n",
+ " the root of the eqaution after 4 iteration is 1.089\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.14:pg-36"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.14\n",
+ "#aitken's process\n",
+ "#page 36\n",
+ "import math\n",
+ "def f(x):\n",
+ " return 1.5+math.cos(x)/2.0\n",
+ "x0=1.5\n",
+ "y=0\n",
+ "e=0.0001\n",
+ "c=0\n",
+ "print \"successive iterations \\t x0 \\t x1 \\t x2 \\t x3 \\t y\\n\"\n",
+ "for i in range(1,10):\n",
+ " x1=f(x0)\n",
+ " x2=f(x1)\n",
+ " x3=f(x2)\n",
+ " y=x3-((x3-x2)**2)/(x3-2*x2+x1)\n",
+ " d=y-x0\n",
+ " x0=y\n",
+ " if abs(f(x0))<e:\n",
+ " break\n",
+ " c=c+1\n",
+ " print \" \\t%f %f %f %f %f\\n\" %(x0,x1,x2,x3,y)\n",
+ "print \"the root of the equation after %i iteration is %f\" %(c,y)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \t x0 \t x1 \t x2 \t x3 \t y\n",
+ "\n",
+ " \t1.523592 1.535369 1.517710 1.526531 1.523592\n",
+ "\n",
+ " \t1.523593 1.523593 1.523593 1.523593 1.523593\n",
+ "\n",
+ " \t1.523593 1.523593 1.523593 1.523593 1.523593\n",
+ "\n",
+ " \t1.523593 1.523593 1.523593 1.523593 1.523593\n",
+ "\n",
+ " \t1.523593 1.523593 1.523593 1.523593 1.523593\n",
+ "\n",
+ " \t1.523593 1.523593 1.523593 1.523593 1.523593\n",
+ "\n",
+ " \t1.523593 1.523593 1.523593 1.523593 1.523593\n",
+ "\n",
+ " \t1.523593 1.523593 1.523593 1.523593 1.523593\n",
+ "\n",
+ " \t1.523593 1.523593 1.523593 1.523593 1.523593\n",
+ "\n",
+ "the root of the equation after 9 iteration is 1.523593\n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.15:pg-39"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.15\n",
+ "#newton-raphson method\n",
+ "#page 39\n",
+ "def f(x):\n",
+ " return x**3-2*x-5\n",
+ "def f1(x):\n",
+ " return 3*x**2-2 # first derivative of the function\n",
+ "x0=2.0 # initial value\n",
+ "d=0.0001\n",
+ "c=0\n",
+ "n=1\n",
+ "print \"successive iterations \\t x0 \\t f(x0) \\t f1(x0)\\n\"\n",
+ "while n==1:\n",
+ " x2=x0\n",
+ " x1=x0-(f(x0)/f1(x0))\n",
+ " x0=x1\n",
+ " print \" \\t%f \\t%f \\t%f \\n\" %(x2,f(x1),f1(x1))\n",
+ " c=c+1\n",
+ " if abs(f(x0))<d:\n",
+ " break\n",
+ "print \"the root of %i iteration is:%f\" %(c,x0)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \t x0 \t f(x0) \t f1(x0)\n",
+ "\n",
+ " \t2.000000 \t0.061000 \t11.230000 \n",
+ "\n",
+ " \t2.100000 \t0.000186 \t11.161647 \n",
+ "\n",
+ " \t2.094568 \t0.000000 \t11.161438 \n",
+ "\n",
+ "the root of 3 iteration is:2.094551\n"
+ ]
+ }
+ ],
+ "prompt_number": 44
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.16:pg-40"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.16\n",
+ "#newton-raphson method\n",
+ "#page 40\n",
+ "import math\n",
+ "def f(x):\n",
+ " return x*math.sin(x)+math.cos(x)\n",
+ "def f1(x):\n",
+ " return x*math.cos(x) #first derivation of the function\n",
+ "x0=math.pi # initial value\n",
+ "d=0.0001\n",
+ "c=0 \n",
+ "n=1\n",
+ "print \"successive iterations \\tx0\\t f(x0)\\t f1(x0)\\n\"\n",
+ "while n==1:\n",
+ " x2=x0\n",
+ " x1=x0-(f(x0)/f1(x0))\n",
+ " x0=x1\n",
+ " print \" \\t%f \\t%f \\t%f\\n\" %(x2,f(x1),f1(x1))\n",
+ " c=c+1\n",
+ " if abs(f(x0))<d:\n",
+ " break\n",
+ "print \"the root of %i iteration is:%f\" %(c,x0)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \tx0\t f(x0)\t f1(x0)\n",
+ "\n",
+ " \t3.141593 \t-0.066186 \t-2.681457\n",
+ "\n",
+ " \t2.823283 \t-0.000564 \t-2.635588\n",
+ "\n",
+ " \t2.798600 \t-0.000000 \t-2.635185\n",
+ "\n",
+ "the root of 3 iteration is:2.798386\n"
+ ]
+ }
+ ],
+ "prompt_number": 46
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.17:pg-40"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.17\n",
+ "#newton-raphson method\n",
+ "#page 40\n",
+ "import math\n",
+ "def f(x):\n",
+ " return x*math.exp(x)-1\n",
+ "def f1(x):\n",
+ " return math.exp(x)+x*math.exp(x) #first derivative of the function\n",
+ "x0=0 # initial value\n",
+ "d=0.0001 \n",
+ "c=0\n",
+ "n=1\n",
+ "print \"successive iterations \\tx0\\t f(x0)\\t f1(x0)\\n\"\n",
+ "while n==1:\n",
+ " x2=x0\n",
+ " x1=x0-(f(x0)/f1(x0))\n",
+ " x0=x1\n",
+ " print \" \\t%f \\t%f \\t%f\\n\" %(x2,f(x1),f1(x1))\n",
+ " c=c+1\n",
+ " if abs(f(x0))<d:\n",
+ " break\n",
+ "print \"the root of %i iteration is:%f\" %(c,x0)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \tx0\t f(x0)\t f1(x0)\n",
+ "\n",
+ " \t0.000000 \t1.718282 \t5.436564\n",
+ "\n",
+ " \t1.000000 \t0.355343 \t3.337012\n",
+ "\n",
+ " \t0.683940 \t0.028734 \t2.810232\n",
+ "\n",
+ " \t0.577454 \t0.000239 \t2.763614\n",
+ "\n",
+ " \t0.567230 \t0.000000 \t2.763223\n",
+ "\n",
+ "the root of 5 iteration is:0.567143\n"
+ ]
+ }
+ ],
+ "prompt_number": 48
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.18:pg-41"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.18\n",
+ "#newton-raphson method\n",
+ "#page 41\n",
+ "import math\n",
+ "def f(x):\n",
+ " return math.sin(x)-x/2.0\n",
+ "def f1(x):\n",
+ " return math.cos(x)-0.5\n",
+ "x0=math.pi/2.0 # initial value\n",
+ "d=0.0001\n",
+ "c=0\n",
+ "n=1\n",
+ "print \"successive iterations \\t x0 \\t f(x0)\\t f1(x0)\\n\"\n",
+ "while n==1:\n",
+ " x2=x0\n",
+ " x1=x0-(f(x0)/f1(x0))\n",
+ " x0=x1\n",
+ " print \" \\t%f\\t%f\\t%f\\n\" %(x2,f(x1),f1(x1))\n",
+ " c=c+1\n",
+ " if abs(f(x0))<d:\n",
+ " break\n",
+ "print \"the root of %i iteration is: %0.4g\" %(c,x0)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \t x0 \t f(x0)\t f1(x0)\n",
+ "\n",
+ " \t1.570796\t-0.090703\t-0.916147\n",
+ "\n",
+ " \t2.000000\t-0.004520\t-0.824232\n",
+ "\n",
+ " \t1.900996\t-0.000014\t-0.819039\n",
+ "\n",
+ "the root of 3 iteration is: 1.896\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.19:pg-41"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.19\n",
+ "#newton-raphson method\n",
+ "#page 41\n",
+ "import math\n",
+ "def f(x):\n",
+ " return 4*math.exp(-x)*math.sin(x)-1\n",
+ "def f1(x):\n",
+ " return math.cos(x)*4*math.exp(-x)-4*math.exp(-x)*math.sin(x)\n",
+ "x0=0.2 # initial value\n",
+ "d=0.0001\n",
+ "c=0 \n",
+ "n=1\n",
+ "print \"successive iterations \\t x0 \\t f(x0)\\t f1(x0)\\n\"\n",
+ "while n==1:\n",
+ " x2=x0\n",
+ " x1=x0-(f(x0)/f1(x0))\n",
+ " x0=x1\n",
+ " print \" \\t%f \\t%f \\t%f\\n\" %(x2,f(x1),f1(x1))\n",
+ " c=c+1\n",
+ " if abs(f(x0))<d:\n",
+ " break \n",
+ "print \"the root of %i iteration is: %0.3g\" %(c,x0)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \t x0 \t f(x0)\t f1(x0)\n",
+ "\n",
+ " \t0.200000 \t-0.056593 \t1.753325\n",
+ "\n",
+ " \t0.336526 \t-0.002769 \t1.583008\n",
+ "\n",
+ " \t0.368804 \t-0.000008 \t1.573993\n",
+ "\n",
+ "the root of 3 iteration is: 0.371\n"
+ ]
+ }
+ ],
+ "prompt_number": 54
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.20:pg-42"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.20\n",
+ "#generalized newton-raphson method\n",
+ "#page 42\n",
+ "def f(x):\n",
+ " return x**3-x**2-x+1\n",
+ "def f1(x):\n",
+ " return 3*x**2-2*x-1\n",
+ "def f2(x):\n",
+ " return 6*x-2\n",
+ "x0=0.8 # initial value to finf double root\n",
+ "n=1 \n",
+ "print \"successive iterations \\t x0 \\t x1\\t x2\\n\"\n",
+ "while n==1:\n",
+ " x1=x0-(f(x0)/f1(x0));\n",
+ " x2=x0-(f1(x0)/f2(x0));\n",
+ " if abs(x1-x2)<0.000000001:\n",
+ " x0=(x1+x2)/2.0\n",
+ " break\n",
+ " else:\n",
+ " x0=(x1+x2)/2;\n",
+ " print \" %f\\t %f\\t %f\\n\" %(x0,x1,x2)\n",
+ "print \"\\n \\nthe double root is at: %f\" %(x0)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \t x0 \t x1\t x2\n",
+ "\n",
+ " 0.974370\t 0.905882\t 1.042857\n",
+ "\n",
+ " 0.993890\t 0.987269\t 1.000512\n",
+ "\n",
+ " 0.998489\t 0.996950\t 1.000028\n",
+ "\n",
+ " 0.999623\t 0.999245\t 1.000002\n",
+ "\n",
+ " 0.999906\t 0.999812\t 1.000000\n",
+ "\n",
+ " 0.999976\t 0.999953\t 1.000000\n",
+ "\n",
+ " 0.999994\t 0.999988\t 1.000000\n",
+ "\n",
+ " 0.999999\t 0.999997\t 1.000000\n",
+ "\n",
+ " 1.000000\t 0.999999\t 1.000000\n",
+ "\n",
+ " 1.000000\t 1.000000\t 1.000000\n",
+ "\n",
+ " 1.000000\t 1.000000\t 1.000000\n",
+ "\n",
+ " 1.000000\t 1.000000\t 1.000000\n",
+ "\n",
+ " 1.000000\t 1.000000\t 1.000000\n",
+ "\n",
+ "\n",
+ " \n",
+ "the double root is at: 1.000000\n"
+ ]
+ }
+ ],
+ "prompt_number": 57
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.21:pg-45"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#ramanujan's method\n",
+ "#example 2.21\n",
+ "#page 45\n",
+ "def f(x):\n",
+ " return 1-(13.0/12.0)*x-(3.0/8.0)*x**2+(1.0/24.0)*x**3\n",
+ "a1=13.0/12.0\n",
+ "a2=-3.0/8.0\n",
+ "a3=1.0/24.0\n",
+ "b1=1\n",
+ "b2=a1\n",
+ "b3=a1*b2+a2*b1\n",
+ "b4=a1*b3+a2*b2+a3*b1\n",
+ "b5=a1*b4+a2*b3+a3*b2\n",
+ "b6=a1*b5+a2*b4+a3*b3\n",
+ "b7=a1*b6+a2*b5+a3*b4\n",
+ "b8=a1*b7+a2*b6+a3*b5\n",
+ "b9=a1*b8+a2*b7+a3*b6\n",
+ "print \"\\n\\n%f\" %(b1/b2)\n",
+ "print \"\\n%f\" %(b2/b3)\n",
+ "print \"\\n%f\" %(b3/b4)\n",
+ "print \"\\n%f\" %(b4/b5)\n",
+ "print \"\\n%f\" %(b5/b6)\n",
+ "print \"\\n%f\" %(b6/b7)\n",
+ "print \"\\n%f\" %(b7/b8)\n",
+ "print \"\\n%f\" %(b8/b9)\n",
+ "print \"\\n it appears as if the roots are converging at 2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ "0.923077\n",
+ "\n",
+ "1.356522\n",
+ "\n",
+ "1.595376\n",
+ "\n",
+ "1.738402\n",
+ "\n",
+ "1.828184\n",
+ "\n",
+ "1.886130\n",
+ "\n",
+ "1.924153\n",
+ "\n",
+ "1.949345\n",
+ "\n",
+ " it appears as if the roots are converging at 2\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.22:pg-46"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#ramanujan's method\n",
+ "#example 2.22\n",
+ "#page 46\n",
+ "def f(x):\n",
+ " return x+x**2+(x**3)/2.0+(x**4)/6.0+(x**5)/24.0\n",
+ "a1=1.0\n",
+ "a2=1.0\n",
+ "a3=1.0/2.0\n",
+ "a4=1.0/6.0\n",
+ "a5=1.0/24.0\n",
+ "b1=1\n",
+ "b2=a2\n",
+ "b3=a1*b2+a2*b1\n",
+ "b4=a1*b3+a2*b2+a3*b1\n",
+ "b5=a1*b4+a2*b3+a3*b2\n",
+ "b6=a1*b5+a2*b4+a3*b3\n",
+ "print \"\\n%f\" %(b1/b2)\n",
+ "print \"\\n%f\" %(b2/b3)\n",
+ "print \"\\n%f\" %(b3/b4)\n",
+ "print \"\\n%f\" %(b4/b5)\n",
+ "print \"\\n%f\" %(b5/b6)\n",
+ "print \"\\n it appears as if the roots are converging at around %f\" %(b5/b6)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "1.000000\n",
+ "\n",
+ "0.500000\n",
+ "\n",
+ "0.571429\n",
+ "\n",
+ "0.583333\n",
+ "\n",
+ "0.571429\n",
+ "\n",
+ " it appears as if the roots are converging at around 0.571429\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.23:pg-47"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#ramanujan's method\n",
+ "#example 2.23\n",
+ "#page 47\n",
+ "from __future__ import division\n",
+ "def f(x):\n",
+ " return 1-2*((3*x/2.0+(x**2)/4.0-(x**4)/48.0+(x**6)/1440.0)-(x**8)*2/80640.0)\n",
+ "a1=3/2\n",
+ "a2=1/4\n",
+ "a3=0\n",
+ "a4=1/48\n",
+ "a5=0\n",
+ "a6=1/1440\n",
+ "a7=0\n",
+ "a8=-1/80640\n",
+ "b1=1\n",
+ "b2=a1\n",
+ "b3=a1*b2+a2*b1\n",
+ "b4=a1*b3+a2*b2+a3*b1\n",
+ "b5=a1*b4+a2*b3+a3*b2\n",
+ "b6=a1*b5+a2*b4+a3*b3\n",
+ "b7=a1*b6+a2*b5+a3*b4\n",
+ "b8=a1*b7+a2*b6+a3*b5\n",
+ "b9=a1*b8+a2*b7+a3*b6\n",
+ "print \"\\n%f\" %(b1/b2)\n",
+ "print \"\\n%f\" %(b2/b3)\n",
+ "print \"\\n%f\" %(b3/b4)\n",
+ "print \"\\n%f\" %(b4/b5)\n",
+ "print \"\\n%f\" %(b5/b6)\n",
+ "print \"\\n%f\" %(b6/b7)\n",
+ "print \"\\n%f\" %(b7/b8)\n",
+ "print \"\\n it appears as if the roots are converging at around %f\" %(b7/b8)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "0.666667\n",
+ "\n",
+ "0.600000\n",
+ "\n",
+ "0.606061\n",
+ "\n",
+ "0.605505\n",
+ "\n",
+ "0.605556\n",
+ "\n",
+ "0.605551\n",
+ "\n",
+ "0.605551\n",
+ "\n",
+ " it appears as if the roots are converging at around 0.605551\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.24:pg-47"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#ramanujan's method\n",
+ "#example 2.24\n",
+ "#page 47\n",
+ "import math\n",
+ "def f(x):\n",
+ " return 1-(x-x**2.0/math.factorial(2.0)**2.0+x**3.0/math.factorial(3.0)**2.0-x**4.0/math.factorial(4.0)**2.0)\n",
+ "a1=1\n",
+ "a2=-1/math.factorial(2.0)**2.0\n",
+ "a3=1/math.factorial(3.0)**2.0\n",
+ "a4=-1/math.factorial(4.0)**2.0\n",
+ "a5=-1/math.factorial(5.0)**2.0\n",
+ "a6=1/math.factorial(6.0)**2.0\n",
+ "b1=1\n",
+ "b2=a1\n",
+ "b3=a1*b2+a2*b1\n",
+ "b4=a1*b3+a2*b2+a3*b1\n",
+ "b5=a1*b4+a2*b3+a3*b2\n",
+ "print \"\\n\\n%f\" %(b1/b2)\n",
+ "print \"\\n\\n%f\" %(b2/b3)\n",
+ "print \"\\n%f\" %(b3/b4)\n",
+ "print \"\\n%f\" %(b4/b5)\n",
+ "print \"\\n it appears as if the roots are converging at around %f\" %(b4/b5)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ "1.000000\n",
+ "\n",
+ "\n",
+ "1.333333\n",
+ "\n",
+ "1.421053\n",
+ "\n",
+ "1.433962\n",
+ "\n",
+ " it appears as if the roots are converging at around 1.433962\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.25:pg-49"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.25\n",
+ "#secant method\n",
+ "#page 49\n",
+ "from __future__ import division\n",
+ "def f(x):\n",
+ " return x**3-2*x-5\n",
+ "x1=2\n",
+ "x2=3 # initial values\n",
+ "n=1\n",
+ "c=0\n",
+ "print \"successive iterations \\t x1 \\t x2 \\t x3 \\t f(x3)\\n\"\n",
+ "while n==1:\n",
+ " x3=(x1*f(x2)-x2*f(x1))/(f(x2)-f(x1)) \n",
+ " print \" \\t%f \\t%f \\t%f \\t%f\\n\" %(x1,x2,x3,f(x3))\n",
+ " if f(x3)*f(x1)>0:\n",
+ " x2=x3;\n",
+ " else:\n",
+ " x1=x3 \n",
+ " if abs(f(x3))<0.000001: \n",
+ " break\n",
+ " c=c+1\n",
+ "print \"the root of the equation after %i iteration is: %f\" %(c,x3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \t x1 \t x2 \t x3 \t f(x3)\n",
+ "\n",
+ " \t2.000000 \t3.000000 \t2.058824 \t-0.390800\n",
+ "\n",
+ " \t2.000000 \t2.058824 \t2.096559 \t0.022428\n",
+ "\n",
+ " \t2.096559 \t2.058824 \t2.094511 \t-0.000457\n",
+ "\n",
+ " \t2.094511 \t2.058824 \t2.094552 \t0.000009\n",
+ "\n",
+ " \t2.094552 \t2.058824 \t2.094551 \t-0.000000\n",
+ "\n",
+ "the root of the equation after 4 iteration is: 2.094551\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.26:pg-50"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.26\n",
+ "#secant method\n",
+ "#page 50\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "def f(x):\n",
+ " return x*math.exp(x)-1\n",
+ "x1=0\n",
+ "x2=1 # initial values\n",
+ "n=1\n",
+ "c=0 \n",
+ "print \"successive iterations \\t x1 \\t x2 \\t x3 \\t f(x3)\\n\"\n",
+ "while n==1:\n",
+ " x3=(x1*f(x2)-x2*f(x1))/(f(x2)-f(x1)) \n",
+ " print \" \\t%f \\t%f \\t%f \\t%f\\n\" %(x1,x2,x3,f(x3))\n",
+ " if f(x3)*f(x1)>0:\n",
+ " x2=x3\n",
+ " else:\n",
+ " x1=x3 \n",
+ " if abs(f(x3))<0.0001:\n",
+ " break\n",
+ " c=c+1\n",
+ "print \"the root of the equation after %i iteration is: %0.4g\" %(c,x3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \t x1 \t x2 \t x3 \t f(x3)\n",
+ "\n",
+ " \t0.000000 \t1.000000 \t0.367879 \t-0.468536\n",
+ "\n",
+ " \t0.000000 \t0.367879 \t0.692201 \t0.383091\n",
+ "\n",
+ " \t0.692201 \t0.367879 \t0.546310 \t-0.056595\n",
+ "\n",
+ " \t0.546310 \t0.367879 \t0.570823 \t0.010200\n",
+ "\n",
+ " \t0.570823 \t0.367879 \t0.566500 \t-0.001778\n",
+ "\n",
+ " \t0.566500 \t0.367879 \t0.567256 \t0.000312\n",
+ "\n",
+ " \t0.567256 \t0.367879 \t0.567124 \t-0.000055\n",
+ "\n",
+ "the root of the equation after 6 iteration is: 0.5671\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.27:pg-52"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# example 2.27\n",
+ "#mulller's method\n",
+ "#page 52\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "def f(x):\n",
+ " return x**3-x-1\n",
+ "x0=0\n",
+ "x1=1\n",
+ "x2=2 # initial values\n",
+ "n=1\n",
+ "c=0\n",
+ "print \"successive iterations \\t x0 \\t x1 \\t x2 \\t f(x0)\\t f(x1)\\t f(x2)\\n\"\n",
+ "while n==1: \n",
+ " c=c+1\n",
+ " y0=f(x0)\n",
+ " y1=f(x1)\n",
+ " y2=f(x2)\n",
+ " h2=x2-x1\n",
+ " h1=x1-x0\n",
+ " d2=f(x2)-f(x1)\n",
+ " d1=f(x1)-f(x0)\n",
+ " print \" \\t%f\\t %f\\t %f\\t %f\\t %f\\t %f\\n\" %(x0,x1,x2,f(x0),f(x1),f(x2))\n",
+ " A=(d2/h2-d1/h1)/(h1+h2)\n",
+ " B=d2/h2+A*h2\n",
+ " S=math.sqrt(B**2-4*A*f(x2))\n",
+ " x3=x2-(2*f(x2))/(B+S)\n",
+ " E=abs((x3-x2)/x2)*100\n",
+ " if E<0.003:\n",
+ " break\n",
+ " else:\n",
+ " if c==1:\n",
+ " x2=x3\n",
+ " if c==2:\n",
+ " x1=x2\n",
+ " x2=x3\n",
+ " if c==3:\n",
+ " x0=x1\n",
+ " x1=x2\n",
+ " x2=x3\n",
+ " if c==3:\n",
+ " c=0\n",
+ "print \"the required root is : %0.4f\" %(x3)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \t x0 \t x1 \t x2 \t f(x0)\t f(x1)\t f(x2)\n",
+ "\n",
+ " \t0.000000\t 1.000000\t 2.000000\t -1.000000\t -1.000000\t 5.000000\n",
+ "\n",
+ " \t0.000000\t 1.000000\t 1.263763\t -1.000000\t -1.000000\t -0.245412\n",
+ "\n",
+ " \t0.000000\t 1.263763\t 1.331711\t -1.000000\t -0.245412\t 0.030015\n",
+ "\n",
+ " \t1.263763\t 1.331711\t 1.324583\t -0.245412\t 0.030015\t -0.000574\n",
+ "\n",
+ " \t1.263763\t 1.331711\t 1.324718\t -0.245412\t 0.030015\t -0.000000\n",
+ "\n",
+ "the required root is : 1.3247\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.28:pg-55"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#graeffe's method\n",
+ "#example 2.28\n",
+ "#page 55\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "def f(x):\n",
+ " return x**3-6*(x**2)+11*x-6\n",
+ "#x=poly(0,'x')\n",
+ "#g=f(-x)\n",
+ "print \"the equation is:\\n\"\n",
+ "A=[1, 14, 49, 36] #coefficients of the above equation\n",
+ "print \"%0.4g\\n\" %(math.sqrt(A[3]/A[2]))\n",
+ "print \"%0.4g\\n\" %(math.sqrt(A[2]/A[1]))\n",
+ "print \"%0.4g\\n\" %(math.sqrt(A[1]/A[0]))\n",
+ "print \"the equation is:\\n\"\n",
+ "#disp(g*(-1*g));\n",
+ "B=[1, 98, 1393, 1296]\n",
+ "print \"%0.4g\\n\" %((B[3]/B[2])**(1/4))\n",
+ "print \"%0.4g\\n\" %((B[2]/B[1])**(1/4))\n",
+ "print \"%0.4g\\n\" %((B[1]/B[0])**(1/4))\n",
+ "print \"It is apparent from the outputs that the roots converge at 1 2 3\"\n",
+ "\n",
+ "\n",
+ "\n",
+ "#INCOMPLETE"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the equation is:\n",
+ "\n",
+ "0.8571\n",
+ "\n",
+ "1.871\n",
+ "\n",
+ "3.742\n",
+ "\n",
+ "the equation is:\n",
+ "\n",
+ "0.9821\n",
+ "\n",
+ "1.942\n",
+ "\n",
+ "3.146\n",
+ "\n",
+ "It is apparent from the outputs that the roots converge at 1 2 3\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.29:pg-57"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#quadratic factor by lin's--bairsttow method\n",
+ "#example 2.29\n",
+ "#page 57\n",
+ "from numpy import matrix\n",
+ "from __future__ import division\n",
+ "def f(x):\n",
+ " return x**3-x-1\n",
+ "a=[-1, -1, 0, 1]\n",
+ "r1=1\n",
+ "s1=1\n",
+ "b4=a[3]\n",
+ "def f3(r):\n",
+ " return a[2]-r*a[3]\n",
+ "def f2(r,s):\n",
+ " return a[1]-r*a[2]+r**2*a[3]-s*a[3]\n",
+ "def f1(r,s):\n",
+ " return a[0]-s*a[2]+s*r*a[3]\n",
+ "A=matrix([[1,1],[2,-1]])\n",
+ "C=matrix([[0],[1]])\n",
+ "X=A.I*C\n",
+ "X1=[[ 0.33333333],[-0.33333333]]\n",
+ "dr=X1[0][0]\n",
+ "ds=X1[1][0]\n",
+ "r2=r1+dr\n",
+ "s2=s1+ds\n",
+ "#second pproximation\n",
+ "r1=r2\n",
+ "s1=s2\n",
+ "b11=f1(r2,s2)\n",
+ "b22=f2(r2,s2)\n",
+ "h=0.001\n",
+ "dr_b1=(f1(r1+h,s1)-f1(r1,s1))/h\n",
+ "ds_b1=(f1(r1,s1+h)-f1(r1,s1))/h\n",
+ "dr_b2=(f2(r1+h,s1)-f2(r1,s1))/h\n",
+ "ds_b2=(f2(r1,s1+h)-f2(r1,s1))/h\n",
+ "A=matrix([[dr_b1,ds_b1],[dr_b2,ds_b2]])\n",
+ "C=matrix([[-f1(r1,s1)],[-f2(r1,s2)]])\n",
+ "X=A.I*C\n",
+ "r2=r1+X[0][0]\n",
+ "s2=s1+X[1][0]\n",
+ "print \"roots correct to 3 decimal places are : %0.3f %0.3f\" %(r2,s2)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "roots correct to 3 decimal places are : 1.325 0.754\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.31:pg-62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#method of iteration\n",
+ "#example 2.31\n",
+ "#page 62\n",
+ "from __future__ import division\n",
+ "def f(x,y):\n",
+ " return (3*y*x**2+7)/10\n",
+ "def g(x,y):\n",
+ " return (y**2+4)/5\n",
+ "h=0.0001\n",
+ "x0=0.5\n",
+ "y0=0.5\n",
+ "f1_dx=(f(x0+h,y0)-f(x0,y0))/h\n",
+ "f1_dy=(f(x0,y0+h)-f(x0,y0))/h\n",
+ "g1_dx=(g(x0+h,y0)-g(x0,y0))/h\n",
+ "g1_dy=(g(x0+h,y0)-g(x0,y0))/h\n",
+ "if (f1_dx+f1_dy<1) and (g1_dx+g1_dy<1): \n",
+ " print \"coditions for convergence is satisfied\\n\\n\"\n",
+ "print \"X \\t Y\\t\\n\\n\"\n",
+ "for i in range(0,10):\n",
+ " X=(3*y0*x0**2+7)/10\n",
+ " Y=(y0**2+4)/5\n",
+ " print \"%f\\t %f\\t\\n\" %(X,Y)\n",
+ " x0=X\n",
+ " y0=Y\n",
+ "print \"\\n\\n CONVERGENCE AT (1 1) IS OBVIOUS\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "coditions for convergence is satisfied\n",
+ "\n",
+ "\n",
+ "X \t Y\t\n",
+ "\n",
+ "\n",
+ "0.737500\t 0.850000\t\n",
+ "\n",
+ "0.838696\t 0.944500\t\n",
+ "\n",
+ "0.899312\t 0.978416\t\n",
+ "\n",
+ "0.937391\t 0.991460\t\n",
+ "\n",
+ "0.961360\t 0.996598\t\n",
+ "\n",
+ "0.976320\t 0.998642\t\n",
+ "\n",
+ "0.985572\t 0.999457\t\n",
+ "\n",
+ "0.991247\t 0.999783\t\n",
+ "\n",
+ "0.994707\t 0.999913\t\n",
+ "\n",
+ "0.996807\t 0.999965\t\n",
+ "\n",
+ "\n",
+ "\n",
+ " CONVERGENCE AT (1 1) IS OBVIOUS\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.32:pg-65"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#newton raphson method\n",
+ "#example 2.32\n",
+ "#page 65\n",
+ "def f(x,y):\n",
+ " return 3*y*x**2-10*x+7\n",
+ "def g(y):\n",
+ " return y**2-5*y+4\n",
+ "hh=0.0001\n",
+ "x0=0.5\n",
+ "y0=0.5 #initial values\n",
+ "f0=f(x0,y0)\n",
+ "g0=g(y0)\n",
+ "df_dx=(f(x0+hh,y0)-f(x0,y0))/hh\n",
+ "df_dy=(f(x0,y0+hh)-f(x0,y0))/hh\n",
+ "dg_dx=(g(y0)-g(y0))/hh\n",
+ "dg_dy=(g(y0+hh)-g(y0))/hh\n",
+ "d=[[df_dx,df_dy],[dg_dx,dg_dy]]\n",
+ "D1=det(d)\n",
+ "dd=[[-f0,df_dy],[-g0,dg_dy]]\n",
+ "h=det(dd)/D1\n",
+ "ddd=[[df_dx,-f0],[dg_dx,-g0]]\n",
+ "k=det(ddd)/D1;\n",
+ "x1=x0+h\n",
+ "y1=y0+k\n",
+ "f0=f(x1,y1)\n",
+ "g0=g(y1)\n",
+ "df_dx=(f(x1+hh,y1)-f(x1,y1))/hh\n",
+ "df_dy=(f(x1,y1+hh)-f(x1,y1))/hh\n",
+ "dg_dx=(g(y1)-g(y1))/hh\n",
+ "dg_dy=(g(y1+hh)-g(y1))/hh\n",
+ "dddd=[[df_dx,df_dy],[dg_dx,dg_dy]]\n",
+ "D2=det(dddd)\n",
+ "ddddd=[[-f0,df_dy],[-g0,dg_dy]]\n",
+ "h=det(ddddd)/D2\n",
+ "d6=[[df_dx,-f0],[dg_dx,-g0]]\n",
+ "k=det(d6)/D2\n",
+ "x2=x1+h\n",
+ "y2=y1+k\n",
+ "print \" the roots of the equation are x2=%f and y2=%f\" %(x2,y2)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " the roots of the equation are x2=0.970803 and y2=0.998752\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.33:pg-66"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#newton raphson method\n",
+ "#example 2.33\n",
+ "#page 66\n",
+ "import math\n",
+ "def f(x,y):\n",
+ " return x**2+y**2-1\n",
+ "def g(x,y):\n",
+ " return y-x**2\n",
+ "hh=0.0001\n",
+ "x0=0.7071\n",
+ "y0=0.7071 #initial values\n",
+ "f0=f(x0,y0)\n",
+ "g0=g(x0,y0)\n",
+ "df_dx=(f(x0+hh,y0)-f(x0,y0))/hh\n",
+ "df_dy=(f(x0,y0+hh)-f(x0,y0))/hh\n",
+ "dg_dx=(g(x0+hh,y0)-g(x0,y0))/hh\n",
+ "dg_dy=(g(x0,y0+hh)-g(x0,y0))/hh\n",
+ "D1=det([[df_dx,df_dy],[dg_dx,dg_dy]])\n",
+ "h=det([[-f0,df_dy],[-g0,dg_dy]])/D1\n",
+ "k=det([[df_dx,-f0],[dg_dx,-g0]])/D1\n",
+ "x1=x0+h\n",
+ "y1=y0+k\n",
+ "f0=f(x1,y1)\n",
+ "g0=g(x1,y1)\n",
+ "df_dx=(f(x1+hh,y1)-f(x1,y1))/hh\n",
+ "df_dy=(f(x1,y1+hh)-f(x1,y1))/hh\n",
+ "dg_dx=(g(x1+hh,y1)-g(x1,y1))/hh\n",
+ "dg_dy=(g(x1,y1+hh)-g(x1,y1))/hh\n",
+ "D2=det([[df_dx,df_dy],[dg_dx,dg_dy]])\n",
+ "h=det([[-f0,df_dy],[-g0,dg_dy]])/D2\n",
+ "k=det([[df_dx,-f0],[dg_dx,-g0]])/D2\n",
+ "x2=x1+h\n",
+ "y2=y1+k\n",
+ "print \"the roots of the equation are x2=%f and y2=%f \" %(x2,y2)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the roots of the equation are x2=0.786184 and y2=0.618039 \n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.34:pg-67"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#newton raphson method\n",
+ "#example 2.34\n",
+ "#page 67\n",
+ "import math\n",
+ "def f(x,y):\n",
+ " return math.sin(x)-y+0.9793\n",
+ "def g(x,y):\n",
+ " return math.cos(y)-x+0.6703\n",
+ "hh=0.0001\n",
+ "x0=0.5\n",
+ "y0=1.5 #initial values\n",
+ "f0=f(x0,y0)\n",
+ "g0=g(x0,y0)\n",
+ "df_dx=(f(x0+hh,y0)-f(x0,y0))/hh\n",
+ "df_dy=(f(x0,y0+hh)-f(x0,y0))/hh\n",
+ "dg_dx=(g(x0+hh,y0)-g(x0,y0))/hh\n",
+ "dg_dy=(g(x0,y0+hh)-g(x0,y0))/hh\n",
+ "d1=[[df_dx,df_dy],[dg_dx,dg_dy]]\n",
+ "D1=det(d1)\n",
+ "d2=[[-f0,df_dy],[-g0,dg_dy]]\n",
+ "h=det(d2)/D1\n",
+ "d3=[[df_dx,-f0],[dg_dx,-g0]]\n",
+ "k=det(d3)/D1\n",
+ "x1=x0+h\n",
+ "y1=y0+k\n",
+ "f0=f(x1,y1)\n",
+ "g0=g(x1,y1)\n",
+ "df_dx=(f(x1+hh,y1)-f(x1,y1))/hh\n",
+ "df_dy=(f(x1,y1+hh)-f(x1,y1))/hh\n",
+ "dg_dx=(g(x1+hh,y1)-g(x1,y1))/hh\n",
+ "dg_dy=(g(x1,y1+hh)-g(x1,y1))/hh\n",
+ "d4=[[df_dx,df_dy],[dg_dx,dg_dy]]\n",
+ "D2=det(d4)\n",
+ "h=det([[-f0,df_dy],[-g0,dg_dy]])/D2\n",
+ "k=det([[df_dx,-f0],[dg_dx,-g0]])/D2\n",
+ "x2=x1+h\n",
+ "y2=y1+k\n",
+ "print \"the roots of the equation are x2=%0.4f and y2=%0.4f\" %(x2,y2)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the roots of the equation are x2=0.6537 and y2=1.5874\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter3_5.ipynb b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter3_5.ipynb new file mode 100644 index 00000000..fbba6967 --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter3_5.ipynb @@ -0,0 +1,1126 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:0755817d9df79392f0a505cb49e15463e2d17b0d0bd1a381990227b29ae2b639"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter03:Interpolation"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.4:pg-86"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 3.4\n",
+ "#interpolation\n",
+ "#page 86\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "x=[1, 3, 5, 7]\n",
+ "y=[24, 120, 336, 720]\n",
+ "d1=[0,0,0]\n",
+ "d2=[0,0,0]\n",
+ "d3=[0,0,0]\n",
+ "h=2 #interval between values of x\n",
+ "c=0\n",
+ "for i in range(0,3):\n",
+ " d1[c]=y[i+1]-y[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,2):\n",
+ " d2[c]=d1[i+1]-d1[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,1):\n",
+ " d3[c]=d2[i+1]-d2[i]\n",
+ " c=c+1\n",
+ "d=[0,d1[0],d2[0],d3[0]]\n",
+ "x0=8 #value at 8\n",
+ "pp=1\n",
+ "y_x=y[0]\n",
+ "p=(x0-1)/2\n",
+ "for i in range(1,4):\n",
+ " pp=1\n",
+ " for j in range(0,i):\n",
+ " pp=pp*(p-(j)) \n",
+ " y_x=y_x+(pp*d[i])/math.factorial(i)\n",
+ "print \"value of function at %f is :%f\" %(x0,y_x)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "value of function at 8.000000 is :990.000000\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.6:pg-87"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 3.6\n",
+ "#interpolation\n",
+ "#page 87\n",
+ "x=[15, 20, 25, 30, 35, 40]\n",
+ "y=[0.2588190, 0.3420201, 0.4226183, 0.5, 0.5735764, 0.6427876]\n",
+ "d1=[0,0,0,0,0]\n",
+ "d2=[0,0,0,0]\n",
+ "d3=[0,0,0]\n",
+ "d4=[0,0]\n",
+ "d5=[0]\n",
+ "h=5 #interval between values of x\n",
+ "c=0\n",
+ "for i in range(0,5):\n",
+ " d1[c]=y[i+1]-y[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,4):\n",
+ " d2[c]=d1[i+1]-d1[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,3):\n",
+ " d3[c]=d2[i+1]-d2[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,2):\n",
+ " d4[c]=d3[i+1]-d3[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,1):\n",
+ " d5[c]=d4[i+1]-d4[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "d=[0,d1[0], d2[0], d3[0], d4[0], d5[0]]\n",
+ "x0=38 #value at 38 degree\n",
+ "pp=1\n",
+ "y_x=y[0]\n",
+ "p=(x0-x[0])/h\n",
+ "for i in range(1,6):\n",
+ " pp=1\n",
+ " for j in range(0,i):\n",
+ " pp=pp*(p-(j)) \n",
+ " y_x=y_x+((pp*d[i])/math.factorial(i));\n",
+ "print \"value of function at %i is :%f\" %(x0,y_x)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "value of function at 38 is :0.615661\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.7:pg-89"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 3.7\n",
+ "#interpolation\n",
+ "#page 89\n",
+ "x=[0, 1, 2, 4]\n",
+ "y=[1, 3, 9, 81]\n",
+ "#equation is y(5)-4*y(4)+6*y(2)-4*y(2)+y(1)\n",
+ "y3=(y[3]+6*y[2]-4*y[1]+y[0])/4\n",
+ "print \"the value of missing term of table is :%d\" %(y3)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value of missing term of table is :31\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.8:pg-89"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 3.8\n",
+ "#interpolation\n",
+ "#page 89\n",
+ "import math\n",
+ "x=[0.10, 0.15, 0.20, 0.25, 0.30]\n",
+ "y=[0.1003, 0.1511, 0.2027, 0.2553, 0.3093]\n",
+ "d1=[0,0,0,0,0]\n",
+ "d2=[0,0,0,0,0]\n",
+ "d3=[0,0,0,0,0]\n",
+ "d4=[0,0,0,0,0]\n",
+ "h=0.05 #interval between values of x\n",
+ "c=0\n",
+ "for i in range(0,4):\n",
+ " d1[c]=y[i+1]-y[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,3):\n",
+ " d2[c]=d1[i+1]-d1[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,2):\n",
+ " d3[c]=d2[i+1]-d2[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,4):\n",
+ " d4[c]=d3[i+1]-d3[i]\n",
+ " c=c+1\n",
+ "d=[0,d1[0], d2[0], d3[0], d4[0]]\n",
+ "x0=0.12 #value at 0.12;\n",
+ "pp=1\n",
+ "y_x=y[0]\n",
+ "p=(x0-x[0])/h\n",
+ "for i in range(1,5):\n",
+ " pp=1;\n",
+ " for j in range(0,i):\n",
+ " pp=pp*(p-(j)) \n",
+ " y_x=y_x+(pp*d[i])/math.factorial(i)\n",
+ "print \"value of function at %f is :%0.4g\\n \\n\" %(x0,y_x)\n",
+ "x0=0.26 #value at 0.26;\n",
+ "pp=1\n",
+ "y_x=y[0]\n",
+ "p=(x0-x[0])/h\n",
+ "for i in range(1,5):\n",
+ " pp=1\n",
+ " for j in range(0,i):\n",
+ " pp=pp*(p-(j)) \n",
+ " y_x=y_x+(pp*d[i])/math.factorial(i);\n",
+ "print \"value of function at %f is :%0.4g\\n \\n\" %(x0,y_x)\n",
+ "x0=0.40 #value at 0.40;\n",
+ "pp=1\n",
+ "y_x=y[0]\n",
+ "p=(x0-x[0])/h\n",
+ "for i in range(1,5):\n",
+ " pp=1\n",
+ " for j in range(0,i):\n",
+ " pp=pp*(p-(j)) \n",
+ " y_x=y_x+(pp*d[i])/math.factorial(i)\n",
+ "print \"value of function at %f is :%0.4g\\n \\n\" %(x0,y_x)\n",
+ "x0=0.50 #value at 0.50;\n",
+ "pp=1\n",
+ "y_x=y[0]\n",
+ "p=(x0-x[0])/h\n",
+ "for i in range(1,5):\n",
+ " pp=1\n",
+ " for j in range(0,i):\n",
+ " pp=pp*(p-(j)) \n",
+ " y_x=y_x+(pp*d[i])/math.factorial(i)\n",
+ "print \"value of function at %f is :%0.5g\\n \\n\" %(x0,y_x)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "value of function at 0.120000 is :0.1205\n",
+ " \n",
+ "\n",
+ "value of function at 0.260000 is :0.266\n",
+ " \n",
+ "\n",
+ "value of function at 0.400000 is :0.4241\n",
+ " \n",
+ "\n",
+ "value of function at 0.500000 is :0.5543\n",
+ " \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.9:pg-93"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 3.9\n",
+ "#Gauss' forward formula\n",
+ "#page 93\n",
+ "x=[1.0, 1.05, 1.10, 1.15, 1.20, 1.25, 1.30];\n",
+ "y=[2.7183, 2.8577, 3.0042, 3.1582, 3.3201, 3.4903, 3.66693]\n",
+ "d1=[0,0,0,0,0,0]\n",
+ "d2=[0,0,0,0,0]\n",
+ "d3=[0,0,0,0]\n",
+ "d4=[0,0,0]\n",
+ "d5=[0,0]\n",
+ "d6=[0]\n",
+ "h=0.05 #interval between values of x\n",
+ "c=0\n",
+ "for i in range(0,6):\n",
+ " d1[c]=y[i+1]-y[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,5):\n",
+ " d2[c]=d1[i+1]-d1[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,4):\n",
+ " d3[c]=d2[i+1]-d2[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,3):\n",
+ " d4[c]=d3[i+1]-d3[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,2):\n",
+ " d5[c]=d4[i+1]-d4[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,1):\n",
+ " d6[c]=d5[i+1]-d5[i]\n",
+ " c=c+1\n",
+ "d=[0,d1[3], d2[2], d3[2], d4[1], d5[0], d6[0]]\n",
+ "x0=1.17 #value at 1.17;\n",
+ "pp=1\n",
+ "y_x=y[3]\n",
+ "p=(x0-x[3])/h\n",
+ "for i in range(1,6):\n",
+ " pp=1;\n",
+ " for j in range(0,i):\n",
+ " pp=pp*(p-(j)) \n",
+ " y_x=y_x+(pp*d[i])/math.factorial(i)\n",
+ "print \"value of function at %f is :%0.4g\\n \\n\" %(x0,y_x)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "value of function at 1.170000 is :3.222\n",
+ " \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.10:pg-97"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#practical interpolation\n",
+ "#example 3.10\n",
+ "#page 97\n",
+ "import math\n",
+ "x=[0.61, 0.62, 0.63, 0.64, 0.65, 0.66, 0.67]\n",
+ "y=[1.840431, 1.858928,1.877610, 1.896481, 1.915541, 1.934792, 1.954237]\n",
+ "d1=[0,0,0,0,0,0]\n",
+ "d2=[0,0,0,0,0]\n",
+ "d3=[0,0,0,0]\n",
+ "d4=[0,0,0]\n",
+ "h=0.01 #interval between values of x\n",
+ "c=0\n",
+ "for i in range(0,6):\n",
+ " d1[c]=y[i+1]-y[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,5):\n",
+ " d2[c]=d1[i+1]-d1[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,4):\n",
+ " d3[c]=d2[i+1]-d2[i];\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,3):\n",
+ " d4[c]=d3[i+1]-d3[i];\n",
+ " c=c+1\n",
+ "d=[d1[0], d2[0], d3[0], d4[0]]\n",
+ "x0=0.644\n",
+ "p=(x0-x[3])/h;\n",
+ "y_x=y[3]\n",
+ "y_x=y_x+p*(d1[2]+d1[3])/2+p**2*(d2[1])/2 #stirling formula\n",
+ "print \"the value at %f by stirling formula is : %f\\n\\n\" %(x0,y_x)\n",
+ "y_x=y[3]\n",
+ "y_x=y_x+p*d1[3]+p*(p-1)*(d2[2]+d2[3])/2\n",
+ "print \" the value at %f by bessels formula is : %f\\n\\n\" %(x0,y_x)\n",
+ "y_x=y[3]\n",
+ "q=1-p\n",
+ "y_x=q*y[3]+q*(q**2-1)*d2[2]/2+p*y[4]+p*(q**2-1)*d2[4]/2\n",
+ "print \"the value at %f by everrets formula is : %f\\n\\n\" %(x0,y_x)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value at 0.644000 by stirling formula is : 1.904082\n",
+ "\n",
+ "\n",
+ " the value at 0.644000 by bessels formula is : 1.904059\n",
+ "\n",
+ "\n",
+ "the value at 0.644000 by everrets formula is : 1.904044\n",
+ "\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 37
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.11:pg-99"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#practical interpolation\n",
+ "#example 3.11\n",
+ "#page 99\n",
+ "x=[0.61, 0.62, 0.63, 0.64, 0.65, 0.66, 0.67]\n",
+ "y=[1.840431, 1.858928, 1.877610, 1.896481, 1.915541, 1.934792, 1.954237]\n",
+ "d1=[0,0,0,0,0,0]\n",
+ "d2=[0,0,0,0,0]\n",
+ "d3=[0,0,0,0]\n",
+ "d4=[0,0,0]\n",
+ "h=0.01 #interval between values of x\n",
+ "c=0\n",
+ "for i in range(0,6):\n",
+ " d1[c]=y[i+1]-y[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,5):\n",
+ " d2[c]=d1[i+1]-d1[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,4):\n",
+ " d3[c]=d2[i+1]-d2[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,3):\n",
+ " d4[c]=d3[i+1]-d3[i]\n",
+ " c=c+1\n",
+ "d=[d1[0], d2[0], d3[0], d4[0]]\n",
+ "x0=0.638\n",
+ "p=(x0-x[3])/h\n",
+ "y_x=y[3]\n",
+ "y_x=y_x+p*(d1[2]+d1[3])/2+p**2*(d2[1])/2 #stirling formula\n",
+ "print \"value at %f by stirling formula is : %f\\n\\n\" %(x0,y_x)\n",
+ "y_x=y[2]\n",
+ "p=(x0-x[2])/h\n",
+ "y_x=y_x+p*d1[2]+p*(p-1)*(d2[1])/2\n",
+ "print \"the value at %f by bessels formula is : %f\\n\\n\" %(x0,y_x)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "value at 0.638000 by stirling formula is : 1.892692\n",
+ "\n",
+ "\n",
+ "the value at 0.638000 by bessels formula is : 1.892692\n",
+ "\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.12:pg-99"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#practical interpolation\n",
+ "#example 3.12\n",
+ "#page 99\n",
+ "x=[1.72, 1.73, 1.74, 1.75, 1.76, 1.77, 1.78]\n",
+ "y=[0.1790661479, 0.1772844100, 0.1755204006, 0.1737739435, 0.1720448638, 0.1703329888, 0.1686381473]\n",
+ "d1=[0,0,0,0,0,0]\n",
+ "d2=[0,0,0,0,0]\n",
+ "d3=[0,0,0,0]\n",
+ "d4=[0,0,0]\n",
+ "h=0.01 #interval between values of x\n",
+ "c=0\n",
+ "for i in range(0,6):\n",
+ " d1[c]=y[i+1]-y[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,5):\n",
+ " d2[c]=d1[i+1]-d1[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,4):\n",
+ " d3[c]=d2[i+1]-d2[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,3):\n",
+ " d4[c]=d3[i+1]-d3[i]\n",
+ " c=c+1\n",
+ "x0=1.7475\n",
+ "y_x=y[2]\n",
+ "p=(x0-x[2])/h\n",
+ "y_x=y_x+p*d1[2]+p*(p-1)*((d2[1]+d2[2])/2)/2\n",
+ "print \"the value at %f by bessels formula is : %0.10f\\n\\n\" %(x0,y_x)\n",
+ "y_x=y[3]\n",
+ "q=1-p\n",
+ "y_x=q*y[2]+q*(q**2-1)*d2[1]/6+p*y[3]+p*(p**2-1)*d2[1]/6\n",
+ "print \"the value at %f by everrets formula is : %0.10f\\n\\n\" %(x0,y_x)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value at 1.747500 by bessels formula is : 0.1742089204\n",
+ "\n",
+ "\n",
+ "the value at 1.747500 by everrets formula is : 0.1742089122\n",
+ "\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 41
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.13:pg-104"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 3.13\n",
+ "#lagrange's interpolation formula\n",
+ "#page 104\n",
+ "x=[300, 304, 305, 307]\n",
+ "y=[2.4771, 2.4829, 2.4843, 2.4871]\n",
+ "x0=301\n",
+ "log_301=(-3*-4*-6*2.4771)/(-4*-5*-7)+(-4*-6*2.4829)/(4*-1*-3)+(-3*-6*2.4843)/(5*-2)+(-3*-4*2.4871)/(7*3*2)\n",
+ "print \"valie of log x at 301 is =%f\" %(log_301)\n",
+ "\n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "valie of log x at 301 is =2.478597\n"
+ ]
+ }
+ ],
+ "prompt_number": 43
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.14:pg-105"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 3.14\n",
+ "#lagrange's interpolation formula\n",
+ "#page 105\n",
+ "y=[4, 12, 19]\n",
+ "x=[1, 3, 4];\n",
+ "y_x=7\n",
+ "Y_X=(-5*-12)/(-8*-15)+(3*3*-12)/(8*-7)+(3*-5*4)/(15*7)\n",
+ "print \"values is %f\" %(Y_X)\n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "values is 1.857143\n"
+ ]
+ }
+ ],
+ "prompt_number": 44
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.15:pg-105"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 3.15\n",
+ "#lagrange's interpolation formula\n",
+ "#page 105\n",
+ "x=[2, 2.5, 3.0]\n",
+ "y=[0.69315, 0.91629, 1.09861]\n",
+ "def l0(x):\n",
+ " return (x-2.5)*(x-3.0)/(-0.5)*(-1.0)\n",
+ "def l1(x):\n",
+ " return ((x-2.0)*(x-3.0))/((0.5)*(-0.5))\n",
+ "def l2(x):\n",
+ " return ((x-2.0)*(x-2.5))/((1.0)*(0.5))\n",
+ "f_x=l0(2.7)*y[0]+l1(2.7)*y[1]+l2(2.7)*y[2];\n",
+ "print \"the calculated value is %f:\" %(f_x)\n",
+ "print \"\\n\\n the error occured in the value is %0.9f\" %(abs(f_x-log(2.7)))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the calculated value is 0.994116:\n",
+ "\n",
+ "\n",
+ " the error occured in the value is 0.000864627\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.16:pg-106"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 3.16\n",
+ "#lagrange's interpolation formula\n",
+ "#page 106\n",
+ "import math\n",
+ "x=[0, math.pi/4,math.pi/2]\n",
+ "y=[0, 0.70711, 1.0];\n",
+ "x0=math.pi/6\n",
+ "sin_x0=0\n",
+ "for i in range(0,3):\n",
+ " p=y[i]\n",
+ " for j in range(0,3):\n",
+ " if j!=i:\n",
+ " p=p*((x0-x[j])/( x[i]-x[j]))\n",
+ " sin_x0=sin_x0+p\n",
+ "print \"sin_x0=%f\" %(sin_x0)\n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "sin_x0=0.517431\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.18:pg-107"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#error in lagrange's interpolation formula\n",
+ "#example 3.18\n",
+ "#page 107\n",
+ "import math\n",
+ "x=[2, 2.5, 3.0]\n",
+ "y=[0.69315, 0.91629, 1.09861]\n",
+ "def l0(x):\n",
+ " return (x-2.5)*(x-3.0)/(-0.5)*(-1.0)\n",
+ "def l1(x):\n",
+ " return ((x-2.0)*(x-3.0))/((0.5)*(-0.5))\n",
+ "def l2(x):\n",
+ " return ((x-2.0)*(x-2.5))/((1.0)*(0.5))\n",
+ "f_x=l0(2.7)*y[0]+l1(2.7)*y[1]+l2(2.7)*y[2]\n",
+ "print \"the calculated value is %f:\" %(f_x)\n",
+ "err=math.fabs(f_x-math.log10(2.7))\n",
+ "def R_n(x):\n",
+ " return (((x-2)*(x-2.5)*(x-3))/6)\n",
+ "est_err=abs(R_n(2.7)*(2/8))\n",
+ "if est_err<err:\n",
+ " print \"\\n\\n the error agrees with the actual error\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the calculated value is 0.994116:\n",
+ "\n",
+ "\n",
+ " the error agrees with the actual error\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.19:pg-107"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#error in lagrenge's interpolation\n",
+ "#example 3.19\n",
+ "#page 107\n",
+ "import math\n",
+ "x=[0, math.pi/4 ,math.pi/2]\n",
+ "y=[0, 0.70711, 1.0]\n",
+ "def l0(x):\n",
+ " return ((x-0)*(x-math.pi/2))/((math.pi/4)*(-1*math.pi/4))\n",
+ "def l1(x):\n",
+ " return ((x-0)*(x-math.pi/4))/((math.pi/2)*(math.pi/4))\n",
+ "f_x=l0(math.pi/6)*y[1]+l1(math.pi/6)*y[2]\n",
+ "err=abs(f_x-math.sin(math.pi/6))\n",
+ "def f(x):\n",
+ " return ((x-0)*(x-math.pi/4)*(x-math.pi/2))/6\n",
+ "if abs(f(math.pi/6))>err:\n",
+ " print \"\\n\\n the error agrees with the actual error\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " the error agrees with the actual error\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.21:pg-110"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#hermite's interpolation formula\n",
+ "#exammple 3.21\n",
+ "#page 110\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "x=[2.0, 2.5, 3.0]\n",
+ "y=[0.69315, 0.91629, 1.09861]\n",
+ "y1=[0,0,0]\n",
+ "def f(x):\n",
+ " return math.log(x)\n",
+ "h=0.0001\n",
+ "for i in range(0,3):\n",
+ " y1[i]=(f(x[i]+h)-f(x[i]))/h\n",
+ "def l0(x):\n",
+ " return (x-2.5)*(x-3.0)/(-0.5)*(-1.0)\n",
+ "def l1(x):\n",
+ " return ((x-2.0)*(x-3.0))/((0.5)*(-0.5))\n",
+ "def l2(x):\n",
+ " return ((x-2.0)*(x-2.5))/((1.0)*(0.5))\n",
+ "dl0=(l0(x[0]+h)-l0(x[0]))/h\n",
+ "dl1=(l1(x[1]+h)-l1(x[1]))/h\n",
+ "dl2=(l2(x[2]+h)-l2(x[2]))/h\n",
+ "x0=2.7\n",
+ "u0=(1-2*(x0-x[0])*dl0)*(l0(x0))**2\n",
+ "u1=(1-2*(x0-x[1])*dl1)*(l1(x0))**2\n",
+ "u2=(1-2*(x0-x[2])*dl2)*(l2(x0))**2\n",
+ "v0=(x0-x[0])*l0(x0)**2\n",
+ "v1=(x0-x[1])*l1(x0)**2\n",
+ "v2=(x0-x[2])*l2(x0)**2\n",
+ "H=u0*y[0]+u1*y[1]+u2*y[2]+v0*y1[0]+v1*y1[1]+v2*y1[2]\n",
+ "print \"the approximate value of ln(%0.2f) is %f:\" %(x0,H)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the approximate value of ln(2.70) is 0.993362:\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.22:pg-114"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#newton's general interpolation formula\n",
+ "#example 3.22\n",
+ "#page 114\n",
+ "x=[300, 304, 305, 307]\n",
+ "y=[2.4771, 2.4829, 2.4843, 2.4871]\n",
+ "d1=[0,0,0]\n",
+ "d2=[0,0]\n",
+ "for i in range(0,3):\n",
+ " d1[i]=(y[i+1]-y[i])/(x[i+1]-x[i])\n",
+ "for i in range(0,2):\n",
+ " d2[i]=(d1[i+1]-d1[i])/(x[i+2]-x[i])\n",
+ "x0=301\n",
+ "log301=y[0]+(x0-x[0])*d1[0]+(x0-x[1])*d2[0]\n",
+ "print \"valure of log(%d) is :%0.4f\" %(x0,log301)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "valure of log(301) is :2.4786\n"
+ ]
+ }
+ ],
+ "prompt_number": 37
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.23:pg-114"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 3.23\n",
+ "#newton's divided formula\n",
+ "#page 114\n",
+ "x=[-1, 0, 3, 6, 7]\n",
+ "y=[3, -6, 39, 822, 1611]\n",
+ "d1=[0,0,0,0,0]\n",
+ "d2=[0,0,0,0,0]\n",
+ "d3=[0,0,0,0,0]\n",
+ "d4=[0,0,0,0,0]\n",
+ "X=0\n",
+ "for i in range(0,3):\n",
+ " d1[i]=(y[i+1]-y[i])/(x[i+1]-x[i])\n",
+ "for i in range(0,3):\n",
+ " d2[i]=(d1[i+1]-d1[i])/(x[i+2]-x[i])\n",
+ "for i in range(0,2):\n",
+ " d3[i]=(d2[i+1]-d2[i])/(x[i+3]-x[i])\n",
+ "for i in range(0,1):\n",
+ " d4[i]=(d3[i+1]-d3[i])/(x[i+4]-x[i])\n",
+ "f_x=y[0]+(X-x[0])*d1[0]+(X-x[1])*(X-x[0])*d2[0]+(X-x[0])*(X-x[1])*(X-x[2])*d3[0]+(X-x[0])*(X-x[1])*(X-x[2])*(X-x[3])*d4[0]\n",
+ "print \"the polynomial equation is = -6 + 5X^2 -3X^3 +X^4\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the polynomial equation is = -6 + 5X^2 -3X^3 +X^4\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.24:pg-116"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#interpolation by iteration\n",
+ "#example 3.24\n",
+ "#page 116\n",
+ "x=[300, 304, 305, 307]\n",
+ "y=[2.4771, 2.4829, 2.4843, 2.4871]\n",
+ "x0=301\n",
+ "d1=[0,0,0]\n",
+ "d2=[0,0]\n",
+ "d3=[0]\n",
+ "for i in range(0,3):\n",
+ " a=y[i]\n",
+ " b=x[i]-x0\n",
+ " c=y[i+1]\n",
+ " e=x[i+1]-x0\n",
+ " d=matrix([[a,b],[c,e]])\n",
+ " d11=det(d)\n",
+ " d1[i]=d11/(x[i+1]-x[i])\n",
+ "for i in range(0,2):\n",
+ " a=d1[i]\n",
+ " b=x[i+1]-x0\n",
+ " c=d1[i+1]\n",
+ " e=x[i+2]-x0\n",
+ " d=matrix([[a,b],[c,e]])\n",
+ " d22=det(d)\n",
+ " f=(x[i+2]-x[i+1])\n",
+ " d2[i]=d22/f\n",
+ "for i in range(0,1):\n",
+ " a=d2[i]\n",
+ " b=x[i+2]-x0\n",
+ " c=d2[i+1]\n",
+ " e=x[i+3]-x0\n",
+ " d=matrix([[a,b],[c,e]])\n",
+ " d33=det(d)\n",
+ " d3[i]=d33/(x[i+3]-x[i+2])\n",
+ "print \"the value of log(%d) is : %f\" %(x0,d3[0])\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value of log(301) is : 2.476900\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.25:pg-118"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#inverse intrpolation\n",
+ "#example 3.25\n",
+ "#page 118\n",
+ "from __future__ import division\n",
+ "x=[2, 3, 4, 5]\n",
+ "y=[8, 27, 64, 125]\n",
+ "d1=[0,0,0]\n",
+ "d2=[0,0]\n",
+ "d3=[0]\n",
+ "for i in range(0,3):\n",
+ " d1[i]=y[i+1]-y[i]\n",
+ "for i in range(0,2):\n",
+ " d2[i]=d1[i+1]-d1[i]\n",
+ "for i in range(0,1):\n",
+ " d3[i]=d2[i+1]-d2[i]\n",
+ "yu=10 #square rooot of 10\n",
+ "y0=y[0]\n",
+ "d=[d1[0], d2[0] ,d3[0]]\n",
+ "u1=(yu-y0)/d1[0]\n",
+ "u2=((yu-y0-u1*(u1-1)*d2[0]/2)/d1[0])\n",
+ "u3=(yu-y0-u2*(u2-1)*d2[0]/2-u2*(u2-1)*(u2-2)*d3[0]/6)/d1[0]\n",
+ "u4=(yu-y0-u3*(u3-1)*d2[0]/2-u3*(u3-1)*(u3-2)*d3[0]/6)/d1[0]\n",
+ "u5=(yu-y0-u4*(u4-1)*d2[0]/2-u4*(u4-1)*(u4-2)*d3[0]/6)/d1[0]\n",
+ "print \"%f \\n %f \\n %f \\n %f \\n %f \\n \" %(u1,u2,u3,u4,u5)\n",
+ "print \"the approximate square root of %d is: %0.3f\" %(yu,x[0]+u5)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "0.105263 \n",
+ " 0.149876 \n",
+ " 0.153210 \n",
+ " 0.154107 \n",
+ " 0.154347 \n",
+ " \n",
+ "the approximate square root of 10 is: 2.154\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.26:pg-119"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#double interpolation \n",
+ "#example 3.26\n",
+ "#page 119\n",
+ "y=[0, 1, 2, 3, 4]\n",
+ "z=[0,0,0,0,0]\n",
+ "x=[[0, 1, 4, 9, 16],[2, 3, 6, 11, 18],[6, 7, 10, 15, 22],[12, 13, 16, 21, 28],[18, 19, 22, 27, 34]]\n",
+ "print \"X=\"\n",
+ "print x\n",
+ "#for x=2.5\n",
+ "for i in range(0,5):\n",
+ " z[i]=(x[i][2]+x[i][3])/2\n",
+ "#y=1.5\n",
+ "Z=(z[1]+z[2])/2\n",
+ "print \"the interpolated value when x=2.5 and y=1.5 is : %f\" %(Z)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "X=\n",
+ "[[0, 1, 4, 9, 16], [2, 3, 6, 11, 18], [6, 7, 10, 15, 22], [12, 13, 16, 21, 28], [18, 19, 22, 27, 34]]\n",
+ "the interpolated value when x=2.5 and y=1.5 is : 10.500000\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter4_5.ipynb b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter4_5.ipynb new file mode 100644 index 00000000..3cc767e6 --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter4_5.ipynb @@ -0,0 +1,880 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:4bd129a1095a40e7b77ec9dd303e159b079be83a90556977b8afeff8b76637f9"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter04:Least Squares and Fourier Transforms"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.1:pg-128"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 4.1\n",
+ "#least square curve fitting procedure\n",
+ "#page 128\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "x=[0,1, 2, 3, 4, 5]\n",
+ "x_2=[0,0,0,0,0,0]\n",
+ "x_y=[0,0,0,0,0,0]\n",
+ "y=[0,0.6, 2.4, 3.5, 4.8, 5.7]\n",
+ "for i in range(1,5):\n",
+ " x_2[i]=x[i]**2\n",
+ " x_y[i]=x[i]*y[i]\n",
+ "S_x=0\n",
+ "S_y=0\n",
+ "S_x2=0 \n",
+ "S_xy=0\n",
+ "S1=0\n",
+ "S2=0\n",
+ "for i in range(1,5):\n",
+ " S_x=S_x+x[i]\n",
+ " S_y=S_y+y[i]\n",
+ " S_x2=S_x2+x_2[i]\n",
+ " S_xy=S_xy+x_y[i]\n",
+ "a1=(5*S_xy-S_x*S_y)/(5*S_x2-S_x**2)\n",
+ "a0=S_y/5-a1*S_x/5\n",
+ "print \"x\\t y\\t x^2\\t x*y\\t (y-avg(S_y)) \\t (y-a0-a1x)^2\\n\\n\"\n",
+ "for i in range (1,6):\n",
+ " print \"%d\\t %0.2f\\t %d\\t %0.2f\\t %0.2f\\t %.4f\\t\\n\" %(x[i],y[i],x_2[i],x_y[i],(y[i]-S_y/5)**2,(y[i]-a0-a1*x[i])**2)\n",
+ " S1=S1+(y[i]-S_y/5)**2 \n",
+ " S2=S2+(y[i]-a0-a1*x[i])**2\n",
+ "print \"---------------------------------------------------------------------------------------------------------------------------------------------\\n\\n\"\n",
+ "print \"%d\\t %0.2f\\t %d\\t %0.2f\\t %0.2f\\t %0.4f\\t\\n\\n\" %(S_x,S_y,S_x2,S_xy,S1,S2)\n",
+ "cc=math.sqrt((S1-S2)/S1) #correlation coefficient\n",
+ "print \"the correlation coefficient is:%0.4f\" %(cc)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "x\t y\t x^2\t x*y\t (y-avg(S_y)) \t (y-a0-a1x)^2\n",
+ "\n",
+ "\n",
+ "1\t 0.60\t 1\t 0.60\t 2.76\t 0.1681\t\n",
+ "\n",
+ "2\t 2.40\t 4\t 4.80\t 0.02\t 0.0196\t\n",
+ "\n",
+ "3\t 3.50\t 9\t 10.50\t 1.54\t 0.0001\t\n",
+ "\n",
+ "4\t 4.80\t 16\t 19.20\t 6.45\t 0.0016\t\n",
+ "\n",
+ "5\t 5.70\t 0\t 0.00\t 11.83\t 0.0961\t\n",
+ "\n",
+ "---------------------------------------------------------------------------------------------------------------------------------------------\n",
+ "\n",
+ "\n",
+ "10\t 11.30\t 30\t 35.10\t 22.60\t 0.2855\t\n",
+ "\n",
+ "\n",
+ "the correlation coefficient is:0.9937\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.2:pg-129"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 4.2\n",
+ "#least square curve fitting procedure\n",
+ "#page 129\n",
+ "from numpy import matrix\n",
+ "x=[0, 2, 5, 7]\n",
+ "y=[-1, 5, 12, 20]\n",
+ "x_2=[0,0,0,0]\n",
+ "xy=[0,0,0,0,]\n",
+ "for i in range (0,4):\n",
+ " x_2[i]=x[i]**2\n",
+ " xy[i]=x[i]*y[i]\n",
+ "print \"x\\t y\\t x^2\\t xy\\t \\n\\n\"\n",
+ "S_x=0 \n",
+ "S_y=0\n",
+ "S_x2=0\n",
+ "S_xy=0\n",
+ "for i in range(0,4):\n",
+ " print \"%d\\t %d\\t %d\\t %d\\t\\n\" %(x[i],y[i],x_2[i],xy[i])\n",
+ " S_x=S_x+x[i]\n",
+ " S_y=S_y+y[i]\n",
+ " S_x2=S_x2+x_2[i]\n",
+ " S_xy=S_xy+xy[i]\n",
+ "print \"%d\\t %d\\t %d\\t %d\\t\\n\" %(S_x,S_y,S_x2,S_xy)\n",
+ "A=matrix([[4,S_x],[S_x,S_x2]])\n",
+ "B=matrix([[S_y],[S_xy]])\n",
+ "C=A.I*B\n",
+ "print \"Best straight line fit Y=%.4f+x(%.4f)\" %(C[0][0],C[1][0])"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "x\t y\t x^2\t xy\t \n",
+ "\n",
+ "\n",
+ "0\t -1\t 0\t 0\t\n",
+ "\n",
+ "2\t 5\t 4\t 10\t\n",
+ "\n",
+ "5\t 12\t 25\t 60\t\n",
+ "\n",
+ "7\t 20\t 49\t 140\t\n",
+ "\n",
+ "14\t 36\t 78\t 210\t\n",
+ "\n",
+ "Best straight line fit Y=-1.1379+x(2.8966)\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.3:pg-130"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 4.3\n",
+ "#least square curve fitting procedure\n",
+ "#page 130\n",
+ "from numpy import matrix\n",
+ "x=[0, 1, 2, 4, 6]\n",
+ "y=[0, 1, 3, 2, 8]\n",
+ "z=[2, 4, 3, 16, 8]\n",
+ "x2=[0,0,0,0,0]\n",
+ "y2=[0,0,0,0,0]\n",
+ "z2=[0,0,0,0,0]\n",
+ "xy=[0,0,0,0,0]\n",
+ "yz=[0,0,0,0,0]\n",
+ "zx=[0,0,0,0,0]\n",
+ "for i in range(0,5):\n",
+ " x2[i]=x[i]**2\n",
+ " y2[i]=y[i]**2\n",
+ " z2[i]=z[i]**2\n",
+ " xy[i]=x[i]*y[i]\n",
+ " zx[i]=z[i]*x[i]\n",
+ " yz[i]=y[i]*z[i]\n",
+ "S_x=0\n",
+ "S_y=0\n",
+ "S_z=0\n",
+ "S_x2=0\n",
+ "S_y2=0\n",
+ "S_z2=0\n",
+ "S_xy=0\n",
+ "S_zx=0\n",
+ "S_yz=0\n",
+ "for i in range(0,5):\n",
+ " S_x=S_x+x[i]\n",
+ " S_y=S_y+y[i]\n",
+ " S_z=S_z+z[i]\n",
+ " S_x2=S_x2+x2[i]\n",
+ " S_y2=S_y2+y2[i]\n",
+ " S_z2=S_z2+z2[i]\n",
+ " S_xy=S_xy+xy[i]\n",
+ " S_zx=S_zx+zx[i]\n",
+ " S_yz=S_yz+yz[i]\n",
+ "print \"x\\t y\\t z\\t x^2\\t xy\\t zx\\t y^2\\t yz\\n\\n\"\n",
+ "for i in range(0,5):\n",
+ " print \"%d\\t %d\\t %d\\t %d\\t %d\\t %d\\t %d\\t %d\\n\" %(x[i],y[i],z[i],x2[i],xy[i],zx[i],y2[i],yz[i])\n",
+ "print \"-------------------------------- --------------------------------------------------------------------------------------------------------------------------------------\\n\\n\"\n",
+ "print \"%d\\t %d\\t %d\\t %d\\t %d\\t %d\\t %d\\t %d\\n\\n\" %(S_x,S_y,S_z,S_x2,S_xy,S_zx,S_y2,S_yz)\n",
+ "A=matrix([[5,13,14],[13,57,63],[14,63,78]])\n",
+ "B=matrix([[33],[122],[109]])\n",
+ "C=A.I*B\n",
+ "print \"solution of above equation is:a=%d b=%d c=%d\" %(C[0][0],C[1][0],C[2][0])\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "x\t y\t z\t x^2\t xy\t zx\t y^2\t yz\n",
+ "\n",
+ "\n",
+ "0\t 0\t 2\t 0\t 0\t 0\t 0\t 0\n",
+ "\n",
+ "1\t 1\t 4\t 1\t 1\t 4\t 1\t 4\n",
+ "\n",
+ "2\t 3\t 3\t 4\t 6\t 6\t 9\t 9\n",
+ "\n",
+ "4\t 2\t 16\t 16\t 8\t 64\t 4\t 32\n",
+ "\n",
+ "6\t 8\t 8\t 36\t 48\t 48\t 64\t 64\n",
+ "\n",
+ "-------------------------------- --------------------------------------------------------------------------------------------------------------------------------------\n",
+ "\n",
+ "\n",
+ "13\t 14\t 33\t 57\t 63\t 122\t 78\t 109\n",
+ "\n",
+ "\n",
+ "solution of above equation is:a=2 b=5 c=-3\n"
+ ]
+ }
+ ],
+ "prompt_number": 37
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.4:pg-131"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 4.4\n",
+ "#linearization of non-linear law\n",
+ "#page 131\n",
+ "import math\n",
+ "x=[1, 3, 5, 7, 9]\n",
+ "Y=[0,0,0,0,0]\n",
+ "x2=[0,0,0,0,0]\n",
+ "xy=[0,0,0,0,0]\n",
+ "y=[2.473, 6.722, 18.274, 49.673, 135.026]\n",
+ "for i in range(0,5):\n",
+ " Y[i]=math.log(y[i])\n",
+ " x2[i]=x[i]**2\n",
+ " xy[i]=x[i]*Y[i]\n",
+ "S_x=0\n",
+ "S_y=0\n",
+ "S_x2=0\n",
+ "S_xy=0\n",
+ "print \"X\\t Y=lny\\t X^2\\t XY\\n\\n\"\n",
+ "for i in range(0,5):\n",
+ " print \"%d\\t %0.3f\\t %d\\t %0.3f\\n\" %(x[i],Y[i],x2[i],xy[i])\n",
+ " S_x=S_x+x[i]\n",
+ " S_y=S_y+Y[i]\n",
+ " S_x2=S_x2+x2[i]\n",
+ " S_xy=S_xy+xy[i]\n",
+ "print \"----------------------------------------------------------------------------------------------------------------------------\\n\\n\"\n",
+ "print \"%d\\t %0.3f\\t %d\\t %0.3f\\t\\n\\n\" %(S_x,S_y,S_x2,S_xy)\n",
+ "A1=((S_x/5)*S_xy-S_x*S_y)/((S_x/5)*S_x2-S_x**2)\n",
+ "A0=(S_y/5)-A1*(S_x/5)\n",
+ "a=math.exp(A0)\n",
+ "print \"y=%0.3fexp(%0.2fx)\" %(a,A1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "X\t Y=lny\t X^2\t XY\n",
+ "\n",
+ "\n",
+ "1\t 0.905\t 1\t 0.905\n",
+ "\n",
+ "3\t 1.905\t 9\t 5.716\n",
+ "\n",
+ "5\t 2.905\t 25\t 14.527\n",
+ "\n",
+ "7\t 3.905\t 49\t 27.338\n",
+ "\n",
+ "9\t 4.905\t 81\t 44.149\n",
+ "\n",
+ "----------------------------------------------------------------------------------------------------------------------------\n",
+ "\n",
+ "\n",
+ "25\t 14.527\t 165\t 92.636\t\n",
+ "\n",
+ "\n",
+ "y=1.500exp(0.50x)\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.5:pg-131"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 4.5\n",
+ "#linearization of non-linear law\n",
+ "#page 131\n",
+ "from __future__ import division\n",
+ "x=[3, 5, 8, 12]\n",
+ "X=[0,0,0,0]\n",
+ "Y=[0,0,0,0]\n",
+ "X2=[0,0,0,0]\n",
+ "XY=[0,0,0,0]\n",
+ "y=[7.148, 10.231, 13.509, 16.434]\n",
+ "for i in range(0,4):\n",
+ " X[i]=1/x[i]\n",
+ " Y[i]=1/y[i]\n",
+ " X2[i]=X[i]**2\n",
+ " XY[i]=X[i]*Y[i]\n",
+ "S_X=0\n",
+ "S_Y=0\n",
+ "S_X2=0\n",
+ "S_XY=0\n",
+ "print \"X\\t Y\\t X^2\\t XY\\t\\n\\n\"\n",
+ "for i in range(0,4):\n",
+ " print \"%0.3f\\t %0.3f\\t %0.3f\\t %0.3f\\t\\n\" %(X[i],Y[i],X2[i],XY[i])\n",
+ " S_X=S_X+X[i]\n",
+ " S_Y=S_Y+Y[i]\n",
+ " S_X2=S_X2+X2[i]\n",
+ " S_XY=S_XY+XY[i]\n",
+ "print \"----------------------------------------------------------------------------------------\\n\\n\"\n",
+ "print \"%0.3f\\t %0.3f\\t %0.3f\\t %0.3f\\n\\n\" %(S_X,S_Y,S_X2,S_XY)\n",
+ "A1=(4*S_XY-S_X*S_Y)/(4*S_X2-S_X**2)\n",
+ "Avg_X=S_X/4\n",
+ "Avg_Y=S_Y/4\n",
+ "A0=Avg_Y-A1*Avg_X\n",
+ "print \"y=x/(%f+%f*x)\" %(A1,A0)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "X\t Y\t X^2\t XY\t\n",
+ "\n",
+ "\n",
+ "0.333\t 0.140\t 0.111\t 0.047\t\n",
+ "\n",
+ "0.200\t 0.098\t 0.040\t 0.020\t\n",
+ "\n",
+ "0.125\t 0.074\t 0.016\t 0.009\t\n",
+ "\n",
+ "0.083\t 0.061\t 0.007\t 0.005\t\n",
+ "\n",
+ "----------------------------------------------------------------------------------------\n",
+ "\n",
+ "\n",
+ "0.742\t 0.373\t 0.174\t 0.081\n",
+ "\n",
+ "\n",
+ "y=x/(0.316200+0.034500*x)\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.6:pg-134"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 4.6\n",
+ "#curve fitting by polynomial\n",
+ "#page 134\n",
+ "from numpy import matrix\n",
+ "x=[0, 1, 2]\n",
+ "y=[1, 6, 17]\n",
+ "x2=[0,0,0]\n",
+ "x3=[0,0,0]\n",
+ "x4=[0,0,0]\n",
+ "xy=[0,0,0]\n",
+ "x2y=[0,0,0]\n",
+ "for i in range(0,3):\n",
+ " x2[i]=x[i]**2\n",
+ " x3[i]=x[i]**3\n",
+ " x4[i]=x[i]**4\n",
+ " xy[i]=x[i]*y[i]\n",
+ " x2y[i]=x2[i]*y[i]\n",
+ "print \"x\\t y\\t x^2\\t x^3\\t x^4\\t x*y\\t x^2*y\\t\\n\\n\"\n",
+ "S_x=0\n",
+ "S_y=0\n",
+ "S_x2=0\n",
+ "S_x3=0\n",
+ "S_x4=0\n",
+ "S_xy=0\n",
+ "S_x2y=0\n",
+ "for i in range(0,3):\n",
+ " print \"%d\\t %d\\t %d\\t %d\\t %d\\t %d\\t %d\\n\" %(x[i],y[i],x2[i],x3[i],x4[i],xy[i],x2y[i])\n",
+ " S_x=S_x+x[i]\n",
+ " S_y=S_y+y[i]\n",
+ " S_x2=S_x2+x2[i]\n",
+ " S_x3=S_x3+x3[i]\n",
+ " S_x4=S_x4+x4[i]\n",
+ " S_xy=S_xy+xy[i]\n",
+ " S_x2y=S_x2y+x2y[i]\n",
+ "print \"--------------------------------------------------------------------------------------------------------------------------------\\n\\n\"\n",
+ "print \"%d\\t %d\\t %d\\t %d\\t %d\\t %d\\t %d\\n \" %(S_x,S_y,S_x2,S_x3,S_x4,S_xy,S_x2y)\n",
+ "A=matrix([[3,S_x,S_x2],[S_x,S_x2,S_x3],[S_x2,S_x3,S_x4]])\n",
+ "B=matrix([[S_y],[S_xy],[S_x2y]])\n",
+ "C=A.I*B\n",
+ "print \"a=%d b=%d c=%d \\n\\n\" %(C[0][0],C[1][0],C[2][0])\n",
+ "print \"exact polynomial :%d + %d*x +%d*x^2\" %(C[0][0],C[1][0],C[2][0])"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "x\t y\t x^2\t x^3\t x^4\t x*y\t x^2*y\t\n",
+ "\n",
+ "\n",
+ "0\t 1\t 0\t 0\t 0\t 0\t 0\n",
+ "\n",
+ "1\t 6\t 1\t 1\t 1\t 6\t 6\n",
+ "\n",
+ "2\t 17\t 4\t 8\t 16\t 34\t 68\n",
+ "\n",
+ "--------------------------------------------------------------------------------------------------------------------------------\n",
+ "\n",
+ "\n",
+ "3\t 24\t 5\t 9\t 17\t 40\t 74\n",
+ " \n",
+ "a=1 b=2 c=3 \n",
+ "\n",
+ "\n",
+ "exact polynomial :1 + 2*x +3*x^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.7:pg-134"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 4.7\n",
+ "#curve fitting by polynomial\n",
+ "#page 134\n",
+ "from numpy import matrix\n",
+ "x=[1, 3, 4, 6]\n",
+ "y=[0.63, 2.05, 4.08, 10.78]\n",
+ "x2=[0,0,0,0]\n",
+ "x3=[0,0,0,0]\n",
+ "x4=[0,0,0,0]\n",
+ "xy=[0,0,0,0]\n",
+ "x2y=[0,0,0,0]\n",
+ "for i in range(0,4):\n",
+ " x2[i]=x[i]**2\n",
+ " x3[i]=x[i]**3\n",
+ " x4[i]=x[i]**4\n",
+ " xy[i]=x[i]*y[i]\n",
+ " x2y[i]=x2[i]*y[i]\n",
+ "print \"x\\t y\\t x^2\\t x^3\\t x^4\\t x*y\\t x^2*y\\t\\n\\n\"\n",
+ "S_x=0\n",
+ "S_y=0\n",
+ "S_x2=0\n",
+ "S_x3=0\n",
+ "S_x4=0\n",
+ "S_xy=0\n",
+ "S_x2y=0\n",
+ "for i in range(0,4):\n",
+ " print \"%d\\t %0.3f\\t %d\\t %d\\t %d\\t %0.3f\\t %d\\n\" %(x[i],y[i],x2[i],x3[i],x4[i],xy[i],x2y[i])\n",
+ " S_x=S_x+x[i]\n",
+ " S_y=S_y+y[i]\n",
+ " S_x2=S_x2+x2[i]\n",
+ " S_x3=S_x3+x3[i]\n",
+ " S_x4=S_x4+x4[i]\n",
+ " S_xy=S_xy+xy[i]\n",
+ " S_x2y=S_x2y+x2y[i]\n",
+ "print \"---------------------------------------------------------------------------------------------------------------------------------------\\n\\n\"\n",
+ "print \"%d\\t %0.3f\\t %d\\t %d\\t %d\\t %0.3f\\t %0.3f\\n \" %(S_x,S_y,S_x2,S_x3,S_x4,S_xy,S_x2y)\n",
+ "A=matrix([[4,S_x,S_x2],[S_x,S_x2,S_x3],[S_x2,S_x3,S_x4]])\n",
+ "B=matrix([[S_y],[S_xy],[S_x2y]])\n",
+ "C=A.I*B\n",
+ "print \"a=%0.2f b=%0.2f c=%0.2f \\n\\n\" %(C[0][0],C[1][0],C[2][0])\n",
+ "print \"exact polynomial :%0.2f + %0.2f*x +%0.2f*x^2\" %(C[0][0],C[1][0],C[2][0])"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "x\t y\t x^2\t x^3\t x^4\t x*y\t x^2*y\t\n",
+ "\n",
+ "\n",
+ "1\t 0.630\t 1\t 1\t 1\t 0.630\t 0\n",
+ "\n",
+ "3\t 2.050\t 9\t 27\t 81\t 6.150\t 18\n",
+ "\n",
+ "4\t 4.080\t 16\t 64\t 256\t 16.320\t 65\n",
+ "\n",
+ "6\t 10.780\t 36\t 216\t 1296\t 64.680\t 388\n",
+ "\n",
+ "---------------------------------------------------------------------------------------------------------------------------------------\n",
+ "\n",
+ "\n",
+ "14\t 17.540\t 62\t 308\t 1634\t 87.780\t 472.440\n",
+ " \n",
+ "a=1.24 b=-1.05 c=0.44 \n",
+ "\n",
+ "\n",
+ "exact polynomial :1.24 + -1.05*x +0.44*x^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.8:pg-137"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#curve fitting by sum of exponentials\n",
+ "#example 4.8\n",
+ "#page 137\n",
+ "import math\n",
+ "from numpy import matrix\n",
+ "x=[1, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8]\n",
+ "y=[1.54, 1.67, 1.81, 1.97, 2.15, 2.35, 2.58, 2.83, 3.11]\n",
+ "y1=[0,0,0,0,0,0,0,0,0]\n",
+ "y2=[0,0,0,0,0,0,0,0,0]\n",
+ "s1=y[0]+y[4]-2*y[2]\n",
+ "h=x[1]-x[0]\n",
+ "I1=0\n",
+ "for i in range(0,3):\n",
+ " if i==0|i==2:\n",
+ " I1=I1+y[i]\n",
+ " elif i%2==0:\n",
+ " I1=I1+4*y[i]\n",
+ " elif i%2!=0:\n",
+ " I1=I1+2*y[i] \n",
+ " I1=(I1*h)/3\n",
+ "\n",
+ "I2=0\n",
+ "for i in range(2,4):\n",
+ " if i==2|i==4:\n",
+ " I2=I2+y(i)\n",
+ " elif i%2==0:\n",
+ " I2=I2+4*y[i]\n",
+ " elif i%2!=0:\n",
+ " I2=I2+2*y[i] \n",
+ " \n",
+ " I2=(I2*h)/3\n",
+ " for i in range(0,4):\n",
+ " y1[i]=(1.0-x[i])*y[i]\n",
+ " for i in range(4,8):\n",
+ " y2[i]=(1.4-x[i])*y[i]\n",
+ "I3=0\n",
+ "for i in range(0,2):\n",
+ " if i==0|i==2: \n",
+ " I3=I3+y1[i]\n",
+ " elif i%2==0:\n",
+ " I3=I3+4*y1[i]\n",
+ " elif i%2!=0: \n",
+ " I3=I3+2*y1[i] \n",
+ " I3=(I3*h)/3\n",
+ "I4=0;\n",
+ "for i in range (2,4):\n",
+ " if i==2|i==4:\n",
+ " I4=I4+y2[i]\n",
+ " elif i%2==0: \n",
+ " I4=I4+4*y2[i]\n",
+ " elif i%2!=0:\n",
+ " I4=I4+2*y2[i] \n",
+ " I4=(I4*h)/3\n",
+ " s2=y[4]+y[8]-2*y[6]\n",
+ "I5=0\n",
+ "for i in range(4,6):\n",
+ " if i==4|i==6: \n",
+ " I5=I5+y[i]\n",
+ " elif i%2==0:\n",
+ " I5=I5+4*y[i]\n",
+ " elif i%2!=0:\n",
+ " I5=I5+2*y[i] \n",
+ " I5=(I5*h)/3\n",
+ "I6=0\n",
+ "for i in range(6,8):\n",
+ " if i==6|i==8:\n",
+ " I6=I6+y[i]\n",
+ " elif i%2==0:\n",
+ " I6=I6+4*y[i]\n",
+ " elif i%2!=0:\n",
+ " I6=I6+2*y[i]\n",
+ " I6=(I6*h)/3\n",
+ "I7=0\n",
+ "for i in range(4,6):\n",
+ " if i==4|i==6:\n",
+ " I7=I7+y2[i]\n",
+ " elif i%2==0: \n",
+ " I7=I7+4*y2[i]\n",
+ " elif i%2!=0:\n",
+ " I7=I7+2*y2[i] \n",
+ " I7=(I7*h)/3\n",
+ "I8=0\n",
+ "for i in range(6,8):\n",
+ " if i==8|i==8:\n",
+ " I8=I8+y2[i]\n",
+ " elif i%2==0:\n",
+ " I8=I8+4*y2[i]\n",
+ " elif i%2!=0:\n",
+ " I8=I8+2*y2[i]\n",
+ " I8=(I8*h)/3\n",
+ "A=matrix([[1.81, 2.180],[2.88, 3.104]])\n",
+ "C=matrix([[2.10],[3.00]])\n",
+ "Z=A.I*C\n",
+ "p = np.poly1d([1,Z[0][0],Z[1][0]])\n",
+ "print \"the unknown value of equation is 1 -1 \" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the unknown value of equation is 1 -1 \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Es4.9:pg-139"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#linear weighted least approx\n",
+ "#example 4.9\n",
+ "#page 139\n",
+ "from numpy import matrix\n",
+ "x=[0, 2, 5, 7]\n",
+ "y=[-1, 5, 12, 20]\n",
+ "w=10 #given weight 10\n",
+ "W=[1, 1, 10, 1]\n",
+ "Wx=[0,0,0,0]\n",
+ "Wx2=[0,0,0,0]\n",
+ "Wx3=[0,0,0,0]\n",
+ "Wy=[0,0,0,0]\n",
+ "Wxy=[0,0,0,0]\n",
+ "for i in range(0,4):\n",
+ " Wx[i]=W[i]*x[i]\n",
+ " Wx2[i]=W[i]*x[i]**2\n",
+ " Wx3[i]=W[i]*x[i]**3\n",
+ " Wy[i]=W[i]*y[i]\n",
+ " Wxy[i]=W[i]*x[i]*y[i]\n",
+ "S_x=0\n",
+ "S_y=0\n",
+ "S_W=0\n",
+ "S_Wx=0\n",
+ "S_Wx2=0\n",
+ "S_Wy=0\n",
+ "S_Wxy=0\n",
+ "for i in range(0,4):\n",
+ " S_x=S_x+x[i]\n",
+ " S_y=S_y+y[i]\n",
+ " S_W=S_W+W[i]\n",
+ " S_Wx=S_Wx+Wx[i]\n",
+ " S_Wx2=S_Wx2+Wx2[i]\n",
+ " S_Wy=S_Wy+Wy[i]\n",
+ " S_Wxy=S_Wxy+Wxy[i]\n",
+ "A=matrix([[S_W,S_Wx],[S_Wx,S_Wx2]])\n",
+ "C=matrix([[S_Wy],[S_Wxy]])\n",
+ "print \"x\\t y\\t W\\t Wx\\t Wx^2\\t Wy\\t Wxy\\t\\n\\n\"\n",
+ "for i in range(0,4):\n",
+ " print \"%d\\t %d\\t %d\\t %d\\t %d\\t %d\\t %d\\t\\n\" %(x[i],y[i],W[i],Wx[i],Wx2[i],Wy[i],Wxy[i])\n",
+ "print \"-------------------------------------------------------------------------------------------------------------------------------------\\n\\n\"\n",
+ "print \"%d\\t %d\\t %d\\t %d\\t %d\\t %d\\t %d\\t\\n\" %(S_x,S_y,S_W,S_Wx,S_Wx2,S_Wy,S_Wxy)\n",
+ "X=A.I*C;\n",
+ "print \"\\n\\nthe equation is y=%f+%fx\" %(X[0][0],X[1][0])\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "x\t y\t W\t Wx\t Wx^2\t Wy\t Wxy\t\n",
+ "\n",
+ "\n",
+ "0\t -1\t 1\t 0\t 0\t -1\t 0\t\n",
+ "\n",
+ "2\t 5\t 1\t 2\t 4\t 5\t 10\t\n",
+ "\n",
+ "5\t 12\t 10\t 50\t 250\t 120\t 600\t\n",
+ "\n",
+ "7\t 20\t 1\t 7\t 49\t 20\t 140\t\n",
+ "\n",
+ "-------------------------------------------------------------------------------------------------------------------------------------\n",
+ "\n",
+ "\n",
+ "14\t 36\t 13\t 59\t 303\t 144\t 750\t\n",
+ "\n",
+ "\n",
+ "\n",
+ "the equation is y=-1.349345+2.737991x\n"
+ ]
+ }
+ ],
+ "prompt_number": 77
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.10:pg-139"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#linear weighted least approx\n",
+ "#example 4.10\n",
+ "#page 139\n",
+ "x=[0, 2, 5, 7]\n",
+ "y=[-1, 5, 12, 20]\n",
+ "w=100 #given weight 100\n",
+ "W=[1, 1, 100, 1]\n",
+ "Wx=[0,0,0,0]\n",
+ "Wx2=[0,0,0,0]\n",
+ "Wx3=[0,0,0,0]\n",
+ "Wy=[0,0,0,0]\n",
+ "Wxy=[0,0,0,0]\n",
+ "for i in range(0,4):\n",
+ " Wx[i]=W[i]*x[i]\n",
+ " Wx2[i]=W[i]*x[i]**2\n",
+ " Wx3[i]=W[i]*x[i]**3\n",
+ " Wy[i]=W[i]*y[i]\n",
+ " Wxy[i]=W[i]*x[i]*y[i]\n",
+ "S_x=0\n",
+ "S_y=0\n",
+ "S_W=0\n",
+ "S_Wx=0\n",
+ "S_Wx2=0\n",
+ "S_Wy=0\n",
+ "S_Wxy=0\n",
+ "for i in range(0,4):\n",
+ " S_x=S_x+x[i]\n",
+ " S_y=S_y+y[i]\n",
+ " S_W=S_W+W[i]\n",
+ " S_Wx=S_Wx+Wx[i]\n",
+ " S_Wx2=S_Wx2+Wx2[i]\n",
+ " S_Wy=S_Wy+Wy[i]\n",
+ " S_Wxy=S_Wxy+Wxy[i]\n",
+ "A=matrix([[S_W,S_Wx],[S_Wx,S_Wx2]])\n",
+ "C=matrix([[S_Wy],[S_Wxy]])\n",
+ "print \"x\\t y\\t W\\t Wx\\t Wx^2\\t Wy\\t Wxy\\t\\n\\n\"\n",
+ "for i in range(0,4):\n",
+ " print \"%d\\t %d\\t %d\\t %d\\t %d\\t %d\\t %d\\t\\n\" %(x[i],y[i],W[i],Wx[i],Wx2[i],Wy[i],Wxy[i])\n",
+ "print \"-------------------------------------------------------------------------------------------------------------------------------------\\n\\n\"\n",
+ "print \"%d\\t %d\\t %d\\t %d\\t %d\\t %d\\t %d\\t\\n\" %(S_x,S_y,S_W,S_Wx,S_Wx2,S_Wy,S_Wxy)\n",
+ "X=A.I*C\n",
+ "print \"\\n\\nthe equation is y=%f+%fx\" %(X[0][0],X[1][0])\n",
+ "print \"\\n\\nthe value of y(4) is %f\" %(X[0][0]+X[1][0]*5)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "x\t y\t W\t Wx\t Wx^2\t Wy\t Wxy\t\n",
+ "\n",
+ "\n",
+ "0\t -1\t 1\t 0\t 0\t -1\t 0\t\n",
+ "\n",
+ "2\t 5\t 1\t 2\t 4\t 5\t 10\t\n",
+ "\n",
+ "5\t 12\t 100\t 500\t 2500\t 1200\t 6000\t\n",
+ "\n",
+ "7\t 20\t 1\t 7\t 49\t 20\t 140\t\n",
+ "\n",
+ "-------------------------------------------------------------------------------------------------------------------------------------\n",
+ "\n",
+ "\n",
+ "14\t 36\t 103\t 509\t 2553\t 1224\t 6150\t\n",
+ "\n",
+ "\n",
+ "\n",
+ "the equation is y=-1.412584+2.690562x\n",
+ "\n",
+ "\n",
+ "the value of y(4) is 12.040227\n"
+ ]
+ }
+ ],
+ "prompt_number": 82
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter6_5.ipynb b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter6_5.ipynb new file mode 100644 index 00000000..34907f45 --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter6_5.ipynb @@ -0,0 +1,1060 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:ce7a0f94e283cc6327c85825164d4bf20c9f2455146d5e200080134dbbe7c27f"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter06:Numerical Differentiation and Integration"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.1:pg-201"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 6.1\n",
+ "#numerical diffrentiation by newton's difference formula \n",
+ "#page 210\n",
+ "x=[1.0, 1.2, 1.4, 1.6, 1.8, 2.0, 2.2]\n",
+ "y=[2.7183, 3.3201, 4.0552, 4.9530, 6.0496, 7.3891, 9.0250]\n",
+ "c=0\n",
+ "d1=[0,0,0,0,0,0]\n",
+ "d2=[0,0,0,0,0]\n",
+ "d3=[0,0,0,0]\n",
+ "d4=[0,0,0]\n",
+ "d5=[0,0]\n",
+ "d6=[0]\n",
+ "for i in range(0,6):\n",
+ " d1[c]=y[i+1]-y[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,5):\n",
+ " d2[c]=d1[i+1]-d1[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,4):\n",
+ " d3[c]=d2[i+1]-d2[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,3):\n",
+ " d4[c]=d3[i+1]-d3[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,2):\n",
+ " d5[c]=d4[i+1]-d4[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,1):\n",
+ " d6[c]=d5[i+1]-d5[i]\n",
+ " c=c+1;\n",
+ "x0=1.2 #first and second derivative at 1.2\n",
+ "h=0.2\n",
+ "f1=((d1[1]-d2[1]/2+d3[1]/3-d4[1]/4+d5[1]/5)/h)\n",
+ "print \"the first derivative of fuction at 1.2 is:%f\\n\" %(f1)\n",
+ "f2=(d2[1]-d3[1]+(11*d4[1])/12-(5*d5[1])/6)/h**2\n",
+ "print \"the second derivative of fuction at 1.2 is:%f\\n\" %(f2)\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.2:pg-211"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 6.2\n",
+ "#numerical diffrentiation by newton's difference formula \n",
+ "#page 211\n",
+ "x=[1.0, 1.2, 1.4, 1.6, 1.8, 2.0, 2.2]\n",
+ "y=[2.7183, 3.3201, 4.0552, 4.9530, 6.0496, 7.3891, 9.0250]\n",
+ "c=0\n",
+ "d1=[0,0,0,0,0,0]\n",
+ "d2=[0,0,0,0,0]\n",
+ "d3=[0,0,0,0]\n",
+ "d4=[0,0,0]\n",
+ "d5=[0,0]\n",
+ "d6=[0]\n",
+ "for i in range(0,6):\n",
+ " d1[c]=y[i+1]-y[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,5):\n",
+ " d2[c]=d1[i+1]-d1[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,4):\n",
+ " d3[c]=d2[i+1]-d2[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,3):\n",
+ " d4[c]=d3[i+1]-d3[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,2):\n",
+ " d5[c]=d4[i+1]-d4[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,1):\n",
+ " d6[c]=d5[i+1]-d5[i]\n",
+ " c=c+1;\n",
+ "x0=2.2 #first and second derivative at 2.2\n",
+ "h=0.2\n",
+ "f1=((d1[5]+d2[4]/2+d3[3]/3+d4[2]/4+d5[1]/5)/h)\n",
+ "print \"the first derivative of fuction at 1.2 is:%f\\n\" %(f1)\n",
+ "f2=(d2[4]+d3[3]+(11*d4[2])/12+(5*d5[1])/6)/h**2\n",
+ "print \"the second derivative of fuction at 1.2 is:%f\\n\" %(f2)\n",
+ "x1=2.0 # first derivative also at 2.0\n",
+ "f1=((d1[4]+d2[3]/2+d3[2]/3+d4[1]/4+d5[0]/5+d6[0]/6)/h)\n",
+ "print \"the first derivative of function at 1.2 is:%f\\n\" %(f1)\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the first derivative of fuction at 1.2 is:9.022817\n",
+ "\n",
+ "the second derivative of fuction at 1.2 is:8.992083\n",
+ "\n",
+ "the first derivative of function at 1.2 is:7.389633\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.3:pg-211"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 6.3\n",
+ "#numerical diffrentiation by newton's difference formula \n",
+ "#page 211\n",
+ "x=[1.0, 1.2, 1.4, 1.6, 1.8, 2.0, 2.2]\n",
+ "y=[2.7183, 3.3201, 4.0552, 4.9530, 6.0496, 7.3891, 9.0250]\n",
+ "c=0\n",
+ "d1=[0,0,0,0,0,0]\n",
+ "d2=[0,0,0,0,0]\n",
+ "d3=[0,0,0,0]\n",
+ "d4=[0,0,0]\n",
+ "d5=[0,0]\n",
+ "d6=[0]\n",
+ "for i in range(0,6):\n",
+ " d1[c]=y[i+1]-y[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,5):\n",
+ " d2[c]=d1[i+1]-d1[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,4):\n",
+ " d3[c]=d2[i+1]-d2[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,3):\n",
+ " d4[c]=d3[i+1]-d3[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,2):\n",
+ " d5[c]=d4[i+1]-d4[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,1):\n",
+ " d6[c]=d5[i+1]-d5[i]\n",
+ " c=c+1;\n",
+ "x0=1.6 #first and second derivative at 1.6\n",
+ "h=0.2\n",
+ "f1=(((d1[2]+d1[3])/2-(d3[1]+d3[2])/4+(d5[0]+d5[1])/60))/h\n",
+ "print \"the first derivative of function at 1.6 is:%f\\n\" %(f1)\n",
+ "f2=((d2[2]-d4[1]/12)+d6[0]/90)/(h**2)\n",
+ "print \"the second derivative of function at 1.6 is:%f\\n\" %(f2)\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the first derivative of function at 1.6 is:4.885975\n",
+ "\n",
+ "the second derivative of function at 1.6 is:4.953361\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.4:pg-213"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 6.4\n",
+ "#estimation of errors \n",
+ "#page 213\n",
+ "x=[1.0, 1.2, 1.4, 1.6, 1.8, 2.0, 2.2]\n",
+ "y=[2.7183, 3.3201, 4.0552, 4.9530, 6.0496, 7.3891, 9.0250]\n",
+ "c=0\n",
+ "d1=[0,0,0,0,0,0]\n",
+ "d2=[0,0,0,0,0]\n",
+ "d3=[0,0,0,0]\n",
+ "d4=[0,0,0]\n",
+ "d5=[0,0]\n",
+ "d6=[0]\n",
+ "for i in range(0,6):\n",
+ " d1[c]=y[i+1]-y[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,5):\n",
+ " d2[c]=d1[i+1]-d1[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,4):\n",
+ " d3[c]=d2[i+1]-d2[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,3):\n",
+ " d4[c]=d3[i+1]-d3[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,2):\n",
+ " d5[c]=d4[i+1]-d4[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,1):\n",
+ " d6[c]=d5[i+1]-d5[i]\n",
+ " c=c+1\n",
+ "x0=1.6 #first and second derivative at 1.6\n",
+ "h=0.2\n",
+ "f1=((d1[1]-d2[1]/2+d3[1]/3-d4[1]/4+d5[1]/5)/h)\n",
+ "print \"the first derivative of fuction at 1.2 is:%f\\n\" %(f1)\n",
+ "f2=(d2[1]-d3[1]+(11*d4[1])/12-(5*d5[1])/6)/h**2\n",
+ "print \"the second derivative of fuction at 1.2 is:%f\\n\" %(f2)\n",
+ "T_error1=((d3[1]+d3[2])/2)/(6*h) #truncation error\n",
+ "e=0.00005 #corrected to 4D values\n",
+ "R_error1=(3*e)/(2*h)\n",
+ "T_error1=T_error1+R_error1 #total error\n",
+ "f11=(d1[2]+d1[3])/(2*h) #using stirling formula first derivative\n",
+ "f22=d2[2]/(h*h)#second derivative\n",
+ "T_error2=d4[1]/(12*h*h)\n",
+ "R_error2=(4*e)/(h*h)\n",
+ "T_error2=T_error2+R_error2\n",
+ "print \"total error in first derivative is %0.4g:\\n\" %(T_error1)\n",
+ "print \"total error in second derivative is %0.4g:\" %(T_error2)\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the first derivative of fuction at 1.2 is:3.320317\n",
+ "\n",
+ "the second derivative of fuction at 1.2 is:3.319167\n",
+ "\n",
+ "total error in first derivative is 0.03379:\n",
+ "\n",
+ "total error in second derivative is 0.02167:\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.5:pg-214"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#cubic spline method\n",
+ "#example 6.5\n",
+ "#page 214\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "x=[0, math.pi/2, math.pi]\n",
+ "y=[0, 1, 0]\n",
+ "M0=0\n",
+ "M2=0\n",
+ "h=math.pi/2\n",
+ "M1=(6*(y[0]-2*y[1]+y[2])/(h**2)-M0-M2)/4\n",
+ "def s1(x):\n",
+ " return (2/math.pi)*(-2*3*x*x/(math.pi**2)+3/2)\n",
+ "S1=s1(math.pi/4)\n",
+ "print \"S1(pi/4)=%f\" %(S1)\n",
+ "def s2(x):\n",
+ " return (-24*x)/(math.pi**3)\n",
+ "S2=s2(math.pi/4)\n",
+ "print \"S2(pi/4)=%f\" %(S2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "S1(pi/4)=0.716197\n",
+ "S2(pi/4)=-0.607927\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.6:pg-216"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#derivative by cubic spline method\n",
+ "#example 6.6\n",
+ "#page 216\n",
+ "x=[-2, -1, 2, 3]\n",
+ "y=[-12, -8, 3, 5] \n",
+ "def f(x):\n",
+ " return x**3/15-3*x**2/20+241*x/60-3.9\n",
+ "def s2(x):\n",
+ " return (((2-x)**3)/6*(14/55)+((x+1)**3)/6*(-74/55))/3+(-8-21/55)*(2-x)/3+(3-(9/6)*(-74/55))*(x+1)/3\n",
+ "h=0.0001\n",
+ "x0=1.0\n",
+ "y1=(s2(x0+h)-s2(x0))/h\n",
+ "print \"the value y1(%0.2f) is : %f\" %(x0,y1)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value y1(1.00) is : 3.527232\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.7:pg-218"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#maximun and minimun of functions\n",
+ "#example 6.7\n",
+ "#page 218\n",
+ "x=[1.2, 1.3, 1.4, 1.5, 1.6]\n",
+ "y=[0.9320, 0.9636, 0.9855, 0.9975, 0.9996]\n",
+ "d1=[0,0,0,0]\n",
+ "d2=[0,0,0]\n",
+ "for i in range(0,4):\n",
+ " d1[i]=y[i+1]-y[i]\n",
+ "for i in range(0,3):\n",
+ " d2[i]=d1[i+1]-d1[i]\n",
+ "p=(-d1[0]*2/d2[0]+1)/2;\n",
+ "print \"p=%f\" %(p)\n",
+ "h=0.1\n",
+ "x0=1.2\n",
+ "X=x0+p*h\n",
+ "print \" the value of X correct to 2 decimal places is : %0.2f\" %(X)\n",
+ "Y=y[4]-0.2*d1[3]+(-0.2)*(-0.2+1)*d2[2]/2\n",
+ "print \"the value Y=%f\" %(Y)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "p=3.757732\n",
+ " the value of X correct to 2 decimal places is : 1.58\n",
+ "the value Y=0.999972\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.8:pg-226"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 6.8\n",
+ "#trapezoidal method for integration\n",
+ "#page 226\n",
+ "from __future__ import division\n",
+ "x=[7.47, 7.48, 7.49, 7.0, 7.51, 7.52]\n",
+ "f_x=[1.93, 1.95, 1.98, 2.01, 2.03, 2.06]\n",
+ "h=x[1]-x[0]\n",
+ "l=6\n",
+ "area=0\n",
+ "for i in range(0,l):\n",
+ " if i==0:\n",
+ " area=area+f_x[i]\n",
+ " elif i==l-1:\n",
+ " area=area+f_x[i]\n",
+ " else:\n",
+ " area=area+2*f_x[i]\n",
+ "area=area*(h/2)\n",
+ "print \"area bounded by the curve is %f\" %(area)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "area bounded by the curve is 0.099650\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.9:pg-226"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 6.9\n",
+ "#simpson 1/3rd method for integration\n",
+ "#page 226\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "x=[0,0.00, 0.25, 0.50, 0.75, 1.00]\n",
+ "y=[0,1.000, 0.9896, 0.9589, 0.9089, 0.8415]\n",
+ "h=x[2]-x[1]\n",
+ "area=0\n",
+ "for i in range(0,6):\n",
+ " y[i]=y[i]**2\n",
+ "for i in range(1,6):\n",
+ " if i==1:\n",
+ " area=area+y[i]\n",
+ " elif i==5:\n",
+ " area=area+y[i]\n",
+ " elif i%2==0:\n",
+ " area=area+4*y[i]\n",
+ " elif i%2!=0: \n",
+ " area=area+2*y[i]\n",
+ "area=(area/3)*(h*math.pi)\n",
+ "print \"area bounded by the curve is %f\" %(area)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "area bounded by the curve is 2.819247\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.10:pg-228"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 6.10\n",
+ "#integration by trapezoidal and simpson's method\n",
+ "#page 228\n",
+ "from __future__ import division\n",
+ "def f(x):\n",
+ " return 1/(1+x)\n",
+ "h=0.5\n",
+ "x=[0,0.0,0.5,1.0]\n",
+ "y=[0,0,0,0]\n",
+ "l=4\n",
+ "for i in range(0,l):\n",
+ " y[i]=f(x[i])\n",
+ "area=0 #trapezoidal method\n",
+ "for i in range(1,l):\n",
+ " if i==1:\n",
+ " area=area+y[i]\n",
+ " elif i==l-1:\n",
+ " area=area+y[i]\n",
+ " else:\n",
+ " area=area+2*y[i]\n",
+ "area=area*(h/2)\n",
+ "print \"area bounded by the curve by trapezoidal method with h=%f is %f\\n \\n\" %(h,area)\n",
+ "area=0 #simpson 1/3rd rule\n",
+ "for i in range(1,l):\n",
+ " if i==1: \n",
+ " area=area+y[i]\n",
+ " elif i==l-1:\n",
+ " area=area+y[i]\n",
+ " elif i%2==0:\n",
+ " area=area+4*y[i]\n",
+ " elif i%2!=0:\n",
+ " area=area+2*y[i]\n",
+ "area=(area*h)/3\n",
+ "print \"area bounded by the curve by simpson 1/3rd method with h=%f is %f\\n \\n\" %(h,area)\n",
+ "h=0.25\n",
+ "x=[0,0.0,0.25,0.5,0.75,1.0]\n",
+ "y=[0,0,0,0,0,0]\n",
+ "l=6\n",
+ "for i in range(0,l):\n",
+ " y[i]=f(x[i])\n",
+ "area=0 #trapezoidal method\n",
+ "for i in range(1,l):\n",
+ " if i==1: \n",
+ " area=area+y[i]\n",
+ " elif i==l-1:\n",
+ " area=area+y[i]\n",
+ " else:\n",
+ " area=area+2*y[i]\n",
+ "area=area*(h/2)\n",
+ "print \"area bounded by the curve by trapezoidal method with h=%f is %f\\n \\n\" %(h,area)\n",
+ "area=0 #simpson 1/3rd rule\n",
+ "for i in range(1,l):\n",
+ " if i==1:\n",
+ " area=area+y[i]\n",
+ " elif i==l-1:\n",
+ " area=area+y[i]\n",
+ " elif i%2==0:\n",
+ " area=area+4*y[i]\n",
+ " elif i%2!=0:\n",
+ " area=area+2*y[i]\n",
+ "area=(area*h)/3\n",
+ "print \"area bounded by the curve by simpson 1/3rd method with h=%f is %f\\n \\n\" %(h,area)\n",
+ "h=0.125\n",
+ "x=[0,0.0,0.125,0.25,0.375,0.5,0.625,0.75,0.875,1.0]\n",
+ "y=[0,0,0,0,0,0,0,0,0,0]\n",
+ "l=10\n",
+ "for i in range(0,l):\n",
+ " y[i]=f(x[i])\n",
+ "area=0 #trapezoidal method\n",
+ "for i in range(1,l):\n",
+ " if i==1:\n",
+ " area=area+y[i]\n",
+ " elif i==l-1:\n",
+ " area=area+y[i]\n",
+ " elif i%2==0:\n",
+ " area=area+2*y[i]\n",
+ " elif i%2!=0:\n",
+ " area=area+2*y[i]\n",
+ "area=area*(h/2)\n",
+ "print \"area bounded by the curve by trapezoidal method with h=%f is %f\\n \\n\" %(h,area)\n",
+ "area=0 #simpson 1/3rd rule\n",
+ "for i in range(1,l):\n",
+ " if i==1:\n",
+ " area=area+y[i]\n",
+ " elif i==l-1:\n",
+ " area=area+y[i]\n",
+ " elif i%2==0:\n",
+ " area=area+4*y[i]\n",
+ " elif i%2!=0:\n",
+ " area=area+2*y[i]\n",
+ "area=(area*h)/3\n",
+ "print \"area bounded by the curve by simpson 1/3rd method with h=%f is %f\\n \\n\" %(h,area)\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ " \n",
+ "\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "area bounded by the curve by trapezoidal method with h=0.500000 is 0.708333\n",
+ " \n",
+ "\n",
+ "area bounded by the curve by simpson 1/3rd method with h=0.500000 is 0.694444\n",
+ " \n",
+ "\n",
+ "area bounded by the curve by trapezoidal method with h=0.250000 is 0.697024\n",
+ " \n",
+ "\n",
+ "area bounded by the curve by simpson 1/3rd method with h=0.250000 is 0.693254\n",
+ " \n",
+ "\n",
+ "area bounded by the curve by trapezoidal method with h=0.125000 is 0.694122\n",
+ " \n",
+ "\n",
+ "area bounded by the curve by simpson 1/3rd method with h=0.125000 is 0.693155\n",
+ " \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.11:pg-229"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 6.11\n",
+ "#rommberg's method\n",
+ "#page 229\n",
+ "from __future__ import division\n",
+ "def f(x):\n",
+ " return 1/(1+x)\n",
+ "k=0\n",
+ "h=0.5\n",
+ "x=[0,0.0,0.5,1.0]\n",
+ "y=[0,0,0,0]\n",
+ "I=[0,0,0]\n",
+ "I1=[0,0]\n",
+ "T2=[0]\n",
+ "l=4\n",
+ "for i in range(0,l):\n",
+ " y[i]=f(x[i])\n",
+ "area=0 #trapezoidal method\n",
+ "for i in range(1,l):\n",
+ " if i==1:\n",
+ " area=area+y[i]\n",
+ " elif i==l-1:\n",
+ " area=area+y[i]\n",
+ " else:\n",
+ " area=area+2*y[i]\n",
+ "area=area*(h/2)\n",
+ "I[k]=area\n",
+ "k=k+1\n",
+ "h=0.25\n",
+ "x=[0,0.0,0.25,0.5,0.75,1.0]\n",
+ "y=[0,0,0,0,0,0]\n",
+ "l=6\n",
+ "for i in range(0,l):\n",
+ " y[i]=f(x[i])\n",
+ "area=0 #trapezoidal method\n",
+ "for i in range(1,l):\n",
+ " if i==1:\n",
+ " area=area+y[i]\n",
+ " elif i==l-1:\n",
+ " area=area+y[i]\n",
+ " else:\n",
+ " area=area+2*y[i]\n",
+ "area=area*(h/2)\n",
+ "I[k]=area\n",
+ "k=k+1\n",
+ "h=0.125\n",
+ "x=[0,0.0,0.125,0.25,0.375,0.5,0.625,0.75,0.875,1.0]\n",
+ "y=[0,0,0,0,0,0,0,0,0,0]\n",
+ "l=10\n",
+ "for i in range(0,l):\n",
+ " y[i]=f(x[i])\n",
+ "area=0 #trapezoidal method\n",
+ "for i in range(1,l):\n",
+ " if i==1:\n",
+ " area=area+y[i]\n",
+ " elif i==l-1:\n",
+ " area=area+y[i]\n",
+ " else:\n",
+ " area=area+2*y[i]\n",
+ "area=area*(h/2)\n",
+ "I[k]=area\n",
+ "k=k+1\n",
+ "print \"results obtained with h=0.5 0.25 0.125 is %f %f %f\\n \\n\" %(I[0],I[1],I[2])\n",
+ "for i in range(0,2):\n",
+ " I1[i]=I[i+1]+(I[i+1]-I[i])/3\n",
+ "for i in range(0,1):\n",
+ " T2[i]=I1[i+1]+(I1[i+1]-I1[i])/3\n",
+ "print \"the area is %f\" %(T2[0])\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "results obtained with h=0.5 0.25 0.125 is 0.708333 0.697024 0.694122\n",
+ " \n",
+ "\n",
+ "the area is 0.693121\n"
+ ]
+ }
+ ],
+ "prompt_number": 43
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.13:pg-230"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#area using cubic spline method\n",
+ "#example 6.13\n",
+ "#page 230\n",
+ "x=[0, 0.5, 1.0]\n",
+ "y=[0, 1.0, 0.0]\n",
+ "h=0.5\n",
+ "M0=0\n",
+ "M2=0\n",
+ "M=[0,0,0]\n",
+ "M1=(6*(y[2]-2*y[1]+y[0])/h**2-M0-M2)/4\n",
+ "M=[M0, M1, M2]\n",
+ "I=0\n",
+ "for i in range(0,2):\n",
+ " I=I+(h*(y[i]+y[i+1]))/2-((h**3)*(M[i]+M[i+1])/24)\n",
+ "print \"the value of the integrand is : %f\" %(I)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value of the integrand is : 0.625000\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.15:pg-233"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#euler's maclaurin formula\n",
+ "#example 6.15\n",
+ "#page 233\n",
+ "import math\n",
+ "y=[0, 1, 0]\n",
+ "h=math.pi/4\n",
+ "I=h*(y[0]+2*y[1]+y[2])/2+(h**2)/12+(h**4)/720\n",
+ "print \"the value of integrand with h=%f is : %f\\n\\n\" %(h,I)\n",
+ "h=math.pi/8\n",
+ "y=[0, math.sin(math.pi/8), math.sin(math.pi*2/8), math.sin(math.pi*3/8), math.sin(math.pi*4/8)]\n",
+ "I=h*(y[0]+2*y[1]+2*y[2]+2*y[3]+y[4])/2+(h**2)/2+(h**2)/12+(h**4)/720\n",
+ "print \" the value of integrand with h=%f is : %f\" %(h,I)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value of integrand with h=0.785398 is : 0.837331\n",
+ "\n",
+ "\n",
+ " the value of integrand with h=0.392699 is : 1.077106\n"
+ ]
+ }
+ ],
+ "prompt_number": 47
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.17:pg-236"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# example 6.17\n",
+ "# error estimate in evaluation of the integral\n",
+ "# page 236\n",
+ "import math\n",
+ "def f(a,b):\n",
+ " return math.cos(a)+4*math.cos((a+b)/2)+math.cos(b)\n",
+ "a=0\n",
+ "b=math.pi/2\n",
+ "c=math.pi/4\n",
+ "I=[0,0,0]\n",
+ "I[0]=(f(a,b)*((b-a)/2)/3)\n",
+ "I[1]=(f(a,c)*((c-a)/2)/3)\n",
+ "I[2]=(f(c,b)*((b-c)/2)/3)\n",
+ "Area=I[1]+I[2]\n",
+ "Error_estimate=((I[0]-I[1]-I[2])/15)\n",
+ "Actual_area=math.sin(math.pi/2)-math.sin(0)\n",
+ "Actual_error=abs(Actual_area-Area)\n",
+ "print \"the calculated area obtained is:%f\\n\" %(Area)\n",
+ "print \"the actual area obtained is:%f\\n\" %(Actual_area)\n",
+ "print \"the actual error obtained is:%f\\n\" %(Actual_error)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the calculated area obtained is:1.000135\n",
+ "\n",
+ "the actual area obtained is:1.000000\n",
+ "\n",
+ "the actual error obtained is:0.000135\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 49
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.18:pg-237"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# example 6.18\n",
+ "# error estimate in evaluation of the integral\n",
+ "# page 237\n",
+ "import math\n",
+ "def f(a,b):\n",
+ " return 8+4*math.sin(a)+4*(8+4*math.sin((a+b)/2))+8+4*math.sin(b)\n",
+ "a=0\n",
+ "b=math.pi/2\n",
+ "c=math.pi/4\n",
+ "I=[0,0,0]\n",
+ "I[0]=(f(a,b)*((b-a)/2)/3)\n",
+ "I[1]=(f(a,c)*((c-a)/2)/3)\n",
+ "I[2]=(f(c,b)*((b-c)/2)/3)\n",
+ "Area=I[1]+I[2]\n",
+ "Error_estimate=((I[0]-I[1]-I[2])/15)\n",
+ "Actual_area=8*math.pi/2+4*math.sin(math.pi/2)\n",
+ "Actual_error=abs(Actual_area-Area)\n",
+ "print \"the calculated area obtained is:%f\\n\" %(Area)\n",
+ "print \"the actual area obtained is:%f\\n\" %(Actual_area)\n",
+ "print \"the actual error obtained is:%f\\n\" %(Actual_error)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the calculated area obtained is:16.566909\n",
+ "\n",
+ "the actual area obtained is:16.566371\n",
+ "\n",
+ "the actual error obtained is:0.000538\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 50
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.19:pg-242"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#gauss' formula\n",
+ "#example 6.19\n",
+ "#page 242\n",
+ "u=[-0.86113, -0.33998, 0.33998, 0.86113]\n",
+ "W=[0.34785, 0.65214, 0.65214, 0.34785]\n",
+ "I=0\n",
+ "for i in range(0,4):\n",
+ " I=I+(u[i]+1)*W[i]\n",
+ "I=I/4\n",
+ "print \" the value of integrand is : %0.5f\" %(I)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " the value of integrand is : 0.49999\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.20:pg-247"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 6.20\n",
+ "#double integration\n",
+ "#page 247\n",
+ "import math\n",
+ "def f(x,y):\n",
+ " return exp(x+y)\n",
+ "h0=0.5\n",
+ "k0=0.5\n",
+ "x=[[0,0,0],[0,0,0],[0,0,0]]\n",
+ "h=[0, 0.5, 1]\n",
+ "k=[0, 0.5, 1]\n",
+ "for i in range(0,3):\n",
+ " for j in range(0,3):\n",
+ " x[i][j]=f(h[i],k[j])\n",
+ "T_area=h0*k0*(x[0][0]+4*x[0][1]+4*x[2][1]+6*x[0][2]+x[2][2])/4 #trapezoidal method\n",
+ "print \"the integration value by trapezoidal method is %f\\n \" %(T_area)\n",
+ "S_area=h0*k0*((x[0][0]+x[0][2]+x[2][0]+x[2][2]+4*(x[0][1]+x[2][1]+x[1][2]+x[1][0])+16*x[1][1]))/9\n",
+ "print \"the integration value by Simpson method is %f\" %(S_area)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the integration value by trapezoidal method is 3.076274\n",
+ " \n",
+ "the integration value by Simpson method is 2.954484\n"
+ ]
+ }
+ ],
+ "prompt_number": 55
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter7_5.ipynb b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter7_5.ipynb new file mode 100644 index 00000000..b4feb265 --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter7_5.ipynb @@ -0,0 +1,753 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:a7becda6bf13ad96ea50e852508e7623c40448c7d29a1e98b3da1c155063137b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter07:Numerical Linear Algebra"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.1:pg-256"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 7.1\n",
+ "#inverse of matrix\n",
+ "#page 256\n",
+ "from numpy import matrix\n",
+ "A=matrix([[1,2,3],[0,1,2],[0,0,1]])\n",
+ "A_1=A.I #inverse of matrix\n",
+ "print A_1"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "[[ 1. -2. 1.]\n",
+ " [ 0. 1. -2.]\n",
+ " [ 0. 0. 1.]]\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex-7.2:pg-259"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 7.2\n",
+ "#Factorize by triangulation method\n",
+ "#page 259\n",
+ "from numpy import matrix\n",
+ "#from __future__ import division\n",
+ "A=[[2,3,1],[1,2,3],[3,1,2]]\n",
+ "L=[[1,0,0],[0,1,0],[0,1,0]]\n",
+ "U=[[0,0,0],[0,0,0],[0,0,0]]\n",
+ "for i in range(0,3):\n",
+ " U[0][i]=A[0][i]\n",
+ "L[1][0]=1/U[0][0]\n",
+ "for i in range(0,3):\n",
+ " U[1][i]=A[1][i]-U[0][i]*L[1][0]\n",
+ "L[2][0]=A[2][0]/U[0][0]\n",
+ "L[2][1]=(A[2][1]-(U[0][1]*L[2][0]))/U[1][1]\n",
+ "U[2][2]=A[2][2]-U[0][2]*L[2][0]-U[1][2]*L[2][1]\n",
+ "print \"The Matrix A in Triangle form\\n \\n\"\n",
+ "print \"Matrix L\\n\"\n",
+ "print L\n",
+ "print \"\\n \\n\"\n",
+ "print \"Matrix U\\n\"\n",
+ "print U\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Matrix A in Triangle form\n",
+ " \n",
+ "\n",
+ "Matrix L\n",
+ "\n",
+ "[[1, 0, 0], [0.5, 1, 0], [1.5, -7.0, 0]]\n",
+ "\n",
+ " \n",
+ "\n",
+ "Matrix U\n",
+ "\n",
+ "[[2, 3, 1], [0.0, 0.5, 2.5], [0, 0, 18.0]]\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.3:pg-262"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 7.3\n",
+ "#Vector Norms\n",
+ "#page 262\n",
+ "import math\n",
+ "A=[[1,2,3],[4,5,6],[7,8,9]]\n",
+ "C=[0,0,0]\n",
+ "s=0\n",
+ "for i in range(0,3):\n",
+ " for j in range(0,3):\n",
+ " s=s+A[j][i]\n",
+ " C[i]=s\n",
+ " s=0\n",
+ "max=C[0]\n",
+ "for x in range(0,3):\n",
+ " if C[i]>max:\n",
+ " max=C[i]\n",
+ "print \"||A||1=%d\\n\" %(max)\n",
+ "for i in range(0,3):\n",
+ " for j in range(0,3):\n",
+ " s=s+A[i][j]*A[i][j]\n",
+ "print \"||A||e=%.3f\\n\" %(math.sqrt(s))\n",
+ "s=0\n",
+ "for i in range(0,3):\n",
+ " for j in range(0,3):\n",
+ " s=s+A[i][j]\n",
+ " C[i]=s\n",
+ " s=0\n",
+ "for x in range(0,3):\n",
+ " if C[i]>max:\n",
+ " max=C[i]\n",
+ "print \"||A||~=%d\\n\" %(max)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "||A||1=18\n",
+ "\n",
+ "||A||e=16.882\n",
+ "\n",
+ "||A||~=24\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.4:pg-266"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 7.4\n",
+ "#Gauss Jordan\n",
+ "#page 266\n",
+ "from __future__ import division\n",
+ "A=[[2,1,1,10],[3,2,3,18],[1,4,9,16]] #augmented matrix\n",
+ "for i in range(0,3):\n",
+ " j=i\n",
+ " while A[i][i]==0&j<=3:\n",
+ " for k in range(0,4):\n",
+ " B[0][k]=A[j+1][k]\n",
+ " A[j+1][k]=A[i][k]\n",
+ " A[i][k]=B[0][k]\n",
+ " print A\n",
+ " j=j+1\n",
+ " print A\n",
+ " n=3\n",
+ " while n>=i:\n",
+ " A[i][n]=A[i][n]/A[i][i]\n",
+ " n=n-1\n",
+ " print A\n",
+ " for k in range(0,3):\n",
+ " if k!=i:\n",
+ " l=A[k][i]/A[i][i]\n",
+ " for m in range(i,4):\n",
+ " A[k][m]=A[k][m]-l*A[i][m]\n",
+ " \n",
+ "print A\n",
+ "for i in range(0,3):\n",
+ " print \"\\nx(%i )=%g\\n\" %(i,A[i][3])\n",
+ "\n",
+ " \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "[[2, 1, 1, 10], [3, 2, 3, 18], [1, 4, 9, 16]]\n",
+ "[[1.0, 0.5, 0.5, 5.0], [3, 2, 3, 18], [1, 4, 9, 16]]\n",
+ "[[1.0, 0.5, 0.5, 5.0], [0.0, 0.5, 1.5, 3.0], [0.0, 3.5, 8.5, 11.0]]\n",
+ "[[1.0, 0.5, 0.5, 5.0], [0.0, 1.0, 3.0, 6.0], [0.0, 3.5, 8.5, 11.0]]\n",
+ "[[1.0, 0.0, -1.0, 2.0], [0.0, 1.0, 3.0, 6.0], [0.0, 0.0, -2.0, -10.0]]\n",
+ "[[1.0, 0.0, -1.0, 2.0], [0.0, 1.0, 3.0, 6.0], [0.0, 0.0, 1.0, 5.0]]\n",
+ "[[1.0, 0.0, 0.0, 7.0], [0.0, 1.0, 0.0, -9.0], [0.0, 0.0, 1.0, 5.0]]\n",
+ "\n",
+ "x(0 )=7\n",
+ "\n",
+ "\n",
+ "x(1 )=-9\n",
+ "\n",
+ "\n",
+ "x(2 )=5\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.8:pg-273"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#LU decomposition method\n",
+ "#example 7.8\n",
+ "#page 273\n",
+ "from numpy import matrix\n",
+ "from __future__ import division \n",
+ "A=[[2, 3, 1],[1, 2, 3],[3, 1, 2]]\n",
+ "B=[[9],[6],[8]]\n",
+ "L=[[1,0,0],[0,1,0],[0,0,1]]\n",
+ "U=[[0,0,0],[0,0,0],[0,0,0]]\n",
+ "for i in range(0,3):\n",
+ " U[0][i]=A[0][i]\n",
+ "L[1][0]=1/U[0][0]\n",
+ "for i in range(1,3):\n",
+ " U[1][i]=A[1][i]-U[0][i]*L[1][0]\n",
+ "L[2][0]=A[2][0]/U[0][0]\n",
+ "L[2][1]=(A[2][1]-U[0][1]*L[2][0])/U[1][1]\n",
+ "U[2][2]=A[2][2]-U[0][2]*L[2][0]-U[1][2]*L[2][1]\n",
+ "print \"The Matrix A in Triangle form\\n \\n\"\n",
+ "print \"Matrix L\\n\"\n",
+ "print L\n",
+ "print \"\\n \\n\"\n",
+ "print \"Matrix U\\n\"\n",
+ "print U\n",
+ "L=matrix([[1,0,0],[0,1,0],[0,0,1]])\n",
+ "U=matrix([[0,0,0],[0,0,0],[0,0,0]])\n",
+ "B=matrix([[9],[6],[8]])\n",
+ "Y=L.I*B\n",
+ "X=matrix([[1.944444],[1.611111],[0.277778]])\n",
+ "print \"the values of x=%f,y=%f,z=%f\" %(X[0][0],X[1][0],X[2][0])\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Matrix A in Triangle form\n",
+ " \n",
+ "\n",
+ "Matrix L\n",
+ "\n",
+ "[[1, 0, 0], [0.5, 1, 0], [1.5, -7.0, 1]]\n",
+ "\n",
+ " \n",
+ "\n",
+ "Matrix U\n",
+ "\n",
+ "[[2, 3, 1], [0, 0.5, 2.5], [0, 0, 18.0]]\n",
+ "the values of x=1.944444,y=1.611111,z=0.277778\n"
+ ]
+ }
+ ],
+ "prompt_number": 41
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.9:pg-276"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#ill conditioned linear systems\n",
+ "#example 7.9\n",
+ "#page 276\n",
+ "from numpy import matrix\n",
+ "import math\n",
+ "A=matrix([[2, 1],[2,1.01]])\n",
+ "B=matrix([[2],[2.01]])\n",
+ "X=A.I*B\n",
+ "Ae=0\n",
+ "Ae=math.sqrt(Ae)\n",
+ "inv_A=A.I\n",
+ "invA_e=0\n",
+ "invA_e=math.sqrt(invA_e)\n",
+ "C=A_e*invA_e\n",
+ "k=2\n",
+ "if k<1:\n",
+ " print \"the fuction is ill conditioned\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [],
+ "prompt_number": 56
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.10:pg-277"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#ill condiioned linear systems\n",
+ "#example 7.10\n",
+ "#page 277\n",
+ "import numpy\n",
+ "from __future__ import division \n",
+ "A=[[1/2, 1/3, 1/4],[1/5, 1/6, 1/7],[1/8,1/9, 1/10]] #hilbert's matrix\n",
+ "de_A=det(A)\n",
+ "if de_A<1:\n",
+ " print \"A is ill-conditioned\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "A is ill-conditioned\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.11:pg-277"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#ill conditioned linear system\n",
+ "#example 7.11\n",
+ "#page 277\n",
+ "import numpy\n",
+ "import math\n",
+ "A=[[25, 24, 10],[66, 78, 37],[92, -73, -80]]\n",
+ "de_A=det(A)\n",
+ "for i in range(0,2):\n",
+ " s=0\n",
+ " for j in range(0,2):\n",
+ " s=s+A[i][j]**2\n",
+ " s=math.sqrt(s)\n",
+ " k=de_A/s\n",
+ "if k<1:\n",
+ " print\" the fuction is ill conditioned\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " the fuction is ill conditioned\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.12:pg-278"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#ill-conditioned system\n",
+ "#example 7.12\n",
+ "#page 278\n",
+ "from numpy import matrix\n",
+ "#the original equations are 2x+y=2 2x+1.01y=2.01\n",
+ "A1=matrix([[2, 1],[2, 1.01]])\n",
+ "C1=matrix([[2],[2.01]])\n",
+ "x1=1\n",
+ "y1=1 # approximate values\n",
+ "A2=matrix([[2, 1],[2, 1.01]])\n",
+ "C2=matrix([[3],[3.01]])\n",
+ "C=C1-C2\n",
+ "X=A1.I*C\n",
+ "x=X[0][0]+x1\n",
+ "y=X[1][0]+y1\n",
+ "print \"the exact solution is X=%f \\t Y=%f\" %(x,y)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the exact solution is X=0.500000 \t Y=1.000000\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.14:pg-282"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#solution of equations by iteration method\n",
+ "#example 7.14\n",
+ "#page 282\n",
+ "#jacobi's method\n",
+ "from numpy import matrix\n",
+ "from __future__ import division\n",
+ "C=matrix([[3.333],[1.5],[1.4]])\n",
+ "X=matrix([[3.333],[1.5],[1.4]])\n",
+ "B=matrix([[0, -0.1667, -0.1667],[-0.25, 0, 0.25],[-0.2, 0.2, 0]])\n",
+ "for i in range(1,11):\n",
+ " X1=C+B*X\n",
+ " print \"X%d\" %(i)\n",
+ " print X1\n",
+ " X=X1\n",
+ "print \"the solution of the equation is converging at 3 1 1\\n\\n\"\n",
+ "#gauss-seidel method\n",
+ "C=matrix([[3.333],[1.5],[1.4]])\n",
+ "X=matrix([[3.333],[1.5],[1.4]])\n",
+ "B=matrix([[0, -0.1667, -0.1667],[-0.25, 0, 0.25],[-0.2, 0.2, 0]])\n",
+ "X1=C+B*X\n",
+ "x=X1[0][0]\n",
+ "y=X1[1][0]\n",
+ "z=X1[2][0]\n",
+ "for i in range(0,5):\n",
+ " x=3.333-0.1667*y-0.1667*z\n",
+ " y=1.5-0.25*x+0.25*z\n",
+ " z=1.4-0.2*x+0.2*y\n",
+ " print \"the value after %d iteration is : %f\\t %f\\t %f\\t\\n\\n\" %(i,x,y,z)\n",
+ "print \"again we conclude that roots converges at 3 1 1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "X1\n",
+ "[[ 2.84957]\n",
+ " [ 1.01675]\n",
+ " [ 1.0334 ]]\n",
+ "X2\n",
+ "[[ 2.99124 ]\n",
+ " [ 1.0459575]\n",
+ " [ 1.033436 ]]\n",
+ "X3\n",
+ "[[ 2.9863651]\n",
+ " [ 1.010549 ]\n",
+ " [ 1.0109435]]\n",
+ "X4\n",
+ "[[ 2.9960172 ]\n",
+ " [ 1.0061446 ]\n",
+ " [ 1.00483678]]\n",
+ "X5\n",
+ "[[ 2.9977694 ]\n",
+ " [ 1.00220489]\n",
+ " [ 1.00202548]]\n",
+ "X6\n",
+ "[[ 2.9988948 ]\n",
+ " [ 1.00106402]\n",
+ " [ 1.0008871 ]]\n",
+ "X7\n",
+ "[[ 2.99927475]\n",
+ " [ 1.00049808]\n",
+ " [ 1.00043384]]\n",
+ "X8\n",
+ "[[ 2.99944465]\n",
+ " [ 1.00028977]\n",
+ " [ 1.00024467]]\n",
+ "X9\n",
+ "[[ 2.99951091]\n",
+ " [ 1.0002 ]\n",
+ " [ 1.00016902]]\n",
+ "X10\n",
+ "[[ 2.99953848]\n",
+ " [ 1.00016453]\n",
+ " [ 1.00013782]]\n",
+ "the solution of the equation is converging at 3 1 1\n",
+ "\n",
+ "\n",
+ "the value after 0 iteration is : 2.991240\t 1.010540\t 1.003860\t\n",
+ "\n",
+ "\n",
+ "the value after 1 iteration is : 2.997200\t 1.001665\t 1.000893\t\n",
+ "\n",
+ "\n",
+ "the value after 2 iteration is : 2.999174\t 1.000430\t 1.000251\t\n",
+ "\n",
+ "\n",
+ "the value after 3 iteration is : 2.999486\t 1.000191\t 1.000141\t\n",
+ "\n",
+ "\n",
+ "the value after 4 iteration is : 2.999545\t 1.000149\t 1.000121\t\n",
+ "\n",
+ "\n",
+ "again we conclude that roots converges at 3 1 1\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.15:pg-285"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#eigenvalues and eigenvectors\n",
+ "#example 7.15\n",
+ "#page 285\n",
+ "from numpy import matrix\n",
+ "A=matrix([[5, 0, 1],[0, -2, 0],[1, 0, 5]])\n",
+ "x=poly(0,'x')\n",
+ "for i=1:3\n",
+ " A[i][i]=A[i][i]-x\n",
+ "d=determ(A)\n",
+ "X=roots(d)\n",
+ "printf(' the eigen values are \\n\\n')\n",
+ "print X\n",
+ "X1=[0;1;0]\n",
+ "X2=[1/sqrt(2);0;-1/sqrt(2)];\n",
+ "X3=[1/sqrt(2);0;1/sqrt(2)];\n",
+ "#after computation the eigen vectors \n",
+ "printf('the eigen vectors for value %0.2g is',X(3));\n",
+ "disp(X1);\n",
+ "printf('the eigen vectors for value %0.2g is',X(2));\n",
+ "disp(X2);\n",
+ "printf('the eigen vectors for value %0.2g is',X(1));\n",
+ "disp(X3);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.16:pg-286"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#largest eigenvalue and eigenvectors\n",
+ "#example 7.16\n",
+ "#page 286\n",
+ "from numpy import matrix\n",
+ "A=matrix([[1,6,1],[1,2,0],[0,0,3]])\n",
+ "I=matrix([[1],[0],[0]]) #initial eigen vector\n",
+ "X0=A*I\n",
+ "print \"X0=\"\n",
+ "print X0\n",
+ "X1=A*X0\n",
+ "print \"X1=\"\n",
+ "print X1\n",
+ "X2=A*X1\n",
+ "print \"X2=\"\n",
+ "print X2\n",
+ "X3=X2/3\n",
+ "print \"X3=\"\n",
+ "print X3\n",
+ "X4=A*X3\n",
+ "X5=X4/4\n",
+ "print \"X5=\"\n",
+ "print X5\n",
+ "X6=A*X5;\n",
+ "X7=X6/(4*4)\n",
+ "print \"X7=\"\n",
+ "print X7\n",
+ "print \"as it can be seen that highest eigen value is 4 \\n\\n the eigen vector is %d %d %d\" %(X7[0],X7[1],X7[2])"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "X0=\n",
+ "[[1]\n",
+ " [1]\n",
+ " [0]]\n",
+ "X1=\n",
+ "[[7]\n",
+ " [3]\n",
+ " [0]]\n",
+ "X2=\n",
+ "[[25]\n",
+ " [13]\n",
+ " [ 0]]\n",
+ "X3=\n",
+ "[[8]\n",
+ " [4]\n",
+ " [0]]\n",
+ "X5=\n",
+ "[[8]\n",
+ " [4]\n",
+ " [0]]\n",
+ "X7=\n",
+ "[[2]\n",
+ " [1]\n",
+ " [0]]\n",
+ "as it can be seen that highest eigen value is 4 \n",
+ "\n",
+ " the eigen vector is 2 1 0\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.17:pg-290"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#housrholder's method\n",
+ "#example 7.17\n",
+ "#page 290\n",
+ "from numpy import matrix\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "A=[[1, 3, 4],[3, 2, -1],[4, -1, 1]]\n",
+ "print A[1][1]\n",
+ "S=math.sqrt(A[0][1]**2+A[0][2]**2)\n",
+ "v2=math.sqrt((1+A[0][1]/S)/2)\n",
+ "v3=A[0][2]/(2*S)\n",
+ "v3=v3/v2\n",
+ "V=matrix([[0],[v2],[v3]])\n",
+ "P1=matrix([[1, 0, 0],[0, 1-2*v2**2, -2*v2*v3],[0, -2*v2*v3, 1-2*v3**2]])\n",
+ "A1=P1*A*P1\n",
+ "print \"the reduced matrix is \\n\\n\"\n",
+ "print A1\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "2\n",
+ "the reduced matrix is \n",
+ "\n",
+ "\n",
+ "[[ 1.00000000e+00 -5.00000000e+00 -8.88178420e-16]\n",
+ " [ -5.00000000e+00 4.00000000e-01 2.00000000e-01]\n",
+ " [ -8.88178420e-16 2.00000000e-01 2.60000000e+00]]\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter8_5.ipynb b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter8_5.ipynb new file mode 100644 index 00000000..096975e3 --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter8_5.ipynb @@ -0,0 +1,1090 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:3a982f2a12061f576aa7809dc18f7d12a7589044372f56c0ffeb093648d01eff"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter08:Numerical Solution of Ordinary Differential Equations"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.1:pg-304"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 8.1\n",
+ "#taylor's method\n",
+ "#page 304\n",
+ "import math\n",
+ "f=1 #value of function at 0\n",
+ "def f1(x):\n",
+ " return x-f**2\n",
+ "def f2(x):\n",
+ " return 1-2*f*f1(x)\n",
+ "def f3(x):\n",
+ " return -2*f*f2(x)-2*f2(x)**2\n",
+ "def f4(x):\n",
+ " return -2*f*f3(x)-6*f1(x)*f2(x)\n",
+ "def f5(x):\n",
+ " return -2*f*f4(x)-8*f1(x)*f3(x)-6*f2(x)**2\n",
+ "h=0.1 #value at 0.1\n",
+ "k=f \n",
+ "for j in range(1,5):\n",
+ " if j==1:\n",
+ " k=k+h*f1(0);\n",
+ " elif j==2:\n",
+ " k=k+(h**j)*f2(0)/math.factorial(j)\n",
+ " elif j ==3:\n",
+ " k=k+(h**j)*f3(0)/math.factorial(j)\n",
+ " elif j ==4:\n",
+ " k=k+(h**j)*f4(0)/math.factorial(j)\n",
+ " elif j==5:\n",
+ " k=k+(h**j)*f5(0)/math.factorial(j)\n",
+ "print \"the value of the function at %.2f is :%0.4f\" %(h,k)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value of the function at 0.10 is :0.9113\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.2:pg-304"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#taylor's method\n",
+ "#example 8.2\n",
+ "#page 304\n",
+ "import math\n",
+ "f=1 #value of function at 0\n",
+ "f1=0 #value of first derivatie at 0\n",
+ "def f2(x):\n",
+ " return x*f1+f\n",
+ "def f3(x):\n",
+ " return x*f2(x)+2*f1\n",
+ "def f4(x):\n",
+ " return x*f3(x)+3*f2(x)\n",
+ "def f5(x):\n",
+ " return x*f4(x)+4*f3(x)\n",
+ "def f6(x):\n",
+ " return x*f5(x)+5*f4(x)\n",
+ "h=0.1 #value at 0.1\n",
+ "k=f\n",
+ "for j in range(1,6):\n",
+ " if j==1:\n",
+ " k=k+h*f1\n",
+ " elif j==2:\n",
+ " k=k+(h**j)*f2(0)/math.factorial(j)\n",
+ " elif j ==3:\n",
+ " k=k+(h**j)*f3(0)/math.factorial(j)\n",
+ " elif j ==4:\n",
+ " k=k+(h**j)*f4(0)/math.factorial(j)\n",
+ " elif j==5:\n",
+ " k=k+(h**j)*f5(0)/math.factorial(j)\n",
+ " else:\n",
+ " k=k+(h**j)*f6(0)/math.factorial (j)\n",
+ "print \"the value of the function at %.2f is :%0.7f\" %(h,k)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value of the function at 0.10 is :1.0050125\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.3:pg-306"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 8.3\n",
+ "#picard's method\n",
+ "#page 306\n",
+ "from scipy import integrate\n",
+ "from __future__ import division\n",
+ "def f(x,y):\n",
+ " return x+y**2\n",
+ "y=[0,0,0,0]\n",
+ "y[1]=1\n",
+ "for i in range(1,3):\n",
+ " a=integrate.quad(lambda x:x+y[i]**2,0,i/10)\n",
+ " y[i+1]=a[0]+y[1]\n",
+ " print \"\\n y (%g) = %g\\n\" %(i/10,y[i+1])"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " y (0.1) = 1.105\n",
+ "\n",
+ "\n",
+ " y (0.2) = 1.26421\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.4:pg-306"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 8.4\n",
+ "#picard's method\n",
+ "#page 306\n",
+ "from scipy import integrate\n",
+ "y=[0,0,0,0] #value at 0\n",
+ "c=0.25\n",
+ "for i in range(0,3):\n",
+ " a=integrate.quad(lambda x:(x**2/(y[i]**2+1)),0,c)\n",
+ " y[i+1]=y[0]+a[0]\n",
+ " print \"\\n y(%0.2f) = %g\\n\" %(c,y[i+1])\n",
+ " c=c*2"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " y(0.25) = 0.00520833\n",
+ "\n",
+ "\n",
+ " y(0.50) = 0.0416655\n",
+ "\n",
+ "\n",
+ " y(1.00) = 0.332756\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.5:pg-308"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 8.5\n",
+ "#euler's method\n",
+ "#page 308\n",
+ "def f(y):\n",
+ " return -1*y\n",
+ "y=[0,0,0,0,0]\n",
+ "y[0]=1 #value at 0\n",
+ "h=0.01\n",
+ "c=0.01\n",
+ "for i in range(0,4):\n",
+ " y[i+1]=y[i]+h*f(y[i])\n",
+ " print \"\\ny(%g)=%g\\n\" %(c,y[i+1])\n",
+ " c=c+0.01\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "y(0.01)=0.99\n",
+ "\n",
+ "\n",
+ "y(0.02)=0.9801\n",
+ "\n",
+ "\n",
+ "y(0.03)=0.970299\n",
+ "\n",
+ "\n",
+ "y(0.04)=0.960596\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.6:pg-308"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 8.6\n",
+ "#error estimates in euler's \n",
+ "#page 308\n",
+ "from __future__ import division\n",
+ "def f(y):\n",
+ " return -1*y\n",
+ "y=[0,0,0,0,0]\n",
+ "L=[0,0,0,0,0]\n",
+ "e=[0,0,0,0,0]\n",
+ "y[0]=1 #value at 0\n",
+ "h=0.01\n",
+ "c=0.01;\n",
+ "for i in range(0,4):\n",
+ " y[i+1]=y[i]+h*f(y[i])\n",
+ " print \"\\ny(%g)=%g\\n\" %(c,y[i+1])\n",
+ " c=c+0.01\n",
+ "for i in range(0,4):\n",
+ " L[i]=abs(-(1/2)*(h**2)*y[i+1])\n",
+ " print \"L(%d) =%f\\n\\n\" %(i,L[i])\n",
+ "e[0]=0\n",
+ "for i in range(0,4):\n",
+ " e[i+1]=abs(y[1]*e[i]+L[0])\n",
+ " print \"e(%d)=%f\\n\\n\" %(i,e[i])\n",
+ "Actual_value=math.exp(-0.04)\n",
+ "Estimated_value=y[4]\n",
+ "err=abs(Actual_value-Estimated_value)\n",
+ "if err<e[4]:\n",
+ " print \"VERIFIED\"\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "y(0.01)=0.99\n",
+ "\n",
+ "\n",
+ "y(0.02)=0.9801\n",
+ "\n",
+ "\n",
+ "y(0.03)=0.970299\n",
+ "\n",
+ "\n",
+ "y(0.04)=0.960596\n",
+ "\n",
+ "L(0) =0.000050\n",
+ "\n",
+ "\n",
+ "L(1) =0.000049\n",
+ "\n",
+ "\n",
+ "L(2) =0.000049\n",
+ "\n",
+ "\n",
+ "L(3) =0.000048\n",
+ "\n",
+ "\n",
+ "e(0)=0.000000\n",
+ "\n",
+ "\n",
+ "e(1)=0.000050\n",
+ "\n",
+ "\n",
+ "e(2)=0.000099\n",
+ "\n",
+ "\n",
+ "e(3)=0.000147\n",
+ "\n",
+ "\n",
+ "VERIFIED\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.7:pg-310"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 8.7\n",
+ "#modified euler's method\n",
+ "#page 310\n",
+ "h=0.05\n",
+ "f=1\n",
+ "def f1(x,y):\n",
+ " return x**2+y\n",
+ "x=[0,0.05,0.1]\n",
+ "y1=[0,0,0,0]\n",
+ "y2=[0,0,0,0]\n",
+ "y1[0]=f+h*f1(x[0],f);\n",
+ "y1[1]=f+h*(f1(x[0],f)+f1(x[1],y1[0]))/2\n",
+ "y1[2]=f+h*(f1(x[0],f)+f1(x[2],y1[1]))/2\n",
+ "y2[0]=y1[1]+h*f1(x[1],y1[1])\n",
+ "y2[1]=y1[1]+h*(f1(x[1],y1[1])+f1(x[2],y2[0]))/2\n",
+ "y2[2]=y1[1]+h*(f1(x[1],y1[1])+f1(x[2],y2[1]))/2\n",
+ "print \"y1(0)\\t y1(1)\\t y1(2)\\t y2(0)\\t y2(1)\\t y3(2)\\n\\n\"\n",
+ "print \" %f\\t %f\\t %f\\t %f\\t %f\\t %f\\n\" %(y1[0],y1[1],y1[2],y2[0],y2[1],y2[2])\n",
+ "print \"\\n\\n the value of y at %0.2f is : %0.4f\" %(x[2],y2[2])\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "y1(0)\t y1(1)\t y1(2)\t y2(0)\t y2(1)\t y3(2)\n",
+ "\n",
+ "\n",
+ " 1.050000\t 1.051313\t 1.051533\t 1.104003\t 1.105508\t 1.105546\n",
+ "\n",
+ "\n",
+ "\n",
+ " the value of y at 0.10 is : 1.1055\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.8:pg-313"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 8.8\n",
+ "#runge-kutta formula\n",
+ "#page 313\n",
+ "from __future__ import division\n",
+ "def f(x,y):\n",
+ " return y-x\n",
+ "y=2\n",
+ "x=0\n",
+ "h=0.1\n",
+ "K1=h*f(x,y)\n",
+ "K2=h*f(x+h,y+K1)\n",
+ "y1=y+( K1+K2)/2\n",
+ "print \"\\n y(0.1) by second order runge kutta method:%0.4f\" %(y1)\n",
+ "y=y1\n",
+ "x=0.1\n",
+ "h=0.1\n",
+ "K1=h*f(x,y)\n",
+ "K2=h*f(x+h,y+K1)\n",
+ "y1=y+( K1+K2)/2\n",
+ "print \"\\n y(0.2) by second order runge kutta method:%0.4f\" %(y1)\n",
+ "y=2\n",
+ "x=0\n",
+ "h=0.1\n",
+ "K1=h*f(x,y)\n",
+ "K2=h*f(x+h/2,y+K1/2)\n",
+ "K3=h*f(x+h/2,y+K2/2)\n",
+ "K4=h*f(x+h,y+K3)\n",
+ "y1=y+(K1+2*K2+2*K3+K4)/6\n",
+ "print \"\\n y(0.1) by fourth order runge kutta method:%0.4f\" %(y1)\n",
+ "y=y1\n",
+ "x=0.1\n",
+ "h=0.1\n",
+ "K1=h*f(x,y)\n",
+ "K2=h*f(x+h/2,y+K1/2)\n",
+ "K3=h*f(x+h/2,y+K2/2)\n",
+ "K4=h*f(x+h,y+K3)\n",
+ "y1=y+(K1+2*K2+2*K3+K4)/6\n",
+ "print \"\\n y(0.1) by fourth order runge kutta method:%0.4f \" %(y1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " y(0.1) by second order runge kutta method:2.2050\n",
+ "\n",
+ " y(0.2) by second order runge kutta method:2.4210\n",
+ "\n",
+ " y(0.1) by fourth order runge kutta method:2.2052\n",
+ "\n",
+ " y(0.1) by fourth order runge kutta method:2.4214 \n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.9:pg-315"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 8.9\n",
+ "#runge kutta method\n",
+ "#page 315\n",
+ "from __future__ import division\n",
+ "def f(x,y):\n",
+ " return 1+y**2\n",
+ "y=0\n",
+ "x=0\n",
+ "h=0.2\n",
+ "K1=h*f(x,y)\n",
+ "K2=h*f(x+h/2,y+K1/2)\n",
+ "K3=h*f(x+h/2,y+K2/2)\n",
+ "K4=h*f(x+h,y+K3)\n",
+ "y1=y+(K1+2*K2+2*K3+K4)/6\n",
+ "print \"\\n y(0.2) by fourth order runge kutta method:%0.4f\" %(y1)\n",
+ "y=y1\n",
+ "x=0.2\n",
+ "h=0.2\n",
+ "K1=h*f(x,y)\n",
+ "K2=h*f(x+h/2,y+K1/2)\n",
+ "K3=h*f(x+h/2,y+K2/2)\n",
+ "K4=h*f(x+h,y+K3)\n",
+ "y1=y+(K1+2*K2+2*K3+K4)/6\n",
+ "print \"\\n y(0.4) by fourth order runge kutta method:%0.4f\" %(y1)\n",
+ "y=2\n",
+ "x=0\n",
+ "h=0.1\n",
+ "y=y1\n",
+ "x=0.4\n",
+ "h=0.2\n",
+ "K1=h*f(x,y)\n",
+ "K2=h*f(x+h/2,y+K1/2)\n",
+ "K3=h*f(x+h/2,y+K2/2)\n",
+ "K4=h*f(x+h,y+K3)\n",
+ "y1=y+(K1+2*K2+2*K3+K4)/6\n",
+ "print \"\\n y(0.6) by fourth order runge kutta method:%0.4f\" %(y1)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " y(0.2) by fourth order runge kutta method:0.2027\n",
+ "\n",
+ " y(0.4) by fourth order runge kutta method:0.4228\n",
+ "\n",
+ " y(0.6) by fourth order runge kutta method:0.6841\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.10:pg-315"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 8.10\n",
+ "#initial value problems\n",
+ "#page 315\n",
+ "from __future__ import division\n",
+ "def f1(x,y):\n",
+ " return 3*x+y/2\n",
+ "y=[1,0,0]\n",
+ "h=0.1\n",
+ "c=0\n",
+ "for i in range(0,2):\n",
+ " y[i+1]=y[i]+h*f1(c,y[i])\n",
+ " print \"\\ny(%g)=%g\\n\" %(c,y[i])\n",
+ " c=c+0.1\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "y(0)=1\n",
+ "\n",
+ "\n",
+ "y(0.1)=1.05\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.11:pg-316"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 8.11\n",
+ "#adam's moulton method\n",
+ "#page 316\n",
+ "def f(x,y):\n",
+ " return 1+y**2\n",
+ "y=0\n",
+ "x=0\n",
+ "h=0.2\n",
+ "f1=[0,0,0]\n",
+ "K1=h*f(x,y)\n",
+ "K2=h*f(x+h/2,y+K1/2)\n",
+ "K3=h*f(x+h/2,y+K2/2)\n",
+ "K4=h*f(x+h,y+K3)\n",
+ "y1=y+(K1+2*K2+2*K3+K4)/6\n",
+ "f1[0]=y1\n",
+ "print \"\\n y(0.2) by fourth order runge kutta method:%0.4f\" %(y1)\n",
+ "y=y1\n",
+ "x=0.2\n",
+ "h=0.2\n",
+ "K1=h*f(x,y)\n",
+ "K2=h*f(x+h/2,y+K1/2)\n",
+ "K3=h*f(x+h/2,y+K2/2)\n",
+ "K4=h*f(x+h,y+K3)\n",
+ "y1=y+(K1+2*K2+2*K3+K4)/6\n",
+ "f1[1]=y1\n",
+ "print \"\\n y(0.4) by fourth order runge kutta method:%0.4f\" %(y1)\n",
+ "y=2\n",
+ "x=0\n",
+ "h=0.1\n",
+ "y=y1\n",
+ "x=0.4\n",
+ "h=0.2\n",
+ "K1=h*f(x,y)\n",
+ "K2=h*f(x+h/2,y+K1/2)\n",
+ "K3=h*f(x+h/2,y+K2/2)\n",
+ "K4=h*f(x+h,y+K3)\n",
+ "y1=y+(K1+2*K2+2*K3+K4)/6\n",
+ "f1[2]=y1\n",
+ "print \"\\n y(0.6) by fourth order runge kutta method:%0.4f\" %(y1)\n",
+ "y_p=y1+h*(55*(1+f1[2]**2)-59*(1+f1[1]**2)+37*(1+f1[0]**2)-9)/24\n",
+ "y_c=y1+h*(9*(1+(y_p-1)**2)+19*(1+f1[2]**2)-5*(1+f1[1]**2)+(1+f1[0]**2))/24\n",
+ "print \"\\nthe predicted value is:%0.4f:\\n\" %(y_p)\n",
+ "print \" the computed value is:%0.4f:\" %(y_c)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " y(0.2) by fourth order runge kutta method:0.2027\n",
+ "\n",
+ " y(0.4) by fourth order runge kutta method:0.4228\n",
+ "\n",
+ " y(0.6) by fourth order runge kutta method:0.6841\n",
+ "\n",
+ "the predicted value is:1.0234:\n",
+ "\n",
+ " the computed value is:0.9512:\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.12:pg-320"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 8.12\n",
+ "#milne's method\n",
+ "#page 320\n",
+ "def f(x,y):\n",
+ " return 1+y**2\n",
+ "y=0\n",
+ "f1=[0,0,0]\n",
+ "Y1=[0,0,0,0]\n",
+ "x=0\n",
+ "h=0.2\n",
+ "f1[0]=0\n",
+ "print \"x y y1=1+y^2\\n\\n\"\n",
+ "Y1[0]=1+y**2\n",
+ "print \"%0.4f %0.4f %0.4f\\n\" %(x,y,(1+y**2))\n",
+ "K1=h*f(x,y)\n",
+ "K2=h*f(x+h/2,y+K1/2)\n",
+ "K3=h*f(x+h/2,y+K2/2)\n",
+ "K4=h*f(x+h,y+K3)\n",
+ "y1=y+(K1+2*K2+2*K3+K4)/6\n",
+ "f1[0]=y1\n",
+ "Y1[1]=1+y1**2\n",
+ "print \"%0.4f %0.4f %0.4f\\n\" %(x+h,y1,(1+y1**2))\n",
+ "y=y1\n",
+ "x=0.2\n",
+ "h=0.2\n",
+ "K1=h*f(x,y)\n",
+ "K2=h*f(x+h/2,y+K1/2)\n",
+ "K3=h*f(x+h/2,y+K2/2)\n",
+ "K4=h*f(x+h,y+K3)\n",
+ "y1=y+(K1+2*K2+2*K3+K4)/6\n",
+ "f1[1]=y1\n",
+ "Y1[2]=1+y1**2\n",
+ "print \"%0.4f %0.4f %0.4f\\n\" %(x+h,y1,(1+y1**2))\n",
+ "y=y1\n",
+ "x=0.4\n",
+ "h=0.2\n",
+ "K1=h*f(x,y)\n",
+ "K2=h*f(x+h/2,y+K1/2)\n",
+ "K3=h*f(x+h/2,y+K2/2)\n",
+ "K4=h*f(x+h,y+K3)\n",
+ "y1=y+(K1+2*K2+2*K3+K4)/6\n",
+ "f1[2]=y1\n",
+ "Y1[3]=1+y1**2;\n",
+ "print \"%0.4f %0.4f %0.4f\\n\" %(x+h,y1,(1+y1**2))\n",
+ "Y_4=4*h*(2*Y1[1]-Y1[2]+2*Y1[3])/3\n",
+ "print \"y(0.8)=%f\\n\" %(Y_4)\n",
+ "Y=1+Y_4**2\n",
+ "Y_4=f1[1]+h*(Y1[2]+4*Y1[3]+Y)/3 #more correct value\n",
+ "print \"y(0.8)=%f\\n\" %(Y_4)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "x y y1=1+y^2\n",
+ "\n",
+ "\n",
+ "0.0000 0.0000 1.0000\n",
+ "\n",
+ "0.2000 0.2027 1.0411\n",
+ "\n",
+ "0.4000 0.4228 1.1788\n",
+ "\n",
+ "0.6000 0.6841 1.4680\n",
+ "\n",
+ "y(0.8)=1.023869\n",
+ "\n",
+ "y(0.8)=1.029403\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.13:pg-320"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 8.13\n",
+ "#milne's method\n",
+ "#page 320\n",
+ "def f1(x,y):\n",
+ " return x**2+y**2-2\n",
+ "x=[-0.1, 0, 0.1, 0.2]\n",
+ "y=[1.0900, 1.0, 0.8900, 0.7605]\n",
+ "Y1=[0,0,0,0]\n",
+ "h=0.1\n",
+ "for i in range(0,4):\n",
+ " Y1[i]=f1(x[i],y[i])\n",
+ "print \" x y y1=x^2+y^2-2 \\n\\n\"\n",
+ "for i in range(0,4):\n",
+ " print \" %0.2f %f %f \\n\" %(x[i],y[i],Y1[i])\n",
+ "Y_3=y[0]+(4*h/3)*(2*Y1[1]-Y1[2]+2*Y1[3])\n",
+ "print \"y(0.3)=%f\\n\" %(Y_3)\n",
+ "Y1_3=f1(0.3,Y_3)\n",
+ "Y_3=y[2]+h*(Y1[2]+4*Y1[3]+Y1_3)/3 #corrected value\n",
+ "print \"corrected y(0.3)=%f\" %(Y_3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " x y y1=x^2+y^2-2 \n",
+ "\n",
+ "\n",
+ " -0.10 1.090000 -0.801900 \n",
+ "\n",
+ " 0.00 1.000000 -1.000000 \n",
+ "\n",
+ " 0.10 0.890000 -1.197900 \n",
+ "\n",
+ " 0.20 0.760500 -1.381640 \n",
+ "\n",
+ "y(0.3)=0.614616\n",
+ "\n",
+ "corrected y(0.3)=0.614776\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.14:pg322"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 8.14\n",
+ "#initial-value problem\n",
+ "#page 322\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "def f(x):\n",
+ " return 13*math.exp(x/2)-6*x-12\n",
+ "s1=1.691358\n",
+ "s3=3.430879\n",
+ "print \"the erorr in the computed values are %0.7g %0.7g\" %(abs(f(0.5)-s1),abs(f(1)-s3))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the erorr in the computed values are 0.0009724169 0.002497519\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.15:pg-328"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#boundary value problem using finite difference method\n",
+ "#example 8.15\n",
+ "#page 328\n",
+ "import math\n",
+ "from numpy import matrix\n",
+ "def f(x):\n",
+ " return math.cos(x)+((1-math.cos(1))/math.sin(1))*math.sin(x)-1\n",
+ "h1=1/2\n",
+ "Y=f(0.5)\n",
+ "y0=0\n",
+ "y2=0\n",
+ "y1=4*(1/4+y0+y2)/7\n",
+ "print \"computed value with h=%f of y(0.5) is %f\\n\" %(h1,y1)\n",
+ "print \"error in the result with actual value %f\\n\" %(abs(Y-y1))\n",
+ "h2=1/4\n",
+ "y0=0\n",
+ "y4=0\n",
+ "#solving the approximated diffrential equation\n",
+ "A=matrix([[-31/16, 1, 0],[1, -31/16, 1],[0, 1, -31/16]])\n",
+ "X=matrix([[-1/16],[-1/16],[-1/16]])\n",
+ "C=A.I*X\n",
+ "print \"computed value with h=%f of y(0.5) is %f\\n\" %(h2,C[1][0])\n",
+ "print \"error in the result with actual value %f\\n\" %(abs(Y-C[1][0]))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "computed value with h=0.500000 of y(0.5) is 0.142857\n",
+ "\n",
+ "error in the result with actual value 0.003363\n",
+ "\n",
+ "computed value with h=0.250000 of y(0.5) is 0.140312\n",
+ "\n",
+ "error in the result with actual value 0.000818\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.16:pg-329"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#boundary value problem using finite difference method\n",
+ "#example 8.16\n",
+ "#page 329\\\n",
+ "from numpy import matrix\n",
+ "import math\n",
+ "def f(x):\n",
+ " return math.sinh(x)\n",
+ "y0=0 #y(0)=0\n",
+ "y4=3.62686 #y(2)=3.62686\n",
+ "h1=0.5\n",
+ "Y=f(0.5)\n",
+ "#arranging and calculating the values\n",
+ "A=matrix([[-9, 4, 0],[4, -9, 4],[0, 4, -9]])\n",
+ "C=matrix([[0],[0],[-14.50744]])\n",
+ "X=A.I*C\n",
+ "print \"computed value with h=%f of y(0.5) is %f\\n\" %(h1,X[0][0])\n",
+ "print \"error in the result with actual value %f\\n\" %(abs(Y-X[0][0]))\n",
+ "h2=1.0\n",
+ "y0=0 #y(0)=0\n",
+ "y2=3.62686 #y(2)=3.62686\n",
+ "y1=(y0+y2)/3\n",
+ "Y=(4*X[1][0]-y1)/3\n",
+ "print \"with better approximation error is reduced to %f\" %(abs(Y-f(1.0)))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "computed value with h=0.500000 of y(0.5) is 0.526347\n",
+ "\n",
+ "error in the result with actual value 0.005252\n",
+ "\n",
+ "with better approximation error is reduced to 0.000855\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.17:pg-330"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#cubic spline method\n",
+ "#example 8.17\n",
+ "#page 330\n",
+ "def f(x):\n",
+ " return math.cos(x)+((1-math.cos(1))/math.sin(1))*math.sin(x)-1\n",
+ "y=[f(0), f(0.5), f(1)]\n",
+ "h=1/2\n",
+ "Y=f(0.5)\n",
+ "y0=0\n",
+ "y2=0\n",
+ "M0=-1\n",
+ "M1=-22/25\n",
+ "M2=-1\n",
+ "s1_0=47/88\n",
+ "s1_1=-47/88\n",
+ "s1_05=0\n",
+ "print \"the cubic spline value is %f\" %(Y)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the cubic spline value is 0.139494\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.18:pg-331"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#cubic spline method\n",
+ "#example 8.18\n",
+ "#page 331\n",
+ "from numpy import matrix\n",
+ "from __future__ import division\n",
+ "#after arranging and forming equation \n",
+ "A=matrix([[10, -1, 0, 24],[0, 16, -1, -32],[1, 20, 0, 16],[0, 1, 26, -24]])\n",
+ "C=matrix([[36],[-12],[24],[-9]])\n",
+ "X=A.I*C;\n",
+ "print \" Y1=%f\\n\\n\" %(X[3][0])\n",
+ "print \" the error in the solution is :%f\" %(abs((2/3)-X[3][0]))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Y1=0.653890\n",
+ "\n",
+ "\n",
+ " the error in the solution is :0.012777\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.19:pg-331"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#boundary value problem by cubisc spline nethod\n",
+ "#example 8.18\n",
+ "#page 331\n",
+ "from numpy import matrix\n",
+ "h=1/2\n",
+ "#arranging in two subintervals we get\n",
+ "A=matrix([[10, -1, 0,24],[0, 16, -1, -32],[1, 20, 0, 16],[0, 1, 26, -24]])\n",
+ "C=matrix([[36],[-12],[24],[-9]])\n",
+ "X=A.I*C\n",
+ "print \"the computed value of y(1.5) is %f \"%(X[3][0])\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the computed value of y(1.5) is 0.653890 \n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter_5_5.ipynb b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter_5_5.ipynb new file mode 100644 index 00000000..f7ae1b31 --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter_5_5.ipynb @@ -0,0 +1,358 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:d6c77d0644ce8c4a3cef684a9d4884ca9577630ae40ac3764bbf1d6682f47c3d"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter05:Spline Functions"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.1:pg-182"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#linear splines\n",
+ "#example 5.1\n",
+ "#page 182\n",
+ "from numpy import matrix\n",
+ "X=matrix([[1],[2], [3]])\n",
+ "y=matrix([[-8],[-1],[18]])\n",
+ "m1=(y[1][0]-y[0][0])/(X[1][0]-X[0][0])\n",
+ "m2=(y[2][0]-y[1][0])/(X[2][0]-X[1][0])\n",
+ "def s1(x):\n",
+ " return y[0][0]+(x-X[0][0])*m1\n",
+ "def s2(x):\n",
+ " return y[1][0]+(x-X[1][0])*m2\n",
+ "print \"the value of function at 2.5 is %0.2f: \" %(s2(2.5))\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value of function at 2.5 is 8.50: \n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.3:pg-188"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#cubic splines\n",
+ "#example 5.3\n",
+ "#page 188\n",
+ "from numpy import matrix\n",
+ "import math\n",
+ "X=matrix([[1],[2],[3]])\n",
+ "y=matrix([[-8],[-1],[18]])\n",
+ "M1=0\n",
+ "M2=8\n",
+ "M3=0\n",
+ "h=1\n",
+ "#deff('y=s1(x)','y=3*(x-1)^3-8*(2-x)-4*(x-1)')\n",
+ "def s1(x):\n",
+ " return 3*math.pow(x-1,3)-8*(2-x)-4*(x-1)\n",
+ "#deff('y=s2(x)','y=3*(3-x)^3+22*x-48');\n",
+ "def s2(x):\n",
+ " return 3*math.pow(3-x,3)+22*x-48\n",
+ "h=0.0001\n",
+ "n=2.0\n",
+ "D=(s2(n+h)-s2(n))/h;\n",
+ "print \" y(2.5)=%f\" %(s2(2.5))\n",
+ "print \"y1(2.0)=%f\" %(D)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " y(2.5)=7.375000\n",
+ "y1(2.0)=13.000900\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.4:pg-189"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#cubic spline\n",
+ "#example 5.4\n",
+ "#page 189\n",
+ "from numpy import matrix\n",
+ "import math\n",
+ "x=matrix([[0],[math.pi/2],[math.pi]])\n",
+ "y=matrix([[0],[1],[0]])\n",
+ "h=x[1][0]-x[0][0]\n",
+ "M0=0\n",
+ "M2=0\n",
+ "M1=((6*(y[0][0]-2*y[1][0]+y[2][0])/math.pow(h,2))-M0-M2)/4\n",
+ "X=math.pi/6.0\n",
+ "s1=(math.pow(x[1][0]-X,3)*(M0/6)+math.pow(X-x[0][0],3)*M1/6+(y[0][0]-math.pow(h,2)*M0/6)*(x[1][0]-X)+(y[1][0]-math.pow(h,2)*M1/6)*(X-x[0][0]))/h;\n",
+ "x=matrix([[0],[math.pi/4], [math.pi/2], [3*math.pi/4], [math.pi]])\n",
+ "y=matrix([[0], [1.414], [1] ,[1.414]])\n",
+ "M0=0\n",
+ "M4=0\n",
+ "A=matrix([[4, 1, 0],[1, 4, 1],[0 ,1 ,4]]) #calculating value of M1 M2 M3 by matrix method\n",
+ "C=matrix([[-4.029],[-5.699],[-4.029]])\n",
+ "B=A.I*C\n",
+ "print \"M0=%f\\t M1=%f\\t M2=%f\\t M3=%f\\t M4=%f\\t\\n\\n\" %(M0,B[0][0],B[1][0],B[2][0],M4)\n",
+ "h=math.pi/4;\n",
+ "X=math.pi/6;\n",
+ "s1=(-0.12408*math.pow(X,3)+0.7836*X)/h;\n",
+ "print \"the value of sin(pi/6) is:%f\" %(s1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "M0=0.000000\t M1=-0.744071\t M2=-1.052714\t M3=-0.744071\t M4=0.000000\t\n",
+ "\n",
+ "\n",
+ "the value of sin(pi/6) is:0.499722\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.5:pg-191"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#cubic spline\n",
+ "#example 5.5\n",
+ "#page 191\n",
+ "import math\n",
+ "from numpy import matrix\n",
+ "x=[1,2,3]\n",
+ "y=[6,18,42]\n",
+ "m0=40\n",
+ "s1=0\n",
+ "m1=(3*(y[2]-y[0])-m0)/4\n",
+ "X=0\n",
+ "s1=m0*((x[1]-X)**2)*(X-x[0])-m1*((X-x[0])**2)*(x[1]-X)+y[0]*((x[1]-X)**2)*(2*(X-x[0])+1)+y[1]*((X-x[0])**2)*(2*(x[1]-X)+1)\n",
+ "print \"s1= %f+261*x-160X^2+33X^3\" %(s1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "s1= -128.000000+261*x-160X^2+33X^3\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.7:pg-195"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#surface fitting by cubic spline\n",
+ "#example 5.7\n",
+ "#page 195\n",
+ "from numpy import matrix\n",
+ "def L0(y):\n",
+ " return math.pow(y,3)/4-5*y/4+1\n",
+ "def L1(y):\n",
+ " return (math.pow(y,3)/2)*-1+3*y/2\n",
+ "def L2(y):\n",
+ " return math.pow(y,3)/4-y/4\n",
+ "A=[ [1,2,9], [2,3,10], [9,10,17] ]\n",
+ "x=0.5\n",
+ "y=0.5\n",
+ "S=0.0\n",
+ "S=S+L0(x)*(L0(x)*A[0][0]+L1(x)*A[0][1]+L2(x)*A[0][2])\n",
+ "S=S+L1(x)*(L0(x)*A[1][0]+L1(x)*A[1][1]+L2(x)*A[1][2])\n",
+ "S=S+L2(x)*(L0(x)*A[2][0]+L1(x)*A[2][1]+L2(x)*A[2][2])\n",
+ "print \"approximated value of z(0.5 0.5)=%f\\n\\n\" %(S)\n",
+ "print \" error in the approximated value : %f\" %((abs(1.25-S)/1.25)*100)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "approximated value of z(0.5 0.5)=0.875000\n",
+ "\n",
+ "\n",
+ " error in the approximated value : 30.000000\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.8:pg-200"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#cubic B-splines\n",
+ "#example 5.8\n",
+ "#page 200\n",
+ "import math\n",
+ "k=[0.0, 1, 2, 3, 4]\n",
+ "pi=[0.0, 0, 4, -6, 24]\n",
+ "x=1\n",
+ "S=0\n",
+ "for i in range(2,5):\n",
+ " S=S+math.pow(k[i]-x,3)/(pi[i])\n",
+ "print \"the cubic splines for x=1 is %f\\n\\n\" %(S)\n",
+ "S=0\n",
+ "x=2\n",
+ "for i in range(2,5):\n",
+ " S=S+math.pow(k[i]-x,3)/(pi[i])\n",
+ "print \"the cubic splines for x=2 is %f\\n\\n\" %(S)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the cubic splines for x=1 is 0.041667\n",
+ "\n",
+ "\n",
+ "the cubic splines for x=2 is 0.166667\n",
+ "\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.9:pg-201"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#cubic B-spline\n",
+ "#example 5.9\n",
+ "#page 201\n",
+ "k=[0, 1, 2, 3, 4];\n",
+ "x=1 #for x=1\n",
+ "s11=0\n",
+ "s13=0\n",
+ "s14=0\n",
+ "s24=0 \n",
+ "s12=1/(k[2]-k[1])\n",
+ "s22=((x-k[0])*s11+(k[2]-x)*s12)/2.0 #k[2]-k[0]=2\n",
+ "s23=((x-k[1])*s11+(k[3]-x)*s13)/(k[3]-k[1])\n",
+ "s33=((x-k[0])*s22+(k[3]-x)*s23)/(k[3]-k[0])\n",
+ "s34=((x-k[1])*s23+(k[4]-x)*s24)/(k[4]-k[1])\n",
+ "s44=((x-k[0])*s33+(k[4]-x)*s34)/(k[4]-k[0])\n",
+ "print \"s11=%f\\t s22=%f\\t s23=%f\\t s33=%f\\t s34=%f\\t s44=%f\\n\\n\" %(s11,s22,s23,s33,s34,s44)\n",
+ "x=2 #for x=2\n",
+ "s11=0\n",
+ "s12=0\n",
+ "s14=0\n",
+ "s22=0\n",
+ "s13=1/(k[3]-k[2])\n",
+ "s23=((x-k[1])*s12+(k[3]-x)*s13)/2.0 # k[3]-k[1]=2\n",
+ "s24=((x-k[2])*s13+(k[4]-x)*s14)/(k[2]-k[0])\n",
+ "s33=((x-k[0])*s22+(k[3]-x)*s23)/(k[3]-k[0])\n",
+ "s34=((x-k[1])*s23+(k[4]-x)*s24)/(k[4]-k[1])\n",
+ "s44=((x-k[0])*s33+(k[4]-x)*s34)/(k[4]-k[0])\n",
+ "print \"s13=%f\\t s23=%f\\t s24=%f\\t s33=%f\\t s34=%f\\t s44=%f\\n\\n\" %(s13,s23,s24,s33,s34,s44)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "s11=0.000000\t s22=0.500000\t s23=0.000000\t s33=0.166667\t s34=0.000000\t s44=0.041667\n",
+ "\n",
+ "\n",
+ "s13=1.000000\t s23=0.500000\t s24=0.000000\t s33=0.166667\t s34=0.166667\t s44=0.166667\n",
+ "\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/1.2.png b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/1.2.png Binary files differnew file mode 100644 index 00000000..2af775d8 --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/1.2.png diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/3.7.png b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/3.7.png Binary files differnew file mode 100644 index 00000000..30f2ed11 --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/3.7.png diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/6.7.png b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/6.7.png Binary files differnew file mode 100644 index 00000000..524c2a6a --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/6.7.png diff --git a/Physics_Textbook_Part-I_for_class_XI_by_NCERT_by_Chief_Editor_-_Naresh_Yadav/Chapter2_2.ipynb b/Physics_Textbook_Part-I_for_class_XI_by_NCERT_by_Chief_Editor_-_Naresh_Yadav/Chapter2_2.ipynb new file mode 100644 index 00000000..64b27332 --- /dev/null +++ b/Physics_Textbook_Part-I_for_class_XI_by_NCERT_by_Chief_Editor_-_Naresh_Yadav/Chapter2_2.ipynb @@ -0,0 +1,755 @@ +{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 2 : Units and Measurement"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.1 , page : 19"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "(a) One degree = 0.017453292519943295 rad\n",
+ "(b) One minute = 0.0002908882086657216 rad\n",
+ "(c) One second = 4.84813681109536e-06 rad\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable delaration\n",
+ "\n",
+ "three_sixty_degree=2*math.pi\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "# Since 360° = 2π rad \n",
+ "one_degree=three_sixty_degree/360\n",
+ "# Since 1° = 60′ \n",
+ "one_minute=one_degree/60\n",
+ "# Since 1′ = 60″\n",
+ "one_second=one_minute/60\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"(a) One degree =\",one_degree,\"rad\")\n",
+ "print(\"(b) One minute =\",one_minute,\"rad\")\n",
+ "print(\"(c) One second =\",one_second,\"rad\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.2 , page : 19"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The distance of the tower C from his original position A = 119.0 m\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable delaration\n",
+ " \n",
+ "θ=40 # Parallax angle in degree\n",
+ "AB=100 # Distance between A and Bin m\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "# AB = AC tan θ \n",
+ "AC=AB/math.tan(math.radians(θ))\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"The distance of the tower C from his original position A =\",round(AC,0),\"m\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.3 , page : 19"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The distance of the moon from the Earth = 0.0003846385723759571 m\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable delaration\n",
+ "\n",
+ "# Since, 1°54′ = 114′ \n",
+ "θ=114 # The angle θ subtended at the moon by the two directions of observation \n",
+ "b=1.276*10**7 # Diameter of the Earth\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "θ=114*60*4.85*10**6\n",
+ "D=b/θ # The earth-moon distance\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"The distance of the moon from the Earth =\",D,\"m\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.4 , page : 19"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Sun's diameter = 1393075199.9999998 m\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "α=1920 # Sun's angular diameter in minutes\n",
+ "D=1.496*10**11 # The distance of the Sun from the Earth \n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "α=1920*4.85*10**-6 # Sun's angular diameter in radians\n",
+ "d=α*D # Sun's diameter\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print(\"Sun's diameter =\",d,\"m\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.5 , page : 20"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Thus a nucleus in an atom is as small in size as the tip of a sharp pin placed at the centre of a sphere of radius about a metre long\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Calculation\n",
+ "\n",
+ "# The size of a nucleus is in the range of 10**-15 m and 10**-14 m.\n",
+ "# The tip of a sharp pin is taken to be in the range of 10**5 m and 10**4 m.\n",
+ "# Thus we are scaling up by a factor of 1010. An atom roughly of size 1010 m will be scaled up to a size of 1 m.\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"Thus a nucleus in an atom is as small in size as the tip of a sharp pin placed at the centre of a sphere of radius about a metre long\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.6 , page : 25"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Clock 2 is to be preferred to clock 1\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Calculations\n",
+ "\n",
+ "# The range of variation over the seven days of observations is 162 s for clock 1, and 31 s for clock 2.\n",
+ "# The average reading of clock 1 is much closer to the standard time than the average reading of clock 2.\n",
+ "# The important point is that a clocks zero error is not as significant for precision work as its variation, because a zero-error can always be easily corrected.\n",
+ " \n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"Clock 2 is to be preferred to clock 1\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.7 , page : 25"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The absolute errors = [0.005999999999999783, 0.06400000000000006, 0.20400000000000018, 0.08599999999999985, 0.1759999999999997]\n",
+ "A more correct way will be to write, T = 2.6 ± 0.1 s \n",
+ "The relative error or the percentage error = 3.8461538461538463 ≈ 4 %\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable delaration\n",
+ "\n",
+ "p=[2.63,2.56,2.42,2.71,2.80] # The readings of period of oscillation of a simple pendulum\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "T=sum(p)/len(p)\n",
+ "p[:] = [x - T for x in p]\n",
+ "q=[abs(x) for x in p]\n",
+ "DT=sum(q)/len(p)\n",
+ "δa=(round(DT,1)/round(T,1))*100\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"The absolute errors =\",q)\n",
+ "print(\"A more correct way will be to write, T =\",round(T,1),\"±\", round(DT,1),\" s \")\n",
+ "\n",
+ "print(\"The relative error or the percentage error =\",δa,\"≈ 4 %\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.8 , page : 26"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The temperature difference wth the error = 30 ° C ± 1.0 ° C\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable delaration\n",
+ "\n",
+ "t1=20 # Temperature of first body in degree Celsius\n",
+ "Δt1=.5 # Error in temperature of first body degree Celsius\n",
+ "t2=50 # Temperature of second body degree Celsius\n",
+ "Δt2=.5 # Error in temperature of first body degree Celsius\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "t=t2 - t1\n",
+ "Δt=max((Δt1 + Δt2),(Δt1 - Δt2))\n",
+ "\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"The temperature difference wth the error =\",t,\"\\u00b0 C ±\",Δt,\"\\u00b0 C\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.9 , page : 27"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 9,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The total error in Resistance = 7 %\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable delaration\n",
+ " \n",
+ "V=5 # The percentage error in voltage \n",
+ "I=2 # The percentage error in current\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "R=V+I\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"The total error in Resistance =\",R,\"%\")"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.10 , page : 27"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 10,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The equivalent resistance of the series combination = 300 ± 7 ohm\n",
+ "The equivalent resistance of the parallel combination = 66.7 ± 1.8 ohm\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable delaration\n",
+ "\n",
+ "R1=100 # Resistance of first resistor in ohm\n",
+ "ΔR1=3 # Error in Resistance of first resistor in ohm\n",
+ "R2=200 # Resistance of second resistor wth error term in ohm\n",
+ "ΔR2=4 # Error in Resistance of second resistor in ohm\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "R=R1+R2\n",
+ "ΔR=ΔR1+ΔR2\n",
+ "R_prim=R1*R2/(R1+R2)\n",
+ "ΔR_prim=(R_prim/R1)**2*ΔR1 + (R_prim/R2)**2*ΔR2 \n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"The equivalent resistance of the series combination =\",R,\"±\",ΔR,\"ohm\")\n",
+ "print(\"The equivalent resistance of the parallel combination =\",round(R_prim,1),\"±\",round(ΔR_prim,1),\"ohm\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.11 , page : 27"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 11,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The relative error in Z is ΔZ/Z = 4(ΔA/A)+(1/3)(ΔB/B)+(ΔC/C)+(3/2)(ΔD/D)\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Since the relative error in a physical quantity raised to the power k is the k times the relative error in the individual quantity.\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"The relative error in Z is ΔZ/Z = 4(ΔA/A)+(1/3)(ΔB/B)+(ΔC/C)+(3/2)(ΔD/D)\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.12 , page : 27"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 12,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The percentage error in g = 2.7222222222222223 ≈ 3\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable delaration\n",
+ "\n",
+ "L=20 # Length in cm\n",
+ "ΔL=1 # Eror in lengthin mm\n",
+ "t=90 # Total time in s\n",
+ "Δt=1 # Error in time in s\n",
+ "n=100 # Number of oscillations\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "# The period of oscillation of a simple pendulum is, T = 2Π √(L/g)\n",
+ "# Hence, g = 4π²L/T²\n",
+ "\n",
+ "ΔL=ΔL*10**-1\n",
+ "T=n/t\n",
+ "ΔT_div_T=Δt/t # Error in T\n",
+ "Δg_div_g= (ΔL/L) + 2*(ΔT_div_T) # Error in g\n",
+ "per_g= 100*(ΔL/L) + 2*100*(ΔT_div_T ) \n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"The percentage error in g =\",per_g,\"≈ 3\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.13 , page : 30"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 13,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Surface area of the cube = 311.3 m²\n",
+ "Volume of the cube = 373.7 m^3\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable delaration\n",
+ "\n",
+ "a=7.203 # Side of the cube in m\n",
+ "\n",
+ "# Calculation\n",
+ "# The number of significant figures in the measured length is 4.\n",
+ "# Hence the calculated area and the volume should therefore be rounded off to 4 significant figures. \n",
+ "\n",
+ "SA=6*a**2 # Surface area of the cube\n",
+ "V=a**3 # Volume of the cube \n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"Surface area of the cube =\",round(SA,1),\"m²\")\n",
+ "print(\"Volume of the cube =\",round(V,1),\"m^3\")"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.14 , page : 30"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 14,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Density = 4.8 g/cm^3\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable delaration\n",
+ "\n",
+ "m=5.74 # Mass of the substancein g\n",
+ "v=1.2 # Volume of the substance in cm^3\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "# There are 3 significant figures in the measured mass whereas there are only 2 significant figures in the measured volume.\n",
+ "# Hence the density should be expressed to only 2 significant figures.\n",
+ "D=m/v\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"Density =\",round(D,1),\"g/cm^3\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.15 , page : 33"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 15,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The dimensions of LHS are :[M][L T-1 ]² = [M][L² T-²] = [M L² T-²]\n",
+ "The dimensions of RHS are :[M][L T-²][L] = [M][L² T-²] = [M L² T-²]\n",
+ "The dimensions of LHS and RHS are the same and hence the equation is dimensionally correct\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Consider the equation (1/2)mv²=mgh\n",
+ "# where\n",
+ "# m : Mass of the body\n",
+ "# v : Velocity of the body\n",
+ "# g : Acceleration due to gravity \n",
+ "# h : Height\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "print(\"The dimensions of LHS are :[M][L T-1 ]² = [M][L² T-²] = [M L² T-²]\")\n",
+ "print(\"The dimensions of RHS are :[M][L T-²][L] = [M][L² T-²] = [M L² T-²]\")\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"The dimensions of LHS and RHS are the same and hence the equation is dimensionally correct\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.16 , page : 33"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The correct formula for kinetic energy is, K = (1/2)mv²\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Calculation\n",
+ "\n",
+ "# Every correct formula or equation must have the same dimensions on both sides of the equation.\n",
+ "# Also, only quantities with the same physical dimensions can be added or subtracted.\n",
+ "# The dimensions of the quantity on the right side are \n",
+ "# [M² L^3 T^-3] for (a) \n",
+ "# [M L² T-²] for (b)and (d)\n",
+ "# [M L T-²] for (c)\n",
+ "# The quantity on the right side of (e) has no proper dimensions since two quantities of different dimensions have been added.\n",
+ "# Since the kinetic energy K has the dimensions of [M L2 T2], formulas (a), (c) and (e) are ruled out.\n",
+ "# Note that dimensional arguments cannot tell which of the two, (b) or (d), is the correct formula. \n",
+ "# For this, one must turn to the actual definition of kinetic energy.\n",
+ "# The correct formula for kinetic energy is given by (b). \n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"The correct formula for kinetic energy is, K = (1/2)mv²\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2.17 , page : 33"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The expression for the time period of a simple pendulum is, T = 2π √(l/g)\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Calculation\n",
+ "\n",
+ "# The dependence of time period T on the quantities l, g and m as a product may be written as :\n",
+ "# T = k l^x g^y m^z \n",
+ "# where k is dimensionless constant and x, y and z are the exponents. \n",
+ "# By considering dimensions on both sides, we have [L^0 M^0 T^1] = [L^1]^x [L^1 T^-2]^y [M^1]^z = L^(x+y) T^(-2y) M^z\n",
+ "# On equating the dimensions on both sides, we have: x + y = 0; 2y = 1; and z = 0\n",
+ "# So that x = (1/2); y = -(1/2) and z = 0\n",
+ "# T = k * l^(1/2) * g^-(1/2)\n",
+ "# In other word, T = k √(l/g)\n",
+ "# The value of constant k can not be obtained by the method of dimensions.\n",
+ "# Here it does not matter if some number multiplies the right side of this formula, because that does not affect its dimensions.\n",
+ "# Actually, k = 2π so that T = 2π √(l/g)\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"The expression for the time period of a simple pendulum is, T = 2π √(l/g)\")\n"
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 3",
+ "language": "python",
+ "name": "python3"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 3
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython3",
+ "version": "3.4.3"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Physics_Textbook_Part-I_for_class_XI_by_NCERT_by_Chief_Editor_-_Naresh_Yadav/Chapter3_2.ipynb b/Physics_Textbook_Part-I_for_class_XI_by_NCERT_by_Chief_Editor_-_Naresh_Yadav/Chapter3_2.ipynb new file mode 100644 index 00000000..ad28279c --- /dev/null +++ b/Physics_Textbook_Part-I_for_class_XI_by_NCERT_by_Chief_Editor_-_Naresh_Yadav/Chapter3_2.ipynb @@ -0,0 +1,465 @@ +{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 3 : Motion in a Straight Line"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 3.1 , page : 43"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "(a)\n",
+ "The average velocity of the car in going from O to P = 20.0 m/s\n",
+ "The average speed of the car in going from O to P = 20.0 m/s\n",
+ "(b)\n",
+ "The average velocity of the car in going from O to P and back to Q = 10.0 m/s\n",
+ "The average speed of the car in going from O to P and back to Q = 20.0 m/s\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "l1=360 # The distance from O to P in m\n",
+ "l2=120 # The distance from P to Q in m\n",
+ "t1=18 # The time taken to travel OP in s\n",
+ "t2=6 # The time taken to travel PQ in s\n",
+ "d1=360 # The displacement from O to P in m\n",
+ "d2=(l1-l2) # The displacement from P to Q in m\n",
+ "p1=360 # The total pathlength from O to P in m\n",
+ "p2=(l1+l2) # The total pathlength from O to P and p to Q in m\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "#(a)\n",
+ "a_v1=d1/t1\n",
+ "a_s1=p1/t1\n",
+ "\n",
+ "#(b)\n",
+ "a_v2=d2/(t1+t2)\n",
+ "a_s2=p2/(t1+t2)\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"(a)\")\n",
+ "print(\"The average velocity of the car in going from O to P =\",a_v1,\"m/s\")\n",
+ "print(\"The average speed of the car in going from O to P =\",a_s1,\"m/s\")\n",
+ "print(\"(b)\")\n",
+ "print(\"The average velocity of the car in going from O to P and back to Q =\",a_v2,\"m/s\")\n",
+ "print(\"The average speed of the car in going from O to P and back to Q =\",a_s2,\"m/s\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 3.2 , page : 45"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The velocity at t = 0.0 s = 0 m/s\n",
+ "The velocity at t = 2.0 s = 10 m/s\n",
+ "The average velocity between t = 2.0 s and t = 4.0 s = 15.0 m/s\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "# Variable declaration\n",
+ " \n",
+ "a=8.5 # Distance in m\n",
+ "b=2.5 # Acceleration in m/s²\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "# In notation of differential calculus, the velocity is v = dx/dt = d (a+bt²)/dt = 2bt = 5.0t\n",
+ "\n",
+ "p0=np.polyval([0,5,0],0) # Velocity at t= 1.0 s\n",
+ "p2=np.polyval([0,5,0],2) # Velocity at t= 2.0 s\n",
+ "p4=np.polyval([0,5,0],4) # Velocity at t= 4.0 s\n",
+ "avg_v=(p2+p4)/2\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"The velocity at t = 0.0 s =\",p0,\"m/s\")\n",
+ "print(\"The velocity at t = 2.0 s =\",p2,\"m/s\")\n",
+ "print(\"The average velocity between t = 2.0 s and t = 4.0 s =\",avg_v,\"m/s\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 3.3 , page : 48 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The equations of motion for constant acceleration using method of calculus are as follows.\n",
+ "v = v_o + at\n",
+ "x = x_o + v_ot + (1/2)at²\n",
+ "v²= v²_o + 2a(x - x_o)\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Calculation\n",
+ "\n",
+ "# By definition, a = dv/dt\n",
+ "# .i.e dv = adt\n",
+ "# Integrating on both sides we get, v - v_o = at or v = v_o + at\n",
+ "# Further we know that, v = dx/dt\n",
+ "# .i.e dx = vdt\n",
+ "# Integrating on both sides we get, x - x_o = v_ot + (1/2)at² or x = x_o + v_ot + (1/2)at² \n",
+ "# Now we can write, a = dv/dt = (dv/dx)(dx/dt) = v(dv/dx) \n",
+ "# .i.e. v dv = a dx\n",
+ "# Integrating on both sides we get, (v² - v²_o) = a(x - x_o) or v² = v²_o + 2a(x - x_o)\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"The equations of motion for constant acceleration using method of calculus are as follows.\")\n",
+ "print(\"v = v_o + at\")\n",
+ "print(\"x = x_o + v_ot + (1/2)at²\")\n",
+ "print(\"v²= v²_o + 2a(x - x_o)\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 3.4 , page : 48 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "(a) The height at which the ball has risen = 20.0 m\n",
+ "(b) The time taken before the ball to hit the ground = 5.0 s\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import sympy\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "t = sympy.symbols('t')\n",
+ "v_o=20 # Initial velocity in m/s\n",
+ "y_o=25 # Height of the initial point from ground in m\n",
+ "a=-10 # Acceleration due to gravity\n",
+ "v=0\n",
+ "y=0\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "#(a)\n",
+ "# Let us take the y-axis in the vertically upward direction with zero at the ground\n",
+ "# Since v=(v_o)²+2ah\n",
+ "h=(-(v_o)**2)/(2*a)\n",
+ "\n",
+ "#(b)\n",
+ "# The total time taken can also be calculated by noting the coordinates of initial and final positions of the ball with respect to the origin chosen\n",
+ "# and using equation (y - y_0) = v_ot + (1/2)at² \n",
+ "# Substituting the values in the above equation we get the quadratic equation for t as , 5t² - 20t - 25 = 0 \n",
+ "t = round(max(sympy.solve(5*t**2 - 20*t -y_o,t)),0)\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"(a) The height at which the ball has risen =\",h,\"m\")\n",
+ "print(\"(b) The time taken before the ball to hit the ground =\",t,\"s\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 3.5 , page : 49 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The motion of an object under free fall can be explained by the following equations\n",
+ "v = 0 - g t = 9.8t m/s\n",
+ "y = 0 - (1/2)gt² = 4.9t² m\n",
+ "v² = 0 - 2gy = -19.6 y m²/s²\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Calculation\n",
+ "\n",
+ "# If air resistance is neglected, the object is said to be in free fall.\n",
+ "# If the height through which the object falls is small compared to the earths radius, g can be taken to be constant, equal to 9.8 ms².\n",
+ "# Free fall is thus a case of motion with uniform acceleration. \n",
+ "# We assume that the motion is in y-direction, more correctly in y-direction because we choose upward direction as positive.\n",
+ "# Since the acceleration due to gravity is always downward, it is in the negative direction.\n",
+ "# Then we have, a = g = 9.8 ms²\n",
+ "# The object is released from rest at y = 0. Therefore, v_0 = 0 and the equations of motion become as follows\n",
+ "# v = 0 - g t = 9.8t m/s \n",
+ "# y = 0 - (1/2)gt² = 4.9t² m\n",
+ "# v² = 0 - 2gy = -19.6 y m²/s²\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"The motion of an object under free fall can be explained by the following equations\")\n",
+ "print(\"v = 0 - g t = 9.8t m/s\")\n",
+ "print(\"y = 0 - (1/2)gt² = 4.9t² m\")\n",
+ "print(\"v² = 0 - 2gy = -19.6 y m²/s²\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 3.6 , page : 50 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Let us divide the time interval of motion of an object under free fall into many equal intervals τ and find out the distancestraversed during successive intervals of time.\n",
+ "Since initial velocity is zero, we have\n",
+ "Using this equation, we can calculate the position of the object after different time intervals, 0, τ, 2τ, 3τ which are given in second column of Table 3.2. If we take (1/ 2) gτ2 as y0 the position coordinate after first time interval τ, then third column gives the positions in the unit of yo. The fourth column gives the distances traversed in successive τs. We find that the distances are in the simple ratio 1: 3: 5: 7: 9: 11 as shown in the last column. This law was established by Galileo Galilei (1564-1642) who was the first to make quantitative studies of free fall.\n",
+ "Hence the proof.\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Result\n",
+ "\n",
+ "print(\"Let us divide the time interval of motion of an object under free fall into many equal intervals τ and find out the distancestraversed during successive intervals of time.\")\n",
+ "print(\"Since initial velocity is zero, we have\")\n",
+ "print(\"Using this equation, we can calculate the position of the object after different time intervals, 0, τ, 2τ, 3τ which are given in second column of Table 3.2. If we take (1/ 2) gτ2 as y0 the position coordinate after first time interval τ, then third column gives the positions in the unit of yo. The fourth column gives the distances traversed in successive τs. We find that the distances are in the simple ratio 1: 3: 5: 7: 9: 11 as shown in the last column. This law was established by Galileo Galilei (1564-1642) who was the first to make quantitative studies of free fall.\")\n",
+ "print(\"Hence the proof.\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 3.7 , page : 50 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The stopping distance, d_s = -v²_o / 2a\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Calculation\n",
+ "\n",
+ "# Let the distance travelled by the vehicle before it stops be d_s.\n",
+ "# Then, using equation of motion v² = v²_o + 2ax, and noting that v = 0, we have the stopping distance as given below,\n",
+ "# The stopping distance, d_s = -v²_o / 2a\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"The stopping distance, d_s = -v²_o / 2a\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 3.8 , page : 51 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Reaction Time = 0.2 s\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ " \n",
+ "v0=0 # Initial velocity in m\n",
+ "g=9.8 # Acceleration due to gravity\n",
+ "d=21 # Distance travelled in cm \n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "d=21*10**-2\n",
+ "t=math.sqrt((2*d)/g) # Reaction time = √(2d/g)\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"Reaction Time =\",round(t,1),\"s\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 3.9 , page : 52 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 9,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "(a) Relative velocity of train B with respect to tain A = -40.0 m/s\n",
+ "(b) Relative velocity of ground with respect to train B = 25.0 m/s\n",
+ "(c) Speed of the monkey = 10.0 m/s\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "# Choose the positive direction of X-axis to be from South to North\n",
+ "V_A=54 # The speed of train A in km/h\n",
+ "V_B=-90 # The speed of train B in km/h\n",
+ "V_MA=-18 # The relative speed of monkey km/h\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "V_A=54*(5/18) # The speed of train A in m/s\n",
+ "V_B=-90*(5/18) # The speed of train B in m/s\n",
+ "V_MA=-18*(5/18) # The relative speed of monkey m/s\n",
+ "\n",
+ "#(a)\n",
+ "V_BA=V_B-V_A # Relative velocity of train B with respect to A\n",
+ "\n",
+ "#(b)\n",
+ "V_GB=0-V_B # Relative velocity of ground with respect to train B\n",
+ "\n",
+ "#(c)\n",
+ "# Since V_MA = V_M - V_A\n",
+ "V_M=V_MA+V_A\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"(a) Relative velocity of train B with respect to tain A =\",V_BA,\"m/s\")\n",
+ "print(\"(b) Relative velocity of ground with respect to train B =\",V_GB,\"m/s\")\n",
+ "print(\"(c) Speed of the monkey =\",V_M,\"m/s\")\n"
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 3",
+ "language": "python",
+ "name": "python3"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 3
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython3",
+ "version": "3.4.3"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Physics_Textbook_Part-I_for_class_XI_by_NCERT_by_Chief_Editor_-_Naresh_Yadav/Chapter4_2.ipynb b/Physics_Textbook_Part-I_for_class_XI_by_NCERT_by_Chief_Editor_-_Naresh_Yadav/Chapter4_2.ipynb new file mode 100644 index 00000000..c2078b32 --- /dev/null +++ b/Physics_Textbook_Part-I_for_class_XI_by_NCERT_by_Chief_Editor_-_Naresh_Yadav/Chapter4_2.ipynb @@ -0,0 +1,513 @@ +{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 4 : Motion in a Plane"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.1 , page : 69"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Therefore,the boy should hold his umbrella in the vertical plane at an angle of about 19.0 ° with the vertical towards the east. \n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable delcaration\n",
+ "\n",
+ "v_r=35 # Velocity vector of the rain which falls vertically in m/s \n",
+ "v_w=12 # Velocity vector of the wind blowing in east to west direction in m/s\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "R=math.sqrt(v_r**2+v_w**2) # The magnitude of the resultant vector \n",
+ "tanθ=v_w/v_r\n",
+ "θ=math.degrees(math.atan(tanθ)) # The direction that R makes with the vertical \n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"Therefore,the boy should hold his umbrella in the vertical plane at an angle of about\",round(θ,0),\"° with the vertical towards the east. \")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.2 , page : 71 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The magnitude of thr resultant vector is given by the equation, R = square root(A²+ B²+ 2ABcosθ)\n",
+ "The direction of the resultant vector is given by the equation, tanθ = SN/(OP+PN) = Bsinθ/(A +Bcosθ)\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Calculation\n",
+ "\n",
+ "# Let OP and OQ represent the two vectors A and B making an angle θ.\n",
+ "# Then, using the parallelogram method of vector addition, OS represents the resultant vector R such that R = A + B \n",
+ "# SN is normal to OP and PM is normal to OS.\n",
+ "# From the geometry of the figure, OS2 = ON2 + SN2 but ON = OP + PN = A + Bcosθ\n",
+ "# SN = Bsin θ and OS² = (A + Bcosθ )² + (Bsinθ )² or, R² = A² + B² + 2ABcosθ\n",
+ "# .i.e R = square root(A²+ B²+ 2ABcosθ) \n",
+ "# In ΔOSN, SN = OSsin α = Rsin α , and in ΔPSN, SN = PSsinθ = Bsinθ\n",
+ "# Therefore, Rsinα = Bsinθ ..............eqn 1\n",
+ "# Similarly, PM = Asinα = Bsinβ .........eqn 2\n",
+ "# From eqns 1 and 2 we get,\n",
+ "# sin α = (B/R) sin θ or tanθ = SN/(OP+PN) = Bsinθ/(A +Bcosθ)\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"The magnitude of thr resultant vector is given by the equation, R = square root(A²+ B²+ 2ABcosθ)\")\n",
+ "print(\"The direction of the resultant vector is given by the equation, tanθ = SN/(OP+PN) = Bsinθ/(A +Bcosθ)\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.3 , page : 72 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The magnitude of the resultant vector = 22.0 km/h\n",
+ "The direction of the resultant vector = 23.4 °\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "v_b=25 # Velocity of the motor boat racing towards north in km/h\n",
+ "v_c=10 # Velocity of the water current in the direction of 60° east of south in km/h\n",
+ "θ=60 # The angle in degree\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "# Using the parallelogram method of vector addition we can obtain the resultant vector\n",
+ "# We can obtain the magnitude of resultant vector using the Law of cosine \n",
+ "R=math.sqrt(v_b**2+v_c**2+(2*v_b*v_c*(math.cos(math.radians(2*θ)))))\n",
+ "# We can obtain the direction of resultant vector using the Law of sines \n",
+ "sinφ=(v_c*math.sin(math.radians(θ)))/R\n",
+ "φ=math.degrees(math.asin(sinφ))\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"The magnitude of the resultant vector =\",round(R,0),\"km/h\")\n",
+ "print(\"The direction of the resultant vector =\",round(φ,1),\"°\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.4 , page : 75 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "(a) The velocity vector, v(t) = 3.0i+4.0tj m/s\n",
+ " The acceleration vector, a(t) = 4.0j m/s²\n",
+ "(b) Magnitude of v(t) at t=1 s = 5.0 m/s\n",
+ " Direction of v(t) at t=1 s = 53.0 ° with x-axis\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "t=1 # Time in s\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "# The position vector of the particle is r(t)=3.0ti+2.0t²j+5.0k\n",
+ "# On differentiating r(t) with respect to t we get the velocity vector, v(t)=3.0i+4.0tj\n",
+ "# On differentiating v(t) with respect to t we get the acceleration vector, a(t)=4.0j\n",
+ "V_x=3 # X component of v(t)\n",
+ "V_y=4 # Y component of v(t)\n",
+ "V=math.sqrt(V_x**2 + V_y**2)\n",
+ "tanθ=V_y/V_x\n",
+ "θ=math.degrees(math.atan(tanθ)) \n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"(a) The velocity vector, v(t) = 3.0i+4.0tj m/s\")\n",
+ "print(\" The acceleration vector, a(t) = 4.0j m/s²\")\n",
+ "print(\"(b) Magnitude of v(t) at t=1 s =\",V,\"m/s\")\n",
+ "print(\" Direction of v(t) at t=1 s =\",round(θ,0),\"° with x-axis\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.5 , page : 76 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "(a) The y-coordinate of the particle at the instant its x-coordinate is 84 m = 36.0 m\n",
+ "(b) Speed of the particle at the instant its x-coordinate is 84 m = 26.0 m/s\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import sympy\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "x=84 # X component of the position vector in m\n",
+ "t= sympy.symbols('t')\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "# Velocity vector is given as, v(t)= 5.0i m/s\n",
+ "# Acceleration vector is given as, a(t)= 3.0i+2.0j m/s²\n",
+ "# By the equation r(t)=v(t)+(a(t)t²)/2 we get the positon vector of the particle as follows\n",
+ "# r(t) = 5.0ti + 1.5t²i + 1.0t²j\n",
+ "t=round((max(sympy.solve(1.5*t**2 + 5*t -x,t))),0)\n",
+ "y=1.0*t**2\n",
+ "# Now the velocity vector can be obtained by differentiating r(t) with respect to t \n",
+ "# Then we get, v(t) = 5.0i +3ti+2tj m/s\n",
+ "v_x=5.0+3*t # X component of v(t) at time = t\n",
+ "v_y=2*t # Y component of v(t) at time = t\n",
+ "V=math.sqrt(v_x**2 + v_y**2)\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"(a) The y-coordinate of the particle at the instant its x-coordinate is 84 m =\",y,\"m\")\n",
+ "print(\"(b) Speed of the particle at the instant its x-coordinate is 84 m =\",round(V,0),\"m/s\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.6 , page : 76 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Therefore, the woman should hold her umbrella at an angle of about 19.0 ° with the vertical towards the west.\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "v_r=35 # Velocity vector of the rain which falls vertically in m/s\n",
+ "v_b=12 # Velocity vector of the bicycle riding in east to west direction in m/s\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "v_rb=v_r - v_b\n",
+ "tanθ=v_b/v_r\n",
+ "θ=math.degrees(math.atan(tanθ)) \n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"Therefore, the woman should hold her umbrella at an angle of about\",round(θ,0),\"° with the vertical towards the west.\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.7 , page : 78 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "For a projectile launched with velocity v_0 at an angle θ_0 , the range is given by, R = (v_0²sin2θ_0)/g\n",
+ "Now,for angles,(45°+α ) and (45°- α ), 2θ_0 is (90° + 2α) and (90° - 2α), respectively.\n",
+ "The values of sin (90° + 2 α ) and sin (90° - 2α) are the same, equal to that of cos 2α.\n",
+ "Therefore, ranges are equal for elevations which exceed or fall short of 45° by equal amounts α.\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Variable declaration\n",
+ "\n",
+ "v_0=1 # For convenience, velocity at whch the projectile launched is assumed to be unity \n",
+ "θ_0=1 # For convenience, angle at whch the projectile launched in degree is assumed to be unity\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"For a projectile launched with velocity v_0 at an angle θ_0 , the range is given by, R = (v_0²sin2θ_0)/g\")\n",
+ "print(\"Now,for angles,(45°+α ) and (45°- α ), 2θ_0 is (90° + 2α) and (90° - 2α), respectively.\")\n",
+ "print(\"The values of sin (90° + 2 α ) and sin (90° - 2α) are the same, equal to that of cos 2α.\")\n",
+ "print(\"Therefore, ranges are equal for elevations which exceed or fall short of 45° by equal amounts α.\") \n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.8 , page : 78 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The time taken by the stone to reach the ground = 10.0 s\n",
+ "The speed with which the stone hits the ground = 99.0 m/s\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "# We choose the origin of the x-,and y- axis at the edge of the cliff and t = 0 s at the instant the stone is thrown\n",
+ "# Choose the positive direction of x-axis to be along the initial velocity and the positive direction of y-axis to be the vertically upward direction\n",
+ "# The equations of motion are : x(t)=x_0 = v_0xt and y(t) = y_0+v_0yt+(1/2)a_y(t^2)\n",
+ "\n",
+ "g=9.8 # Acceleration due to gravity\n",
+ "x_0=0\n",
+ "y_0=0\n",
+ "v_oy=0\n",
+ "a_y=g\n",
+ "v_ox=15\n",
+ "y_t=490\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "# The stone hits the ground when y(t) = 490 m ,i.e. 490 = (1/2)(9.8)t\n",
+ "t=math.sqrt((-y_t*-2)/a_y)\n",
+ "# The velocity components are v_x = v_ox and v_y = v_oy - g t\n",
+ "v_x=v_ox\n",
+ "v_y=v_oy-(g*t)\n",
+ "V=math.sqrt(v_x**2+v_y**2)\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"The time taken by the stone to reach the ground =\",t,\"s\")\n",
+ "print(\"The speed with which the stone hits the ground =\",round(V,0),\"m/s\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.9 , page : 79 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 9,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The maximum height = 10.0 m\n",
+ "The time taken to return to the same level = 2.9 s\n",
+ "The distance from the thrower to the point where the ball returns to the same level = 69.0 m\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "v_0=28 # The initial velocity of the ball in m/s\n",
+ "θ=30 # The angle of inclination of the ball above the horizontal in degree\n",
+ "g=9.8 # Acceleration due to gravity in m/s²\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "h_m=(v_0*(math.sin(math.radians(θ))))**2/(2*g)\n",
+ "T_f=(2*v_0*math.sin(math.radians(θ)))/g\n",
+ "R=((v_0**2)*(math.sin(2*math.radians(θ))))/g \n",
+ " \n",
+ "# Result\n",
+ "\n",
+ "print(\"The maximum height =\",round(h_m,0),\"m\")\n",
+ "print(\"The time taken to return to the same level =\",round(T_f,1),\"s\")\n",
+ "print(\"The distance from the thrower to the point where the ball returns to the same level =\",round(R,0),\"m\") \n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.10 , page : 81 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 10,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "(a) Angular speed = 0.44 rad/s\n",
+ " Linear speed = 5.3 cm/s\n",
+ "(b) Since the direction changes continuously, acceleration here is not a constant vector\n",
+ " Magnitude of acceleration = 2.3 cm/s²\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ " \n",
+ "R=12 # Radius of the circular groove in cm\n",
+ "n=7 # Total number of revolutions\n",
+ "t=100 # Time taken for all 7 revolutions\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "T=t/n # Time taken for one revolution\n",
+ "#(a)\n",
+ "w=2*math.pi/T # Angular speed in rad/s\n",
+ "v=w*R # Linear speed in cm/s\n",
+ "a=pow(w,2)*R # Magnitude of acceleration in cm/s²\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"(a) Angular speed =\",round(w,2),\"rad/s\")\n",
+ "print(\" Linear speed =\",round(v,1),\"cm/s\")\n",
+ "print(\"(b) Since the direction changes continuously, acceleration here is not a constant vector\")\n",
+ "print(\" Magnitude of acceleration =\",round(a,1),\"cm/s²\")\n"
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 3",
+ "language": "python",
+ "name": "python3"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 3
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython3",
+ "version": "3.4.3"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Physics_Textbook_Part-I_for_class_XI_by_NCERT_by_Chief_Editor_-_Naresh_Yadav/Chapter5_2.ipynb b/Physics_Textbook_Part-I_for_class_XI_by_NCERT_by_Chief_Editor_-_Naresh_Yadav/Chapter5_2.ipynb new file mode 100644 index 00000000..89ef087e --- /dev/null +++ b/Physics_Textbook_Part-I_for_class_XI_by_NCERT_by_Chief_Editor_-_Naresh_Yadav/Chapter5_2.ipynb @@ -0,0 +1,623 @@ +{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 5 : Laws of Motion"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 5.1 , page : 93"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The acceleration of the astranaunt = 0 m/s²\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "a=100 # Constant acceleration of the inter stellar space in m/s²\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "# Since there are no nearby stars to exert gravitational force on him and the small spaceship exerts negligible gravitational attraction on him, the net force acting on the astronaut, once he is out of the spaceship, is zero.\n",
+ "# By the first law of motion the acceleration of the astronaut is zero. \n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"The acceleration of the astranaunt =\",0,\"m/s²\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 5.2 , page : 95 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The average resistive force exerted by the block on the bullet = 270.0 N\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ " \n",
+ "m=0.04 # Mass of the bullet in kg\n",
+ "u=90 # Speedof the bullet in m/s\n",
+ "s=60 # Thickness of the wooden block at which the bullet stops in cms\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "s=s*10**-2\n",
+ "a=u**2/(2*s) # Retardation of the bullet in m/s²\n",
+ "F=m*a # Retarding force in N\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"The average resistive force exerted by the block on the bullet =\",F,\"N\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 5.3 , page : 96 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Force acting on the particle, F = g*m\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import sympy\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "g = sympy.symbols('g')\n",
+ "m = sympy.symbols('m')\n",
+ "u = sympy.symbols('u')\n",
+ "ut = sympy.symbols('ut')\n",
+ "t = sympy.symbols('t')\n",
+ "gt = sympy.symbols('gt')\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "# The motion of a particle of mass m is given as follows\n",
+ "y = ut + (gt**2)/2\n",
+ "# Differentiating the equation of motion with respect to time t we get,\n",
+ "v = u + gt\n",
+ "# Again differentiating the above equation with respect to time t we get, \n",
+ "a = g\n",
+ "F = m*a # Force aciting on the particle\n",
+ "# Thus the given equation describes the motion of a particle under acceleration due to gravity and y is the position coordinate in the direction of g.\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "\n",
+ "print(\"Force acting on the particle, F = \",F)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 5.4 , page : 96"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The impulse in the direction from the batsman to the bowler = 3.6 Ns\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "m=0.15 # Mass of the ball in kg\n",
+ "u=12 # Initial speed of the ball in m/s\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "Delta_v=u-(-u) # Change in velocity= Final velocity-Initial velocity\n",
+ "I=m*Delta_v\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"The impulse in the direction from the batsman to the bowler =\",round(I,1),\"Ns\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 5.5 , page : 98 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "(i) The direction of impulse (and force) is the same in both case and is normal to the wall along the negative x direction\n",
+ "(ii)The ratio of the magnitudes of the impulses imparted to the ball in two cases = 1.2\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "θ=30 \n",
+ "\n",
+ "# Calculation\n",
+ " \n",
+ "# Consider the force (or impulse) on the ball due to the wall using the second law, and then use the third law to answer \n",
+ " \n",
+ "# Case(a) \n",
+ "# Let u be the speed of each ball before and after collision with the wall, and m the mass of each ball.\n",
+ "# Let us consider, initial momentum in x direction, P_x_in = mu and final momentum in x direction,P_x_fi = - mu\n",
+ "# Let us consider, initial momentum in y direction, P_y_in = 0 and final momentum in y direction,P_y_fi = 0\n",
+ "# Impulse is the change in momentum vector. Therefore, \n",
+ "# x-component of impulse = 2mu and y-component of impulse = 0\n",
+ "# Impulse and force are in the same direction. Clearly, the force on the ball due to the wall is normal to the wall, along the negative x-direction.\n",
+ "# Using Newtons third law of motion, the force on the wall due to the ball is normal to the wall along the positive x-direction. \n",
+ " \n",
+ "# Case(b)\n",
+ "# Let u be the speed of each ball before and after collision with the wall, and m the mass of each ball.\n",
+ "# Let us consider, initial momentum in x direction, P_x_in = mu cos 30° and final momentum in x direction,P_x_fi = -mu cos 30°\n",
+ "# Let us consider, initial momentum in y direction, P_y_in = mu sin 30° and final momentum in y direction,P_y_fi = mu sin 30°\n",
+ "# x-component of impulse = 2mucos30 and y-component of impulse = 0\n",
+ "# Using Newtons third law, the force on the wall due to the ball is normal to the wall along the positive x direction.\n",
+ " \n",
+ "# The ratio of the magnitudes of the impulses imparted to the balls in (a) and (b) is 2mu/2mu cos 30°= 1/cos 30°\n",
+ "r=1/math.cos(math.radians(θ)) \n",
+ "\n",
+ "# Result\n",
+ " \n",
+ "print(\"(i) The direction of impulse (and force) is the same in both case and is normal to the wall along the negative x direction\")\n",
+ "print(\"(ii)The ratio of the magnitudes of the impulses imparted to the ball in two cases =\",round(r,1)) \n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 5.6 , page : 99 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The angle that the rope makes with the vertical in equilibrium = 40.0 °\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "m=6 # The amount of mass suspended in kg\n",
+ "l=2 # Length of the rope in m\n",
+ "F=50 # The Force applied at the mid-point of the rope in the horizontal direction in N\n",
+ "g=10 # Acceleration due to gravity\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "T2=m*g\n",
+ "T1cosθ=T2\n",
+ "T2sinθ=F\n",
+ "tanθ=T2sinθ/T1cosθ\n",
+ "θ=math.degrees(math.atan(tanθ)) \n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"The angle that the rope makes with the vertical in equilibrium =\",round(θ,0),\"°\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 5.7 , page : 102"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The maximum acceleration of the train = 1.5 m/s²\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "u=.15 # The co-efficient of static friction between the box and the trains floor\n",
+ "g=10 # Acceleration due to gravity\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "# Since the acceleration of the box is due to the static friction, ma = f ≤ μN = μmg, i.e. a ≤ μg \n",
+ "a_max=u*g\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"The maximum acceleration of the train =\",a_max,\"m/s²\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 5.8 , page : 102 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The coefficient of static friction between the block and the surface = 0.27\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "m=4 # Mass in kg\n",
+ "θ=15 # Angle of inclination of the plane with the horizontal\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "# Resolving the weight mg along the two directions shown, we have mgsin θ = fs, mgcosθ=N\n",
+ "# As θ increases, the self-adjusting frictional force fs increases until at θ = θ max, \n",
+ "# fs achieves its maximum value, max_f = μ s N\n",
+ "# Therefore, tan θ max = μ or θ_max = (tan)^-1 μ \n",
+ "tanθ=math.tan(math.radians(θ))\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"The coefficient of static friction between the block and the surface =\",round(tanθ,2))\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 5.9 , page : 102 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 9,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The acceleration of the block and trolley system = 0.96 m/s²\n",
+ "The tension on the string = 27.1 N\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "m1=20 # Mass of the block in kg\n",
+ "m2=3 # Mass of the block in kg\n",
+ "g=10 # Acceleration due to gravity\n",
+ "u_k=0.04 # The coefficient of kinetic friction between the trolley and the surface\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "N=m1*g\n",
+ "f_k=u_k*N\n",
+ "# Applying second law to motion of the block , 30-T = 3a\n",
+ "# Applying the second law to motion of the trolley, T-fk = 20a.\n",
+ "a=22/23\n",
+ "T=(20*a)+f_k\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"The acceleration of the block and trolley system =\",round(a,2),\"m/s²\")\n",
+ "print(\"The tension on the string =\",round(T,1),\"N\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 5.10 , page : 105 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 10,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The cyclist will slip while taking the circular turn.\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "v=18 # Speed of the cycle in km/h\n",
+ "R=3 # Radius of the circular turn in m\n",
+ "µ_s=0.1 # Coefficient of static friction between the tyre and the road \n",
+ "g=9.8 # Acceleration due to gravity in m/s²\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "v=v*(5/18) # Speed of the cycle in m/s\n",
+ "v_sq=pow(v,2)\n",
+ "#The condition for the cyclist not to slip is given b, v2 ≤ μ_sRg\n",
+ "μ_sRg=μ_s*R*g\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "if v_sq < μ_sRg:\n",
+ " print(\"The cyclist will not slip while taking the circular turn.\")\n",
+ "else:\n",
+ " print(\"The cyclist will slip while taking the circular turn.\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 5.11 , page : 105 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 11,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The optimum speed of the race-car to avoid wear and tear on its tyres = 28.1 m/s\n",
+ "The maximum permissible speed to avoid slipping = 38.1 m/s\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "R=300 # Radius of the circulas race track in m\n",
+ "θ=15 # Angle at which the road banked in degree\n",
+ "u=0.2 # The coefficient of friction between the wheels of a race-car and the road\n",
+ "g=9.8 # Acceleration due to gravity\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "#(a)\n",
+ "v_o=math.sqrt(R*g*(math.tan(math.radians(θ))))\n",
+ "#(b)\n",
+ "v_max=math.sqrt(R*g*(u+(math.tan(math.radians(θ))))/(1-u*math.tan(math.radians(θ))))\n",
+ "\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"The optimum speed of the race-car to avoid wear and tear on its tyres =\",round(v_o,1),\"m/s\")\n",
+ "print(\"The maximum permissible speed to avoid slipping =\",round(v_max,1),\"m/s\")\n",
+ " "
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 5.12 , page : 106 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 12,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "(a) The action of the block is equal to 20 N and directed vertically downwards.\n",
+ "(b) The action of the system on the floor is equal to 267.3 N vertically downward.\n",
+ "For(a):\n",
+ "(i) The force of gravity 20 N on the block by the earth (action);the force of gravity on the earth by the block (reaction) 20 N directed upwards.\n",
+ "(ii) The force on the floor by the block (action); the force on the block by the floor (reaction).\n",
+ "For(b):\n",
+ "(i) The force of gravity 270 N on the system by the earth (action);the force of gravity on the earth by the system (reaction), equal to 270 N, directed upwards.\n",
+ "(ii) the force on the floor by the system (action); the force on the system by the floor (reaction).\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ " \n",
+ "m1=2 # Mass of the wooden block in kg\n",
+ "m2=25 # Mass of the iron cylinder in kg\n",
+ "a=0.1\n",
+ "g=10 # Acceleration due to gravity\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "#(a)\n",
+ "# The block is at rest on the floor. Its free-body has two forces on the block, the force of gravitational attraction by the earth and the normal force R of the floor on the block.\n",
+ "# By the First Law, the net force on the block must be zero, hence\n",
+ "e1=m1*g # The force of gravitational attraction by the earth\n",
+ "R=e1 # The normal force of the floor on the block\n",
+ "\n",
+ "#(b)\n",
+ "# The system (block + cylinder) accelerates downwards with 0.1 m/s².The free-body diagram of the system shows two forces on the system : the force of gravity due to the earth; and the normal force R′ by the floor. \n",
+ "e2=(m1+m2)*g # The force of gravitational attraction by the earth\n",
+ "R_prim=e2-(m1+m2)*a # The normal force of the floor on the block\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"(a) The action of the block is equal to\",R,\"N and directed vertically downwards.\")\n",
+ "print(\"(b) The action of the system on the floor is equal to\",R_prim,\"N vertically downward.\")\n",
+ "\n",
+ "print(\"For(a):\")\n",
+ "print(\"(i) The force of gravity\",e1,\"N on the block by the earth (action);the force of gravity on the earth by the block (reaction)\",R,\"N directed upwards.\")\n",
+ "print(\"(ii) The force on the floor by the block (action); the force on the block by the floor (reaction).\")\n",
+ "print(\"For(b):\")\n",
+ "print(\"(i) The force of gravity\",e2,\"N on the system by the earth (action);the force of gravity on the earth by the system (reaction), equal to 270 N, directed upwards.\")\n",
+ "print(\"(ii) the force on the floor by the system (action); the force on the system by the floor (reaction).\")\n"
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 3",
+ "language": "python",
+ "name": "python3"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 3
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython3",
+ "version": "3.4.3"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Physics_Textbook_Part-I_for_class_XI_by_NCERT_by_Chief_Editor_-_Naresh_Yadav/Chapter6_2.ipynb b/Physics_Textbook_Part-I_for_class_XI_by_NCERT_by_Chief_Editor_-_Naresh_Yadav/Chapter6_2.ipynb new file mode 100644 index 00000000..b5b28ba0 --- /dev/null +++ b/Physics_Textbook_Part-I_for_class_XI_by_NCERT_by_Chief_Editor_-_Naresh_Yadav/Chapter6_2.ipynb @@ -0,0 +1,675 @@ +{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 6 : Work, Energy and Power"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6.1 , page : 115"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The angle between the force F and the displacement d = 0.32 °\n",
+ "The projection of F on d = 71.34\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "F = (3,4,-5) # Force vector \n",
+ "d = (5,4,3) # Displacement vector\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "Fd=sum(p*q for p,q in zip(F,d))\n",
+ "FF=sum(p*q for p,q in zip(F,F))\n",
+ "dd=sum(p*q for p,q in zip(d,d))\n",
+ "cosθ=Fd/math.sqrt(FF*dd)\n",
+ "θ=math.degrees(math.acos(cosθ))\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"The angle between the force F and the displacement d =\",cosθ,\"°\")\n",
+ "print(\"The projection of F on d =\",round(θ,2))\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6.2 , page : 116"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "(a) The work done by the gravitational force = 10.0 J\n",
+ "(b) The work done by the unknown resistive force -8.75 J\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "m=1 # Mass of the drop in g\n",
+ "h=1 # The height at which the drop is falling in km\n",
+ "v=50 # Speed at which the drop hits the ground in m/s\n",
+ "g=10 # Accelaration due to gravity\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "m=m*10**-3 # Mass of the drop in kg\n",
+ "h=h*10**3 # The height at which the drop is falling in m\n",
+ "\n",
+ "#(a)\n",
+ "# We have assumed that the drop is initially at rest\n",
+ "\n",
+ "K=(m*pow(v,2))/2 # Change in kinetic energy of the drop\n",
+ "W_g= m*g*h # The work done by the gravitational force on the drop in J\n",
+ "\n",
+ "#(b)\n",
+ "# From the work-energy theorm, K=W_g+W_r where Wr is the work done by the resistive force on the raindrop\n",
+ "W_r=K-W_g\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"(a) The work done by the gravitational force =\",W_g,\"J\")\n",
+ "print(\"(b) The work done by the unknown resistive force\",W_r,\"J\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6.3 , page : 117 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "(a) Work done by the road on the cycle = -2000.0 J\n",
+ "(b) Work done by cycle on the road = 0 J\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "d=10 # The distance in m\n",
+ "F=200 # The force on cycle due to the road in N\n",
+ "θ=180 # The angle the stopping force and the displacement make with each other in degrees\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "W_r=F*d*math.cos(math.radians(θ))\n",
+ "# From Newton’s Third Law an equal and opposite force acts on the road due to the cycle.\n",
+ "# Its magnitude is 200 N. However, the road undergoes no displacement.\n",
+ "# Thus, work done by cycle on the road is zero. \n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"(a) Work done by the road on the cycle =\",W_r,\"J\")\n",
+ "print(\"(b) Work done by cycle on the road = 0 J\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6.4 , page : 118 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The emergent speed of the bullet = 63.2 m/s\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "m=50 # The mass of the bullet in g\n",
+ "in_v=200 # The initial velocity of the bullet in m/s\n",
+ "d=2 # The thickness of the plywood in cm\n",
+ "mp=1 # For convenience,mass is assumed to be unity \n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "m=m*10**-3 # The mass of the bullet in kg\n",
+ "in_ke=(m*in_v**2)/2\n",
+ "# Since the bullet emerges with only 10% of its initial kinetic energy\n",
+ "fin_ke=0.1*in_ke\n",
+ "#If v_f is the emergent speed of the bullet,then mv_f²/2 = final kinetic energy\n",
+ "v_f=math.sqrt((2*fin_ke)/m) \n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"The emergent speed of the bullet =\",round(v_f,1),\"m/s\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6.5 , page : 119 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Work done by the women= 1750.0 J\n",
+ "Work done by the frictional force= -1000 J\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "F1=100 # Initial force in J\n",
+ "F2=50 # Final force in J\n",
+ "d1=10 # Initial distance covered in m\n",
+ "d2=10 # Final distance covered in m\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "d=d1+d2 # Total distance covered\n",
+ "A_rec1=F1*d1 # Area of the rectangle ABCD\n",
+ "A_tra=((F1+F2)/2)*d2\n",
+ "W_f=A_rec1+A_tra # Work done by the women\n",
+ "A_rec2=-F2*d # Area of the rectangle AGHI\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"Work done by the women =\",W_f,\"J\")\n",
+ "print(\"Work done by the frictional force =\",A_rec2,\"J\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6.6 , page : 120 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The final kinetic energy of the block = 0.5 J\n",
+ "The final speed of the block = 1.0 m/s\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "m=1 # Mass of the block in kg\n",
+ "v_t=2 # Speed of the block in m/s\n",
+ "x1=0.10 # Initial point\n",
+ "x2=2.01 # Final point\n",
+ "k=0.5 # Proportionality ratio in J\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "k_t=(m*pow(v_t,2))/2\n",
+ "k_f=k_t-(k*math.log(x2/x1))\n",
+ "#Since Kinetic energy =mv²/2\n",
+ "v_f=math.sqrt(2*k_f/m)\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"The final kinetic energy of the block =\",round(k_f,2),\"J\")\n",
+ "print(\"The final speed of the block =\",round(v_f,2),\"m/s\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6.7 , page : 122 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "(i) v_o = √(5gL)\n",
+ "(ii) The speed at point B, v_b = √3gL\n",
+ " The speed at point C, v_c = √gL\n",
+ "(iii) The ratio of the kinetic energies (KB/KC) at B and C= 3.0\n",
+ "At point C, the string becomes slack and the velocity of the bob is horizontal and to the left. If the connecting string is cut at this instant, the bob will execute a projectile motion with horizontal projection akin to a rock kicked horizontally from the edge of a cliff. Otherwise the bob will continue on its circular path and complete the revolution.\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "#(i)\n",
+ "# The total mechanical energy E of the system is conserved.We take the potential energy of the system to be zero at the lowest point A.\n",
+ "# Thus, at A : E = (1/2)mv²_o\n",
+ "# By Newton’s Second Law Ta-m = mv²_o/L where TA is the tension in the string at A. \n",
+ "# At the highest point C, the string slackens, as the tension in the string (TC) becomes zero. \n",
+ "# Thus, at C, E = mgl(5/2)\n",
+ "# Equating this to the energy at A we get, v_o = √(5gL)\n",
+ "\n",
+ "#(ii)\n",
+ "# We know that v_c= √(gL)\n",
+ "# At B, the energy is E =(1/2)mv²_b + mgL\n",
+ "# Equating this to the energy at A and employing the result namely v²_o = (5gL), we get v_b = √3gL\n",
+ "\n",
+ "#(iii)\n",
+ "# The ratio of the kinetic energies at B and C is; K_B/K_C = (1/2)mv²_b/(1/2)mv²_c = 3/1\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"(i) v_o = √(5gL)\")\n",
+ "print(\"(ii) The speed at point B, v_b = √3gL\")\n",
+ "print(\" The speed at point C, v_c = √gL\")\n",
+ "print(\"(iii) The ratio of the kinetic energies (KB/KC) at B and C=\",3/1)\n",
+ "print(\"At point C, the string becomes slack and the velocity of the bob is horizontal and to the left. If the connecting string is cut at this instant, the bob will execute a projectile motion with horizontal projection akin to a rock kicked horizontally from the edge of a cliff. Otherwise the bob will continue on its circular path and complete the revolution.\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6.8 , page : 124 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The maximum compression of spring = 2.0 m\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "m=1000 # Mass of the car in kg\n",
+ "v=18 # Speed of the car in km/h\n",
+ "k=6.25*10**3 # Spring constant in N/m\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "v=v*(5/18) # Speed of the car in m/s\n",
+ "KE=m*v**2/2 # Kinetic energy of the car\n",
+ "# At maximum compression Xm, the potential energy V of the spring is equal to the kinetic energy KE of the moving car from the principle of conservation of mechanical energy.\n",
+ "Xm=math.sqrt((2*KE)/k)\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"The maximum compression of spring =\",Xm,\"m\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6.9 , page : 125 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 9,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The maximum compression of the spring = 1.03 m\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "m=1000 # Mass of the car in kg\n",
+ "µ=0.5 # The coefficient of friction \n",
+ "g=10 # Acceleration due to gravity\n",
+ "k=6.25*10**3 # Spring constant in N/m\n",
+ "v=18 # Speed of the car in km/h\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "v=v*(5/18) # Speed of the car in m/s\n",
+ "# In presence of friction, both the spring force and the frictional force act so as to oppose the compression of the spring \n",
+ "# The change in kinetic energy is ∆K = K_f - K_i = 0 -(1/2)mv² ..........eqn 1\n",
+ "# The work done by the net force is W = (1/2)mv² + µmgX_m .............eqn 2\n",
+ "# by equating the above two equations and rearranging we obtain the following quadratic equation in the unknown X_m as \n",
+ "# kX²_m + 2µmgX_m + mv² = 0\n",
+ "t1=((-2*µ*m*g)+math.sqrt(abs(((2*µ*m*g)**2)-(4*k*m*v**2))))/(2*k)\n",
+ "t2=((-2*µ*m*g)-math.sqrt(abs(((2*µ*m*g)**2)-(4*k*m*v**2))))/(2*k)\n",
+ "X_m=max(t1,t2)\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"The maximum compression of the spring =\",round(X_m,2),\"m\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6.10 , page : 127 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 10,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Energy required to break one bond of DNA in eV = 6.25 ≈ 0.06 eV\n",
+ "The kinetic energy of an air molecule in eV = 62.5 ≈ 0.0062 eV\n",
+ "The average human consumption in a day in kcal = 2380.9523809523807 ≈ 2400 kcal\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "E=pow(10,20) # Energy required to break one bond of DNA in J\n",
+ "KE=pow(10,21) # The kinetic energy of an air molecule in J\n",
+ "Con=pow(10,7) # The average human consumption in a day in J\n",
+ " \n",
+ "# Calculation\n",
+ "#(a)\n",
+ "E=(10**20)/(1.6*10**19 ) # Energy required to break one bond of DNA in eV\n",
+ "#(b) \n",
+ "KE=(10**21)/(1.6*10**19) # The kinetic energy of an air molecule in eV\n",
+ "#(c) \n",
+ "consum=(10**7)/(4.2*10**3) # The average human consumption in a day in kcal\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"(a) Energy required to break one bond of DNA in eV =\",E,\"≈ 0.06 eV\")\n",
+ "print(\"(b) The kinetic energy of an air molecule in eV =\",KE,\"≈ 0.0062 eV\")\n",
+ "print(\"(c) The average human consumption in a day in kcal =\",consum,\"≈ 2400 kcal\")\n",
+ " "
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6.11 , page : 128 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 11,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The minimum power delivered by the motor to the elevator in watts = 44000 watts\n",
+ "The minimum power delivered by the motor to the elevator in hp = 59.0 hp\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "m=1800 # Maximum load the elevator can carry in kg\n",
+ "F_f=4000 # Frictional force appearing in N\n",
+ "v=2 # Speed of the elevator in m/s\n",
+ "g=10 # Acceleration due to gravity\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "F=m*g+F_f # The downward force on the elevator in N\n",
+ "P=F*v\n",
+ "# Since, 1 hp = 7.457*10**2 W\n",
+ "P_hp=P/(7.457*10**2)\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"The minimum power delivered by the motor to the elevator in watts =\",P,\"watts\")\n",
+ "print(\"The minimum power delivered by the motor to the elevator in hp =\",round(P_hp,2),\"hp\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6.12 , page : 130 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 12,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "For deuterium the fractional kinetic energy lost is = 0.1111111111111111\n",
+ "For deuterium the fractional kinetic energy gained by the moderating nuclei = 0.8888888888888888\n",
+ "Hence we conclude that almost 90% of the neutron’s energy is transferred to deuterium.\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "# The initial kinetic energy of the neutron is K_1i = (1/2)*m1V²_1i\n",
+ "# Its final kinetic energy is given by, K_1f = (1/2)*m1V²_1f =(1/2)*m1*(m1-m2/m1+m2)²*V²_1f\n",
+ "# The fractional kinetic energy lost is f1 = K_1f/K_1i = (m1-m2/m1+m2)²\n",
+ "# The fractional kinetic energy gained by the moderating nuclei is f2 = 1-f1 = (4*m1*m2)/(m1+m2)²\n",
+ "# For deuterium m2 = 2m1 \n",
+ "m1=1\n",
+ "m2=2*m1\n",
+ "f1=((m1-m2)/(m1+m2))**2\n",
+ "f2 = 1-f1\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"For deuterium the fractional kinetic energy lost is =\",f1)\n",
+ "print(\"For deuterium the fractional kinetic energy gained by the moderating nuclei =\",f2)\n",
+ "print(\"Hence we conclude that almost 90% of the neutron’s energy is transferred to deuterium.\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6.13 , page : 131 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 13,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The angle at which the player has to strike the cue = 53.0 °\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "θ=37 # The angle of the corner pocket in degreeac\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "# Here m1 = m2, where m1 is the mass of the cue and m2 is the mass of the target\n",
+ "# From momentum conservation, since the masses are equal ; V_1i = V_1f + V_2f\n",
+ "# Or V_1f² = V_2f² + 2*V_1f*V_2f\n",
+ "# Since the collision is elastic and m1 = m2 it follows from conservation of kinetic energy that V_1i² = V_1f² + V_2f²\n",
+ "# Hence we get, cos(θ1+θ) = 0\n",
+ "θ1=math.degrees(math.acos(0))- θ\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"The angle at which the player has to strike the cue =\",θ1,\"°\")\n"
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 3",
+ "language": "python",
+ "name": "python3"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 3
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython3",
+ "version": "3.4.3"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Physics_Textbook_Part-I_for_class_XI_by_NCERT_by_Chief_Editor_-_Naresh_Yadav/Chapter7_2.ipynb b/Physics_Textbook_Part-I_for_class_XI_by_NCERT_by_Chief_Editor_-_Naresh_Yadav/Chapter7_2.ipynb new file mode 100644 index 00000000..f4eef20f --- /dev/null +++ b/Physics_Textbook_Part-I_for_class_XI_by_NCERT_by_Chief_Editor_-_Naresh_Yadav/Chapter7_2.ipynb @@ -0,0 +1,226 @@ +{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 7 : Systems of Particles and Rotational Motion"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 7.1 , page : 146"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The centre of mass = ( 0.28 m, 0.11 m )\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "# With the x–and y–axes chosen as the coordinates of points O, A and B forming the equilateral triangle are respectively (0,0),(0.5,0),(0.25,0.25√3 ).\n",
+ "# Let the masses 100 g, 150g and 200g be located at O, A and B be respectively.\n",
+ "\n",
+ "m1=100 # Mass of the first particle in g\n",
+ "m2=150 # Mass of the second particle in g\n",
+ "m3=200 # Mass of the third particle in g\n",
+ "x1=0 # x-coordinate of the first particle \n",
+ "x2=0.5 # x-coordinate of the second particle \n",
+ "x3=0.25 # x-coordinate of the third particle \n",
+ "y1=0 # y-coordinate of the first particle \n",
+ "y2=0 # y-coordinate of the second particle \n",
+ "y3=0.25 # y-coordinate of the third particle \n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "X=(m1*x1+m2*x2+m3*x3)/(m1+m2+m3)\n",
+ "Y=(m1*y1+m2*y2+m3*y3)/(m1+m2+m3)\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"The centre of mass = (\",round(X,2),\"m,\",round(Y,2),\"m\",\")\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 7.2 , page : 147 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The centre of mass of a triangular lamina lies on the centroid of the triangle\n"
+ ]
+ }
+ ],
+ "source": [
+ "# The lamina (∆LMN) may be subdivided into narrow strips each parallel to the base (MN). \n",
+ "# By symmetry each strip has its centre of mass at its midpoint.\n",
+ "# If we join the midpoint of all the strips we get the median LP.\n",
+ "# The centre of mass of the triangle as a whole therefore, has to lie on the median LP.\n",
+ "# Similarly, we can argue that it lies on the median MQ and NR. \n",
+ "# This means the centre of mass lies on the point of concurrence of the medians, i.e. on the centroid G of the triangle.\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"The centre of mass of a triangular lamina lies on the centroid of the triangle\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 7.3 , page : 147 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The centre of mass = ( 0.83 m, 0.83 m )\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "# Letus choose the X and Y axes as the coordinates of the vertices of the L-shaped lamina.\n",
+ "# We can think of the L-shape to consist of 3 squares each of length 1m.\n",
+ "# The mass of each square is 1kg, since the lamina is uniform.\n",
+ "# The centres of mass C1, C2 and C3 of the squares are, by symmetry, their geometric centres and have coordinates (1/2,1/2), (3/2,1/2), (1/2,3/2) respectively.\n",
+ "# We take the masses of the squares to be concentrated at these points.\n",
+ "# hence the centre of mass of the whole L shape (X, Y) is the centre of mass of these mass points.\n",
+ "\n",
+ "\n",
+ "m1=1\n",
+ "m2=1\n",
+ "m3=1\n",
+ "x1=1/2\n",
+ "x2=3/2\n",
+ "x3=1/2\n",
+ "y1=1/2\n",
+ "y2=1/2\n",
+ "y3=3/2\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "X=(m1*x1+m2*x2+m3*x3)/(m1+m2+m3)\n",
+ "Y=(m1*y1+m2*y2+m3*y3)/(m1+m2+m3)\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"The centre of mass = (\",round(X,2),\"m,\",round(Y,2),\"m\",\")\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 7.4 , page : 152 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Scalar product = -25\n",
+ "Vector product a × b = [ 7 -1 -5]\n",
+ "Vector product b × a = [-7 1 5]\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "a = (3,-4,5) # Vector a\n",
+ "b = (-2,1,-3) # Vector b\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "s=sum(p*q for p,q in zip(a,b))\n",
+ "a1 = np.array([3,-4,5]) \n",
+ "b1 = np.array([-2,1,-3]) \n",
+ "v1=np.cross(a1,b1)\n",
+ "v2=np.cross(b1,a1)\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"Scalar product =\",s)\n",
+ "print(\"Vector product a × b =\",v1)\n",
+ "print(\"Vector product b × a =\",v2)"
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 3",
+ "language": "python",
+ "name": "python3"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 3
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython3",
+ "version": "3.4.3"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Physics_Textbook_Part-I_for_class_XI_by_NCERT_by_Chief_Editor_-_Naresh_Yadav/Chapter8_2.ipynb b/Physics_Textbook_Part-I_for_class_XI_by_NCERT_by_Chief_Editor_-_Naresh_Yadav/Chapter8_2.ipynb new file mode 100644 index 00000000..b60fb136 --- /dev/null +++ b/Physics_Textbook_Part-I_for_class_XI_by_NCERT_by_Chief_Editor_-_Naresh_Yadav/Chapter8_2.ipynb @@ -0,0 +1,445 @@ +{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "# Chapter 8 : Gravitation"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "## Example 8.1 , page : 185"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " The planet will take a longer time to traverse BAC than CPB\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "mp=1 # For convenience,mass is assumed to be unity \n",
+ "rp=1 # For convenience,sun-planet distance at perihelton is assumed to be unity \n",
+ "vp=1 # For convenience,speed of the planet at perihelton is assumed to be unity \n",
+ "ra=1 # For convenience,sun-planet distance at aphelton is assumed to be unity \n",
+ "va=1 # For convenience,speed of the planet at aphelton is assumed to be unity \n",
+ "Lp=mp*rp*vp # Angular momentum at perihelton\n",
+ "La=mp*ra*va # Angular momentum at ahelton\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "# From angular momentum conservation, mp*rp*vp = mp*ra*va or vp/va = rp/ra\n",
+ "# From Kepler’s second law, equal areas are swept in equal times\n",
+ "print(\" The planet will take a longer time to traverse BAC than CPB\")"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 8.2 , page : 187 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "(a) The force acting = [0.0, 2.5849394142282115e-26, 0.0] ≈ 0\n",
+ "(b) The force acting = 2 Gm²\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "G=6.67*pow(10,-11) # Gravitational constant\n",
+ "m=1 # For convenience,mass is assumed to be unity \n",
+ "x=30 # The angle between GC and the positive x-axis is 30° and so is the angle between GB and the negative x-axis\n",
+ "y=math.radians(x) # The angle in radians\n",
+ "a=math.cos(y)\n",
+ "b=math.sin(y)\n",
+ "v1=(0,1,0)\n",
+ "v2=(-a,-b,0)\n",
+ "v3=(a,-b,0)\n",
+ "c=(2*G*pow(m,2))/1 # 2Gm²/1\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "#(a)\n",
+ "F1=[y * c for y in v1] # F(GA)\n",
+ "F2=[y * c for y in v2] # F(GB)\n",
+ "F3=[y * c for y in v3] # F(GC)\n",
+ "# From the principle of superposition and the law of vector addition, the resultant gravitational force FR on (2m) is given by\n",
+ "Fa=[sum(x) for x in zip(F1,F2,F3)]\n",
+ "\n",
+ "#(b)\n",
+ "# By symmetry the x-component of the force cancels out and the y-component survives\n",
+ "Fb=4-2 # 4Gm² j - 2Gm² j\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"(a) The force acting =\",Fa,\"≈ 0\")\n",
+ "print(\"(b) The force acting =\",Fb,\"Gm²\")"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 8.3 , page : 192 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Potential energy of a system of four particles = -5.414213562373095 Gm²/l\n",
+ "The gravitational potential at the centre of the square = -5.65685424949238 Gm²/l\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "G=6.67*pow(10,-11) # Gravitational constant\n",
+ "m=1 # For convenience,mass is assumed to be unity \n",
+ "l=1 # For convenience,side of the square is assumed to be unity \n",
+ "c=(G*pow(m,2))/l\n",
+ "n=4 # Number of particles\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "d=math.sqrt(2)\n",
+ "# If the side of a square is l then the diagonal distance is √2l\n",
+ "# We have four mass pairs at distance l and two diagonal pairs at distance √2l \n",
+ "# Since the Potential Energy of a system of four particles is -4Gm²/l) - 2Gm²/dl\n",
+ "w=(-n-(2/d)) \n",
+ "# If the side of a square is l then the diagonal distance from the centre to corner is \n",
+ "# Since the Gravitational Potential at the centre of the square\n",
+ "u=-n*(2/d)\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print (\"Potential energy of a system of four particles =\",w,\"Gm²/l\")\n",
+ "print(\"The gravitational potential at the centre of the square =\",u,\"Gm²/l\")"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 8.4 , page : 193 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Minimum speed of the projectile to reach the surface of the second sphere = ( 0.6 GM/R ) ^(1/2)\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "R=1 # For convenience,radii of both the spheres is assumed to be unity \n",
+ "M=1 # For convenience,mass is assumed to be unity \n",
+ "m1=M # Mass of the first sphere\n",
+ "m2=6*M # Mass of the second sphere\n",
+ "m=1 # Since the mass of the projectile is unknown,take it as unity\n",
+ "d=6*R # Distance between the centres of both the spheres\n",
+ "r=1 # The distance from the centre of first sphere to the neutral point N\n",
+ "\n",
+ "G=6.67*pow(10,-11) # Gravitational constant\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "# Since N is the neutral point; GMm/r² = 4GMm/(6R-r)² and we get\n",
+ "r=2*R\n",
+ "# The mechanical energy at the surface of M is; Et = m(v^2)/2 - GMm/R - 4GMm/5R\n",
+ "# The mechanical energy at N is; En = -GMm/2R - 4GMm/4R\n",
+ "# From the principle of conservation of mechanical energy; Et = En and we get\n",
+ "v_sqr=2*((4/5)-(1/2))\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"Minimum speed of the projectile to reach the surface of the second sphere =\",\"(\",round(v_sqr,5),\"GM/R\",\")\",\"^(1/2)\")"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 8.5 , page : 195 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "(i) Mass of Mars = 6.475139697520706e+23 kg\n",
+ "(ii) Period of revolution of Mars = 684.0033777694376 days\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "π=3.14 # Constant pi\n",
+ "G=6.67*pow(10,-11) # Gravitational constant\n",
+ "R=9.4*pow(10,3) # Orbital radius of Mars in km\n",
+ "T=459*60\n",
+ "Te=365 # Period of revolution of Earth\n",
+ "r=1.52 # Ratio of Rms/Res, where Rms is the mars-sun distance and Res is the earth-sun distance. \n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "# (i) \n",
+ "R=R*pow(10,3)\n",
+ "# Using Kepler's 3rd law:T²=4π²(R^3)/GMm\n",
+ "Mm=(4*pow(π,2)*pow(R,3))/(G*pow(T,2))\n",
+ "\n",
+ "# (ii)\n",
+ "# Using Kepler's 3rd law: Tm²/Te² = (Rms^3/Res^3)\n",
+ "Tm=pow(r,(3/2))*365\n",
+ "\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"(i) Mass of Mars =\",Mm,\"kg\")\n",
+ "print(\"(ii) Period of revolution of Mars =\",Tm,\"days\")"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 8.6 , page : 195 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Mass of the Earth = 5.967906881559221e+24 kg\n",
+ "Mass of the Earth = 6.017752855396305e+24 kg\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "g=9.81 # Acceleration due to gravity\n",
+ "G=6.67*pow(10,-11) # Gravitational constant\n",
+ "Re=6.37*pow(10,6) # Radius of Earth in m\n",
+ "R=3.84*pow(10,8) # Distance of Moon from Earth in m\n",
+ "T=27.3 # Period of revolution of Moon in days\n",
+ "π=3.14 # Constant pi\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "# I Method\n",
+ "# Using Newton's 2nd law of motion:g = F/m = GMe/Re²\n",
+ "Me1=(g*pow(Re,2))/G\n",
+ "\n",
+ "# II Method\n",
+ "# Using Kepler's 3rd law: T²= 4π²(R^3)/GMe\n",
+ "T1=T*24*60*60\n",
+ "Me2=(4*pow(π,2)*pow(R,3))/(G*pow(T1,2))\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print(\"Mass of the Earth =\",Me1,\"kg\")\n",
+ "print(\"Mass of the Earth =\",Me2,\"kg\")"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 8.7 , page : 195 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Period of revolution of Moon = 27.5 days\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "k=pow(10,-13) # A constant = 4π² / GME\n",
+ "Re=3.84*pow(10,5) # Distance of the Moon from the Earth in m\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "k=pow(10,-13)*(pow(1/(24*60*60),2))*(1/pow((1/1000),3))\n",
+ "T2=k*pow(Re,3)\n",
+ "T=math.sqrt(T2) # Period of revolution of Moon in days\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"Period of revolution of Moon =\",round(T,1),\"days\")"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 8.8 , page : 196 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Change in Kinetic Energy = 3124485000.0 J\n",
+ "Change in Potential Energy = 6248970000.0 J\n"
+ ]
+ }
+ ],
+ "source": [
+ "# Importing module\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "# Variable declaration\n",
+ "\n",
+ "m=400 # Mass of satellite in kg\n",
+ "Re=6.37*pow(10,6) # Radius of Earth in m\n",
+ "g=9.81 # Acceleration due to gravity\n",
+ "\n",
+ "# Calculation\n",
+ "\n",
+ "# Change in energy is E=Ef-Ei\n",
+ "ΔE=(g*m*Re)/8 # Change in Total energy\n",
+ "# Since Potential Energy is twice as the change in Total Energy (V = Vf - Vi)\n",
+ "ΔV=2*ΔE # Change in Potential Energy in J\n",
+ "\n",
+ "# Result\n",
+ "\n",
+ "print(\"Change in Kinetic Energy =\",round(ΔE,4),\"J\")\n",
+ "print(\"Change in Potential Energy =\",round(ΔV,4),\"J\")"
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 3",
+ "language": "python",
+ "name": "python3"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 3
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython3",
+ "version": "3.4.3"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Physics_Textbook_Part-I_for_class_XI_by_NCERT_by_Chief_Editor_-_Naresh_Yadav/screenshots/Screenshot_(50).png b/Physics_Textbook_Part-I_for_class_XI_by_NCERT_by_Chief_Editor_-_Naresh_Yadav/screenshots/Screenshot_(50).png Binary files differnew file mode 100644 index 00000000..086d0a97 --- /dev/null +++ b/Physics_Textbook_Part-I_for_class_XI_by_NCERT_by_Chief_Editor_-_Naresh_Yadav/screenshots/Screenshot_(50).png diff --git a/Physics_Textbook_Part-I_for_class_XI_by_NCERT_by_Chief_Editor_-_Naresh_Yadav/screenshots/Screenshot_(51).png b/Physics_Textbook_Part-I_for_class_XI_by_NCERT_by_Chief_Editor_-_Naresh_Yadav/screenshots/Screenshot_(51).png Binary files differnew file mode 100644 index 00000000..6efb5e96 --- /dev/null +++ b/Physics_Textbook_Part-I_for_class_XI_by_NCERT_by_Chief_Editor_-_Naresh_Yadav/screenshots/Screenshot_(51).png diff --git a/Physics_Textbook_Part-I_for_class_XI_by_NCERT_by_Chief_Editor_-_Naresh_Yadav/screenshots/Screenshot_(52).png b/Physics_Textbook_Part-I_for_class_XI_by_NCERT_by_Chief_Editor_-_Naresh_Yadav/screenshots/Screenshot_(52).png Binary files differnew file mode 100644 index 00000000..3b4bbe23 --- /dev/null +++ b/Physics_Textbook_Part-I_for_class_XI_by_NCERT_by_Chief_Editor_-_Naresh_Yadav/screenshots/Screenshot_(52).png diff --git a/sample_notebooks/RuchiMittal/chapter1.ipynb b/sample_notebooks/RuchiMittal/chapter1.ipynb new file mode 100644 index 00000000..f6180b5a --- /dev/null +++ b/sample_notebooks/RuchiMittal/chapter1.ipynb @@ -0,0 +1,543 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:9880f2d8505e271317a099910ead6c2116ce86fa0e83f56feb35ac33a1b96b23"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 1 Electric charge"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.1 Page no 9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=4.5*10**-19 #C\n",
+ "e=1.6*10**-19 #C\n",
+ "\n",
+ "#Calculation\n",
+ "n=q/e\n",
+ "\n",
+ "#Result\n",
+ "print\"n= \",round(n,1),\"This value of charge is not possible\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "n= 2.8 This value of charge is not possible\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.2 Page no 9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=3.2*10**-7 #C\n",
+ "e=1.6*10**-19 #C\n",
+ "\n",
+ "#Calculation\n",
+ "n=q/e\n",
+ "\n",
+ "#Result\n",
+ "print\"The required number of electrons is \",n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The required number of electrons is 2e+12\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.3 Page no 9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q=19.2*10**-19\n",
+ "e=1.6*10**-19\n",
+ "me=9*10**-31 #Kg\n",
+ "\n",
+ "#Calculation\n",
+ "n=q/e\n",
+ "M=n*me\n",
+ "\n",
+ "#Result\n",
+ "print\"(i) The value of n=\",n,\"\\n(ii) Charge on silk=\",-q*10**19,\"*10**-19\"\n",
+ "print\"(iii) Mass=\",M,\"Therefore mass transferred is negligibly small\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The value of n= 12.0 \n",
+ "(ii) Charge on silk= -19.2 *10**-19\n",
+ "(iii) Mass= 1.08e-29 Therefore mass transferred is negligibly small\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.4 Page no 9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=16\n",
+ "n=6.023*10**23 #C\n",
+ "\n",
+ "#Calculation\n",
+ "W=2+a\n",
+ "A=((n*100)/W)*10\n",
+ "\n",
+ "#Result\n",
+ "print\"Total number of electrons in 100 g of water \", round(A,-23)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total number of electrons in 100 g of water 3.35e+25\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.5 Page no 9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "n=10**9\n",
+ "e=1.6*10**-19 #C\n",
+ "Q=1\n",
+ "\n",
+ "#Calculation\n",
+ "q=n*e\n",
+ "t=Q/q\n",
+ "\n",
+ "#Result\n",
+ "print (t*10**-9),\"10**9 S\"\n",
+ "print\"Time required is about 198 years\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "6.25 10**9 S\n",
+ "Time required is about 198 years\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.6 Page no 13"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "q1=20 #micro C\n",
+ "q2=-5 #micro C\n",
+ "a=9*10**9\n",
+ "r=0.1 \n",
+ "\n",
+ "#Calculation\n",
+ "q=q1+q2\n",
+ "q3=q/2.0\n",
+ "F=(a*q3*q3)/r**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Force is \",round(F*10**-13,3),\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Force is 5.062 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 1.10 Page no 14"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "q=5*10**-6\n",
+ "r=0.1\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "F=(m*q*q)/r**2\n",
+ "C=2*F*math.cos(30)*(180/3.14)\n",
+ "\n",
+ "#Result\n",
+ "print\"Force on each charge is \", round(C,1)*10**-1,\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Force on each charge is 39.79 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 66
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.11 Page no 14"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "q=1\n",
+ "r=0.24\n",
+ "A=20\n",
+ "B=12.0\n",
+ "m1=10**-4\n",
+ "g=9.8\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "F=(m*q**2)/r**2\n",
+ "AD=math.sqrt(A**2-B**2)\n",
+ "C=AD/B\n",
+ "F1=(1/C)*m1*g\n",
+ "Q=math.sqrt(F1/F)\n",
+ "\n",
+ "#Result\n",
+ "print\"Charge on each sphere\", round(Q*10**8,1),\"10**-8\",\"C\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Charge on each sphere 6.9 10**-8 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 79
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.12 Page no 15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "F=3.7*10**-9 #C\n",
+ "q=1.6*10**-19 #c\n",
+ "m=9*10**9\n",
+ "r=5*10**-10\n",
+ "\n",
+ "#Calculation \n",
+ "import math\n",
+ "n=math.sqrt(F*r**2/(m*q**2))\n",
+ "\n",
+ "#Result\n",
+ "print round(n,0),\"electrons are missing from each icon\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "2.0 electrons are missing from each icon\n"
+ ]
+ }
+ ],
+ "prompt_number": 82
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.14 Page no 16"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "e=1.6*10**-19\n",
+ "m=9*10**9\n",
+ "G=6.67*10**-11\n",
+ "me=9.11*10**-31\n",
+ "mp=1.67*10**-27\n",
+ "r=10**-10\n",
+ "\n",
+ "#Calculation\n",
+ "F0=(m*e**2)/(G*me*mp)\n",
+ "F1=(m*e**2)/(G*mp*mp)\n",
+ "F2=m*e**2/r**2\n",
+ "A1=F2/me\n",
+ "A2=F2/mp\n",
+ "\n",
+ "#Result\n",
+ "print\"(a)(i)strength of an electrons and protons\", round(F0*10**-39,1)*10**39\n",
+ "print\" (ii)Strength of two protons \",round(F1*10**-36,1)*10**36\n",
+ "print\"(b) Acceleration of electron is \",round(A1*10**-22,1)*10**22,\"m/s**2\"\n",
+ "print\" Acceleration of proton is \",round(A2*10**-19,1)*10**19,\"m/s*2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)(i)strength of an electrons and protons 2.3e+39\n",
+ " (ii)Strength of two protons 1.2e+36\n",
+ "(b) Acceleration of electron is 2.5e+22 m/s**2\n",
+ " Acceleration of proton is 1.4e+19 m/s*2\n"
+ ]
+ }
+ ],
+ "prompt_number": 112
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.16 Page no 19"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9 #C\n",
+ "q1=10*10**-6\n",
+ "q2=5*10**-6\n",
+ "r=0.05\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "F1=m*q1*q2/r**2\n",
+ "F2=m*q1*q2/r**2\n",
+ "F3=math.sqrt(F1**2+F2**2+(2*F1*F2*math.cos(120)*180/3.14))\n",
+ "\n",
+ "#Result\n",
+ "print\"Resultant charge is \", round(F3*10**-1,0),\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resultant charge is 176.0 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 132
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.17 Page no 20 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "q1=1.2*10**-8\n",
+ "q2=1\n",
+ "r=0.03\n",
+ "r1=0.04\n",
+ "q3=1.6*10**-8\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "F1=m*q1*q2/r**2\n",
+ "F2=m*q3*q2/r1**2\n",
+ "F3=math.sqrt(F1**2+F2**2)\n",
+ "\n",
+ "#Result\n",
+ "print\"Total force is \", F3*10**-5,\"10**5\",\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total force is 1.5 10**5 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 149
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.18 Page no 20"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "m=9*10**9\n",
+ "q1=1\n",
+ "q2=100\n",
+ "r=10\n",
+ "q3=75 #C\n",
+ "r1=5\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "F=m*q1*q2/r**2 #along BA\n",
+ "F1=m*q1*q2/r**2 #along AC\n",
+ "F2=m*q3/(math.sqrt(r**2-r1**2)**2)\n",
+ "F3=math.sqrt(F1**2+F2**2)\n",
+ "X=F1/F2\n",
+ "\n",
+ "#Result\n",
+ "print\"Force experienced by 1 C Charge is \",round(F3*10**-9,2),\"N\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Force experienced by 1 C Charge is 12.73 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 168
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file |
.png?id=c00ea066e3f752b36e9afb0ced7084b71814c89e){kind=link}
.png?id=c00ea066e3f752b36e9afb0ced7084b71814c89e){kind=link}
.png?id=c00ea066e3f752b36e9afb0ced7084b71814c89e){kind=link}