From a7fe2ac0177fd9c64dc1dfdfb11e85d4095a6501 Mon Sep 17 00:00:00 2001 From: Trupti Kini Date: Wed, 23 Dec 2015 23:30:09 +0600 Subject: Added(A)/Deleted(D) following books A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/Chapter9_2.ipynb A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter1_2.ipynb A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter2_2.ipynb A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter3_2.ipynb A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter4_2.ipynb A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter6_2.ipynb A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter7_2.ipynb A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter8_2.ipynb A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter_5_2.ipynb A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/ex1.2_2.png A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/ex3.13_2.png A Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/ex6.7_2.png A _Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/chapter1.ipynb A _Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/chapter10.ipynb A _Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/chapter11.ipynb A _Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/chapter2.ipynb A _Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/chapter3.ipynb A _Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/chapter4.ipynb A _Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/chapter5.ipynb A _Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/chapter6.ipynb A _Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/chapter7.ipynb A _Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/chapter8.ipynb A _Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/chapter9.ipynb A _Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/screenshots/Screenshot_(56).png A _Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/screenshots/Screenshot_(57).png A _Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/screenshots/Screenshot_(58).png A sample_notebooks/VarunSakpal/chapter_2.ipynb --- .../Chapter9_2.ipynb | 427 ++++ .../chapter1_2.ipynb | 633 ++++++ .../chapter2_2.ipynb | 2194 ++++++++++++++++++++ .../chapter3_2.ipynb | 1113 ++++++++++ .../chapter4_2.ipynb | 888 ++++++++ .../chapter6_2.ipynb | 1068 ++++++++++ .../chapter7_2.ipynb | 761 +++++++ .../chapter8_2.ipynb | 1098 ++++++++++ .../chapter_5_2.ipynb | 366 ++++ .../screenshots/ex1.2_2.png | Bin 0 -> 19571 bytes .../screenshots/ex3.13_2.png | Bin 0 -> 16157 bytes 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_Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/chapter9.ipynb create mode 100644 _Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/screenshots/Screenshot_(56).png create mode 100644 _Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/screenshots/Screenshot_(57).png create mode 100644 _Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/screenshots/Screenshot_(58).png create mode 100644 sample_notebooks/VarunSakpal/chapter_2.ipynb diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/Chapter9_2.ipynb b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/Chapter9_2.ipynb new file mode 100644 index 00000000..ed702744 --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/Chapter9_2.ipynb @@ -0,0 +1,427 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:355541e90dba9a8d43b842e0fe28ab8938c872697589fddf3fc55c333074b753" + }, + "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 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter1_2.ipynb b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter1_2.ipynb new file mode 100644 index 00000000..4bf3cff1 --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter1_2.ipynb @@ -0,0 +1,633 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:659bf40fd24dcbea56efd9142b8ac03b3b827f2e9c5f83005d90a9bfbf26cc13" + }, + "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 E1d:\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))0:\n", + " b=x1\n", + " else:\n", + " a=x1\n", + " if abs(f(x1))0:\n", + " b=x1\n", + " else:\n", + " a=x1\n", + " if abs(f(x1))0:\n", + " b=x1\n", + " else:\n", + " a=x1\n", + " if abs(f(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)" + ], + "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))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 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter3_2.ipynb b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter3_2.ipynb new file mode 100644 index 00000000..77d8f79f --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter3_2.ipynb @@ -0,0 +1,1113 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:58def12f7e424e92e928d020c21b40714eff26275c7ce87aa5600004fbc92a49" + }, + "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_errerr:\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", + "for in range(0,4):\n", + " d1[i]=(y[i+1]-y[i])/(x[i+1]-x[i])\n", + "for in range(0,3):\n", + " d2[i]=(d1[i+1]-d1[i])/(x[i+2]-x[i])\n", + "for in range(0,2):\n", + " d3[i]=(d2[i+1]-d2[i])/(x[i+3]-x[i])\n", + "for iin range(0,1):\n", + " d4[i]=(d3[i+1]-d3[i])/(x[i+4]-x[i])\n", + "X=poly(0,'X')\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", + "disp(f_x,'the polynomial equation is =')" + ], + "language": "python", + "metadata": {}, + "outputs": [] + }, + { + "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": [ + "Ex2.26: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_2.ipynb b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter4_2.ipynb new file mode 100644 index 00000000..0ffbc728 --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter4_2.ipynb @@ -0,0 +1,888 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:491578e93659e656ccfdeaf1d5ad50cd42b24dd8a73c6ca4fafe521f6ad022c8" + }, + "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": 16 + }, + { + "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 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter6_2.ipynb b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter6_2.ipynb new file mode 100644 index 00000000..7faef1ea --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter6_2.ipynb @@ -0,0 +1,1068 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:f89b91b8c25ad887942affb48e813afa9449ad54d5e7ab1cf115acac70bf5f26" + }, + "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 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter7_2.ipynb b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter7_2.ipynb new file mode 100644 index 00000000..235166ee --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter7_2.ipynb @@ -0,0 +1,761 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:5bfd3c7c4de8a3c81a889273c59a61a612b4e62a4dc20f2d2db4b70b66a7b2dc" + }, + "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 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter8_2.ipynb b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter8_2.ipynb new file mode 100644 index 00000000..001e1904 --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter8_2.ipynb @@ -0,0 +1,1098 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:9931680bf490e72be52e25ee57fcec17ccbbf7d8a2c5f86513644cb4dda66cab" + }, + "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 err0:\n", + "\tprint\"\\n Case (a) Fission is possible since m1=\",m1\n", + "else:\n", + "\tprint\"\\n Case (a) Fission is not possiblesince m1=\",m1\n", + "\n", + "\n", + "if m2>0:\n", + " print\"\\n Case (b) Fission is possible since m2=\",m2\n", + "else:\n", + " print\"\\n Case (b) Fission is not possible since m2=\",m2\n", + "\n", + " \n", + "if m3>0:\n", + " print\"\\n Case (c) Fission is possible since m3=\",m3\n", + "else:\n", + "\tprint\"\\n Case (c) Fission is not possible since m3=\",m3" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 2.3;pg no:21" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 2.3, Page:21 \n", + " \n", + "\n", + "\n", + " Nuclear spin of the nucleus is 2.5\n", + " (-) \n", + "\n", + " Magnetic moment is = in nuclear magneton 0.8621\n" + ] + } + ], + "source": [ + "#cal of Nuclear spin and Magnetic moment\n", + "#intiation of all variables\n", + "# Chapter 2\n", + "print\"Example 2.3, Page:21 \\n \\n\"\n", + "# Given:\n", + "l=3.;\n", + "f=3.;\n", + "s=1./2.;\n", + "\n", + "# Solution:\n", + "I=l-s;# total angulr momentum\n", + "P=(-1)**(l);#nuclear parity\n", + "mm=(I-2.293*(I/(I+1)));# in nuclear magneton\n", + "\n", + "print\"\\n Nuclear spin of the nucleus is \",I\n", + "if P>0:\n", + " print\" (+) \"\n", + "else:\n", + " print\" (-) \"\n", + "\n", + "print\"\\n Magnetic moment is = in nuclear magneton\",round(mm,4)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 2.4;pg no:21" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 2.4, Page:21 \n", + " \n", + "\n", + "\n", + " Nuclear spin of the nucleus is 4.5\n", + " (+) \n", + "\n", + " Magnetic moment is =in nuclear magneton 6.793\n" + ] + } + ], + "source": [ + "#cal of Nuclear spin and Magnetic moment\n", + "#intiation of all variables\n", + "# Chapter 2\n", + "print\"Example 2.4, Page:21 \\n \\n\"\n", + "# Given:\n", + "\n", + "l=4.;\n", + "g=4.;\n", + "s=1./2.;\n", + "\n", + "# Solution:\n", + "I=l+s;# total angulr momentum\n", + "P=(-1)**(l);#nuclear parity\n", + "mm=(I+2.293);# for odd proton and I=l+s in nuclear magneton\n", + "\n", + "print\"\\n Nuclear spin of the nucleus is\",I\n", + "if P>0:\n", + " print\" (+) \"\n", + "else:\n", + " print\" (-) \"\n", + "print\"\\n Magnetic moment is =in nuclear magneton\",mm" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 2.5;pg no:22" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 2.5, Page:22 \n", + " \n", + "\n", + " The odd nucleon is 55 proton in level 2 d 5/2 (+). Hence spin is 5/2 and parity (+)\n", + "\n", + " Magnetic moment is =in nuclear magneton 4.793\n", + "\n", + " \n", + " The odd nucleon is 47 neutron in level 1 g 9/2 (+). Hence spin is 9/2 and parity (+)\n", + "Magnetic moment for neutron is m=-1.913 in nuclear magneton\n" + ] + } + ], + "source": [ + "#cal of Magnetic moment\n", + "#intiation of all variables\n", + "# Chapter 2\n", + "print\"Example 2.5, Page:22 \\n \\n\"\n", + "# Given:(a) Cs(55) (b) Kr(36)\n", + "\n", + "# Solution: Part(a)\n", + "print\" The odd nucleon is 55 proton in level 2 d 5/2 (+). Hence spin is 5/2 and parity (+)\"\n", + "mm=((5./2.)+2.293);\n", + "print\"\\n Magnetic moment is =in nuclear magneton\",mm\n", + "# Solution: Part(b)\n", + "print\"\\n \\n The odd nucleon is 47 neutron in level 1 g 9/2 (+). Hence spin is 9/2 and parity (+)\"\n", + "print\"Magnetic moment for neutron is m=-1.913 in nuclear magneton\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 2.7;pg no:22" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 2.7, Page:22 \n", + " \n", + "\n", + "The nuclear g factor for P is = 2.262\n" + ] + } + ], + "source": [ + "#cal of The nuclear g factor for P\n", + "#intiation of all variables\n", + "# Chapter 2\n", + "print\"Example 2.7, Page:22 \\n \\n\"\n", + "# Given:\n", + "h=6.6262*10**-34;# in J.s\n", + "f=17.24*10**6;# in Hz/T\n", + "m=5.05*10**-27;# in J/T\n", + "# Solution:\n", + "\n", + "E=h*f;\n", + "g=E/(m)\n", + "print\"The nuclear g factor for P is =\",round(g,3)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 2.8;pg no:23" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 2.8, Page:23 \n", + " \n", + "\n", + "The NMR frequency in MHz 10.71\n" + ] + } + ], + "source": [ + "#cal of The NMR frequency\n", + "#intiation of all variables\n", + "# Chapter 2\n", + "print\"Example 2.8, Page:23 \\n \\n\"\n", + "# Given:\n", + "\n", + "h=6.6262*10**-34;# in J.s\n", + "f=17.24*10**6;# in Hz/T\n", + "m=5.05*10**-27;# in J/T\n", + "g=1.405;\n", + "\n", + "# Solution:\n", + "\n", + "E=g*m;\n", + "f=E/(h*10**6);# NMR frequency\n", + "\n", + "print\"The NMR frequency in MHz\",round(f,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 2.9;pg no:23" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 2.9, Page:23 \n", + "Magnetic field required for a proton in T= 0.711\n", + "Magnetic field required for C 13 in T= 2.826\n" + ] + } + ], + "source": [ + "#cal of Magnetic field required\n", + "#intiation of all variables\n", + "# Chapter 2\n", + "print\"Example 2.9, Page:23 \"\n", + "# Given:\n", + "\n", + "h=6.6262*10**-34;# in J.s\n", + "f=30.256*10**6;# in Hz/T\n", + "m=5.05*10**-27;# in J/T\n", + "g1=5.585;\n", + "g2=1.405;\n", + "\n", + "# Solution:\n", + "H1=(h*f)/(g1*m);\n", + "H2=(h*f)/(g2*m);\n", + "print\"Magnetic field required for a proton in T=\",round(H1,3)\n", + "print\"Magnetic field required for C 13 in T=\",round(H2,3)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 2.10;pg no:23" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 2.10, Page:23\n", + "Magnetic field required for a proton in T= 0.3316\n" + ] + } + ], + "source": [ + "#cal of Magnetic field required\n", + "#intiation of all variables\n", + "# Chapter 2\n", + "print\"Example 2.10, Page:23\"\n", + "# Given:\n", + "\n", + "h=6.62*10**-34;# in J.s\n", + "f=9.302*10**9;# in Hz/T\n", + "m=9.2741*10**-24;# in J/T\n", + "g1=2.0025;\n", + "\n", + "\n", + "# Solution:\n", + "\n", + "H1=(h*f)/(g1*m);\n", + "print\"Magnetic field required for a proton in T=\",round(H1,4)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 2.11;pg no:24" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 2.11, Page:24 \n", + " \n", + "\n", + "The frequency needed to bring in resonance in GHz= 33.66\n" + ] + } + ], + "source": [ + "#cal of frequency needed\n", + "#intiation of all variables\n", + "# Chapter 2\n", + "print\"Example 2.11, Page:24 \\n \\n\"\n", + "# Given:\n", + "mf=1.201;# In T\n", + "h=6.6262*10**-34;# in J.s\n", + "m=9.2741*10**-24;# in J/T\n", + "g=2.0025;\n", + "# Solution:\n", + "v=(g*m*mf)/(h*10**9);\n", + "print\"The frequency needed to bring in resonance in GHz=\",round(v,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 2.12;pg no:24" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 2.12, Page:24 \n", + " \n", + "\n", + "\n", + " The frequency needed to bring proton spin resonance is=MHz 63.86\n", + "\n", + " The frequency needed to bring electron spin resonance in GHz 4.2\n" + ] + } + ], + "source": [ + "#cal of The frequency needed to bring proton and electron\n", + "#intiation of all variables\n", + "# Chapter 2\n", + "print\"Example 2.12, Page:24 \\n \\n\"\n", + "# Given:\n", + "mf=1.5;# In T\n", + "h=6.6262*10**-34;# in J.s\n", + "mb=9.2741*10**-24;# in J/T\n", + "mn=5.0504*10**-27;#in J/T\n", + "ge=2.002;\n", + "gp=5.5854;\n", + "# Solution: Part(a)\n", + "v1=(gp*mn*mf)/(h*10**6);\n", + "print\"\\n The frequency needed to bring proton spin resonance is=MHz\",round(v1,2)\n", + "# Solution: Part(b)\n", + "v2=(ge*mb*mf)/(h*10**10);\n", + "print\"\\n The frequency needed to bring electron spin resonance in GHz\",round(v2,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 2.13;pg no:24" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 2.13, Page:24 \n", + " \n", + "\n", + "\n", + " Magnetic field required for causing resonance in T= 1.004\n" + ] + } + ], + "source": [ + "#cal of Magnetic field required\n", + "#intiation of all variables\n", + "# Chapter 2\n", + "print\"Example 2.13, Page:24 \\n \\n\"\n", + "# Given:\n", + "h=6.6262*10**-34;# in J.s\n", + "f=40.2*10**6;# in Hz/T\n", + "m=5.05*10**-27;# in J/T\n", + "g1=5.256;\n", + "\n", + "\n", + "# Solution:\n", + "H1=(h*f)/(g1*m);\n", + "print\"\\n Magnetic field required for causing resonance in T=\",round(H1,3)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 2.14;pg no:25" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 2.14, Page: 25 \n", + " \n", + "\n", + "\n", + " The ratio of NMR frequencies of B/N is = 1.35\n" + ] + } + ], + "source": [ + "#cal of ratio of NMR frequencies\n", + "#intiation of all variables\n", + "# Chapter 2\n", + "print\"Example 2.14, Page: 25 \\n \\n\"\n", + "#Given:\n", + "mf=1.0;# In T\n", + "h=6.6262*10**-34;# in J.s\n", + "mn=5.0504*10**-27;#in J/T\n", + "gB=5.4;\n", + "gN=4.01;\n", + "# Solution:\n", + "v1=(gB*mn*mf)/(h*10**6);\n", + "\n", + "v2=(gN*mn*mf)/(h*10**6);\n", + "\n", + "print\"\\n The ratio of NMR frequencies of B/N is =\",round(v1/v2,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 2.15;pg no:25" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 2.15, Page:25 \n", + " \n", + "\n", + "The recoil energy in (meV)= 1.95\n" + ] + } + ], + "source": [ + "#cal of recoil energy\n", + "#intiation of all variables\n", + "# Chapter 2\n", + "print\"Example 2.15, Page:25 \\n \\n\"\n", + "# Given:\n", + "E=14.4*10**-3;# in MeV\n", + "m=57;\n", + "# Solution:\n", + "Er=(536*(E)**2)/(m*10**-3);\n", + "print\"The recoil energy in (meV)=\",round(Er,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 2.16;pg no:25" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 2.16, Page:25 \n", + " \n", + "\n", + "The energy emtted by the nucleus in (KeV)= 23.8\n" + ] + } + ], + "source": [ + "#cal of energy emtted by the nucleus\n", + "#intiation of all variables\n", + "# Chapter 2\n", + "print\"Example 2.16, Page:25 \\n \\n\"\n", + "import math\n", + "# Given:\n", + "Er=2.551;# in meV\n", + "m=119;# atomic wt of Sn\n", + "\n", + "# Solution:\n", + "E=math.sqrt(2.551*10**-3*119/536);# energy emitted by nucleus\n", + "print\"The energy emtted by the nucleus in (KeV)=\",round(E*10**3,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 2.17;pg no:26" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 2.17, Page:26 \n", + " \n", + "\n", + "The doppler shift frequency in 10^11 Hz= 3.97\n" + ] + } + ], + "source": [ + "#cal of doppler shift frequency\n", + "#intiation of all variables\n", + "# Chapter 2\n", + "print\"Example 2.17, Page:26 \\n \\n\"\n", + "# Given:\n", + "l=10**-10;# in m\n", + "m=100;# in u\n", + "h=6.6262*10**-34;# in J.s\n", + "\n", + "\n", + "# Solution:\n", + "v=h/(m*l*1.67*10**-27);# velocity\n", + "f=v/l;# frequency\n", + "\n", + "print\"The doppler shift frequency in 10^11 Hz=\",round(f/10**11,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 2.18;pg no:26" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 2.18, Page:26 \n", + " \n", + "\n", + "The frequency in 10**8 cm^-1 is = 1.161\n", + "The frequency in 10^12 MHz is = 3.48\n" + ] + } + ], + "source": [ + "#cal of frequency\n", + "#intiation of all variables\n", + "# Chapter 2\n", + "print\"Example 2.18, Page:26 \\n \\n\"\n", + "#Given:\n", + "# 1 ev=8065 cm^-1\n", + "E=14.4*10**3;# in eV\n", + "# Solution:\n", + "f1=E*8065;# frequency in cm^-1\n", + "print\"The frequency in 10**8 cm^-1 is =\",round(f1/10**8,3)\n", + "fr=f1*3*10**8*100;\n", + "print\"The frequency in 10^12 MHz is =\",round(fr/10**18,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 2.19;pg no:26" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 2.19, Page:26 \n", + " \n", + "\n", + "The shift in frequency between the source and the sample in MHz= 25.55\n" + ] + } + ], + "source": [ + "#cal of shift in frequency between the source and the sample\n", + "#intiation of all variables\n", + "# Chapter 2\n", + "print\"Example 2.19, Page:26 \\n \\n\"\n", + "# Given:\n", + "#Given:\n", + "# 1 ev=8065 cm^-1\n", + "E=14.4*10**3;# in eV\n", + "v1=2.2*10**-3;# in m/s\n", + "# Solution:\n", + "f1=E*8065;# frequency in cm^-1\n", + "fr=f1*3*10**8*100;\n", + "fr1=(fr*v1)/(3*10**8);\n", + "print\"The shift in frequency between the source and the sample in MHz=\",round(fr1/10**6,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 2.20;pg no:27" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 2.20, Page:27 \n", + " \n", + "\n", + "The recoil velocity of free atom is =(m/s) 84.21\n", + "The recoil velocity of atom that is part of crystal in 10^-20 (m/s) 7.68\n", + "The doppler shift for free atom in 10^10 Hz= 97.6\n", + "The doppler shift of atom that is part of crystal in 10^-10 Hz is = 8.9\n" + ] + } + ], + "source": [ + "#cal of recoil velocity and doppler shift for an atom\n", + "#intiation of all variables\n", + "# Chapter 2\n", + "print\"Example 2.20, Page:27 \\n \\n\"\n", + "# Given:\n", + "E=1.6*14.4*10**3*10**-19;# energy in J\n", + "c=3*10**8;# in m/s\n", + "m=57*1.6*10**-27;\n", + "M=10**-4;\n", + "h=6.6262*10**-34;# in J.s\n", + "# Solution:\n", + "p=E/c;\n", + "v=p/m;\n", + "v1=(v*m)/(M);\n", + "v2=(v*m)/(M*10**-20);\n", + "f1=(E*v)/(h*c);\n", + "f2=(E*v1)/(h*c*10**-10);\n", + "print\"The recoil velocity of free atom is =(m/s)\",round(v,2)\n", + "print\"The recoil velocity of atom that is part of crystal in 10^-20 (m/s)\",round(v2,2)\n", + "print\"The doppler shift for free atom in 10^10 Hz=\",round(f1/10**10,2)\n", + "print\"The doppler shift of atom that is part of crystal in 10^-10 Hz is =\",round(f2,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 2.21;pg no:28" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 2.21, Page:28 \n", + " \n", + "\n", + "Ellipticity is = 0.17\n", + "b/a is = 1.185\n" + ] + } + ], + "source": [ + "#cal of Ellipticity\n", + "#intiation of all variables\n", + "# Chapter 2\n", + "print\"Example 2.21, Page:28 \\n \\n\"\n", + "# Given:\n", + "A=175.;\n", + "R=1.4*10**-15*((A)**(1./3.));\n", + "# Soluiton:\n", + "#Part a\n", + "sqrBMinusSqrA = (5.9 * (10** (-28))) * 5. /(2. * 71.);\n", + "BMinusA = sqrBMinusSqrA / (2. * R);\n", + "ellipticity = 2 * (BMinusA) / (2. * R);\n", + "print\"Ellipticity is =\",round(ellipticity,2)\n", + "#Part B\n", + "b = (BMinusA + (2 * R)) /2;\n", + "a = (-BMinusA + (2 * R)) /2;\n", + "print\"b/a is =\",round(b/a,3)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 2.22;pg no:28" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 2.22, Page: \n", + " \n", + "\n", + "Ellipticity is = 0.2\n", + "b/a is = 1.222\n", + "Ellipticity is = -0.06\n", + "b/a is = 0.946\n" + ] + } + ], + "source": [ + "#cal of Ellipticity\n", + "#intiation of all variables\n", + "# Chapter 2\n", + "print\"Example 2.22, Page: \\n \\n\"\n", + "# Given:\n", + "A1=176.;\n", + "A2=233.;\n", + "R1=1.4*10**-15*((A1)**(1./3.));\n", + "R2=1.4*10**-15*((A2)**(1./3.));\n", + "# Soluiton:\n", + "#Part a\n", + "sqrBMinusSqrA = (5. * 7. * (10** (-28))) /(2. * 71.);\n", + "BMinusA = sqrBMinusSqrA / (2. * R1);\n", + "ellipticity = 2. * (BMinusA) / (2. * R1);\n", + "print\"Ellipticity is =\",round(ellipticity,2)\n", + "b = (BMinusA + (2. * R1)) /2.;\n", + "a = (-BMinusA + (2. * R1)) /2.;\n", + "print\"b/a is =\",round(b/a,3)\n", + "#Part B\n", + "sqrBMinusSqrA = -(5. * 3. * (10** (-28))) /(2. * 91.);\n", + "BMinusA = sqrBMinusSqrA / (2. * R2);\n", + "ellipticity = 2. * (BMinusA) / (2. * R2);\n", + "print\"Ellipticity is =\",round(ellipticity,2)\n", + "b = (BMinusA + (2. * R2)) /2.;\n", + "a = (-BMinusA + (2. * R2)) /2.;\n", + "print\"b/a is =\",round(b/a,3)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 2.23;pg no:29" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 2.23, Page:29 \n", + " \n", + "\n", + "Quadrapole moment in barns= 0.2761\n" + ] + } + ], + "source": [ + "#cal of Quadrapole moment\n", + "#intiation of all variables\n", + "# Chapter 2\n", + "print\"Example 2.23, Page:29 \\n \\n\"\n", + "# Given:\n", + "e = 0.03;\n", + "A=75.;\n", + "R=1.4*10**-15*((A)**(1./3.));\n", + "# Soluiton:\n", + "BPlusA = 2. * R;\n", + "BMinusA = e * R;\n", + "sqrBMinusSqrA = BPlusA * BMinusA;\n", + "BMinusA = sqrBMinusSqrA / (2. * R);\n", + "Q = (2.*33.*sqrBMinusSqrA)/5.;\n", + "q1=Q/10**-28;\n", + "print\"Quadrapole moment in barns=\",round(q1,4)" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/_Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/chapter3.ipynb b/_Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/chapter3.ipynb new file mode 100644 index 00000000..761cfb0c --- /dev/null +++ b/_Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/chapter3.ipynb @@ -0,0 +1,787 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "#chapter 3:nuclear models" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 3.1;pg no:38" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 3.1, Page:38 \n", + " \n", + "\n", + "The coulomb barrier for the penetration of Th by proton in MeV= 12.96\n", + "The coulomb barrier for the penetration of Th by alpha particle in MeV= 23.94\n" + ] + } + ], + "source": [ + "#cal of coulomb barrier for the penetration of Th by alpha particle and proton\n", + "#intiation of all variables\n", + "# Chapter 3\n", + "print\"Example 3.1, Page:38 \\n \\n\"\n", + "#Given in cgs units\n", + "m1=232.;\n", + "m2=1.;\n", + "m3=4.;\n", + "z1=90.;\n", + "z2=1.;\n", + "z3=2.;\n", + "e=4.8*10**-10;# in ergs\n", + "c=1.4;# nuclear radius constant\n", + "\n", + "#Formula: E=(z1*z2*e^2)/(r1+r2)\n", + "r1=(m1)**(1./3.);\n", + "r2=(m2)**(1./3.);\n", + "r3=(m3)**(1./3.);\n", + "E1=(z1*z2*e*e)/(c*(r1+r2)*10**-13*(1.6*10**-6));\n", + "print\"The coulomb barrier for the penetration of Th by proton in MeV=\",round(E1,2)\n", + "E2=(z1*z3*e*e)/(c*(r1+r3)*10**-13*(1.6*10**-6));\n", + "print\"The coulomb barrier for the penetration of Th by alpha particle in MeV=\",round(E2,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 3.2;pg no:38" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 3.2, Page38 \n", + " \n", + "\n", + "\n", + " The coulomb barrier for the penetration of Th by proton in MeV= 8.84\n", + "\n", + " \n", + " The coulomb barrier for the penetration of Th by alpha particle in MeV= 10.96\n" + ] + } + ], + "source": [ + "#cal of coulomb barrier for the penetration of Th by alpha particle and proton\n", + "#intiation of all variables\n", + "# Chapter 3\n", + "print\"Example 3.2, Page38 \\n \\n\"\n", + "#Given in cgs units\n", + "m1=112;\n", + "m2=1;\n", + "m3=4;\n", + "m4=66;\n", + "z1=50;\n", + "z2=1;\n", + "z3=2;\n", + "z4=30;\n", + "e=4.8*10**-10;# in ergs\n", + "c=1.4;# nuclear radius constant\n", + "\n", + "#Formula: E=(z1*z2*e^2)/(r1+r2)\n", + "r1=(m1)**(1./3.);\n", + "r2=(m2)**(1./3.);\n", + "r3=(m3)**(1./3.);\n", + "r4=(m4)**(1./3.);\n", + "E1=(z1*z2*e*e)/(c*(r1+r2)*10**-13*(1.6*10**-6));\n", + "print\"\\n The coulomb barrier for the penetration of Th by proton in MeV=\",round(E1,2)\n", + "E2=(z4*z3*e*e)/(c*(r4+r3)*10**-13*(1.6*10**-6));\n", + "print\"\\n \\n The coulomb barrier for the penetration of Th by alpha particle in MeV=\",round(E2,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 3.3;pg no:39" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 3.3, Page:39 \n", + " \n", + "\n", + "The closest distance of approach in fm= 37.9\n" + ] + } + ], + "source": [ + "#cal of closest distance of approach\n", + "#intiation of all variables\n", + "# Chapter 3\n", + "print\"Example 3.3, Page:39 \\n \\n\"\n", + "# Given:\n", + "E=6;# in MeV\n", + "z1=79;\n", + "z2=2;\n", + "q=4.8*10**-10;\n", + "# Solution:\n", + "\n", + "# At the closest distance of approach, the kineic energy of the alpha particle balances the columb barrier energy.\n", + "\n", + "r1=(z1*z2*q*q)/(E*1.6*10**-6);# distance in cm\n", + "r=r1*10**13;# distance in fm\n", + "\n", + "print\"The closest distance of approach in fm=\",round(r,1)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 3.4;pg no:39" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 3.4, Page:39 \n", + " \n", + "\n", + "The stable nuclied of the isobaric series is Hf atomic no. = 72.6\n" + ] + } + ], + "source": [ + "#cal of stable nuclied of the isobaric series\n", + "#intiation of all variables\n", + "# Chapter 3\n", + "print\"Example 3.4, Page:39 \\n \\n\"\n", + "# Given:\n", + "A=180.;\n", + "# Solution:\n", + "z=(40.*A)/(0.6*(A**(2./3.))+80.);\n", + "print\"The stable nuclied of the isobaric series is Hf atomic no. =\",round(z,1)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 3.5;pg no:39" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 3.5, Page:39 \n", + " \n", + "\n", + "The stable nuclied of the isobaric series is Sr atomic no. = 38.0\n", + "\n", + " Hence the nuclides of z<38 fall on the left of the limb of B vs Z parabola while the nuclides of z>38 fall on the right limb of the parabola.\n" + ] + } + ], + "source": [ + "#cal of stable nuclied of the isobaric series is Sr atomic no\n", + "#intiation of all variables\n", + "# Chapter 3\n", + "print\"Example 3.5, Page:39 \\n \\n\"\n", + "# Given:\n", + "A=87.;\n", + "\n", + "# Solution:\n", + "z=(40.*A)/(0.6*(A**(2./3.))+80.);\n", + "print\"The stable nuclied of the isobaric series is Sr atomic no. =\",round(z)\n", + "# nereast integer is 38\n", + "print\"\\n Hence the nuclides of z<38 fall on the left of the limb of B vs Z parabola while the nuclides of z>38 fall on the right limb of the parabola.\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 3.7;pg no:40" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 3.7, Page:40 \n", + " \n", + "\n", + "\n", + " The binding energy for the last proton in 12C in (MeV)= 15.92\n", + "\n", + " The binding energy for the last neutron in 12C in (MeV)= 9.43\n", + "\n", + " The binding energy for the last proton in 28Si in (MeV)= 11.55\n", + "\n", + " The binding energy for the last neutron in 28Si in (MeV)= 17.2\n" + ] + } + ], + "source": [ + "#cal of binding energy for the last proton,neutron\n", + "#intiation of all variables\n", + "# Chapter 3\n", + "print\"Example 3.7, Page:40 \\n \\n\"\n", + "# Given:\n", + "B=11.009305;\n", + "C1=12;\n", + "C2=11.001433;\n", + "p=1.0078;\n", + "n=1.0087;\n", + "Al=26.981535;\n", + "Si1=27.976927;\n", + "Si2=26.986705;\n", + "# Solution:\n", + "m1=(B+p-C1);#(a)\n", + "E1=m1*931;# of last proton in C in MeV\n", + "print\"\\n The binding energy for the last proton in 12C in (MeV)=\",round(E1,2)\n", + "\n", + "m2=(C2+n-C1);#(b)\n", + "E2=m2*931;# of last neutron in C in MeV\n", + "print\"\\n The binding energy for the last neutron in 12C in (MeV)=\",round(E2,2)\n", + "\n", + "m3=(Al+p-Si1);#(c)\n", + "E3=m3*931;# of last proton in Si in MeV\n", + "print\"\\n The binding energy for the last proton in 28Si in (MeV)=\",round(E3,2)\n", + "\n", + "m4=(Si2+n-Si1);#(d)\n", + "E4=m4*931;# of last neutron in Si in MeV\n", + "print\"\\n The binding energy for the last neutron in 28Si in (MeV)=\",round(E4,2)\n", + "\n", + "# Note: There is a calculation error in the textbook for the (b) part." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 3.8;pg no:41" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 3.8, Page: 41 \n", + " \n", + "\n", + "\n", + " The binding energy for N(14) in (MeV)= 10.27\n", + "\n", + " The binding energy for O(16) in (MeV)= 7.16\n" + ] + } + ], + "source": [ + "#cal of binding energy for N,O\n", + "#intiation of all variables\n", + "# Chapter 3\n", + "print\"Example 3.8, Page: 41 \\n \\n\"\n", + "# Given:\n", + "d=2.014102;\n", + "C=12;\n", + "a=4.002603;\n", + "N=14.003074;\n", + "O=15.994915;\n", + "\n", + "# Solution:\n", + "m1=(C+d-N);\n", + "E1=m1*931;# The binding energy for N(14)\n", + "print\"\\n The binding energy for N(14) in (MeV)=\",round(E1,2)\n", + "\n", + "m2=(C+a-O);\n", + "E2=m2*931;#The binding energy for O(16) \n", + "print\"\\n The binding energy for O(16) in (MeV)=\",round(E2,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 3.9;pg no:41" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 3.9, Page:41 \n", + " \n", + "\n", + "\n", + " The binding energy in (MeV)= 18.95\n" + ] + } + ], + "source": [ + "#cal of binding energy\n", + "#intiation of all variables\n", + "# Chapter 3\n", + "print\"Example 3.9, Page:41 \\n \\n\"\n", + "# Given:\n", + "D=-1.997042;\n", + "n=1.0087;\n", + "# Solution:\n", + "m=(D+2.*n);\n", + "E=m*931.;\n", + "print\"\\n The binding energy in (MeV)=\",round(E,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 3.10;pg no:41" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 3.10, Page:41 \n", + " \n", + "\n", + "\n", + " The neutron seperation energy in (MeV)= 7.3596\n", + "\n", + " The proton seperation energy in (MeV)= 7.9647\n" + ] + } + ], + "source": [ + "#cal of seperation energy\n", + "#intiation of all variables\n", + "# Chapter 3\n", + "print\"Example 3.10, Page:41 \\n \\n\"\n", + "# Given:\n", + "mH=1.007825;\n", + "mn=1.008665;\n", + "M1=207.97666;# mass of Pb 208\n", + "M2=206.97590;# mass of Pb 207\n", + "M3=206.97739;# mass of Tl 207\n", + "\n", + "# Solution:\n", + "\n", + "B1=((82*1.007825+126*1.008665)-207.97666)*931;# binding energy for Pb 208\n", + "B2=((82*1.007825+125*1.008665)-206.97590)*931;# binding energy for Pb 207\n", + "B3=((81*1.007825+126*1.008665)-206.97739)*931;# binding energy for Tl 207\n", + "Sn=B1-B2;# neutron seperation energy\n", + "Sp=B1-B3;# proton seperation energy\n", + "\n", + "print\"\\n The neutron seperation energy in (MeV)=\",round(Sn,4)\n", + "print\"\\n The proton seperation energy in (MeV)=\",round(Sp,4)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 3.11;pg no:42" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 3.11, Page:42 \n", + " \n", + "\n", + " The neutron seperation energy in (MeV)= 12.41\n", + " The proton seperation energy in (MeV)= 8.79\n" + ] + } + ], + "source": [ + "#cal of seperation energy\n", + "#intiation of all variables\n", + "# Chapter 3\n", + "print\"Example 3.11, Page:42 \\n \\n\"\n", + "# Given:\n", + "mH=1.007825;\n", + "mn=1.008665;\n", + "M1=22.98977;# mass of Na 23\n", + "M2=21.994435;# mass of Na 22\n", + "M3=21.991385;# mass of Ne 22\n", + "# Solution:\n", + "\n", + "m1=((11.*1.007825+12.*1.008665)-M1);\n", + "m2=((11.*1.007825+11.*1.008665)-M2);\n", + "m3=((10.*1.007825+12.*1.008665)-M3);\n", + "Sn=(m1-m2)*931.;# neutron seperation energy\n", + "Sp=(m1-m3)*931.;# proton seperation energy\n", + "\n", + "print\" The neutron seperation energy in (MeV)=\",round(Sn,2)\n", + "print\" The proton seperation energy in (MeV)=\",round(Sp,2)\n", + "\n", + "# Note: The answers are given in the form of atomic mass units where as in the question its asked for energies." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 3.12;pg no:43" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 3.12, Page:43 \n", + " \n", + "\n", + "\n", + " The excited level density in (MeV)= 894.29\n", + "\n", + " The level spacing in (keV)= 1.12\n", + "\n", + " The nuclear temperature in (MeV)= 2.0\n" + ] + } + ], + "source": [ + "#cal of excited level density,level spacing and nuclear temperature\n", + "#intiation of all variables\n", + "# Chapter 3\n", + "print\"Example 3.12, Page:43 \\n \\n\"\n", + "# Given:\n", + "C=0.3;# in MeV^-1\n", + "a=2.0;# in MeV\n", + "E=8; # in MeV\n", + "import math\n", + "# Solution:\n", + "d=C*(math.exp(2*((2*8)**(0.5))));# excited level density\n", + "s=(1/d)*1000;# level spacing\n", + "nT=(E/a)**(0.5);# nuclear temperature\n", + "print\"\\n The excited level density in (MeV)=\",round(d,2)\n", + "print\"\\n The level spacing in (keV)=\",round(s,2)\n", + "print\"\\n The nuclear temperature in (MeV)=\",round(nT,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 3.13;pg no:43" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 3.13, Page:43 \n", + " \n", + "\n", + "The expression for 1st, 2nd, and 3rd excited states are K times respectively. 5.0 12.0 21.0\n" + ] + } + ], + "source": [ + "#cal of expression for 1st, 2nd, and 3rd excited states\n", + "#intiation of all variables\n", + "# Chapter 3\n", + "print\"Example 3.13, Page:43 \\n \\n\"\n", + "# Given:\n", + "I0=3./2.;# ground state of spin\n", + "\n", + "# Solution:\n", + "I1=I0+1.;\n", + "I2=I0+2.;\n", + "I3=I0+3.;\n", + "K=1.;# Assumed as some constant\n", + "# Formula: E=(h^2/(2*I))*((I*(I+1))-I0*(I0+1))\n", + "# Consider K=(h^2/(2*I))=1\n", + "\n", + "E1=K*((I1*(I1+1.))-(I0*(I0+1.)));# For 1 excited state\n", + "\n", + "E2=K*((I2*(I2+1.))-(I0*(I0+1.)));# For 2 excited state\n", + "\n", + "E3=K*((I3*(I3+1.))-(I0*(I0+1.)));# For 3 excited state\n", + "\n", + "print\"The expression for 1st, 2nd, and 3rd excited states are K times respectively.\",E1,E2,E3" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 3.14;pg no:44" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 3.14, Page:44 \n", + " \n", + "\n", + "\n", + " The energy of state 4 (+) in (keV)= 146.67\n", + "\n", + " The energy of state 6 (+) in (keV)= 308.0\n", + "\n", + " The energy of state 8 (+) in (keV)= 528.0\n", + "\n", + " The energy of state 10 (+) in (keV)= 806.67\n" + ] + } + ], + "source": [ + "#cal of energy of state 4,6,8,10\n", + "#intiation of all variables\n", + "# Chapter 3\n", + "print\"Example 3.14, Page:44 \\n \\n\"\n", + "# Given:\n", + "E2=44;# in keV\n", + "\n", + "# Solution:\n", + "E4=E2*((4.*5.)/(2.*3.));# for part (a)\n", + "E6=E2*((6.*7.)/(2.*3.));# for part (b)\n", + "E8=E2*((8.*9.)/(2.*3.));# for part (c)\n", + "E10=E2*((10.*11.)/(2.*3.));# for part (d)\n", + "\n", + "print\"\\n The energy of state 4 (+) in (keV)=\",round(E4,2)\n", + "print\"\\n The energy of state 6 (+) in (keV)=\",round(E6,2)\n", + "print\"\\n The energy of state 8 (+) in (keV)=\",round(E8,2)\n", + "print\"\\n The energy of state 10 (+) in (keV)=\",round(E10,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 3.15;pg no:44" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 3.15, Page:44 \n", + " \n", + "\n", + "n= 11\n", + "\n", + " For the required level of energy 525 keV nearest even integer is = & spin is (+) 12.0\n" + ] + } + ], + "source": [ + "#cal of energy\n", + "#intiation of all variables\n", + "# Chapter 3\n", + "print\"Example 3.15, Page:44 \\n \\n\"\n", + "# Given:\n", + "E2=44;# in keV\n", + "En=525;# in keV\n", + "\n", + "# Solution:\n", + "n=(En)/E2;\n", + "# \n", + "print\"n=\",n\n", + "print\"\\n For the required level of energy 525 keV nearest even integer is = & spin is (+)\",round(n+1,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 3.16;pg no:45" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 3.16, Page:45 \n", + " \n", + "\n", + "n1= 3.31818181818\n", + "\n", + " For the required level of energy 146 keV nearest even integer is =& spin is (+) 4.32\n", + "\n", + " \n", + " n2= 6.90909090909\n", + "\n", + " For the required level of energy 304 keV nearest even integer is =& spin is (+) 6.91\n", + "\n", + " \n", + " n3= 11.6818181818\n", + "\n", + " For the required level of energy 514 keV nearest even integer is =& spin is (+) 12.68\n" + ] + } + ], + "source": [ + "#cal of energy\n", + "#intiation of all variables\n", + "# Chapter 3\n", + "print\"Example 3.16, Page:45 \\n \\n\"\n", + "# Given:\n", + "E2=44.;# in keV\n", + "En1=146.;# in keV\n", + "En2=304.;# in keV\n", + "En3=514.;# in keV\n", + "# Solution:\n", + "n1=(En1)/E2;\n", + "n2=(En2)/E2;\n", + "n3=(En3)/E2;\n", + "print\"n1=\",n1\n", + "print\"\\n For the required level of energy 146 keV nearest even integer is =& spin is (+)\",round(n1+1.,2)\n", + "print\"\\n \\n n2=\",n2\n", + "print\"\\n For the required level of energy 304 keV nearest even integer is =& spin is (+)\",round(n2,2)\n", + "print\"\\n \\n n3=\",n3\n", + "print\"\\n For the required level of energy 514 keV nearest even integer is =& spin is (+)\",round(n3+1.,2)\n", + "\n", + "#Note: In the last part (c) the answer given in the textbook is 8(+). But the correct answer is 12(+)" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/_Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/chapter4.ipynb b/_Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/chapter4.ipynb new file mode 100644 index 00000000..3f17707e --- /dev/null +++ b/_Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/chapter4.ipynb @@ -0,0 +1,1632 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# chapter 4:radio activity" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 4.1;pg no:55" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.1, Page:55 \n", + " \n", + "\n", + "The value of avagadro constant in 10^23 atoms per mole= 6.05\n" + ] + } + ], + "source": [ + "#cal of value of avagadro constant\n", + "#intiation of all variables\n", + "# Chapter 4\n", + "print\"Example 4.1, Page:55 \\n \\n\"\n", + "#Given:\n", + "t1=1600;# in year\n", + "a=11.6*10**17;# atoms\n", + "# Solution:\n", + "k=0.693/t1;# year^-1\n", + "L=(a*226)/k;# atomic mass of Radon is 226\n", + "print\"The value of avagadro constant in 10^23 atoms per mole=\",round(L/10**23,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 4.2;pg no:55" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.2, Page:55 \n", + " \n", + "\n", + "The specific activity in dis min^-1 g^-1= 1813.0\n" + ] + } + ], + "source": [ + "#cal of specific activity\n", + "#intiation of all variables\n", + "# Chapter 4\n", + "print\"Example 4.2, Page:55 \\n \\n\"\n", + "# Given:\n", + "t1=1.3*10**9;# in years\n", + "w=0.0119;# wt %\n", + "\n", + "# Solution:\n", + "N=(w*6.022*10**23)/(40*100);\n", + "k=(0.693*60)/(t1*3.16*10**7);\n", + "sa=N*k;# specific activity\n", + "print\"The specific activity in dis min^-1 g^-1=\",round(sa)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 4.3;pg no:55" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.3, Page:55 \n", + " \n", + "\n", + " The no. of atoms of Na(24) are = 2.88311688312e+12\n", + " The mass of Na(24) is * 10^-10 g 1.15\n", + "The no. of atoms of P(32) are = 6.59657142857e+13\n", + "The mass of P(32) is * 10^-9 g 3.51\n", + "The no. of atoms of Ra(226) are = 2.69945165945e+18\n", + "The mass of Ra(226) is * 10^-3 g 1.01\n" + ] + } + ], + "source": [ + "#cal of no. of atoms,mass of Na,P,Ra \n", + "#intiation of all variables\n", + "# Chapter 4\n", + "print\"Example 4.3, Page:55 \\n \\n\"\n", + "# Given:\n", + "L=6.022*10**23;\n", + "# Solution:\n", + "# 1 mCi= 3.7*10^7 dis/s\n", + "k1=0.693/(15*3600);\n", + "N1=3.7*10**7/k1;\n", + "m1=(24*N1*10**10)/L;\n", + "print\" The no. of atoms of Na(24) are =\",round(N1,2)\n", + "print\" The mass of Na(24) is * 10^-10 g\",round(m1,2)\n", + "k2=0.693/(14.3*24*3600);\n", + "N2=3.7*10**7/k2;\n", + "m2=(32*N2*10**9)/L;\n", + "print\"The no. of atoms of P(32) are =\",round(N2,2)\n", + "print\"The mass of P(32) is * 10^-9 g\",round(m2,2)\n", + "k3=0.693/(1600*3.16*10**7);\n", + "N3=3.7*10**7/k3;\n", + "m3=(226*N3*10**3)/L;\n", + "print\"The no. of atoms of Ra(226) are =\",round(N3,2)\n", + "print\"The mass of Ra(226) is * 10^-3 g\",round(m3,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 4.4;pg no:56" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.4, Page:56 \n", + " \n", + "\n", + "The activity in Ci/cm^3 = 2.59\n" + ] + } + ], + "source": [ + "#cal of activity\n", + "#intiation of all variables\n", + "# Chapter 4\n", + "print\"Example 4.4, Page:56 \\n \\n\"\n", + "# Given:\n", + "t1=12.3;# in yrs\n", + "L=6.022*10**23;\n", + "# Solution:\n", + "k=.693/(t1*3.16*10**7);# in s^-1\n", + "A=(2*L)/(2.24*10**4);# no. of atoms\n", + "a1=A*k;# dis per s\n", + "a=a1/(3.7*10**10);# activity in Ci/cm^3\n", + "print\"The activity in Ci/cm^3 =\",round(a,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 4.5;pg no:56" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.5, Page:56 \n", + " \n", + "\n", + "The ratio of C14/C12 in atmosphere in 10^-12 is = 1.4\n" + ] + } + ], + "source": [ + "#cal of ratio of C14/C12 in atmosphere\n", + "#intiation of all variables\n", + "# Chapter 4\n", + "print\"Example 4.5, Page:56 \\n \\n\"\n", + "# Given:\n", + "t=5736;# in years\n", + "Nk=16.1;# dis/min\n", + "L=6.022*10**23;\n", + "# Solution:\n", + "k=(0.693*60)/(t*3.16*10**7);\n", + "N1=Nk/k;# atoms per g for C14\n", + "N2=L/12;#\n", + "r=(N1*10**12)/N2;# ratio of C14/C12 in atmosphere\n", + "print\"The ratio of C14/C12 in atmosphere in 10^-12 is =\",round(r,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 4.7;pg no:57" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.7, Page:57 \n", + " \n", + "\n", + "Number of alpha decays = and number of beta decays = 8 6\n" + ] + } + ], + "source": [ + "#cal of Number of alpha decays and number of beta decays\n", + "#intiation of all variables\n", + "# Chapter 4\n", + "print\"Example 4.7, Page:57 \\n \\n\"\n", + "# Given:\n", + "dA = 206-238;\n", + "dA_Beta=0;\n", + "dA_Alpha = -4;\n", + "\n", + "dZ_Alpha = -2;\n", + "dZ_Beta = 1;\n", + "nBeta=0; #random initialisation\n", + "dZ = 82 -92;\n", + "# Solution:\n", + "nAlpha = (dA- (dA_Beta* nBeta))/dA_Alpha;\n", + "\n", + "nBeta = (dZ- (dZ_Alpha * nAlpha))/dZ_Beta;\n", + "\n", + "print\"Number of alpha decays = and number of beta decays = \",nAlpha,nBeta" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 4.8;pg no:57" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.8, Page:57 \n", + " \n", + "\n", + "\n", + " For first photon, dE1= MeV, dI1=, since EL/ML1= & (L+PI+PF) is odd, M3 0.059 -3 5\n", + "\n", + " For second photon, dE2= MeV, dI2=,since EL/ML2= & (L+PI+PF) is even, E2 1.17 2 4\n", + "\n", + " For third photon, dE3= MeV, d3I=, since EL/ML3= & (L+PI+PF) is even, E2 1.33 2 4\n" + ] + } + ], + "source": [ + "#cal of energy\n", + "#intiation of all variables\n", + "# Chapter 4\n", + "print\"Example 4.8, Page:57 \\n \\n\"\n", + "# Given:\n", + "E1=0.059;\n", + "E2=2.5;\n", + "E3=1.33;\n", + "Ei=0;\n", + "Ef=0;\n", + "\n", + "# Solution:\n", + "# delta E for 1,2 & 3 photon \n", + "dE1=E1-Ei;\n", + "dE2=E2-E3;\n", + "dE3=E3-Ef;\n", + "# delta I for 1,2 & 3 photon \n", + "dI1=2-5;\n", + "dI2=4-2;\n", + "dI3=2-0;\n", + "# EL/ML for 1,2 & 3 photon \n", + "ELML1=3+1+1;\n", + "ELML2=2+1+1;\n", + "ELML3=2+1+1;\n", + "print\"\\n For first photon, dE1= MeV, dI1=, since EL/ML1= & (L+PI+PF) is odd, M3\",dE1,dI1,ELML1\n", + "print\"\\n For second photon, dE2= MeV, dI2=,since EL/ML2= & (L+PI+PF) is even, E2\",dE2,dI2,ELML2\n", + "print\"\\n For third photon, dE3= MeV, d3I=, since EL/ML3= & (L+PI+PF) is even, E2\",dE3,dI3,ELML3" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 4.9;pg no:58" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.9, Page:58 \n", + " \n", + "\n", + "The energy that would be released for 0.1 mole of Co will be =(GJ) 24.08\n" + ] + } + ], + "source": [ + "#cal of energy\n", + "#intiation of all variables\n", + "# Chapter 4\n", + "print\"Example 4.9, Page:58 \\n \\n\"\n", + "# Given:\n", + "E=2.5; # in MeV\n", + "# Solution:\n", + "# 1 Mev/atom=96.32GJ/mole\n", + "E1=E*96.32# GJ/mole\n", + "E2=0.1*E1;# for 0.1 mole\n", + "print\"The energy that would be released for 0.1 mole of Co will be =(GJ)\",E2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 4.10;pg no:58" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.10, Page: \n", + " \n", + "\n", + "The total energy dissipate per hour is = (KJ) 360.860029304\n" + ] + } + ], + "source": [ + "#cal of total energy dissipated\n", + "#intiation of all variables\n", + "# Chapter 4\n", + "print\"Example 4.10, Page: \\n \\n\"\n", + "# Given\n", + "E=2.5;# in MeV\n", + "# Solution:\n", + "k=0.693/(5.27*3.16*10**7);# decay constant\n", + "A=k*0.1*6.022*10**23;# atoms/s\n", + "A1=3.6*10**3*A;# atoms /hr\n", + "\n", + "E1=A1*E*1.6*10**-13*10**-3;#Energy in KJ/hr\n", + "\n", + "print\"The total energy dissipate per hour is = (KJ)\",E1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 4.11;pg no:59" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.11, Page:59 \n", + " \n", + "\n", + "The net mass loss is = u 0.0044\n" + ] + } + ], + "source": [ + "#cal of net mass loss\n", + "#intiation of all variables\n", + "# Chapter 4\n", + "print\"Example 4.11, Page:59 \\n \\n\"\n", + "# Given:\n", + "Ma=4.;#mass of alpha particle\n", + "Mr=228.;# mass of Th\n", + "Ea=4.; #in MeV\n", + "\n", + "\n", + "# Solution:\n", + "Er=(Ma/Mr)*Ea;# energy of recoil\n", + "Et=Ea+Er;# total energy of transition\n", + "\n", + "dM=Et/931.;# net mass loss in u\n", + "\n", + "print\"The net mass loss is = u\",round(dM,4)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 4.12;pg no:59" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.12, Page:59 \n", + " \n", + "\n", + "For Ra emitting alpha\n", + "\tEnergy of recoil is (MeV) 0.0876\n", + "\tTotal transition energy is (MeV) 4.9506\n", + "For Bi emitting alpha\n", + "\tEnergy of recoil is (MeV) 0.117\n", + "\tTotal transition energy is (MeV) 6.199\n" + ] + } + ], + "source": [ + "#cal of value of avagadro constant\n", + "#intiation of all variables\n", + "# Chapter 4\n", + "print\"Example 4.12, Page:59 \\n \\n\"\n", + "# Given:\n", + "Ma=4.;#mass of alpha particle\n", + "Mr1=222.;# mass of \n", + "Mr2=208.;\n", + "Ea1=4.863;\n", + "Ea2=6.082;\n", + "# Solution:\n", + "\n", + "Er1=(Ma/Mr1)*Ea1;\n", + "Et1=Ea1+Er1;\n", + "print\"For Ra emitting alpha\"\n", + "print\"\\tEnergy of recoil is (MeV)\",round(Er1,4)\n", + "print\"\\tTotal transition energy is (MeV)\",round(Et1,4)\n", + "Er2=(Ma/Mr2)*Ea2;\n", + "Et2=Ea2+Er2;\n", + "print\"For Bi emitting alpha\"\n", + "print\"\\tEnergy of recoil is (MeV)\",round(Er2,4)\n", + "print\"\\tTotal transition energy is (MeV)\",round(Et2,4)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 4.13;pg no:60" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.13, Page:60 \n", + " \n", + "\n", + "The Kinetic Energy and velocity are MeV and 10^5 m/s respectively 0.0966 2.79\n" + ] + } + ], + "source": [ + "#cal of Kinetic Energy and velocity\n", + "#intiation of all variables\n", + "# Chapter 4\n", + "print\"Example 4.13, Page:60 \\n \\n\"\n", + "# Given:\n", + "import math\n", + "dm=0.006332;# in u\n", + "ma=4.;\n", + "mCm=244.;\n", + "\n", + "# Solution:\n", + "\n", + "E=dm*931.;# in MeV\n", + "KE=E*(ma/mCm); # in MeV\n", + "v=math.sqrt((2.*KE*1.6*10.**-13.)/(240.*1.6605*10.**-27.));\n", + "print\"The Kinetic Energy and velocity are MeV and 10^5 m/s respectively\",round(KE,4),round(v/10**5,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 4.14;pg no:60" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.14, Page:60 \n", + " \n", + "\n", + "The range in Al for beta radiation is mg/cm^2 using Katz and Penfold empirical equation and mg/cm^2 using feathers relation. 784.8 795.0\n" + ] + } + ], + "source": [ + "#cal of rate of energy emission and Penfold empirical equation\n", + "#intiation of all variables\n", + "# Chapter 4\n", + "print\"Example 4.14, Page:60 \\n \\n\"\n", + "import math\n", + "# Given:\n", + "E0=1.7;# in MeV\n", + "# Solution:\n", + "# For E0<2.5 MeV; using Katz and Penfold empirical equation we have\n", + "R1=412*((E0)**(1.265-0.0954*math.log(E0)));# mg/cm^2\n", + "# Using feather's relation we have\n", + "R2=530*E0-106;# mg/cm^2\n", + "\n", + "print\"The range in Al for beta radiation is mg/cm^2 using Katz and Penfold empirical equation and mg/cm^2 using feathers relation.\",round(R1,1),round(R2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 4.15;pg no:60" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.15, Page:61 \n", + " \n", + "\n", + "The order for Case 1 is and emission is type M2,Case 2 is and emission is type E2,Case 3 is and emission is type E1,Case 4 is not possible. 2.0 3 2 4 1.0 2\n" + ] + } + ], + "source": [ + "#cal of emissions \n", + "#intiation of all variables\n", + "# Chapter 4\n", + "print\"Example 4.15, Page:61 \\n \\n\"\n", + "# Solution:\n", + "L1=(5.5-3.5);# Case 1\n", + "L2=2-0;# Case 2\n", + "L3=1.5-.5;# Case 3\n", + "ELML1=1+0+2;\n", + "ELML2=1+1+2;\n", + "ELML3=0+1+1;\n", + "print\"The order for Case 1 is and emission is type M2,Case 2 is and emission is type E2,Case 3 is and emission is type E1,Case 4 is not possible.\",L1,ELML1,L2,ELML2,L3,ELML3\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 4.16;pg no:61" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [ + "#cal of rate of energy emission\n", + "#intiation of all variables\n", + "# Chapter 4\n", + "print\"Example 4.16, Page:61 \\n \\n\"\n", + "# Given:\n", + "m=4*10**-3;# in gms\n", + "M=210;\n", + "E=0.34;# in MeV\n", + "# Solution:\n", + "\n", + "N=(m*6.022*10**23)/M;\n", + "k=0.693/(5*24*3600);# in s^-1\n", + "A=N*k;# in dis/s\n", + "# Energy released at 0.34 MeV per dis/s will be\n", + "E1=E*A;# in MeV/s\n", + "E2=E1*1.6*10**-13;# watts\n", + "\n", + "print\"The rate of energy emission is W\",round(E2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 4.17;pg no:61" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.17, Page:61 \n", + " \n", + "\n", + "The strength in KCi is 6.77\n" + ] + } + ], + "source": [ + "#cal of strength in KCi\n", + "#intiation of all variables\n", + "# Chapter 4\n", + "print\"Example 4.17, Page:61 \\n \\n\"\n", + "# Given:\n", + "A=0.2506*10**15;# atoms/s re: Ex4_10\n", + "# Solution:\n", + "Strength=A/(3.7*10**10);# in kCi\n", + "S1=Strength*10**-3;# in KCi\n", + "print\"The strength in KCi is \",round(S1,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example4.18;pg no:62" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.18, Page:62 \n", + " \n", + "\n", + "\n", + " The 1st half life, 2nd half life, 3rd half life are respectively. 9.0 4.5 2.25\n", + "\n", + " The 1st half life, 2nd half life, 3rd half life are respectively. 13 6 3\n", + "\n", + " The 1st half life, 2nd half life, 3rd half life are respectively. 500 250 125\n" + ] + } + ], + "source": [ + "#cal of half life periods\n", + "#intiation of all variables\n", + "# Chapter 4\n", + "print\"Example 4.18, Page:62 \\n \\n\"\n", + "# Given:\n", + "N1=10^24;# atoms\n", + "N2=10^16;# atoms\n", + "N3=1000;# atoms\n", + "N4=80;# atoms\n", + "\n", + "# Solution:\n", + "N11=N1*0.5;# 1st half life\n", + "N12=N11/2;# 2nd half life\n", + "N13=N12/2;# 3rd half life\n", + "print\"\\n The 1st half life, 2nd half life, 3rd half life are respectively.\",N11,N12,N13\n", + "N21=N2/2;# 1st half life\n", + "N22=N21/2;# 2nd half life\n", + "N23=N22/2;# 3rd half life\n", + "print\"\\n The 1st half life, 2nd half life, 3rd half life are respectively.\",N21,N22,N23\n", + "N31=N3/2;# 1st half life\n", + "N32=N31/2;# 2nd half life\n", + "N33=N32/2;# 3rd half life\n", + "print\"\\n The 1st half life, 2nd half life, 3rd half life are respectively.\",N31,N32,N33\n", + "#Radiactivity is a statistical property. Decay kinetics are reliable only when initial number is large" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 4.20;pg no:62" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.20, Page:62 \n", + " \n", + "\n", + "The partial half life for beta decay in Gy and partial half life for EC decay in Gy. 1.441 11.429\n" + ] + } + ], + "source": [ + "#cal of partial half life for beta and EC decay\n", + "#intiation of all variables\n", + "# Chapter 4\n", + "print\"Example 4.20, Page:62 \\n \\n\"\n", + "# Given:\n", + "t1=1.28*10**9;# in years\n", + "# Solution:\n", + "k=0.693/(1.28*10**9);\n", + "# beta deay is 88.8%\n", + "k1=0.888*k;\n", + "# EC decay is 11.2%\n", + "k2=0.112*k;\n", + "tbeta=(0.693*10**-9)/(k1);# partial half life for beta decay in Gy\n", + "tEC=(0.693*10**-9)/(k2);# partial half life for EC decay in Gy\n", + "print\"The partial half life for beta decay in Gy and partial half life for EC decay in Gy.\",round(tbeta,3),round(tEC,3)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 4.21;pg no:63" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.21, Page:63 \n", + " \n", + "\n", + "The time required will be h 150.0\n" + ] + } + ], + "source": [ + "#cal of time\n", + "#intiation of all variables\n", + "# Chapter 4\n", + "print\"Example 4.21, Page:63 \\n \\n\"\n", + "# Given:\n", + "import math\n", + "t=15.02;# in hours\n", + "# Solution:\n", + "ar=1000;# activity ratio given that 0.1% of intial activity\n", + "k=0.693/t;\n", + "t1=(math.log(ar))/k;\n", + "print\"The time required will be h\",round(t1)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 4.22;pg no:63" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.22, Page:63 \n", + " \n", + "\n", + "The time required will be h 26.0\n" + ] + } + ], + "source": [ + "#cal of time required\n", + "#intiation of all variables\n", + "# Chapter 4\n", + "print\"Example 4.22, Page:63 \\n \\n\"\n", + "# Given:\n", + "import math\n", + "t=6.01;# in hours\n", + "# Solution:\n", + "ar=100/5;# activity ratio given that 5% of intial activity\n", + "k=0.693/t;\n", + "\n", + "t1=(math.log(ar))/k;\n", + "print\"The time required will be h\",round(t1)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 4.23;pg no:64" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.23, Page:64 \n", + " \n", + "\n", + "\n", + " The overall decay constant will be *10^-17 s^-1 1.2\n", + "\n", + " The activity for k(40) in 10^5 beta/s 1.804\n", + "\n", + " The activity for k(41) in beta/s 21.12\n" + ] + } + ], + "source": [ + "#cal of activity for k\n", + "#intiation of all variables\n", + "# Chapter 4\n", + "print\"Example 4.23, Page:64 \\n \\n\"\n", + "# Given:\n", + "\n", + "t=1.83*10**9;# in years\n", + "# Solution:\n", + "# Part (a)\n", + "k=(0.693)/(t*3.16*10**7);\n", + "k1=(0.693*10**17)/(t*3.16*10**7);# in 10^-17 s^-1\n", + "print\"\\n The overall decay constant will be *10^-17 s^-1\",round(k1,2)\n", + "# Part (b)\n", + "a=(6.022*10**23)/40; # atoms of K(40)\n", + "A=a*k;# activity\n", + "print\"\\n The activity for k(40) in 10^5 beta/s\",round(A/10**5,3)\n", + "\n", + "# Part (c)\n", + "a1=(6.022*10**23*1.2*10**-4)/41; # atoms of K(41)\n", + "A1=a1*k;# activity\n", + "print\"\\n The activity for k(41) in beta/s\",round(A1,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 4.24;pg no:65" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.24, Page:65 \n", + " \n", + "\n", + "The decay constant is min^-1 and the half life is min 0.1511 4.5879\n" + ] + } + ], + "source": [ + "#cal of decay constant and half life\n", + "#intiation of all variables\n", + "# Chapter 4\n", + "print\"Example 4.24, Page:65 \\n \\n\"\n", + "# Given:\n", + "import math\n", + "a1=6520.;# c/min\n", + "a2=4820.;#c/min\n", + "t=2.;#min\n", + "# Solution:\n", + "k=math.log(a1/a2)/t;\n", + "t1=0.693/k;# half life\n", + "print\"The decay constant is min^-1 and the half life is min\",round(k,4),round(t1,4)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 4.25;pg no:65" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.25, Page:65 \n", + " \n", + "\n", + "The half life for case (a)is h and case(b) is min 1.5 12.89\n" + ] + } + ], + "source": [ + "#cal of value of avagadro constant\n", + "#intiation of all variables\n", + "# Chapter 4\n", + "print\"Example 4.25, Page:65 \\n \\n\"\n", + "# Given:\n", + "import math\n", + "a=(1./32.);# activity drop of its initial value\n", + "t1=7.5;#in h case(a)\n", + "t2=64.45;# in min case(b)\n", + "# Solution:\n", + "n=math.log(a)/math.log(0.5);\n", + "t11=t1/n;# half life\n", + "t12=t2/n;# half life\n", + "print\"The half life for case (a)is h and case(b) is min\",t11,t12" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 4.26;pg no:66" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.26, Page:66 \n", + " \n", + "\n", + "The proportion of U235 704 million years back in 10^-2 is 1.3\n" + ] + } + ], + "source": [ + "#cal of proportion of U235 704 million years back\n", + "#intiation of all variables\n", + "# Chapter 4\n", + "print\"Example 4.26, Page:66 \\n \\n\"\n", + "# Given:\n", + "import math\n", + "t238=4.5*10**9;# in y\n", + "t235=7.04*10**8;# in y\n", + "a0=0.72;# atoms per cent\n", + "t=7.04*10**8;\n", + "# Solution:\n", + "\n", + "k1=0.693/(t238);#decay constant for U 238\n", + "N1=(100-a0)*math.exp(k1*t);\n", + "\n", + "k2=0.693/(t235);#decay constant for U 235\n", + "N2=(a0)*math.exp(k2*t);\n", + "\n", + "proportion=N2/N1;\n", + "print\"The proportion of U235 704 million years back in 10^-2 is \" ,round(proportion*100,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 4.27;pg no:67" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.27, Page:67 \n", + " \n", + "\n", + "The no. of atoms produced is and its mass is *10^-20 grams 1.5873 4.74\n" + ] + } + ], + "source": [ + "#cal of The no. of atoms produced is and its mass\n", + "#intiation of all variables\n", + "# Chapter 4\n", + "print\"Example 4.27, Page:67 \\n \\n\"\n", + "# Given:\n", + "t=110;# in min\n", + "a=10;#dpmg^-1\n", + "# Solution:\n", + "k=0.693/t;\n", + "N=a/k;# atoms of F18\n", + "mass=(N*18)/6.022*10**23;\n", + "mass1=(N*18*10**20)/(6.022*10**23);# in 10^-20 grams\n", + "print\"The no. of atoms produced is and its mass is *10^-20 grams\",round(N/10**3,4),round(mass1,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 4.28;pgno:67" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.28, Page:67 \n", + " \n", + "\n", + "The mass in micro-grams is 0.166\n" + ] + } + ], + "source": [ + "#cal of mass in micro-grams\n", + "#intiation of all variables\n", + "# Chapter 4\n", + "print\"Example 4.28, Page:67 \\n \\n\"\n", + "# Given:\n", + "import math\n", + "t=14.3;# half life in days\n", + "# Solution:\n", + "k=0.693/(t*24*3600);\n", + "N=(3.7*10**10)/(k);# No. of atoms in 1 Ci\n", + "N1=N*(1-(math.exp(-0.693/14.3)));# atoms of S32 produced\n", + "mass=(N1*32)/(6.022*10**23);\n", + "m1=mass*10**6;# in micro grams\n", + "print\"The mass in micro-grams is \",round(m1,3)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 4.29;pg no:68" + ] + }, + { + "cell_type": "code", + "execution_count": 28, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.29, Page:68 \n", + " \n", + "\n", + "\n", + " The fraction decayed in 1 day will be 16.59\n", + "\n", + " The fraction decayed in 5 days will be 59.63\n", + "\n", + " The fraction decayed in 10 days will be . 83.7\n", + "\n", + " The fraction decayed in 6 half lives will be . 98.44\n", + "\n", + " Time needed for the decay of 99.9 percent is half lives i.e.days. 9.97 38.069\n" + ] + } + ], + "source": [ + "#cal of fraction decays and half lives\n", + "#intiation of all variables\n", + "# Chapter 4\n", + "print\"Example 4.29, Page:68 \\n \\n\"\n", + "# Given:\n", + "import math\n", + "t=3.82;# in days\n", + "# Solution:\n", + "# part(a)\n", + "days=1;\n", + "D1=(1-(math.exp(-0.693*days/t)))*100;\n", + "print\"\\n The fraction decayed in 1 day will be \",round(D1,2)\n", + "# part(b)\n", + "days=5;\n", + "D1=(1-(math.exp(-0.693*days/t)))*100;\n", + "print\"\\n The fraction decayed in 5 days will be \",round(D1,2)\n", + "# part(c)\n", + "days=10;\n", + "D1=(1-(math.exp(-0.693*days/t)))*100;\n", + "print\"\\n The fraction decayed in 10 days will be .\",round(D1,2)\n", + "# part(d)\n", + "days=6*t;\n", + "D1=(1-(math.exp(-0.693*days/t)))*100;\n", + "print\"\\n The fraction decayed in 6 half lives will be .\",round(D1,2)\n", + "# part(e)\n", + "n=math.log(0.001)/math.log(0.5);\n", + "print\"\\n Time needed for the decay of 99.9 percent is half lives i.e.days.\",round(n,2),round(t*n,3)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 4.30;pg no:68" + ] + }, + { + "cell_type": "code", + "execution_count": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.30, Page:68 \n", + " \n", + "\n", + "\n", + " The decay constant is y^-1 0.24\n", + "\n", + " The decay constant is y^-1 0.03\n", + "\n", + " The half life is y 2.92\n", + "\n", + " The half life is y 23.64\n" + ] + } + ], + "source": [ + "#cal of decay constant and half life\n", + "#intiation of all variables\n", + "# Chapter 4\n", + "print\"Example 4.30, Page:68 \\n \\n\"\n", + "# Given:\n", + "t=2.6;# years\n", + "# Solution:\n", + "k=0.693/t;# decay constant\n", + "#part(a)\n", + "kbeta=0.89*k;\n", + "print\"\\n The decay constant is y^-1\",round(kbeta,2)\n", + "kEC=0.11*k;\n", + "print\"\\n The decay constant is y^-1\",round(kEC,2)\n", + "#part(b)\n", + "tbeta=0.693/kbeta;\n", + "print\"\\n The half life is y\",round(tbeta,2)\n", + "tEC=0.693/kEC;\n", + "print\"\\n The half life is y\",round(tEC,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 4.31;pg no:69" + ] + }, + { + "cell_type": "code", + "execution_count": 30, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.31, Page:69 \n", + " \n", + "\n", + "Decay by EC\n", + "\n", + "\t The decay constant is h^-1 0.02\n", + "\n", + "\t The half life is h 30.48\n", + "\n", + "Decay by beta+\n", + "\n", + "\t The decay constant is h^-1 0.01\n", + "\n", + "\t The half life is h 67.37\n", + "\n", + "Decay by beta-\n", + "\n", + "\t The decay constant is h^-1 0.02\n", + "\n", + "\t The half life is h 32.82\n" + ] + } + ], + "source": [ + "#cal of Decay by beta\n", + "#intiation of all variables\n", + "# Chapter 4\n", + "print\"Example 4.31, Page:69 \\n \\n\"\n", + "# Given:\n", + "t=12.8;# hours\n", + "# Solution:\n", + "k=0.693/t;# decay constant\n", + "#part(a)by EC\n", + "kEC=0.42*k;\n", + "print\"Decay by EC\"\n", + "print\"\\n\\t The decay constant is h^-1\",round(kEC,2)\n", + "tEC=0.693/kEC;\n", + "print\"\\n\\t The half life is h\",round(tEC,2)\n", + "\n", + "#part(b)by beta+\n", + "kbeta1=0.19*k;\n", + "print\"\\nDecay by beta+\"\n", + "print\"\\n\\t The decay constant is h^-1\",round(kbeta1,2)\n", + "tbeta1=0.693/kbeta1;\n", + "print\"\\n\\t The half life is h\",round(tbeta1,2)\n", + "\n", + "#part(b)by beta+\n", + "kbeta2=0.39*k;\n", + "print\"\\nDecay by beta-\"\n", + "print\"\\n\\t The decay constant is h^-1\",round(kbeta2,2)\n", + "tbeta2=0.693/kbeta2;\n", + "print\"\\n\\t The half life is h\",round(tbeta2,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 4.32;pg no:69" + ] + }, + { + "cell_type": "code", + "execution_count": 31, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.32, Page:69 \n", + " \n", + "\n", + "The proportion of U,Ra,Rn is 1: 152879.581152 4.27107329843e+11\n" + ] + } + ], + "source": [ + "#cal of proportion of U,Ra,Rn\n", + "#intiation of all variables\n", + "# Chapter 4\n", + "print\"Example 4.32, Page:69 \\n \\n\"\n", + "# Given:\n", + "tU=4.47*10**9;# y\n", + "tRa=1600;# y\n", + "tRn=3.82;# days\n", + "nU=1;\n", + "# Solution:\n", + "#under secular equilibrium we have\n", + "nRa=(tRa*365/tRn)*nU;\n", + "nRn=(tU*365/tRn)*nU;\n", + "print\"The proportion of U,Ra,Rn is 1:\",nRa,nRn" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 4.33;pg no:70" + ] + }, + { + "cell_type": "code", + "execution_count": 32, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.33, Page:70 \n", + " \n", + "\n", + "The proportion of ay4 at the end of 4 hrs is 63.6\n" + ] + } + ], + "source": [ + "#cal of proportion of ay4\n", + "#intiation of all variables\n", + "# Chapter 4\n", + "print\"Example 4.33, Page:70 \\n \\n\"\n", + "# Given:\n", + "ax0 =1.; #assume\n", + "tx = 2.; #hrs\n", + "ty = 1.; #hrs\n", + "# Solution:\n", + "# general equation connecting Ax and Ay is\n", + "# Ax(n) = (ky * Ax(0) * (exp(-kx * t) - exp(-ky * t))/ (ky - kx)) + Ay(0) * exp(-ky * t)\n", + "ax0 = 1.;\n", + "ay4 = (ax0 * (0.693/1.) * ((1./4.)-(1./16.)))/((0.693/1.)-(0.693/2.)) + ax0 * (1./16.);\n", + "ax4 = (1.* ax0)/4.;\n", + "proportion = (ay4 * 100)/(ay4+ax4)\n", + "print\"The proportion of ay4 at the end of 4 hrs is\",round(proportion,1)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 4.34;pg no:71" + ] + }, + { + "cell_type": "code", + "execution_count": 33, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.34, Page:71 \n", + " \n", + "\n", + "Activity due to La(140) at the end of 12 hrs will be dps 1111.0\n", + "Activity due to La(140) at the end of 24 d will be dps 518.0\n" + ] + } + ], + "source": [ + "#cal of Activity due to La(140)\n", + "#intiation of all variables\n", + "# Chapter 4\n", + "print\"Example 4.34, Page:71 \\n \\n\"\n", + "#Given:\n", + "Ax0 = 2000.; #dps\n", + "#Solution:\n", + "#part a\n", + "ky = 0.693/10.;\n", + "kx = 0.693/288.;\n", + "# general equation connecting Ax and Ay is\n", + "Ax12 = (ky * Ax0 * (0.5**(1./24.) - 0.5**(1.2)))/ (ky - kx)\n", + "\n", + "print\"Activity due to La(140) at the end of 12 hrs will be dps\",round(Ax12)\n", + "#part b\n", + "ky = 0.693/10.;\n", + "kx = 0.693/288.;\n", + "# general equation connecting Ax and Ay is\n", + "Ax24 = (ky * Ax0 * (0.5**(2) - 0.5**(57.6)))/ (ky - kx)\n", + "print\"Activity due to La(140) at the end of 24 d will be dps\",round(Ax24)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 4.35;pg no:71" + ] + }, + { + "cell_type": "code", + "execution_count": 34, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.35, Page:71 \n", + " \n", + "\n", + "The time when daughter activity reaches maximum is and this is same when activities of both are equal. 4.48\n" + ] + } + ], + "source": [ + "#cal of time\n", + "#intiation of all variables\n", + "# Chapter 4\n", + "print\"Example 4.35, Page:71 \\n \\n\"\n", + "# Given:\n", + "import math\n", + "t1=2.7;# h\n", + "t2=3.6;# h\n", + "# Solution:\n", + "k1=0.693/t1;\n", + "k2=.693/t2;\n", + "tmax=(math.log(k2/k1))/(k2-k1);\n", + "print\"The time when daughter activity reaches maximum is and this is same when activities of both are equal.\",round(tmax,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 4.36;pg no:71" + ] + }, + { + "cell_type": "code", + "execution_count": 35, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.36, Page:71 \n", + " \n", + "\n", + "The half life of Bi is days 3.84\n" + ] + } + ], + "source": [ + "#cal of half life of Bi\n", + "#intiation of all variables\n", + "# Chapter 4\n", + "print\"Example 4.36, Page:71 \\n \\n\"\n", + "# Given:\n", + "tPo=138;# days\n", + "n=24.86;# days\n", + "# Solution:\n", + "kPo = 0.693/tPo;\n", + "# using simplification logx=2(x-1)/(x+1)\n", + "kBi=((2 * 2.303)-(n*kPo))/n;\n", + "tBi=0.693/kBi;\n", + "print\"The half life of Bi is days\",round(tBi,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 4.37;pg no:72" + ] + }, + { + "cell_type": "code", + "execution_count": 37, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 4.37, Page:72 \n", + " \n", + "\n", + "The residual activity in the sample in 10^5 dps= 93.75\n" + ] + } + ], + "source": [ + "#cal of residual activity in the sample\n", + "#intiation of all variables\n", + "# Chapter 4\n", + "print\"Example 4.37, Page:72 \\n \\n\"\n", + "# Given:\n", + "a=10*10**7;# rate\n", + "t=15;# h\n", + "# Soution:\n", + "A30=a*(1-(0.5)**(2));# dps\n", + "A45=A30*((0.5)**(3));# dps\n", + "print\"The residual activity in the sample in 10^5 dps=\",A45/10**5" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/_Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/chapter5.ipynb b/_Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/chapter5.ipynb new file mode 100644 index 00000000..9a9ccc52 --- /dev/null +++ b/_Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/chapter5.ipynb @@ -0,0 +1,1096 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# chapter 5:nuclear reactions" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 5.1;pg no:80" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 5.1, Page:80 \n", + " \n", + "\n", + "The excitation energy for Al*(26) in MeV= and that of Na*(22) in MeV= 18.79 65.4993\n" + ] + } + ], + "source": [ + "#cal of excitation energy for Al and Na\n", + "#intiation of all variables\n", + "# Chapter 5\n", + "print\"Example 5.1, Page:80 \\n \\n\"\n", + "# Given:\n", + "mMg=23.985045;\n", + "md=2.014102;\n", + "mAl=25.986900;\n", + "mNe=19.99244;\n", + "mNa=21.944;\n", + "# Solution:\n", + "# for compound nucleus Al*(26)\n", + "KE1=(24./26.)*8.;\n", + "BE1=(mMg+md-mAl)*931.;# in MeV\n", + "EE1=BE1+KE1;\n", + "# for compound nucleus Na*(22)\n", + "KE2=(20./22.)*8.;\n", + "BE2=(mNe+md-mNa)*931.;# in MeV\n", + "EE2=BE2+KE2;\n", + "print\"The excitation energy for Al*(26) in MeV= and that of Na*(22) in MeV=\",round(EE1,2),round(EE2,4)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 5.2;pg no:81" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 5.2, Page:81 \n", + " \n", + "\n", + "The energy of protons scattered through an angle of 90 deg. in MeV= 3.91\n", + "\n", + " \n", + " The energy of proton observed at 90 deg. after they have excited the lithium to a level of 0.48 MeV is = MeV 4.52\n" + ] + } + ], + "source": [ + "#cal of energy of protons\n", + "#intiation of all variables\n", + "# Chapter 5\n", + "print\"Example 5.2, Page:81 \\n \\n\"\n", + "# Given:\n", + "import math\n", + "E0=5.;# in MeV\n", + "m=1.;\n", + "M=7.\n", + "\n", + "# Solution:\n", + "\n", + "Erecoil=(4*5*m*M*((math.sin(45*3.14/180))**(2)))/((m+M)**2);\n", + "Escat=E0-Erecoil;\n", + "\n", + "print\"The energy of protons scattered through an angle of 90 deg. in MeV=\",round(Escat,2)\n", + "Eresi=E0-0.48;\n", + "\n", + "Erecoil2=(14/64)*Eresi;\n", + "Escat2=Eresi-Erecoil2;\n", + "print\"\\n \\n The energy of proton observed at 90 deg. after they have excited the lithium to a level of 0.48 MeV is = MeV\",Escat2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 5.3;pg no:81" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 5.3, Page:81 \n", + " \n", + "\n", + "The mass difference between A & B in Uu= 1678.0\n" + ] + } + ], + "source": [ + "#cal of mass difference between A & B\n", + "#intiation of all variables\n", + "# Chapter 5\n", + "print\"Example 5.3, Page:81 \\n \\n\"\n", + "# Given:\n", + "th=2.4;# in Mev\n", + "z=0.0009;# mp-mn in atomic mass unit\n", + "# Solution:\n", + "x=-(2.4/931.);#assuming no barrier operates\n", + "y=x+z;\n", + "print\"The mass difference between A & B in Uu=\", round(-y*10**6)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 5.4;pg no:82" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 5.4, Page:82 \n", + " \n", + "\n", + "The energy required for (a) & (b) are = respectively in MeV 1875.55 937.77\n" + ] + } + ], + "source": [ + "#cal of energy required\n", + "#intiation of all variables\n", + "# Chapter 5\n", + "print\"Example 5.4, Page:82 \\n \\n\"\n", + "# Given:\n", + "mp=1.007277;\n", + "# Solution:\n", + "E1=2*mp*931;# in MeV / part(a)\n", + "E2=E1/2;# in MeV / part (b)\n", + "print\"The energy required for (a) & (b) are = respectively in MeV\",round(E1,2),round(E2,2) " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 5.5;pg no:82" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 5.5, Page:82 \n", + " \n", + "\n", + "The Q-values are in MeV for B(9) & F(19) reactions respectively -1.853 -2.45\n" + ] + } + ], + "source": [ + "#cal of Q-values\n", + "#intiation of all variables\n", + "# Chapter 5\n", + "print\"Example 5.5, Page:82 \\n \\n\"\n", + "#Given\n", + "E1=2.059;\n", + "E2=2.59;\n", + "M1=9.;\n", + "M2=18.;\n", + "m=1.;\n", + "# Solution\n", + "Q1=-E1*(M1/(m+M1)); # part(a)\n", + "Q2=-E2*(M2/(m+M2)); # part(b)\n", + "print\"The Q-values are in MeV for B(9) & F(19) reactions respectively\",round(Q1,3),round(Q2,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 5.6;pg no:82" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 5.6, Page:82 \n", + " \n", + "\n", + "The thershold energies are in MeV for B(12) & N(13) reactions respectively 1.343 3.485\n" + ] + } + ], + "source": [ + "#cal of thershold energies\n", + "#intiation of all variables\n", + "# Chapter 5\n", + "print\"Example 5.6, Page:82 \\n \\n\"\n", + "# Given:\n", + "Q1=1.136;\n", + "Q2=3.236;\n", + "M1=11.;\n", + "M2=13.;\n", + "m1=2.;\n", + "m2=1.;\n", + "# Solution\n", + "E1=Q1*((m1+M1)/M1); # part(a)\n", + "E2=Q2*((m2+M2)/M2); # part(b)\n", + "print\"The thershold energies are in MeV for B(12) & N(13) reactions respectively\",round(E1,3),round(E2,3)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 5.7;pg no:82" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 5.7, Page:82 \n", + " \n", + "\n", + "The binding energy of last 2 neutron in part(a) and part(b) are = in MeV respectively 20.37 20.5\n" + ] + } + ], + "source": [ + "#cal of The binding energy of last 2 neutron\n", + "#intiation of all variables\n", + "# Chapter 5\n", + "print\"Example 5.7, Page:82 \\n \\n\"\n", + "# Given:\n", + "\n", + "Q1=9.28;# in Mev\n", + "Q2=0.21;# in Mev\n", + "Q3=7.25;# in Mev\n", + "Q4=3.63;# in Mev\n", + "mn=1.008665;\n", + "md=1.995311;# mass difference between Fe(56) & Fe(54)\n", + "# Solution:\n", + "E1=Q1+Q2+Q3+Q4;# part (a)\n", + "E2=(2*mn-md)*931;# part (b)\n", + "print\"The binding energy of last 2 neutron in part(a) and part(b) are = in MeV respectively\",round(E1,2),round(E2,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 5.8;pg no:83" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 5.8, Page:83 \n", + " \n", + "\n", + "The threshold energy is in MeV for O(17) reaction 1.543\n" + ] + } + ], + "source": [ + "#cal of threshold energy\n", + "#intiation of all variables\n", + "# Chapter 5\n", + "print\"Example 5.8, Page:83 \\n \\n\"\n", + "# Given:\n", + "Q1=1.2;\n", + "M1=14.;\n", + "m1=4.;\n", + "# Solution:\n", + "E1=Q1*((m1+M1)/M1);\n", + "print\"The threshold energy is in MeV for O(17) reaction\",round(E1,3)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 5.9;pg no:83" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 5.9, Page:83 \n", + " \n", + "\n", + "The threshold energy is in MeV for C(13) reaction 3.485\n" + ] + } + ], + "source": [ + "#cal of threshold energy\n", + "#intiation of all variables\n", + "# Chapter 5\n", + "print\"Example 5.9, Page:83 \\n \\n\"\n", + "# Given:\n", + "Q1=3.236;\n", + "M1=13.;\n", + "m1=1.;\n", + "# Solution:\n", + "E1=Q1*((m1+M1)/M1);\n", + "print\"The threshold energy is in MeV for C(13) reaction\",round(E1,3)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 5.10;pg no:84" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 5.10, Page:84 \n", + " \n", + "\n", + "The cs area required will be in b= 3.836\n" + ] + } + ], + "source": [ + "#cal of cs area required\n", + "#intiation of all variables\n", + "# Chapter 5\n", + "print\"Example 5.10, Page:84 \\n \\n\"\n", + "# Given:\n", + "import math\n", + "A1=3836;#in barns\n", + "E1=1;# in eV\n", + "E2=10**6# in eV\n", + "\n", + "# Solution:\n", + "vr=math.sqrt(E2/E1);\n", + "A2=A1/vr;\n", + "print\"The cs area required will be in b=\", A2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 5.11;pg no:84" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 5.11, Page:84 \n", + " \n", + "\n", + "The activity in 10^7 dis s^-1 g^-1 NaCl= 6.13\n" + ] + } + ], + "source": [ + "#cal of activity\n", + "#intiation of all variables\n", + "# Chapter 5\n", + "print\"Example 5.11, Page:84 \\n \\n\"\n", + "# Given:\n", + "a=0.56*10.**-24.; # area\n", + "flux=10.**13.;\n", + "# Solution:\n", + "A=6.022*10.**23.*10.**-3.*2.5/(58.5);\n", + "k=A*flux*0.56*10.**-24.;\n", + "y=(0.5)**(4./5.);\n", + "activity=k*(1-y);\n", + "print\"The activity in 10^7 dis s^-1 g^-1 NaCl=\",round(activity/10**7,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 5.12;pg no:84" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 5.12, Page:84 \n", + " \n", + "\n", + "The cros section area in (b)= 3762.66\n" + ] + } + ], + "source": [ + "#cal of cros section area\n", + "#intiation of all variables\n", + "# Chapter 5\n", + "print\"Example 5.12, Page:84 \\n \\n\"\n", + "# Given:\n", + "w=0.1189;\n", + "flux=10**16;\n", + "# Solution:\n", + "A=w/(flux*3.16*10**7);# in m^2\n", + "A1=A*10000/(10**-24);# in Barns\n", + "print\"The cros section area in (b)=\",round(A1,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 5.13;pg no:85" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 5.13, Page:85 \n", + " \n", + "\n", + "The cros section area in (b)= 98.61\n", + "\n", + " \n", + " The saturation activity possible in 10^13 dis s^-1 g^-1= 3.01\n", + "\n", + " \n", + " The activity in 10^12 dis s^-1 g^-1= 6.83\n" + ] + } + ], + "source": [ + "#cal of cros section area,activity,saturation activity\n", + "#intiation of all variables\n", + "# Chapter 5\n", + "print\"Example 5.13, Page:85 \\n \\n\"\n", + "# Given:\n", + "w=8.52*10**-4;\n", + "flux=10**18;\n", + "# Solution:\n", + "\n", + "A=w/(flux*24*3600);# in m^2\n", + "A1=A*10000/(10**-24);# in Barns\n", + "print\"The cros section area in (b)=\", round(A1,2)\n", + "k=flux*A*6.022*10**23/197;\n", + "print\"\\n \\n The saturation activity possible in 10^13 dis s^-1 g^-1=\",round(k/10**13,2)\n", + "y=(0.5)**(0.3704);\n", + "activity=k*(1-y);\n", + "print\"\\n \\n The activity in 10^12 dis s^-1 g^-1= \",round(activity/10**12,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 5.14;pg no:85" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 5.14, Page:85 \n", + " \n", + "\n", + "The activity in 10^12 dis s^-1 g^-1= 1.037\n", + "The activity in Ci/cm^3 of foil is = 1402.0\n" + ] + } + ], + "source": [ + "#cal of activity\n", + "#intiation of all variables\n", + "# Chapter 5\n", + "print\"Example 5.14, Page:85 \\n \\n\"\n", + "# Given:\n", + "import math\n", + "a=98.7*10**-24; # area in cm ^2\n", + "flux=10**16;\n", + "d=19.3;# density\n", + "l=0.02;# thickness in cm\n", + "area=1;# in cm^2\n", + "# Solution:\n", + "V=area*l;\n", + "m=V*d;\n", + "A=(6.022*10**23*m)/(197);\n", + "k=A*flux*a;\n", + "y=math.exp((-0.693*5)/(2.7*24*60));\n", + "activity=k*(1-y);\n", + "print\"The activity in 10^12 dis s^-1 g^-1=\",round(activity/10**12,3)\n", + "# specific activity in Ci/cm^3\n", + "a1=activity/(3.7*10**10);# in Ci/gold foil\n", + "a2=a1/V;# in Ci/cm^3 of foil\n", + "print\"The activity in Ci/cm^3 of foil is = \",round(a2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 5.15;pg no:86" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 5.15, Page:86 \n", + " \n", + "\n", + "The geometric cross-section area are for Sr(88), Sr(87), Xe(136) & Xe(135) respectively 1.218 1.208 1.628 1.62\n" + ] + } + ], + "source": [ + "#cal of geometric cross-section area\n", + "#intiation of all variables\n", + "# Chapter 5\n", + "print\"Example 5.15, Page:86 \\n \\n\"\n", + "# Given:\n", + "r0=1.4*10**-15;# in m\n", + "A1=88;\n", + "A2=87;\n", + "A3=136;\n", + "A4=135;\n", + "# Solution:\n", + "\n", + "rSr1=(3.14*(r0*(A1)**(0.33333))**2)/10**-28;# in barns\n", + "rSr2=(3.14*(r0*(A2)**(0.33333))**2)/10**-28;# in barns\n", + "rXe1=(3.14*(r0*(A3)**(0.33333))**2)/10**-28;# in barns\n", + "rXe2=(3.14*(r0*(A4)**(0.33333))**2)/10**-28;# in barns\n", + "print\"The geometric cross-section area are for Sr(88), Sr(87), Xe(136) & Xe(135) respectively\",round(rSr1,3),round(rSr2,3),round(rXe1,3),round(rXe2,3)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 5.16;pg no:87" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 5.16, Page:87 \n", + " \n", + "\n", + " The activity in mCi is = 0.042\n", + " The activity in M Bq is = 1.564\n" + ] + } + ], + "source": [ + "#cal of activity in mCi,Bq\n", + "#intiation of all variables\n", + "# Chapter 5\n", + "print\"Example 5.16, Page:87 \\n \\n\"\n", + "# Given:\n", + "m=4*10**-3;# in gms\n", + "flux=1.3*10**14;\n", + "a=19.6*10**-24;# in cm^2\n", + "# Solution:\n", + "N=(m/59)*6.022*10**23;\n", + "A=N*flux*a*3600;# atoms\n", + "k=0.693/(5.25*3.16*10**7);# s^-1\n", + "A1=k*A;# Activity in dps\n", + "A2=(A1)/(3.7*10**10);# in Ci\n", + "A3=(A1*10**3)/(3.7*10**10);# in mCi\n", + "A4=A2*37*10**8;# in Bq\n", + "print\" The activity in mCi is =\",round(A3,3)\n", + "print\" The activity in M Bq is = \",round(A4/10**5,3)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 5.17;pg no:88" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 5.17, Page:88 \n", + " \n", + "\n", + "The thickness of Cd foil in (cm)= 0.0406\n" + ] + } + ], + "source": [ + "#cal of thickness of Cd foil\n", + "#intiation of all variables\n", + "# Chapter 5\n", + "print\"Example 5.17, Page:88 \\n \\n\"\n", + "# Given:\n", + "import math\n", + "a=2.44*1000*10**-24;# in barns\n", + "d=8.64;# g/cm^3\n", + "# Solution:\n", + "n=(d*6.02*10**23)/112;# atoms/cm^2\n", + "x=(math.log(100))/(n*a);# in cm\n", + "print\"The thickness of Cd foil in (cm)=\",round(x,4)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 5.18;pg no:89" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 5.18, Page:89 \n", + " \n", + "\n", + "The thickness of B foil in (cm)= 0.0095\n" + ] + } + ], + "source": [ + "#cal of excitation energy for Al and Na\n", + "#intiation of all variables\n", + "# Chapter 5\n", + "print\"Example 5.18, Page:89 \\n \\n\"\n", + "# Given:\n", + "import math\n", + "a=3.8*1000*10**-24;# in barns\n", + "Ir=0.004;# I0/Ix\n", + "d=2.55;# g/cm^3\n", + "\n", + "# Solution:\n", + "n=(d*6.02*10**23)/10;# atoms/cm^2\n", + "y=(Ir)**-1;\n", + "x=math.log(y)/(n*a);# in cm\n", + "print\"The thickness of B foil in (cm)=\",round(x,4)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 5.19;pg no:89" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 5.19, Page:89 \n", + " \n", + "\n", + "\n", + " The excitation energy of compound nucleus in (MeV)= 9.66\n" + ] + } + ], + "source": [ + "#cal of excitation energy \n", + "#intiation of all variables\n", + "# Chapter 5\n", + "print\"Example 5.19, Page:89 \\n \\n\"\n", + "# Given:\n", + "E1=6;# MeV\n", + "mAl=26.981535;\n", + "malpha=4.002604;\n", + "mP=30.973763;\n", + "# Solution:\n", + "KE=E1*(27/31);# in MeV\n", + "BE=(mAl+malpha-mP)*931;# in MeV\n", + "Ex=KE+BE;\n", + "print\"\\n The excitation energy of compound nucleus in (MeV)=\",round(Ex,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 5.20;pg no:90" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 5.20, Page:90 \n", + " \n", + "\n", + "The excitation energy of compound nucleus in (MeV)= 4.63\n", + "The excitation energy of compound nucleus in (MeV)= 6.022\n" + ] + } + ], + "source": [ + "#cal of excitation energy \n", + "#intiation of all variables\n", + "# Chapter 5\n", + "print\"Example 5.20, Page:90 \\n \\n\"\n", + "# Given:\n", + "E1=1.4;# MeV\n", + "mBi=208.980417;\n", + "mn=1.008665;\n", + "mBI=209.984110;\n", + "# Solution:part(a)\n", + "\n", + "KE1=0.;# in MeV\n", + "BE1=(mBi+mn-mBI)*931.;# in MeV\n", + "Ex1=KE1+BE1;\n", + "print\"The excitation energy of compound nucleus in (MeV)=\",round(Ex1,2)\n", + "# Solution:part(b)\n", + "KE2=E1*(209./210.);# in MeV\n", + "BE2=(mBi+mn-mBI)*931.;# in MeV\n", + "Ex2=KE2+BE2;\n", + "print\"The excitation energy of compound nucleus in (MeV)=\",round(Ex2,3)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 5.21;pg no:90" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 5.21, Page:90 \n", + " \n", + "\n", + "The resonance in part(a) will occur at =(MeV) 1.668\n", + "The resonance in part(b) will occur at (MeV)= 2.955\n" + ] + } + ], + "source": [ + "#cal of resonance\n", + "#intiation of all variables\n", + "# Chapter 5\n", + "print\"Example 5.21, Page:90 \\n \\n\"\n", + "Ex=12.8;# MeV\n", + "mB=10.012939;\n", + "malpha=4.002604;\n", + "mN=14.003074;\n", + "mC=12.00;\n", + "md=2.014102;\n", + "# Solution:part(a)\n", + "\n", + "BE1=(mB+malpha-mN)*931;# in MeV\n", + "KE1=Ex-BE1;\n", + "E1=KE1*(14./10.);\n", + "print\"The resonance in part(a) will occur at =(MeV)\",round(E1,3)\n", + "\n", + "# Solution:part(b)\n", + "BE2=(mC+md-mN)*931.;# in MeV\n", + "KE2=Ex-BE2;\n", + "E2=KE2*(14./12.);\n", + "print\"The resonance in part(b) will occur at (MeV)=\",round(E2,3)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 5.22;pg no:91" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 5.22, Page:91 \n", + " \n", + "\n", + "The resonance frequency in (MHz)= 19.2\n" + ] + } + ], + "source": [ + "#cal of resonance frequency\n", + "#intiation of all variables\n", + "# Chapter 5\n", + "print\"Example 5.22, Page:91 \\n \\n\"\n", + "# Given:\n", + "B=2.5;# tesla\n", + "q=1.6*10**-19;\n", + "m=1.66*10**-27;\n", + "# Solution:\n", + "\n", + "f=(B*q*10**-6)/(2*3.14*2*m);\n", + "print\"The resonance frequency in (MHz)=\",round(f,1)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 5.23;pg no:91" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 5.23, Page:91 \n", + " \n", + "\n", + "The magnetic field needed to accelerate protons in T= 0.56\n", + "The magnetic field needed to accelerate N(14) ions in T= 1.31\n" + ] + } + ], + "source": [ + "#cal of magnetic field needed to accelerate\n", + "#intiation of all variables\n", + "# Chapter 5\n", + "print\"Example 5.23, Page:91 \\n \\n\"\n", + "# Given:\n", + "f=8.6*10**6;# in Hz\n", + "q1=1.6*10**-19;\n", + "q2=6*1.6*10**-19;\n", + "m1=1.66*10**-27;\n", + "m2=14*1.66*10**-27;\n", + "# Solution:\n", + "# for proton\n", + "B1=2*3.14*f*m1/q1;\n", + "print\"The magnetic field needed to accelerate protons in T=\",round(B1,2)\n", + "# for N(14) ions\n", + "B2=2*3.14*f*m2/q2;\n", + "print\"The magnetic field needed to accelerate N(14) ions in T=\",round(B2,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 5.24;pg no:91" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 5.24, Page:91 \n", + " \n", + "\n", + "The excitation energy of compound nucleus Si* in MeV= 10.5622\n", + "The excitation energy of compound nucleus Rb* in MeV= 12.0756\n" + ] + } + ], + "source": [ + "#cal of excitation energy of compound nucleus Si and Rb\n", + "#intiation of all variables\n", + "# Chapter 5\n", + "print\"Example 5.24, Page:91 \\n \\n\"\n", + "E1=2.75;# MeV\n", + "E2=14;# in MeV\n", + "mMg=23.985045;\n", + "malpha=4.00260;\n", + "mSi=27.9763;\n", + "mNe=19.99244;\n", + "mCo=58.93320;\n", + "mRb=78.9239\n", + "# Solution:\n", + "\n", + "KE1=E1*(24/28);# in MeV\n", + "BE1=(mMg+malpha-mSi)*931;# in MeV\n", + "Ex1=KE1+BE1;\n", + "print\"The excitation energy of compound nucleus Si* in MeV=\",round(Ex1,4)\n", + "KE2=E2*(59./79.);# in MeV\n", + "BE2=(mNe+mCo-mRb)*931.;# in MeV\n", + "Ex2=KE2+BE2;\n", + "print\"The excitation energy of compound nucleus Rb* in MeV=\",round(Ex2,4)" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/_Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/chapter6.ipynb b/_Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/chapter6.ipynb new file mode 100644 index 00000000..569a9390 --- /dev/null +++ b/_Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/chapter6.ipynb @@ -0,0 +1,923 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "#chapter 6:nuclear fission " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 6.1;pg no:98" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 6.1, Page:98 \n", + " \n", + "\n", + "The half-thickness of In is times more than of Gd. 789.0\n" + ] + } + ], + "source": [ + "#cal of excitation energy for Al and Na\n", + "#intiation of all variables\n", + "# Chapter 6\n", + "print\"Example 6.1, Page:98 \\n \\n\"\n", + "# Given:\n", + "import math\n", + "d1=7.9;# density of Gd\n", + "d2=2.31;# Density of In\n", + "a1=49.;# in Kb\n", + "a2=155.;# in b\n", + "m1=157.25;\n", + "m2=114.8;\n", + "Na=6.02*10.**23.;\n", + "# Solution:\n", + "x1=math.log(1./(1./2.))/((d1*Na*a1*10.**-24.*10.**3.)/m1);# half-thickness for Gd\n", + "x2=math.log(1./(1./2.))/((d2*Na*a2*10.**-24.)/m2);# half-thickness for In\n", + "r=x2/x1;\n", + "print\"The half-thickness of In is times more than of Gd.\",round(r)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 6.2;pg no:98" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 6.2, Page:99 \n", + " \n", + "\n", + "The thickness of Pd foil which would reduce the intensity of a beam to excatly 1/1000 of its initial value in (cm)= 2.36\n" + ] + } + ], + "source": [ + "#cal of thickness of Pd foil\n", + "#intiation of all variables\n", + "# Chapter 6\n", + "print\"Example 6.2, Page:99 \\n \\n\"\n", + "# Given:\n", + "import math\n", + "d1=12.;# density of Gd\n", + "a1=43.11;# in b\n", + "m1=106.4;\n", + "Na=6.02*10**23;\n", + "i1=1.;\n", + "i2=1./1000.;\n", + "# Solution\n", + "x=math.log(i1/i2)/((d1*Na*a1*10.**-24.)/m1);# thickness for Pd foil\n", + "print\"The thickness of Pd foil which would reduce the intensity of a beam to excatly 1/1000 of its initial value in (cm)=\",round(x,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 6.3;pg no:99" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 6.3, Page:99 \n", + " \n", + "\n", + "The fission energy of Te(130)in MeV is = 47.6\n", + "The barrier energy of Te(130) in MeV is = 80.7\n", + "The activation energy of Te(130) in MeV is = 33.1\n" + ] + } + ], + "source": [ + "#cal of fission,barrier,activation\n", + "#intiation of all variables\n", + "# Chapter 6\n", + "print\"Example 6.3, Page:99 \\n \\n\"\n", + "# Given:\n", + "mTe=129.9067; # mol wt. of Te(52)\n", + "mCu=64.9278;# mol wt of Cu(29)\n", + "mFe=65;# mol wt of Fe(26)\n", + "# Solution\n", + "E1=(mTe-2*mCu)*931; # Fission Energy in MeV\n", + "print\"The fission energy of Te(130)in MeV is =\",round(E1,1)\n", + "r=((65)**0.33333);\n", + "E2=(26*26*4.8*4.8*10**-20)/(2*1.5*1.6*10**-13*10**-6*r);# Barrier energy in MeV\n", + "print\"The barrier energy of Te(130) in MeV is =\",round(E2,1)\n", + "E3=E2-E1;# Activation Energy in MeV\n", + "print\"The activation energy of Te(130) in MeV is = \",round(E3,1)\n", + "# Since barrier energy is greater than fission energy, spontaneous fission is not possible unless the activation energy is provided." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 6.4;pg no:99" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 6.4, Page:99 \n", + " \n", + "\n", + "\n", + " The fission energy in MeV is = 24.5784\n", + "\n", + " The barrier energy in MeV is = 77.95\n", + "\n", + " The activation energy in MeV is = 53.38\n" + ] + } + ], + "source": [ + "#cal of fission,barrier,activation energies\n", + "#intiation of all variables\n", + "# Chapter 6\n", + "print\"Example 6.4, Page:99 \\n \\n\"\n", + "# Given:\n", + "mSn=113.903; # mol wt. of Sn(50)\n", + "mMn=56.9383;# mol wt of Mn(25)\n", + "mFe=57;#mol wt of Fe(26)\n", + "# Solution\n", + "E1=(mSn-2*mMn)*931; # Fission Energy in MeV\n", + "print\"\\n The fission energy in MeV is =\",E1\n", + "r=((mFe)**0.33333);\n", + "E2=(25*25*4.8*4.8*10**-20)/(2*1.5*1.6*10**-13*10**-6*r);# Barrier energy in MeV\n", + "print\"\\n The barrier energy in MeV is = \",round(E2,2)\n", + "E3=E2-E1;# Activation Energy in MeV\n", + "print\"\\n The activation energy in MeV is =\",round(E3,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 6.5;pg no: 100" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 6.5, Page:100 \n", + " \n", + "\n", + "\n", + " z1= 40.0\n", + "\n", + " z2= 54.0\n" + ] + } + ], + "source": [ + "#intiation of all variables\n", + "# Chapter 6\n", + "print\"Example 6.5, Page:100 \\n \\n\"\n", + "# Given:\n", + "a1=94;# atomic no. of Pu\n", + "a2=42;#atomic no. of Mo\n", + "a3=56;# atomic no. of Ba\n", + "# Solution:\n", + "# By principle of equal charge displacement\n", + "z1=0.5*(a1+a2-a3);\n", + "print\"\\n z1=\",z1\n", + "z2=0.5*(a1-a2+a3);\n", + "print\"\\n z2=\",z2\n", + "\n", + "#From z1 and z2 we have the primary fragments as Zr(40), atomic mass(100) and Xe(54), atomic mass (138)." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 6.6;pg no: 100" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 6.6, Page:100 \n", + " \n", + "\n", + "\n", + " z1= 37.0\n", + "\n", + " z2= 55.0\n" + ] + } + ], + "source": [ + "#intiation of all variables\n", + "# Chapter 6\n", + "print\"Example 6.6, Page:100 \\n \\n\"\n", + "# Given:\n", + "a1=92;# atomic no. of U\n", + "a2=40;#atomic no. of Zr\n", + "a3=58;# atomic no. of Ce\n", + "# Solution\n", + "# By principle of equal charge displacement\n", + "z1=0.5*(a1+a2-a3);\n", + "print\"\\n z1=\",z1\n", + "z2=0.5*(a1-a2+a3);\n", + "print\"\\n z2=\",z2\n", + "#From z1 and z2 we have the primary fragments are Rb(37), atomic mass(94) and Cs(55), atomic mass (140)." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 6.7;pg no: 100" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 6.7, Page:100 \n", + " \n", + "\n", + "\n", + " z1= 39.0\n", + "\n", + " z2= 51.0\n" + ] + } + ], + "source": [ + "#intiation of all variables\n", + "# Chapter 6\n", + "print\"Example 6.7, Page:100 \\n \\n\"\n", + "# Given:\n", + "a1=92;# atomic no. of U\n", + "a2=42;#atomic no. of Mo\n", + "a3=56;# atomic no. of Ba\n", + "# Solution\n", + "# By principle of equal charge displacement\n", + "z1=0.5*(a1+a2-a3);\n", + "print\"\\n z1=\",z1\n", + "z2=0.5*(a1-a2+a3-4);\n", + "print\"\\n z2=\",z2\n", + "\n", + "# From z1 and z2 we have the primary fragments are Y(39), atomic mass(95) and Sb(51), atomic mass (137)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 6.8;pg no: 101" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 6.8, Page:101 \n", + " \n", + "\n", + "The fission energy in MeV is = 168.24\n", + "\n", + " The barrier energy in MeV is = 196.54\n", + "\n", + " The activation energy in MeV is = 28.3\n", + "\n", + " The fission by thermal neutrons is not possible since excitation energy is less than activation energy. 6.4239\n" + ] + } + ], + "source": [ + "#cal of fission,barrier,activation energies\n", + "#intiation of all variables\n", + "# Chapter 6\n", + "print\"Example 6.8, Page:101 \\n \\n\"\n", + "# Given:\n", + "mU=236.04533;\n", + "mU1=236.045733;\n", + "mU2=235.043933;\n", + "mY=94.912;\n", + "mSb=136.91782;\n", + "mn=1.0087;\n", + "Na=6.02*10**23;\n", + "\n", + "# Solution:\n", + "E1=(mU-mY-mSb-4*mn)*931; # Fission Energy in MeV\n", + "print\"The fission energy in MeV is = \",round(E1,2)\n", + "r1=((mY)**0.33333);\n", + "r2=((mSb)**0.33333);\n", + "E2=(39*51*4.8*4.8*10**-20)/(1.5*10**-13*(r1+r2)*1.6*10**-6);# Barrier energy in MeV\n", + "print\"\\n The barrier energy in MeV is = \",round(E2,2)\n", + "E3=E2-E1;# Activation Energy in MeV\n", + "print\"\\n The activation energy in MeV is = \",round(E3,2)\n", + "# Note : There is discrepancy in the final answer.\n", + "E4=(mU2+mn-mU1)*931; # Fission Energy in MeV\n", + "print\"\\n The fission by thermal neutrons is not possible since excitation energy is less than activation energy.\",E4" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 6.9;pg no: 101" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 6.9, Page:101 \n", + " \n", + "\n", + "The independent fractional chain yield of Sr is = 0.01\n", + "\n", + " The independent fractional chain yield of Y is = 1.54\n" + ] + } + ], + "source": [ + "#cal of independent fractional chain yield of Sr,Y\n", + "#intiation of all variables\n", + "# Chapter 6\n", + "print\"Example 6.9, Page:101 \\n \\n\"\n", + "# Given:\n", + "import math\n", + "nSr=38.;\n", + "nY=39.;\n", + "# Solution:\n", + "PSr=0.565*math.exp(-((nSr-40)**2)); #independent fractional chain yield of Sr\n", + "PY=0.565*math.exp(((nY-40.)**2.)); #independent fractional chain yield of Y\n", + "print\"The independent fractional chain yield of Sr is =\",round(PSr,2)\n", + "print\"\\n The independent fractional chain yield of Y is =\",round(PY,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 6.10;pg no: 102" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 6.10, Page:102 \n", + " \n", + "\n", + "The energy for the given fission is =(MeV) 184.7\n" + ] + } + ], + "source": [ + "#cal of energy for the given fission\n", + "#intiation of all variables\n", + "# Chapter 6\n", + "print\"Example 6.10, Page:102 \\n \\n\"\n", + "# Given:\n", + "mU=235.043091;\n", + "mn=1.0087;\n", + "mXe=138.9187;\n", + "mSn=94.919;\n", + "# Solution:\n", + "dm=(235.04309+1.0087-138.917-94.919-2.0174);# delta m\n", + "E=dm*931;# energy of given fission in MeV\n", + "print\"The energy for the given fission is =(MeV)\",round(E,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 6.11;pg no: 102" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 6.11, Page:102 \n", + " \n", + "\n", + "The energy for the Pd(108)+Xe(129)+3n fission is =(MeV) 210.46\n", + "\n", + " The energy for the Gd(155)+Br(81)+4n fission is =(MeV) 174.76\n" + ] + } + ], + "source": [ + "#cal of energy for the Gd(155)+Br(81)+4n,Pd(108)+Xe(129)+3n\n", + "#intiation of all variables\n", + "# Chapter 6\n", + "print\"Example 6.11, Page:102 \\n \\n\"\n", + "# Given:\n", + "mPu=239.052161;\n", + "mPd=107.903920;\n", + "mXe=128.904784;\n", + "mn=1.0087;\n", + "mGd=154.922010;\n", + "mBr=80.916344;\n", + "\n", + "# Solution: Part (a)\n", + "\n", + "dm1=(mPu-(mPd+mXe+2*mn));# delta m\n", + "E1=dm1*931;# energy of given fission in MeV\n", + "print\"The energy for the Pd(108)+Xe(129)+3n fission is =(MeV)\",round(E1,2)\n", + "\n", + "dm2=(mPu-(mGd+mBr+3*mn));# delta m\n", + "E2=dm2*931;# energy of given fission in MeV\n", + "print\"\\n The energy for the Gd(155)+Br(81)+4n fission is =(MeV)\",round(E2,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 6.12;pg no: 103" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 6.12, Page:103 \n", + " \n", + "\n", + "The fission energy in MeV is = 245.41\n", + "\n", + " The barrier energy in MeV is = 240.0\n" + ] + } + ], + "source": [ + "#cal of barrier,fission energy\n", + "#intiation of all variables\n", + "# Chapter 6\n", + "print\"Example 6.12, Page:103 \\n \\n\"\n", + "# Given:\n", + "\n", + "mFm=250.079;\n", + "mSn=124.9077;\n", + "Na=6.02*10**23;\n", + "\n", + "# Solution:\n", + "E1=(mFm-2*mSn)*931; # Fission Energy in MeV\n", + "print\"The fission energy in MeV is = \",round(E1,2)\n", + "r=((mSn)**0.33333);\n", + "E2=(50*50*4.8*4.8*10**-20)/(2*1.5*10**-13*(r)*1.6*10**-6);# Barrier energy in MeV\n", + "print\"\\n The barrier energy in MeV is =\",round(E2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 6.13;pg no: 103" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 6.13, Page:103 \n", + " \n", + "\n", + "\n", + " (a) Excitation energy is =(MeV) 4.93\n", + "\n", + " Fission is not possible\n", + "\n", + " (b) Excitation energy is =(MeV) 5.93\n", + "\n", + " Fission is not possible\n", + "\n", + " (c) Excitation energy is =(MeV) 13.93\n", + "\n", + " Fission is possible\n" + ] + } + ], + "source": [ + "#cal of thickness of Pd foil\n", + "#intiation of all variables\n", + "# Chapter 6\n", + "print\"Example 6.13, Page:103 \\n \\n\"\n", + "# Given:\n", + "mTh1=232;\n", + "mTh2=233;\n", + "ETh1=6.4;# in MeV\n", + "ETh2=4.93;# in MeV\n", + "E=6.5;# fission barrier energy in MeV\n", + "\n", + "# Solution: Part(a)\n", + "E1=0*mTh1/mTh2;\n", + "Ex1=E1+ETh2;\n", + "print\"\\n (a) Excitation energy is =(MeV)\",Ex1\n", + "if (Ex1>E):\n", + "\tprint\"\\n Fission is possible\"\n", + "else:\n", + "\tprint\"\\n Fission is not possible\"\n", + "\n", + "# Solution: Part(b)\n", + "E2=2*mTh1/mTh2;\n", + "Ex2=E2+ETh2;\n", + "print\"\\n (b) Excitation energy is =(MeV)\",Ex2\n", + "if (Ex2>E):\n", + "\tprint\"\\n Fission is possible\"\n", + "else:\n", + "\tprint\"\\n Fission is not possible\"\n", + "\n", + "# Solution: Part (c)\n", + "E3=10*mTh1/mTh2;\n", + "Ex3=E3+ETh2;\n", + "print\"\\n (c) Excitation energy is =(MeV)\",Ex3\n", + "if (Ex3>E):\n", + "\tprint\"\\n Fission is possible\"\n", + "else:\n", + "\tprint\"\\n Fission is not possible\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 6.14;pg no: 104" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 6.14, Page:104 \n", + " \n", + "\n", + "The energy for the given fission is =(MeV) 201.9\n" + ] + } + ], + "source": [ + "#cal of energy for the given fission\n", + "#intiation of all variables\n", + "# Chapter 6\n", + "print\"Example 6.14, Page:104 \\n \\n\"\n", + "\n", + "# Given:\n", + "\n", + "mEs=249.0762;\n", + "mn=1.0087;\n", + "mGd=160.9286;\n", + "mBr=86.922;\n", + "\n", + "# Solution:\n", + "dm=(mEs-(mGd+mBr+mn));# delta m\n", + "E=dm*931;# energy of given fission in MeV\n", + "print\"The energy for the given fission is =(MeV)\",round(E,1)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 6.15;pg no: 104" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 6.15, Page:104 \n", + " \n", + "\n", + "The velocity will be related as and Energy will be related as 6.0 6.0\n" + ] + } + ], + "source": [ + "#cal of velocity will be related as and Energy will be related\n", + "#intiation of all variables\n", + "# Chapter 6\n", + "print\"Example 6.15, Page:104 \\n \\n\"\n", + "\n", + "# Given:\n", + "\n", + "m=1./6.;# mass ratio of pair of fission product\n", + "\n", + "# Solution:\n", + "# Velocities as well as energies are in inverse ratio of their masses.\n", + "\n", + "v=(m)**(-1);# Velocity ratio\n", + "e=(m)**(-1);# Energy ratio\n", + "print\"The velocity will be related as and Energy will be related as \",v,e" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 6.16;pg no: 104" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 6.16, Page:104 \n", + " \n", + "\n", + "The no. of fissions produced per second will be = 3.125e+12\n" + ] + } + ], + "source": [ + "#cal of no. of fissions produced per second\n", + "#intiation of all variables\n", + "# Chapter 6\n", + "print\"Example 6.16, Page:104 \\n \\n\"\n", + "\n", + "# Given:\n", + "P=100;# in watts\n", + "\n", + "# Solution:\n", + "P1=P*10**7;# in erg/s\n", + "P2=P1/(1.6*10**-6);# in MeV/s\n", + "# 1 ifssion generates 200 MeV of energy\n", + "f=P2/200;# no. of fissions\n", + "print\"The no. of fissions produced per second will be =\",f" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 6.17;pg no: 104" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 6.17, Page:104 \n", + " \n", + "\n", + "The fission barrier energy is = (MeV) 221.78\n" + ] + } + ], + "source": [ + "#cal of fission barrier energy\n", + "#intiation of all variables\n", + "# Chapter 6\n", + "print\"Example 6.17, Page:104 \\n \\n\"\n", + "# Given:\n", + "r0=1.4*10**-15;# nuclear radius constant in m\n", + "p=8.85*10**-12;# permittivity of free space in J^-1*C^2*m^-1\n", + "A=92;\n", + "e=1.6*10**-19;\n", + "mPd=118;\n", + "# Solution:\n", + "r=((mPd)**0.33333);\n", + "Eb=((A/2)**2)*(e**2)/(2*r0*r*4*3.14*p*1.6*10**-13);\n", + "print\"The fission barrier energy is = (MeV)\",round(Eb,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 6.18;pg no: 105" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 6.18, Page:105 \n", + " \n", + "\n", + "The fission barrier energy is =(MeV) 224.0\n" + ] + } + ], + "source": [ + "#cal of fission barrier energy\n", + "#intiation of all variables\n", + "# Chapter 6\n", + "print\"Example 6.18, Page:105 \\n \\n\"\n", + "# Given:\n", + "r0=1.37*10**-15;# nuclear radius constant in m\n", + "p=8.85*10**-12;# permittivity of free space in J^-1*C^2*m^-1\n", + "A=92;\n", + "e=1.6*10**-19;\n", + "mTe=140;\n", + "mZr=95;\n", + "# Solution:\n", + "r1=((mTe)**0.33333);\n", + "r2=((mZr)**0.33333);\n", + "Eb=(52*40)*(e**2)/(r0*(r1+r2)*4*3.14*p*1.6*10**-13)\n", + "print\"The fission barrier energy is =(MeV)\",round(Eb)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 6.19;pg no: 105" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 6.19, Page:105 \n", + " \n", + "\n", + "The critical deformation energy for the fission is =(MeV) 6.237\n" + ] + } + ], + "source": [ + "#cal of critical deformation energy for the fission\n", + "#intiation of all variables\n", + "# Chapter 6\n", + "print\"Example 6.19, Page:105 \\n \\n\"\n", + "# Given:\n", + "A=240.;\n", + "Z=94.;\n", + "#Solution:\n", + "Ecr=(0.89*(A**(2./3.)))-(0.02*(Z*(Z-1.)))/(A**(1./3.));\n", + "print\"The critical deformation energy for the fission is =(MeV)\",round(Ecr,3)" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/_Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/chapter7.ipynb b/_Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/chapter7.ipynb new file mode 100644 index 00000000..8731741e --- /dev/null +++ b/_Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/chapter7.ipynb @@ -0,0 +1,298 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "#chapter 7:nuclear reactions" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 7.1;pg no: 111" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 7.1, Page:111 \n", + " \n", + "\n", + "The reproduction factor for the reactor is = 1.053\n" + ] + } + ], + "source": [ + "#cal of reproduction factor for the reactor\n", + "#intiation of all variables\n", + "# Chapter 7\n", + "print\"Example 7.1, Page:111 \\n \\n\"\n", + "# Given:\n", + "\n", + "f=1.03;# fast fission factor\n", + "n=1.32;# no. of fast neutrons generated per thermal radiations\n", + "ref=0.89;# resonance escape factor\n", + "tuf=0.87;# thermal utilization factor\n", + "\n", + "# Solution\n", + "rf=f*n*ref*tuf;#reproduction factor\n", + "print\"The reproduction factor for the reactor is = \",round(rf,3)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 7.2;pg no: 111" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 7.2, Page:111 \n", + " \n", + "\n", + "\n", + " The approximate critical dimensionsof a Pu 239 in (m)= 4.86\n", + "\n", + " \n", + " The radius of the reactor in (m)= 2.81\n" + ] + } + ], + "source": [ + "#cal of approximate critical dimensionsof a Pu 239 and radius of the reactor\n", + "#intiation of all variables\n", + "# Chapter 7\n", + "print\"Example 7.2, Page:111 \\n \\n\"\n", + "# Given:\n", + "import math\n", + "k=1.04;\n", + "m=0.032;# in m^2 i.e., migration area M^2\n", + "# Solution: (a) Cubical reactor\n", + "a=3.14*math.sqrt(3*m/(k-1));\n", + "print\"\\n The approximate critical dimensionsof a Pu 239 in (m)=\",round(a,2)\n", + "# Solution: (a) Spherical reactor\n", + "r=a/math.sqrt(3);\n", + "print\"\\n \\n The radius of the reactor in (m)=\",round(r,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 7.3;pg no: 111" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 7.3, Page:111 \n", + " \n", + "\n", + "\n", + " The design parameter for H2O is = 1114.0\n", + "\n", + " \n", + " The design parameter for D2O is = 79042.0\n", + "\n", + " \n", + " The design parameter for C is = 163353.0\n", + "Example 7.3, Page:111 \n", + " \n", + "\n", + "\n", + " The design parameter for H2O is = 1114.0\n", + "\n", + " \n", + " The design parameter for D2O is = 79042.0\n", + "\n", + " \n", + " The design parameter for C is = 163353.0\n" + ] + } + ], + "source": [ + "#cal of design parameter for H2O,D2O,C\n", + "#intiation of all variables\n", + "# Chapter 7\n", + "print\"Example 7.3, Page:111 \\n \\n\"\n", + "# Given:\n", + "a1=687; # neutron absorption cross section for U 235 in barns\n", + "a2=0.66 # neutron absorption cross section for H2O in barns\n", + "a3=0.0093; # neutron absorption cross section for D2O in barns\n", + "a4=0.0045; # neutron absorption cross section for C in barns\n", + "\n", + "#Solution:\n", + "\n", + "F1=1.07*a1/a2;#design parameter for H2O part(a)\n", + "print\"\\n The design parameter for H2O is =\",round(F1)\n", + "\n", + "F2=1.07*a1/a3;#design parameter for D2O part(b)\n", + "print\"\\n \\n The design parameter for D2O is =\",round(F2)\n", + "\n", + "F3=1.07*a1/a4;#design parameter for C part(c)\n", + "print\"\\n \\n The design parameter for C is =\",round(F3)#cal of design parameter for H2O,D2O,C\n", + "#intiation of all variables\n", + "# Chapter 7\n", + "print\"Example 7.3, Page:111 \\n \\n\"\n", + "# Given:\n", + "a1=687; # neutron absorption cross section for U 235 in barns\n", + "a2=0.66 # neutron absorption cross section for H2O in barns\n", + "a3=0.0093; # neutron absorption cross section for D2O in barns\n", + "a4=0.0045; # neutron absorption cross section for C in barns\n", + "\n", + "#Solution:\n", + "\n", + "F1=1.07*a1/a2;#design parameter for H2O part(a)\n", + "print\"\\n The design parameter for H2O is =\",round(F1)\n", + "\n", + "F2=1.07*a1/a3;#design parameter for D2O part(b)\n", + "print\"\\n \\n The design parameter for D2O is =\",round(F2)\n", + "\n", + "F3=1.07*a1/a4;#design parameter for C part(c)\n", + "print\"\\n \\n The design parameter for C is =\",round(F3)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 7.4;pg no: 112" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 7.4, Page:112 \n", + " \n", + "\n", + "The no. of fissions per second are = 3.125e+17\n" + ] + } + ], + "source": [ + "#cal of no. of fissions per second\n", + "#intiation of all variables\n", + "# Chapter 7\n", + "print\"Example 7.4, Page:112 \\n \\n\"\n", + "# Given:\n", + "P=10*10**6; # power in watts\n", + "E=200*10**6; # in eV\n", + "\n", + "# Solution:\n", + "e=E*1.6*10**-19;# in joules\n", + "# Thus for 1 fission occurs per second, rate of power generation is e\n", + "n=(P)/e;# no. of fissions\n", + "print\"The no. of fissions per second are =\",n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 7.5;pg no: 112" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 7.5, Page:112 \n", + " \n", + "\n", + "\n", + " The thermal energy generated is =(J) 153.85\n" + ] + } + ], + "source": [ + "#cal of thermal energy generated\n", + "#intiation of all variables\n", + "# Chapter 7\n", + "print\"Example 7.5, Page:112 \\n \\n\"\n", + "# Given:\n", + "density=19;# in g/cc\n", + "E1=200*10**6*(1.6*10**-19); # energy released per fission in J\n", + "flux1=10**12;# in cm^2/s\n", + "a1=590*10**-24;#fission cross-section in cm^2\n", + "Na1=6.02*10**23;\n", + "\n", + "# Solution:\n", + "\n", + "#Ntgt=volume of target*No.of atoms per cm^3\n", + "\n", + "Ni=(30*((0.5)**2)*3.14*density*Na1*(0.72*10**-2))/238;\n", + "\n", + "Np=Ni*a1*flux1;\n", + "\n", + "E2=E1*Np;# Thermal energy generated in J\n", + "\n", + "print\"\\n The thermal energy generated is =(J)\",round(E2,2)\n", + "# Note: There is discrepancy in answer given in the textbook. After calculations the answer comes out to be 153.850366 " + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/_Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/chapter8.ipynb b/_Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/chapter8.ipynb new file mode 100644 index 00000000..444ef624 --- /dev/null +++ b/_Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/chapter8.ipynb @@ -0,0 +1,398 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "#chapter 8:direction and measurment of activity" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 8.1;pg no: 120" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 8.1, Page:120 \n", + " \n", + "\n", + "The potential signal recorded will be (mV)= 1.33\n" + ] + } + ], + "source": [ + "#cal of potential signal\n", + "#intiation of all variables\n", + "# Chapter 8\n", + "print\"Example 8.1, Page:120 \\n \\n\"\n", + "#Given:\n", + "e=1.6*10**-19;# electron charge\n", + "C=6*10**-12;# in F\n", + "N=10**5;# # electron multiplication\n", + "#Solution:\n", + "e1=N*e;\n", + "v=e1/(2*C);\n", + "v1=1000*v;\n", + "print\"The potential signal recorded will be (mV)=\",round(v1,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 8.2;pg no: 120" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 8.2, Page:120 \n", + " \n", + "\n", + "The capacitance that would be required in (nF)= 8.0\n" + ] + } + ], + "source": [ + "#cal of capacitance that would be required\n", + "#intiation of all variables\n", + "# Chapter 8\n", + "print\"Example 8.2, Page:120 \\n \\n\"\n", + "# Given:\n", + "N=10**5;# electron multiplication\n", + "v=10**-6;# in V\n", + "e=1.6*10**-19;# electron charge\n", + "\n", + "# Solution:\n", + "e1=N*e;\n", + "C=e1/(2*v);\n", + "C1=C*10**9;\n", + "print\"The capacitance that would be required in (nF)=\",C1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 8.3;pg no: 120" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 8.3, Page:120 \n", + " \n", + "\n", + "The alpha coefficient in electrons electron^-1 cm^-1= 4.4\n" + ] + } + ], + "source": [ + "#cal of alpha coefficient\n", + "#intiation of all variables\n", + "# Chapter 8\n", + "import math\n", + "print\"Example 8.3, Page:120 \\n \\n\"\n", + "# Given:\n", + "n0=1;# initial primary electrons\n", + "n=1.6*10**4;\n", + "x=2.2; #distance in cm\n", + "\n", + "# Solution:\n", + "a=math.log(n/n0)/(x);\n", + "\n", + "print\"The alpha coefficient in electrons electron^-1 cm^-1=\",round(a,2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 8.4;pg no: 121" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 8.4, Page:121 \n", + " \n", + "\n", + "\n", + " The potential at the inner surface is = (V) 1600.0\n", + "\n", + " The field at the inner surface is =(V/cm) 411.8\n", + "\n", + " \n", + " The potential at the outer surface is =(V) 0.0\n", + "\n", + " The field at the outer surface in (V/cm)= 30.88\n", + "\n", + " \n", + " The potential in mid-way between the cylinder is =(V) 383.482616627\n", + "\n", + " The field in mid-way between the cylinderis =(V/cm) 57.46\n" + ] + } + ], + "source": [ + "#cal of potential,field\n", + "#intiation of all variables\n", + "# Chapter 8\n", + "print\"Example 8.4, Page:121 \\n \\n\"\n", + "# Given:\n", + "import math\n", + "V=1600.;# potential across the electrodes\n", + "di=3.;# inner diameter\n", + "do=40.;# outer diameter\n", + "a=1.5;#in mm\n", + "A=20.;#in mm\n", + "\n", + "# Solution:\n", + "# Part(a)At the inner surface\n", + "r1=1.5;# in mm\n", + "V1=V*(math.log(A/r1)/math.log(A/a));\n", + "X1=V/(r1*(math.log(A/a)));\n", + "print\"\\n The potential at the inner surface is = (V)\",V1\n", + "print\"\\n The field at the inner surface is =(V/cm)\",round(X1,2)\n", + "# Part(b)At the outer surface\n", + "r2=20.;# in mm\n", + "V2=V*(math.log(A/r2)/math.log(A/a));\n", + "X2=V/(r2*(math.log(A/a)));\n", + "print\"\\n \\n The potential at the outer surface is =(V)\",V2\n", + "print\"\\n The field at the outer surface in (V/cm)=\",round(X2,2)\n", + "# Part(c)In mid-way between the cylinder\n", + "r3=(A+a)/2.;# in mm\n", + "V3=V*(math.log(A/r3)/math.log(A/a));\n", + "X3=V/(r3*(math.log(A/a)));\n", + "print\"\\n \\n The potential in mid-way between the cylinder is =(V)\",V3\n", + "print\"\\n The field in mid-way between the cylinderis =(V/cm)\",round(X3,3)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 8.5;pg no: 122" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 8.5, Page:122 \n", + " \n", + "\n", + "The resolving time of the given system in microseconds is = 1711.0\n" + ] + } + ], + "source": [ + "#cal of resolving time\n", + "#intiation of all variables\n", + "# Chapter 8\n", + "print\"Example 8.5, Page:122 \\n \\n\"\n", + "# Given:\n", + "ma1=3600.;# counts in 3 min\n", + "mb1=2400.;# counts in 5 min\n", + "mab1=9900.;# counts in 6 min\n", + "\n", + "# Solution:\n", + "ma=ma1/3;\n", + "mb=mb1/5;\n", + "mab=mab1/6;\n", + "\n", + "t1=(ma+mb-mab)/(mab**2-ma**2-mb**2);\n", + "t2=t1*60;# in seconds\n", + "t=t2*1000000;# in microseconds\n", + "print\"The resolving time of the given system in microseconds is =\",round(t)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 8.6;pg no: 122" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 8.6, Page:122 \n", + " \n", + "\n", + "The resolving time of the given system in microseconds is = 2490.0\n", + "\n", + " The true count rate of unknown sample in (cpm)= 2868.0\n" + ] + } + ], + "source": [ + "#cal of The resolving time,true count rate\n", + "#intiation of all variables\n", + "# Chapter 8\n", + "print\"Example 8.6, Page:122 \\n \\n\"\n", + "# Given:\n", + "ma1=3321.;# counts in 3 min\n", + "mb1=2862.;# counts in 2 min\n", + "mab1=4798.;# counts in 2 min\n", + "m=1080.;# counts in 30 min\n", + "muk1=5126.;# counts in 2 min\n", + "# Solution:\n", + "ma=ma1/3.;\n", + "mb=mb1/2.;\n", + "mab=mab1/2.;\n", + "mbc=m/30.;\n", + "muk=muk1/2.;\n", + "t1=(ma+mb-mab-mbc)/(mab**2.-ma**2.-mb**2.);# in min\n", + "t2=t1*60.;# in seconds\n", + "t=t2*1000000.;# in microseconds\n", + "print\"The resolving time of the given system in microseconds is =\",round(t)\n", + "n=muk/(1-muk*t1);# true count rate\n", + "print\"\\n The true count rate of unknown sample in (cpm)=\",round(n)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 8.7;pg no: 123" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 8.7, Page:123 \n", + " \n", + "\n", + "\n", + " The resolving time of the given system in microseconds is = 227.74452861\n", + "\n", + " The counting loss of sample A in (cpm)= 359.0\n", + "\n", + " The counting loss of sample B in (cpm)= 460.0\n", + "\n", + " The counting loss of sample AB in (cpm)= 1523.0\n" + ] + } + ], + "source": [ + "#cal of counting loss of sample A,B,C\n", + "#intiation of all variables\n", + "# Chapter 8\n", + "print\"Example 8.7, Page:123 \\n \\n\"\n", + "# Given:\n", + "\n", + "ma=9728.;# cpm\n", + "mb=11008.;# cpm\n", + "mab=20032.;# cpm\n", + "\n", + "# Solution:\n", + "\n", + "t1=(ma+mb-mab)/(mab**2-ma**2-mb**2);# in min\n", + "\n", + "t2=t1*60;# in seconds\n", + "t=t2*1000000;# in microseconds\n", + "print\"\\n The resolving time of the given system in microseconds is =\",t\n", + "\n", + "#From true count rate equation we have, n=muk/(1-muk*t).\n", + "# This implies, n-m=m^2*t where n-m corresponds to counting loss\n", + "na=ma**2*t1;# For sample A\n", + "nb=mb**2*t1;# For sample B\n", + "nab=mab**2*t1;# For sample AB\n", + "print\"\\n The counting loss of sample A in (cpm)=\",round(na)\n", + "print\"\\n The counting loss of sample B in (cpm)=\",round(nb)\n", + "print\"\\n The counting loss of sample AB in (cpm)=\",round(nab)\n", + "# NOTE: The resolving time of the given system in microseconds is give 222.7. This is a calculation error in the textbook." + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/_Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/chapter9.ipynb b/_Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/chapter9.ipynb new file mode 100644 index 00000000..ae812b40 --- /dev/null +++ b/_Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/chapter9.ipynb @@ -0,0 +1,356 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "#chapter 9:isotopes for nuclear reactions" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 9.1;pg no: 126" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 9.1, Page:126 \n", + " \n", + "\n", + "The value of numerical constant is = 0.0373\n" + ] + } + ], + "source": [ + "#cal of value of numerical constant\n", + "#intiation of all variables\n", + "# Chapter 9\n", + "print\"Example 9.1, Page:126 \\n \\n\"\n", + "#Give:\n", + "t=3600.;# in seconds\n", + "F=96500.;# in columbs\n", + "# Formula: m=0.0373fMit, Faraday's law: m=(itE)/F\n", + "#Solution:\n", + "constant=t/F;\n", + "print\"The value of numerical constant is =\",round(constant,4)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 9.2;pg no: 126" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 9.2, Page:126 \n", + " \n", + "\n", + "The ion current should be in (mA)= 0.4167\n" + ] + } + ], + "source": [ + "#cal of ion current\n", + "#intiation of all variables\n", + "# Chapter 9\n", + "print\"Example 9.2, Page:126 \\n \\n\"\n", + "# Given:\n", + "m1=24*10**-6;# g per day\n", + "m2=10**-2;# g per day\n", + "i1=10**-6;# in A\n", + "#Formula: i1*m2=m1*i2\n", + "#Solution:\n", + "i2=(i1*m2)/m1;\n", + "i=i2/10**-3;# in mA\n", + "print\"The ion current should be in (mA)=\",round(i,4)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 9.3;pg no: 126" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 9.3, Page:126 \n", + " \n", + "\n", + "The net yield is = 6.0337\n" + ] + } + ], + "source": [ + "#cal of net yield\n", + "#intiation of all variables\n", + "# Chapter 9\n", + "print\"Example 9.3, Page:126 \\n \\n\"\n", + "#Given:\n", + "f=1.0014;# seperation factor\n", + "s=4;# series\n", + "p=6;# parallel\n", + "# Note: The global yield for s stages in series is(f)^s and each parallel stages simply multiplies the yield of the stage, Hence overall yield with p parallel stages (each with s stages in series) will be Y=p*(f)^s\n", + "#Solution:\n", + "Y=p*(f)**s;\n", + "print\"The net yield is =\",round(Y,4)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 9.4;pg no: 127" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 9.14, Page:127 \n", + " \n", + "\n", + "\n", + " The no. of stages will be = 1130.0\n", + "\n", + " The no. of stages will be = 3451.0\n" + ] + } + ], + "source": [ + "#cal of no. of stages\n", + "#intiation of all variables\n", + "# Chapter 9\n", + "print\"Example 9.14, Page:127 \\n \\n\"\n", + "# Given:\n", + "import math\n", + "f=1.0014;# seperation factor\n", + "#Solution: Part (a)\n", + "A1=3.5/0.72;# total enrichment\n", + "n1=math.log(A1)/math.log(f);\n", + "print\"\\n The no. of stages will be = \",round(n1)\n", + "\n", + "#Solution: Part (b)\n", + "A2=90/0.72;# total enrichment\n", + "n2=math.log(A2)/math.log(f);\n", + "print\"\\n The no. of stages will be =\",round(n2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 9.5;pg no: 127" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 9.5, Page:127 \n", + " \n", + "\n", + "The overall seperation factor is = 1.105\n" + ] + } + ], + "source": [ + "#cal of overall seperation factor\n", + "#intiation of all variables\n", + "# Chapter 9\n", + "print\"Example 9.5, Page:127 \\n \\n\"\n", + "# Given:\n", + "f=1.01;# seperation factor\n", + "n=10;# plates\n", + "#Solution: \n", + "A=f**n;\n", + "print\"The overall seperation factor is =\",round(A,3)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 9.6;pg no: 127" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 9.6, Page:127 \n", + " \n", + "\n", + "The overall seperation factor is = 1.173\n" + ] + } + ], + "source": [ + "#cal of overall seperation factor\n", + "#intiation of all variables\n", + "# Chapter 9\n", + "print\"Example 9.6, Page:127 \\n \\n\"\n", + "# Given:\n", + "f=1.01;# seperation factor\n", + "n=16;# plates\n", + "#Solution: \n", + "A=f**n;\n", + "print\"The overall seperation factor is =\",round(A,3)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 9.7;pg no: 128" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 9.7, Page:128 \n", + " \n", + "\n", + "The enthalpy for the exchange reaction in (J)= 4829.008\n" + ] + } + ], + "source": [ + "#cal of enthalpy for the exchange reaction\n", + "#intiation of all variables\n", + "# Chapter 9\n", + "print\"Example 9.7, Page:128 \\n \\n\"\n", + "# Given:\n", + "import math\n", + "k1=3.78;\n", + "k2=2.79;\n", + "t1=298.;# in K\n", + "t2=353.;# in K\n", + "R=8.314# Gas constant\n", + "# Formula: log(k1/k2)=(H/R)*((t2-t1)t1*t2)\n", + "# Solution:\n", + "H=R*math.log(k1/k2)/((t2-t1)/(t1*t2));\n", + "print\"The enthalpy for the exchange reaction in (J)=\",round(H,3)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##example 9.8;pg no: 128" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 9.8, Page:128 \n", + " \n", + "\n", + "The single stage seperation factor is = 2.67\n" + ] + } + ], + "source": [ + "#cal of single stage seperation factor\n", + "#intiation of all variables\n", + "# Chapter 9\n", + "print\"Example 9.8, Page:128 \\n \\n\"\n", + "# Given:\n", + "a1=0.015;\n", + "a2=0.04;\n", + "# Solution: Defining the seperation factor f as approximately equal to (a2/a1) where a1, a2 are the relative abundances of the isotope of interest in the initial and final fractions, we have\n", + "f=(a2/a1);\n", + "print\"The single stage seperation factor is = \",round(f,2)" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/_Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/screenshots/Screenshot_(56).png b/_Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/screenshots/Screenshot_(56).png new file mode 100644 index 00000000..d9f1e926 Binary files /dev/null and b/_Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/screenshots/Screenshot_(56).png differ diff --git a/_Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/screenshots/Screenshot_(57).png b/_Nuclear_Chemistry_through_Problems_by__H._J._Arnikar_and_N._S._Rajurkar/screenshots/Screenshot_(57).png new file mode 100644 index 00000000..8c463815 Binary files /dev/null and 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of Antennas" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.1, Page 37" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from scipy.integrate import quad,dblquad\n", + "\n", + "#formula for beam solid angle theta_a=double_integration of d_omega\n", + "theta_a=quad(lambda x:1,0,2*pi)[0]*quad(lambda x:sin(x),0,pi/6)[0]\n", + "print 'Exact Beam Solid Angle:',theta_a,'steradians'\n", + "\n", + "#formula for approx angle=delta1*delta2\n", + "delta1=pi/3\n", + "delta2=pi/3\n", + "theta_a1=delta1*delta2\n", + "theta_a1=delta1**2\n", + "print 'Approximate Beam Solid Angle:',theta_a1,'steradians'" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Exact Beam Solid Angle: 0.841787214477 steradians\n", + "Approximate Beam Solid Angle: 1.09662271123 steradians\n" + ] + } + ], + "prompt_number": 31 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.7, Page 52" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import scipy\n", + "\n", + "#The half power point of the pattern occurs at 60 degrees. Therefore theta_1r=2*pi/3\n", + "theta_1r=(2*pi)/3\n", + "theta_2r=(2*pi)/3\n", + "\n", + "#Given U=B0*cos(theta)\n", + "exact_theta_a=dblquad(lambda x,y:cos(x)*sin(x), 0, (2*pi), lambda x:0, lambda x:(pi/2))\n", + "print 'Exact Beam Solid Angle:',exact_theta_a[0],'steradians'\n", + "\n", + "#Formula for approx theta = theta_1r*theta_2r\n", + "approx_theta_a=theta_1r*theta_2r\n", + "print 'Approximate Beam Solid Angle:',approx_theta_a,'steradians'\n", + "\n", + "#formula for exact directivity=4*pi/exact_beam_angle\n", + "exact_direct=((4*pi)/(exact_theta_a[0]))\n", + "\n", + "#formula for approx directivity=4*pi/approx_beam_angle\n", + "approx_direct=((4*pi)/(approx_theta_a))\n", + "\n", + "#exact directivity in dB\n", + "exact_direct_db=10*log10(exact_direct)\n", + "\n", + "#approx directivity in dB\n", + "approx_direct_db=10*log10(approx_direct)\n", + "\n", + "print 'Exact directivity:',exact_direct_db,'dB'\n", + "print 'Approx. directivity:',approx_direct_db,'dB'" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Exact Beam Solid Angle: 3.14159265359 steradians\n", + "Approximate Beam Solid Angle: 4.38649084493 steradians\n", + "Exact directivity: 6.02059991328 dB\n", + "Approx. directivity: 4.57092636745 dB\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.8, Page 58" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import scipy\n", + "\n", + "#Maximum intensity\n", + "u_max=1\n", + "\n", + "#Calculation of radiated power\n", + "p_rad=dblquad(lambda x,y:(sin(x)**2)*sin(x),0,2*pi,lambda x:0,lambda x:pi)\n", + "print 'Radiated Power:',p_rad[0],'W'\n", + "\n", + "#Calulation of maximum directivity\n", + "D0=(4*pi)/(p_rad[0])\n", + "\n", + "#Directivity in dB\n", + "D0_db=10*log10(D0)\n", + "print 'Directivity:',D0_db,'dB'\n", + "\n", + "deg=90\n", + "\n", + "#Calculation od directivity\n", + "D0_1=101/(deg-0.0027*deg**2)\n", + "D0_1_db=10*log10(D0_1)\n", + "print 'Directivity:',D0_1_db,'dB'\n", + "\n", + "#Calculation of directivity\n", + "D0_2=(-172.4)+(191*sqrt((0.818+(1/deg))))\n", + "D0_2_db=10*log10(D0_2)\n", + "print 'Directivity:',D0_2,'dB'" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Radiated Power: 8.37758040957 W\n", + "Directivity: 1.76091259056 dB\n", + "Directivity: 1.70982984843 dB\n", + "Directivity: 0.346803154212 dB\n" + ] + } + ], + "prompt_number": 34 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.9(a), Page 61" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import scipy\n", + "\n", + "B0=1\n", + "#Maximum intensity\n", + "u_max=1\n", + "\n", + "#Array containing angles in radians\n", + "a=sin(array([10,20,30,40,50,60,70,80,90,100,110,120,130,140,150,160,170,180])*pi/180)**2\n", + "\n", + "#Calculation of radiated power\n", + "p_rad1=B0*((pi/18)**2)*sum(a)*sum(a)\n", + "print 'Power Radiated:',p_rad1,'W'\n", + "\n", + "#Calculation of directivity\n", + "D0=(4*pi)/(p_rad1)\n", + "\n", + "print 'Directivity using numerical techniques:',D0\n", + "\n", + "#Calu=culation of radiated power\n", + "a=quad(lambda x:sin(x)**2,0,pi)\n", + "b=quad(lambda x:sin(x)**2,0,pi)\n", + "p_rad2=a[0]*b[0]\n", + "\n", + "#Directivity\n", + "D01=(4*pi)/(p_rad2)\n", + "\n", + "print 'Directivity:',D01" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Power Radiated: 2.46740110027 W\n", + "Directivity using numerical techniques: 5.09295817894\n", + "Directivity: 5.09295817894\n" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.9(b), Page 63" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import scipy\n", + "\n", + "\n", + "B0=1\n", + "\n", + "#Maximum intensity\n", + "u_max=1\n", + "\n", + "#Arrays containing angles in radians\n", + "a=sin(array([5,15,25,35,45,55,65,75,85])*pi/180)**2\n", + "b=sin(array([5,15,25,35,45,55,65,75,85])*pi/180)**2\n", + "\n", + "#Calculation of radiated power\n", + "p_rad=B0*((pi/18)**2)*(2*sum(a))*(2*sum(b))\n", + "\n", + "#Calculation of directivity\n", + "D0=(4*pi*u_max)/(p_rad)\n", + "\n", + "print 'Directivity using 18 divisions:',D0" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Directivity using 18 divisions: 5.09295817894\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.10, Page 68" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import scipy\n", + "\n", + "#maximum intensuty\n", + "u_max=1\n", + "B0=1\n", + "\n", + "#Input impedance in Ohms\n", + "inp_imp=73\n", + "#Characteristic impedance in Ohms\n", + "char_imp=50\n", + "\n", + "#Calculation of radiated power\n", + "p_rad=B0*quad(lambda x:1,0,2*pi)[0]*quad(lambda x:sin(x)**4,0,pi)[0]\n", + "\n", + "#Calulation of directivity\n", + "D0=(4*pi*u_max)/(p_rad)\n", + "\n", + "#conduction & dielectric efficiency ecd=1 since antenna is loseless\n", + "ecd=1\n", + "\n", + "#Maximum Gain\n", + "G0=ecd*D0\n", + "G0_db=10*log10(G0)\n", + "\n", + "#Reflection Coefficient Tau\n", + "tau=float(inp_imp-char_imp)/float(inp_imp+char_imp)\n", + "\n", + "#Reflection efficiency=1-tau**2\n", + "er=1-tau**2\n", + "er_db=10*log10(er)\n", + "\n", + "#Total efficiency\n", + "e0=er*ecd\n", + "e0_db=10*log10(e0)\n", + "\n", + "#Absolute Gain\n", + "G0_abs=e0*D0\n", + "G0abs_db=10*log10(G0_abs)\n", + "\n", + "print 'Maximum Gain:',G0_db\n", + "\n", + "print 'Reflection efficiency:',er_db\n", + "\n", + "print 'Total efficiency:',e0_db\n", + "\n", + "print 'Absolute Gain:',G0abs_db" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Maximum Gain: 2.29848855242\n", + "Reflection efficiency: -0.154573670944\n", + "Total efficiency: -0.154573670944\n", + "Absolute Gain: 2.14391488148\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.11, Page 77" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import scipy\n", + "\n", + "#unit vector of the wave\n", + "rho_w=array([1,0])\n", + "\n", + "#unit vector of the electric field\n", + "rho_a=array([1/sqrt(2),1/sqrt(2)])\n", + "\n", + "#Polarization factor\n", + "PLF=abs(dot(rho_w,rho_a))**2\n", + "print 'Polarization Factor:',PLF" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "0.5\n" + ] + } + ], + "prompt_number": 56 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.12, Page 78" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import scipy\n", + "\n", + "#unit vector of the wave\n", + "rho_w=array([1/sqrt(2),1/sqrt(2)])\n", + "\n", + "#unit vector of the electric field\n", + "rho_a=array([1/sqrt(2),-1/sqrt(2)])\n", + "\n", + "#Polarization Factor\n", + "PLF=abs(dot(rho_w,rho_a))**2\n", + "\n", + "print 'Polarization Factor:',PLF" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "0.0\n" + ] + } + ], + "prompt_number": 57 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.13, Page 86" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import scipy\n", + "\n", + "#Radiation Resistance\n", + "rad_res=73\n", + "\n", + "#Frequency of antenna\n", + "f=10**8\n", + "\n", + "#Velocity\n", + "v=3*10**8\n", + "\n", + "#Wavelength\n", + "lamda=v/f\n", + "\n", + "#Length of antenna\n", + "l=lamda/2\n", + "\n", + "#Perimeter of the antenna\n", + "b=(3*10**-4)*lamda\n", + "C=2*pi*b\n", + "\n", + "#value of omega\n", + "w=2*pi*f\n", + "\n", + "#Constant\n", + "mu0=4*pi*10**-7\n", + "\n", + "#Conductivity\n", + "sigma=5.7*10**7\n", + "\n", + "#High frequency resistance\n", + "Rhf=(l/C)*(sqrt((w*mu0)/(2*sigma)))\n", + "\n", + "#Load resistance\n", + "Rl=Rhf/2\n", + "\n", + "#calculation of conduction & dielectric efficiency\n", + "ecd=(rad_res)/(rad_res+Rl)\n", + "ecd_db=10*log10(ecd)\n", + "\n", + "print 'Conduction-dielectric efficiency:',ecd_db" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Conduction-dielectric efficiency: -0.0138216614754\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.16, Page 98 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import scipy\n", + "\n", + "lamda=1\n", + "\n", + "#Maximum directivity of transmitter\n", + "D0_t_db=16\n", + "D0_t=10**(float(D0_t_db)/10)\n", + "\n", + "#Maximum directivity of receiver\n", + "D0_r_db=20\n", + "D0_r=10**(D0_r_db/10)\n", + "\n", + "#Reflection coeficients of transmitter and receiver\n", + "tau_r=0.1\n", + "tau_t=0.2\n", + "\n", + "#Power at transmitter\n", + "P_t=2\n", + "\n", + "#Calculation of Power to the receiver\n", + "P_r=(1-tau_r**2)*(1-tau_t**2)*((lamda/(4*pi*100*lamda))**2)*D0_t*D0_r*P_t\n", + "print 'Power delivered to the load of receiver:',P_r,'W'" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Power delivered to the load of receiver: 0.00479199874075 W\n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.18, Page 108" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import scipy\n", + "\n", + "#antenna temp at receiver terminals\n", + "Ta=150\n", + "\n", + "#physical temp of transmission line\n", + "T0=300\n", + "\n", + "#thermal efficiency of the antennna\n", + "eA=0.99\n", + "\n", + "#antenna physical temperature\n", + "Tp=300\n", + "l=1\n", + "\n", + "#antenna temp at antenna terminals due to physical temperature\n", + "T_ap=Tp*(1/eA-1)\n", + "\n", + "#Loss of waveguide in dB/m\n", + "alpha_db=0.13\n", + "\n", + "#Loss of waveguide in Np/m\n", + "alpha_np=alpha_db/0.868\n", + "\n", + "#Calulation of effective temperature\n", + "T_A=Ta*exp(-l*alpha_np*2)+T_ap*exp(-l*alpha_np*2)+T0*(1-exp(-l*alpha_np*2))\n", + "print 'Effective temperature:',T_A,'K'" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Effective temperature: 191.071984919 K\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit