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Rogawski problems contributed by publisher WHFreeman. These are a subset of the problems available to instructors who use the Rogawski textbook. The remainder can be obtained from the publisher.
1 ## DBsubject('Calculus') 2 ## DBchapter('Limits and Derivatives') 3 ## DBsection('Definition of the Derivative') 4 ## KEYWORDS('calculus', 'derivatives', 'slope') 5 ## TitleText1('Calculus: Early Transcendentals') 6 ## EditionText1('2') 7 ## AuthorText1('Rogawski') 8 ## Section1('10.2') 9 ## Problem1('11') 10 ## Author('LA Danielson') 11 ## Institution('The College of Idaho') 12 13 DOCUMENT(); 14 loadMacros("PG.pl","PGbasicmacros.pl","PGanswermacros.pl"); 15 loadMacros("Parser.pl"); 16 loadMacros("freemanMacros.pl"); 17 #loadMacros("PGauxiliaryFunctions.pl"); 18 #loadMacros("PGgraphmacros.pl"); 19 20 #$showPartialCorrectAnswers=1; 21 22 $n0 = Real(random(2,9,1)); 23 24 sub partial_sum { 25 $n = shift; #number of terms 26 $sum = 0; 27 28 for($i=$n0; $i<($n0+$n); $i++){ 29 $sum += (1/($i+1))-(1/($i+2)); 30 31 } 32 33 return $sum; 34 } 35 36 37 $ans1=partial_sum(3); 38 $ans2=partial_sum(4); 39 $ans3=partial_sum(5); 40 $ans4=1/($n0+1); 41 #for solution formatting 42 $ns[0] = $n0; 43 for($i=1; $i<=6; $i++){ 44 $ns[$i]= $ns[$i-1]+1; 45 } 46 47 $num1=$ns[4]-$ns[1]; 48 $den1=$ns[4]*$ns[1]; 49 50 $num2=$ns[5]-$ns[1]; 51 $den2=$ns[5]*$ns[1]; 52 53 $num3=$ns[6]-$ns[1]; 54 $den3=$ns[6]*$ns[1]; 55 56 Context()->texStrings; 57 BEGIN_TEXT 58 \{ beginproblem() \} 59 \{ textbook_ref_exact("Rogawski ET 2e", "10.2","11") \} 60 $PAR 61 62 Calculate \(S_3,S_{4}\), and \(S_{5}\) and then find the sum for the telescoping series 63 \[S=\sum_{n=$n0}^\infty \left(\frac{1}{n+1}-\frac{1}{n+2}\right)\] 64 65 $PAR \(S_3\) = \{ans_rule()\} 66 $PAR \(S_{4}\) = \{ans_rule()\} 67 $PAR \(S_{5}\) = \{ans_rule()\} 68 $PAR \(S\) = \{ans_rule()\} 69 END_TEXT 70 71 Context()->normalStrings; 72 73 ANS(Real($ans1)->cmp); 74 ANS(Real($ans2)->cmp); 75 ANS(Real($ans3)->cmp); 76 ANS(Real($ans4)->cmp); 77 78 Context()->texStrings; 79 SOLUTION(EV3(<<'END_SOLUTION')); 80 $PAR 81 $SOL 82 \[S_3=\left(\frac{1}{$ns[1]}-\frac{1}{$ns[2]}\right)+\left(\frac{1}{$ns[2]}-\frac{1}{$ns[3]}\right)+\left(\frac{1}{$ns[3]}-\frac{1}{$ns[4]}\right)=\frac{1}{$ns[1]}-\frac{1}{$ns[4]}=\frac{$num1}{$den1};\] 83 84 \[S_4=S_3+\left(\frac{1}{$ns[4]}-\frac{1}{$ns[5]}\right)=\frac{1}{$ns[1]}-\frac{1}{$ns[5]}=\frac{$num2}{$den2};\] 85 86 \[S_5=S_4+\left(\frac{1}{$ns[5]}-\frac{1}{$ns[6]}\right)=\frac{1}{$ns[1]}-\frac{1}{$ns[6]}=\frac{$num3}{$den3}.\] 87 88 The general term in the sequence of partial sums is 89 \[S_N=\left(\frac{1}{$ns[1]}-\frac{1}{$ns[2]}\right)+\left(\frac{1}{$ns[2]}-\frac{1}{$ns[3]}\right)+\left(\frac{1}{$ns[3]}-\frac{1}{$ns[4]}\right)+\cdots +\left(\frac{1}{N+$ns[0]}-\frac{1}{N+$ns[1]}\right)=\frac{1}{$ns[1]}-\frac{1}{N+$ns[1]};\] 90 thus, 91 \[S=\lim_{N\rightarrow \infty} S_N=\lim_{N\rightarrow \infty}\left(\frac{1}{$ns[1]}-\frac{1}{N+$ns[1]}\right)=\frac{1}{$ns[1]}.\] 92 $PAR Thus the sum of the telescoping series is \(\frac{1}{$ns[1]}\). 93 END_SOLUTION 94 95 ENDDOCUMENT();
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