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Tue Nov 8 15:17:41 2011 UTC (2 years, 5 months ago) by aubreyja
File size: 2770 byte(s)
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();
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();