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Tue Nov 8 15:17:41 2011 UTC (2 years, 1 month ago) by aubreyja
File size: 2319 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('Infinite Series and Sequences')
3 # DBsection('Absolute Convergence and the Root and Ratio Tests')
4 # KEYWORDS('calculus', 'series', 'sequences', 'convergence', 'root test', 'ratio test')
5 # TitleText1('Calculus: Early Transcendentals')
6 # EditionText1('2')
7 # AuthorText1('Rogawski')
8 # Section1('10.5')
9 # Problem1('48')
10 # Author('Emily Price')
11 # Institution('W.H.Freeman')
12 DOCUMENT();
13
14
15
17
21
22
24
25 #Book Values
26 #numerator = n^2 + 4n
27 #denominator = 3n^4 + 9
28
29 #building the numerator - n^2 + bn + c
30 $b1 = random(0, 9); 31$c1 = random(1, 9);
32 $numerator = Formula("n^2 +$b1*n + $c1")->reduce; 33 34 #building the denominator - an^4 + bn^3 + cn^2 + dn + e 35$a2 = random(2, 9);
36 $b2 = random(0, 9); 37$c2 = random(0, 9);
38 $d2 = random(1, 9); 39$e2 = random(0, 9);
40 $denominator = Formula("$a2 n^4 + $b2 n^3 +$c2 n^2 + $d2 n +$c2")->reduce;
41
42 #Let's try to make a multiple choice question
43 $question = new_multiple_choice(); 44$question->qa(' $$\sum\limits_{n=1}^{\infty} \frac{numerator}{denominator}$$ is:', 'convergent');
45 $question->makeLast( 'convergent', 'divergent'); 46 47 48 Context()->texStrings; 49 50 BEGIN_TEXT 51 \{ beginproblem() \} 52 \{ textbook_ref_exact("Rogawski ET 2e", "10.5", "48") \} 53$PAR
54 Determine convergence or divergence using any method covered so far.
55 $PAR 56 \{$question->print_q() \}
57 \{ $question->print_a() \} 58 END_TEXT 59 60 Context()->normalStrings; 61 62 #Answer Check Time! 63 ANS(radio_cmp($question->correct_ans));
64
65 Context()->texStrings;
66 SOLUTION(EV3(<<'END_SOLUTION'));
67 $PAR 68$SOL
69
70 This series is similar to a $$p$$-series; because
71 $\frac{numerator}{denominator} \approx \frac{n^2}{a2 n^4} = \frac{1}{a2 n^2}$
72 for large $$n$$, we will apply the Limit Comparison Test comparing with the $$p$$-series with $$p = 2$$.  Now,
73 $L = \lim_{n \to \infty} \frac{\frac{numerator}{denominator}}{\frac{1}{n^2}} = \lim_{n \to \infty} \frac{n^2(numerator)}{denominator} = \frac{1}{a2}.$
74 The $$p$$-series with $$p = 2$$ converges and $$L$$ exists; therefore, the series also converges.
75
76 END_SOLUTION
77
78 ENDDOCUMENT()