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Revision 2584 - (download) (annotate)
Tue Nov 8 15:17:41 2011 UTC (2 years, 5 months ago) by aubreyja
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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('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('47')
   10 # Author('Emily Price')
   11 # Institution('W.H.Freeman')
   12 DOCUMENT();
   13 
   14 
   15 
   16 #Load Necessary Macros
   17 
   18 loadMacros("PG.pl", "PGbasicmacros.pl", "PGchoicemacros.pl", "PGanswermacros.pl", );
   19 loadMacros("Parser.pl");
   20 loadMacros("freemanMacros.pl");
   21 
   22 
   23 Context()->variables->add(n=>'Real');
   24 
   25 #Book Values
   26 #exp1 = 3
   27 #exp2 = 2
   28 
   29 $exp2 = random(2, 5);
   30 $exp1 = $exp2 + random(1, 3);
   31 
   32 ($an, $approx, $frac, $term, $answer, $trueanswer,) = @{list_random(
   33   [ "\frac{1}{\sqrt[$exp1]{n^{$exp2}-n}}", "\frac{1}{\sqrt[$exp1]{n^{$exp2}}}", "\frac{$exp2}{$exp1}", "\sqrt[$exp1]{\frac{n^{$exp2}}{n^{$exp2} - n}}", "diverges", 'divergent'],
   34   [ "\frac{1}{\sqrt{n^{$exp1} - n^{$exp2}}}", "\frac{1}{\sqrt{n^{$exp1}}}", "\frac{$exp1}{2}",  "\sqrt{\frac{n^{$exp1}}{n^{$exp1} - n^{$exp2}}}", "converges", 'convergent'] )};
   35 
   36 
   37 
   38 #Let's try to make a multiple choice question
   39 $question = new_multiple_choice();
   40 $question->qa(' \( \sum\limits_{n=2}^{\infty} $an \) is:', $trueanswer);
   41 $question->makeLast( 'convergent', 'divergent');
   42 
   43 
   44 Context()->texStrings;
   45 
   46 BEGIN_TEXT
   47 \{ beginproblem() \}
   48 \{ textbook_ref_exact("Rogawski ET 2e", "10.5", "47") \}
   49 $PAR
   50 Determine convergence or divergence using any method covered so far.
   51 $PAR
   52 \{ $question->print_q() \}
   53 \{ $question->print_a() \}
   54 END_TEXT
   55 
   56 Context()->normalStrings;
   57 
   58 #Answer Check Time!
   59 ANS(radio_cmp($question->correct_ans));
   60 
   61 Context()->texStrings;
   62 SOLUTION(EV3(<<'END_SOLUTION'));
   63 $PAR
   64 $SOL
   65 This series is similar to a \( p \)-series; because
   66 \[ $an \approx $approx = \frac{1}{n^{$frac}}\]
   67 for large \( n \), we will apply the Limit Comparison Test comparing with the \( p\)-series with \( p = $frac\).  Now,
   68 \[ L = \lim_{n \to \infty} \frac{$an}{\frac{1}{n^{$frac}}} = \lim_{n \to \infty} $term = 1. \]
   69 The \( p \)-series with \( p = $frac \) $answer and \( L>0 \) exists; therefore, the series \( \sum\limits_{n=2}^{\infty} $an \) also $answer.
   70 
   71 
   72 
   73 END_SOLUTION
   74 
   75 ENDDOCUMENT()

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