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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('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('30')
   10 ## Author('Keith Thompson')
   11 ## Institution('W.H.Freeman')
   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 = random(2,9,1);
   23 $base=random(3,7,2);
   24 $mult=random(2,6,2);
   25 $netb=$base-$n0*$mult;
   26 $p=$netb+$mult;
   27 
   28 $ans1=exp($netb)*exp($mult)/(exp($mult)-1);
   29 Context()->texStrings;
   30 BEGIN_TEXT
   31 \{ beginproblem() \}
   32 \{ textbook_ref_exact("Rogawski ET 2e", "10.2","30") \}
   33 $PAR
   34 
   35 Use the formula for the sum of a geometric series to find the sum or state that the series diverges (enter DIV for a divergent series).
   36 
   37 \[\sum_{n=$n0}^\infty e^{$base-$mult n}\]
   38 
   39 \(S=\) \{ans_rule()\}
   40 END_TEXT
   41 
   42 Context()->normalStrings;
   43 
   44 ANS(std_num_str_cmp($ans1,['DIV'])).
   45 
   46 Context()->texStrings;
   47 SOLUTION(EV3(<<'END_SOLUTION'));
   48 $PAR
   49 $SOL
   50 Re-write the series as
   51 \[\sum_{n=$n0}^\infty e^{$base}e^{-$mult n}=\sum_{n=$n0}^\infty e^{$base}\left(\frac{1}{e^{$mult}}\right)^n\]
   52 
   53 to recognize it as a geometric series with \(c=e^{$base}\left(\frac{1}{e^{$mult}}\right)^{$n0}=e^{$netb}\) and \(0<r=\frac{1}{e^{$mult}}<1\). Thus
   54 
   55 \[\sum_{n=$n0}^\infty e^{$base-$mult n}=e^{$netb}\frac{1}{1-\frac{1}{e^{$mult}}}=e^{$netb}\frac{e^{$mult}}{e^{$mult}-1}=\frac{e^{$p}}{e^{$mult}-1}.\]
   56 END_SOLUTION
   57 ENDDOCUMENT();

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