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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('')
    3 # DBsection('')
    4 # KEYWORDS('')
    5 # TitleText1('Calculus: Early Transcendentals')
    6 # EditionText1('2')
    7 # AuthorText1('Rogawski')
    8 # Section1('10.7')
    9 # Problem1('14')
   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 #$power1 = 2
   27 #$power2 = 1
   28 
   29 $power1 = random(2, 9);
   30 $power2 = random(2, 4);
   31 
   32 Context()->variables->set(x=>{limits=>[.1,10]});
   33 Context()->variables->set(n=>{limits=>[1,5]});
   34 
   35 $finalpower = Formula("2*$power1*n - $power2")->reduce;
   36 $interimpower = Formula("2*$power1")->reduce;
   37 $mcterms = Formula("(-1)^(n+1)*x^($finalpower)/(2*n)!")->reduce;
   38 $denominator = Formula("x^($power2)")->reduce;
   39 $interval = Interval("(-inf,inf)"); #added LAD
   40 
   41 $point1=.5;
   42 $point2=1.5;
   43 $point3=2.5;
   44 $mcterms->{test_points} = [[1,$point1],[2,$point2],[3,$point3]];
   45 
   46 Context()->texStrings;
   47 
   48 BEGIN_TEXT
   49 \{ beginproblem() \}
   50 \{ textbook_ref_exact("Rogawski ET 2e", "10.7", "14") \}
   51 $PAR
   52 Find the Maclaurin series and corresponding interval of convergence of the following function.
   53 $PAR
   54 \[f(x) = \frac{1 - \cos (x^{$power1})}{$denominator} \]
   55 $PAR
   56 \( f(x) = \sum\limits_{n=1}^{\infty} \) \{ ans_rule() \}
   57 $PAR
   58 The interval of convergence for this power series is: \{ans_rule() \}
   59 END_TEXT
   60 Context()->normalStrings;
   61 
   62 #Answer Check Time!
   63 ANS($mcterms->cmp);
   64 ANS($interval->cmp);
   65 
   66 Context()->texStrings;
   67 SOLUTION(EV3(<<'END_SOLUTION'));
   68 $PAR
   69 $SOL
   70 
   71 Substituting \( x^{$power1} \) for \( x \) in the Maclaurin series for \( \cos x\) gives
   72 \[ \cos (x^{$power1}) = \sum_{n=0}^{\infty} (-1)^n \frac{(x^{$power1})^{2n}}{(2n)!} = 1 + \sum_{n=1}^{\infty} (-1)^n \frac{x^{$interimpower n}}{(2n)!} \]
   73 Thus,
   74 \[ 1 - \cos (x^{$power1}) = 1 - \left( 1 + \sum_{n=1}^{\infty} (-1)^n \frac{x^{$interimpower n}}{(2n)!} \right) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^{$interimpower n}}{(2n)!} \]
   75 and
   76 \[ \frac{1 - \cos (x^{$power1})}{$denominator} = \frac{1}{$denominator} \sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^{$interimpower n}}{(2n)!} = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^{$finalpower}}{(2n)!} .\]
   77 This series is valid for all \(x\ne0\).  Thus, the interval of convergence is \((-\infty,0)\cup (0,\infty) \).
   78 
   79 
   80 
   81 END_SOLUTION
   82 
   83 ENDDOCUMENT()

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