View of /trunk/NationalProblemLibrary/WHFreeman/Rogawski_Calculus_Early_Transcendentals_Second_Edition/10_Infinite_Series/10.4_Absolute_and_Conditional_Convergence/10.4.13.pg

    1 # DBsubject('Calculus')
    2 # DBchapter('Infinite Series and Sequences')
    3 # DBsection('Alternating Series')
    4 # KEYWORDS('calculus', 'series', 'sequences', 'alternating series', 'convergence')
    5 # TitleText1('Calculus: Early Transcendentals')
    6 # EditionText1('2')
    7 # AuthorText1('Rogawski')
    8 # Section1('10.4')
    9 # Problem1('13')
   10 # Author('LA Danielson')
   11 # Institution('The College of Idaho')
   12 DOCUMENT();
   13 
   14 #Load Necessary Macros
   15 
   16 loadMacros("PG.pl", "PGbasicmacros.pl", "PGchoicemacros.pl", "PGanswermacros.pl", );
   17 loadMacros("Parser.pl");
   18 loadMacros("freemanMacros.pl");
   19 
   20 Context()->variables->add(n=>'Real');
   21 
   22 #Book Values
   23 #$series = \sum (-1)^(n+1)/n^4
   24 
   25 $errorp = random(4,5,1);
   26 $plus = random(1,2,1);
   27 $exponent = $errorp +$plus;
   28 $tolerance =10**(-$errorp);
   29 
   30 
   31 $terms = Formula("(-1)^(n+1)/(n^($exponent))");
   32 
   33 #$error = 10**($errorp/$exponent)-1;
   34 $error = sprintf "%.2f",10**($errorp/$exponent)-1;
   35 $N = Real(int($error) + 1);
   36 $answer=0;
   37 $sum = "1";
   38 @sign = ("-","+");
   39 
   40 for($i=1;$i<=$N;$i++){
   41   $answer+=$terms->eval(n=>$i);
   42   if($i>1){
   43       $sum.="$sign[$i%2]\frac{1}{$i^{$exponent}}";
   44   }
   45 }
   46 
   47 Context()->texStrings;
   48 
   49 BEGIN_TEXT
   50 \{ beginproblem() \}
   51 \{ textbook_ref_exact("Rogawski ET 2e", "10.4", "13") \}
   52 $PAR
   53 Approximate the value of the series to within an error of at most \(10^{-$errorp}\).
   54 
   55 
   56 \[ \sum_{n=1}^{\infty} \frac{ (-1)^{n+1}}{n^{$exponent}} \]
   57 According to Equation (2):
   58 \[ \left| S_N-S \right|\le a_{N+1} \]
   59 what is the smallest value of \(N\) that approximates \(S\) to within an error of at most \(10^{-$errorp}\)?
   60 $BR
   61 \(N = \) \{ ans_rule() \}
   62 $PAR
   63 \(S\approx\) \{ ans_rule() \}
   64 END_TEXT
   65 
   66 Context()->normalStrings;
   67 Context()->flags->set(tolerance=>$tolerance);
   68 
   69 #Answer Check Time!
   70 ANS($N->cmp);
   71 ANS(Real("$answer")->cmp);
   72 
   73 Context()->texStrings;
   74 SOLUTION(EV3(<<'END_SOLUTION'));
   75 $PAR
   76 $SOL
   77 Let \( S =  \sum _{n=1}^{\infty} \frac{ (-1)^{n+1}}{n^{$exponent}} \), so that \( a_n = \frac{1}{n^{$exponent}} \).  By Equation (2),
   78 
   79 \[ | S_N - S | \le a_{N+1} = \frac{1}{(N+1)^{$exponent}}. \]
   80 
   81 To make the error less than \(10^{-$errorp}\), we must choose \( N \) so that
   82 \[ \frac{1}{(N+1)^{$exponent}} < 10^{-$errorp} \qquad\textrm{or}\qquad N > 10^\frac{$errorp}{$exponent}-1\approx $error. \]
   83 
   84 The smallest value that satisfies the required inequality is then \( N=$N \).  Thus,
   85 \[ S \approx S_{$N} = $sum = $answer \]
   86 
   87 
   88 END_SOLUTION
   89 
   90 ENDDOCUMENT()