View of /trunk/NationalProblemLibrary/WHFreeman/Rogawski_Calculus_Early_Transcendentals_Second_Edition/10_Infinite_Series/10.4_Absolute_and_Conditional_Convergence/10.4.15.pg

    1 # DBsubject('Calculus')
    2 # DBchapter('')
    3 # DBsection('')
    4 # KEYWORDS('')
    5 # TitleText1('Calculus: Early Transcendentals')
    6 # EditionText1('2')
    7 # AuthorText1('Rogawski')
    8 # Section1('10.4')
    9 # Problem1('15')
   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/(n)(n+3)(n+4)
   24 
   25 $errorp = random(3,4,1);
   26 $a=Real(random(1,10,1));
   27 do{$b=Real(random(1,10,1));}until($b!=$a);
   28 if($errorp>3){#modify constants so magnitude of N similar for all
   29    $a+=70;
   30    $b+=70;
   31 }
   32 $ap1 = $a+1;
   33 $bp1 = $b+1;
   34 
   35 $tolerance =10**(-$errorp);
   36 
   37 
   38 $term = Formula("(-1)^(n+1)/((n+$a)(n+$b))");
   39 
   40 $error = sprintf "%.2f",(sqrt(($ap1+$bp1)**2-4*($ap1*$bp1-10**$errorp))-$ap1-$bp1)/2;
   41 $N = Real(int($error)+1);
   42 
   43 $answer=0;
   44 
   45 for($i=1;$i<=$N;$i++){
   46   $answer+=$term->eval(n=>$i);
   47 }
   48 
   49 Context()->texStrings;
   50 
   51 BEGIN_TEXT
   52 \{ beginproblem() \}
   53 \{ textbook_ref_exact("Rogawski ET 2e", "10.4", "15") \}
   54 $PAR
   55 Approximate the value of the series to within an error of at most \(10^{-$errorp}\).
   56 
   57 \[ \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{(n+$a)(n+$b)} \]
   58 $PAR
   59 According to Equation (2):
   60 \[ \left| S_N-S \right|\le a_{N+1} \]
   61 what is the smallest value of \(N\) that approximates \(S\) to within an error of at most \(10^{-$errorp}\)?
   62 $BR
   63 \(N = \) \{ ans_rule() \}
   64 $PAR
   65 \(S\approx\) \{ ans_rule() \}
   66 END_TEXT
   67 
   68 Context()->normalStrings;
   69 Context()->flags->set(tolerance=>$tolerance);
   70 
   71 #Answer Check Time!
   72 ANS($N->cmp);
   73 ANS(Real("$answer")->cmp);
   74 
   75 Context()->texStrings;
   76 SOLUTION(EV3(<<'END_SOLUTION'));
   77 $PAR
   78 $SOL
   79 Let \( S =  \sum _{n=1}^{\infty} \frac{ (-1)^{n+1}}{(n+$a)(n+$b)} \), so that \( a_n = \frac{1}{(n+$a)(n+$b)} \).  By Equation (2),
   80 
   81 
   82 \[ | S_N - S | \le a_{N+1} = \frac{1}{(N+$ap1)(N+$bp1)}. \]
   83 We must choose \( N \) so that
   84 \[ \frac{1}{(N+$ap1)(N+$bp1)} \le 10^{-$errorp} \quad \textrm{or} \quad (N+$ap1)(N+$bp1) \ge 10^{$errorp}. \]
   85 
   86 Solving this quadratic inequality yields \( N \ge $error\).   The smallest value that satisfies the required inequality is then \( N=$N \).
   87 
   88 Thus \[S \approx S_{$N} = \sum_{n=1}^{$N} \frac{ (-1)^{n+1}}{(n+$a)(n+$b)} =  $answer \]
   89 
   90 
   91 END_SOLUTION
   92 
   93 ENDDOCUMENT()