# DBsubject('Calculus')
## DBchapter('Infinite Sequences and Series')
# DBsection('Alternating Series')
# KEYWORDS('calculus', 'series', 'sequences', 'alternating series', 'convergence')
# TitleText1('Calculus: Early Transcendentals')
# EditionText1('2')
# AuthorText1('Rogawski')
# Section1('10.4')
# Problem1('13')
# Author('LA Danielson')
# Institution('The College of Idaho')
DOCUMENT();

#Load Necessary Macros

loadMacros("PG.pl", "PGbasicmacros.pl", "PGchoicemacros.pl", "PGanswermacros.pl", );
loadMacros("Parser.pl");
loadMacros("freemanMacros.pl");

Context()->variables->add(n=>'Real');

#Book Values
#$series = \sum (-1)^(n+1)/n^4

$errorp = random(4,5,1);
$plus = random(1,2,1);
$exponent = $errorp +$plus;
$tolerance =10**(-$errorp);


$terms = Formula("(-1)^(n+1)/(n^($exponent))");

#$error = 10**($errorp/$exponent)-1;
$error = sprintf "%.2f",10**($errorp/$exponent)-1;
$N = Real(int($error) + 1);
$answer=0;
$sum = "1";
@sign = ("-","+");

for($i=1;$i<=$N;$i++){
  $answer+=$terms->eval(n=>$i);
  if($i>1){
      $sum.="$sign[$i%2]\frac{1}{$i^{$exponent}}";
  }
}

Context()->texStrings;

BEGIN_TEXT
\{ beginproblem() \}
\{ textbook_ref_exact("Rogawski ET 2e", "10.4", "13") \}
$PAR
Approximate the value of the series to within an error of at most \(10^{-$errorp}\).


\[ \sum_{n=1}^{\infty} \frac{ (-1)^{n+1}}{n^{$exponent}} \]
According to Equation (2):
\[ \left| S_N-S \right|\le a_{N+1} \]
what is the smallest value of \(N\) that approximates \(S\) to within an error of at most \(10^{-$errorp}\)?
$BR
\(N = \) \{ ans_rule() \}
$PAR
\(S\approx\) \{ ans_rule() \}
END_TEXT

Context()->normalStrings;
Context()->flags->set(tolerance=>$tolerance);

#Answer Check Time!
ANS($N->cmp);
ANS(Real("$answer")->cmp);

Context()->texStrings;
SOLUTION(EV3(<<'END_SOLUTION'));
$PAR
$SOL
Let \( S =  \sum _{n=1}^{\infty} \frac{ (-1)^{n+1}}{n^{$exponent}} \), so that \( a_n = \frac{1}{n^{$exponent}} \).  By Equation (2),

\[ | S_N - S | \le a_{N+1} = \frac{1}{(N+1)^{$exponent}}. \]

To make the error less than \(10^{-$errorp}\), we must choose \( N \) so that
\[ \frac{1}{(N+1)^{$exponent}} < 10^{-$errorp} \qquad\textrm{or}\qquad N > 10^\frac{$errorp}{$exponent}-1\approx $error. \]

The smallest value that satisfies the required inequality is then \( N=$N \).  Thus, 
\[ S \approx S_{$N} = $sum = $answer \]


END_SOLUTION

ENDDOCUMENT()