View of /trunk/NationalProblemLibrary/WHFreeman/Rogawski_Calculus_Early_Transcendentals_Second_Edition/10_Infinite_Series/10.7_Taylor_Series/10.7.29.pg

    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('29')
   10 # Author('Nick Hamblet')
   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 
   18 Context()->variables->add(n=>'Real');
   19 
   20 $c = random(2,7,1);
   21 
   22 $f = Compute("1/x");
   23 
   24 $series = Formula("(-1)^n/($c)^(n+1) * (x-$c)^n");
   25 $series->{test_points} = [ [ 1, random(.5,9,.5) ],
   26                            [ 2, random(.5,9,.5) ],
   27                            [ 3, random(.5,9,.5) ],
   28                            [ 4, random(.5,9,.5) ],
   29                            [ 5, random(.5,9,.5) ],
   30                            [ 6, random(.5,9,.5) ],
   31                            [ 7, random(.5,9,.5) ],
   32                            [ 8, random(.5,9,.5) ],
   33                            [ 9, random(.5,9,.5) ] ];
   34 
   35 $interval = Interval("(0,$c+$c)");
   36 
   37 Context()->texStrings;
   38 BEGIN_TEXT
   39 \{ beginproblem() \}
   40 \{ textbook_ref_exact("Rogawski ET 2e", "10.7","29") \}
   41 $PAR
   42 Find the Taylor series, centered at \(c=$c\), for the function
   43 \[ f(x) = $f. \]
   44 $PAR
   45 \(\displaystyle f(x)=\sum_{n=0}^{\infty}\) \{ans_rule()\}. $BR
   46 The interval of convergence is: \{ans_rule()\}.
   47 END_TEXT
   48 Context()->normalStrings;
   49 
   50 ANS($series->cmp, $interval->cmp);
   51 
   52 Context()->texStrings;
   53 SOLUTION(EV3(<<'END_SOLUTION'));
   54 $PAR
   55 $SOL
   56 $PAR
   57 Write
   58 \[ $f = \frac{1}{$c+(x-$c)}=\frac{1}{$c}\cdot \frac{1}{1+\left(\frac{x-$c}{$c}\right)} \]
   59 and then substitute \(-\left(\frac{x-$c}{$c}\right)\) for \(x\) in the Maclaurin series for \(\frac{1}{1-x}\) to obtain
   60 \[ $f = \frac{1}{$c}\sum_{n=0}^{\infty} \left[-\left(\frac{x-$c}{$c}\right)\right]^n = \sum_{n=0}^{\infty} \frac{(-1)^n}{$c^{n+1}}(x-$c)^n. \]
   61 This series is valid for \(\left|-\left(\frac{x-$c}{$c}\right)\right|<1\), or \(|x-$c|<$c\).
   62 
   63 END_SOLUTION
   64 
   65 ENDDOCUMENT();