View of /trunk/NationalProblemLibrary/WHFreeman/Rogawski_Calculus_Early_Transcendentals_Second_Edition/10_Infinite_Series/10.7_Taylor_Series/10.7.37.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('37')
   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 $k = random(2,4,1);
   21 $k = 2;
   22 $nk = -$k;
   23 $kmo = $k - 1;
   24 if($kmo != 1){ $kmocdot = "$kmo\cdot"; }
   25 $ksq = $k**2;
   26 $c = random($k + 1, $ksq - 1, 1); # $k >= 2, so this range is non-empty
   27 $b = $c - $k;
   28 $a = $ksq - $c;
   29 $apb = $a + $b;
   30 
   31 if($a == $b){
   32   my $s = $a**2; # = a*b
   33   $f = Formula("1/($s - x^2)");
   34 } else {
   35   $f = Formula("1/(($a + x)*($b - x))");
   36 }
   37 
   38 $series = Formula("(1/$kmo) * (-1)^(n+1)/($k)^(2*n+3) * (($k)^(n+1)-1) * (x-$c)^n");
   39 $lend = $c - $k; # = $b
   40 $rend = $c + $k;
   41 $le = $lend + .5;
   42 $re = $rend - .5;
   43 $series->{test_points} = [ [ 1, random($le,$re,.5) ],
   44                            [ 2, random($le,$re,.5) ],
   45                            [ 3, random($le,$re,.5) ],
   46                            [ 4, random($le,$re,.5) ],
   47                            [ 5, random($le,$re,.5) ],
   48                            [ 6, random($le,$re,.5) ],
   49                            [ 7, random($le,$re,.5) ],
   50                            [ 8, random($le,$re,.5) ],
   51                            [ 9, random($le,$re,.5) ] ];
   52 
   53 $interval = Interval("($lend, $rend)");
   54 
   55 Context()->texStrings;
   56 BEGIN_TEXT
   57 \{ beginproblem() \}
   58 \{ textbook_ref_exact("Rogawski ET 2e", "10.7","37") \}
   59 $PAR
   60 Find the Taylor series, centered at \(c=$c\), for the function
   61 \[ f(x) = $f. \]
   62 $PAR
   63 \(\displaystyle f(x)=\sum_{n=0}^{\infty}\) \{ans_rule()\}. $BR
   64 The interval of convergence is: \{ans_rule()\}.
   65 END_TEXT
   66 Context()->normalStrings;
   67 
   68 ANS($series->cmp, $interval->cmp);
   69 
   70 Context()->texStrings;
   71 SOLUTION(EV3(<<'END_SOLUTION'));
   72 $PAR
   73 $SOL
   74 $PAR
   75 By partial fraction decomposition
   76 \[ $f = \frac{1/$apb}{$a+x} + \frac{1/$apb}{$b-x}, \]
   77 so
   78 \[
   79 \begin{array}{rcl}
   80 \displaystyle $f
   81 &=& \displaystyle \frac{1}{$apb}\left(\frac{1}{$ksq+(x-$c)}+\frac{1}{$nk - (x-$c)}\right) \\
   82 &=& \displaystyle \frac{1}{$apb}\left(\frac{1}{$ksq}\cdot \frac{1}{1+\frac{x-$c}{$ksq}} - \frac{1}{$k}\cdot \frac{1}{1+\frac{x-$c}{$k}}\right)
   83 \end{array}
   84 \]
   85 
   86 $PAR
   87 
   88 By making substitutions into the Maclaurin series for \(\frac{1}{1-x}\), we write
   89 \[
   90 \begin{array}{rcl}
   91 \displaystyle \frac{1}{$ksq}\cdot \frac{1}{1+\frac{x-$c}{$ksq}}
   92 &=& \displaystyle \frac{1}{$ksq}\sum_{n=0}^{\infty} \left(-\,\frac{x-$c}{$ksq}\right)^n \\
   93 &=& \displaystyle \frac{1}{$ksq}\sum_{n=0}^{\infty} \frac{(-1)^n}{$ksq^n}(x-$c)^n \\
   94 &=& \displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n}{$ksq^{n+1}}(x-$c)^n \\
   95 &=& \displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n}{$k^{2n+2}}(x-$c)^n
   96 \end{array}
   97 \]
   98 (valid for \(|x-$c|<$ksq\)) and
   99 \[
  100 \begin{array}{rcl}
  101 \displaystyle \frac{1}{$k}\cdot \frac{1}{1+\frac{x-$c}{$k}}
  102 &=& \displaystyle \frac{1}{$k}\sum_{n=0}^{\infty} \left(-\,\frac{x-$c}{$k}\right)^n \\
  103 &=& \displaystyle \frac{1}{$k}\sum_{n=0}^{\infty} \frac{(-1)^n}{$k^n}(x-$c)^n \\
  104 &=& \displaystyle \sum_{n=0}^{\infty}\frac{(-1)^n}{$k^{n+1}}(x-$c)^n
  105 \end{array}
  106 \]
  107 (valid for \(|x-$c|<$k\)).
  108 
  109 $PAR
  110 
  111 Thus,
  112 \[
  113 \begin{array}{rcl}
  114 \displaystyle $f
  115 &=& \displaystyle \frac{1}{$apb}\left(\sum_{n=0}^{\infty} \frac{(-1)^n}{$k^{2n+2}}(x-$c)^n + \sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{$k^{n+1}}(x-$c)^n\right) \\
  116 &=& \displaystyle \frac{1}{$apb}\left(\sum_{n=0}^{\infty} \left(\frac{(-1)^{n+1}}{$k^{n+1}} + \frac{(-1)^n}{$k^{2n+2}}\right)(x-$c)^n\right) \\
  117 &=& \displaystyle \frac{1}{$apb}\left(\sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{$k^{2n+2}}($k^{n+1}-1)(x-$c)^n\right) \\
  118 &=& \displaystyle \sum_{n=0}^{\infty}\frac{(-1)^{n+1}($k^{n+1}-1)}{$kmocdot $k^{2n+3}}(x-$c)^n.
  119 \end{array}
  120 \]
  121 
  122 $PAR
  123 
  124 This is valid when both of the intermediate series are valid, so when \(|x-$c|<$k\), which is the interval \(($lend,$rend)\).
  125 
  126 END_SOLUTION
  127 
  128 ENDDOCUMENT();