View of /trunk/NationalProblemLibrary/WHFreeman/Rogawski_Calculus_Early_Transcendentals_Second_Edition/10_Infinite_Series/10.1_Sequences/10.1.13.pg

    1 ## DBsubject('Calculus')
    2 ## DBchapter('Infinite Sequences and Series')
    3 ## DBsection('Infinite Sequences and Series')
    4 ## KEYWORDS('calculus', 'series')
    5 ## TitleText1('Calculus: Early Transcendentals')
    6 ## EditionText1('2')
    7 ## AuthorText1('Rogawski')
    8 ## Section1('10.1')
    9 ## Problem1('13')
   10 ## Author('Danny Glin')
   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()->texStrings;
   19 Context()->flags->set(reduceConstants=>0);
   20 Context()->variables->are(n=>'Real');
   21 
   22 #$k=3
   23 $k=random(2,4,1);
   24 
   25 $andenom=Formula("n^($k)");
   26 
   27 foreach $i (1..3) {
   28 $denom[$i]=$andenom->eval(n=>$i);
   29 $numer[$i]=(-1)**($i+1);
   30 $a[$i]=Formula("$numer[$i]/$denom[$i]");
   31 }
   32 
   33 #$a1=Formula("1/1");
   34 #
   35 #$d2=2**$k;
   36 #$a2=Formula("-1/$d2");
   37 #
   38 #$d3=3**$k;
   39 #$a3=Formula("1/$d3");
   40 
   41 $an=Formula("(-1)^(n+1)/n^($k)");
   42 $an->{test_points} = [[1],[2],[3],[4],[5],[6]];
   43 
   44 BEGIN_TEXT
   45 \{ beginproblem() \}
   46 \{ textbook_ref_exact("Rogawski ET 2e", "10.1","13") \}
   47 $PAR
   48 Find a formula for the \(n\)th term of the following sequences:
   49 
   50 $PAR
   51 (a) \($a[1],$a[2],$a[3],\ldots\)
   52 $PAR
   53 \( a_n = \) \{ans_rule()\}$BR
   54 END_TEXT
   55 Context()->normalStrings;
   56 
   57 ANS($an->cmp);
   58 
   59 Context()->texStrings;
   60 @ord=('second','third','fourth');
   61 $m=$k-2;
   62 $neg=(-1)**$k;
   63 $antrans=Formula("$neg/n^($k)");
   64 
   65 SOLUTION(EV3(<<'END_SOLUTION'));
   66 $PAR
   67 $SOL
   68 The denominators are the $ord[$m] powers of the positive integers starting with \(n=1\).  Also, the sign of the terms is alternating with the sign of the first term being positive.  Thus,
   69 \[
   70 a_1=\{$antrans->substitute(n=>1)\}=\{$an->substitute(n=>1)\},\
   71 a_2=\{$antrans->substitute(n=>2)\}=\{$an->substitute(n=>2)\},\
   72 a_3=\{$antrans->substitute(n=>3)\}=\{$an->substitute(n=>3)\}
   73 \]
   74 This rule leads to the following formula for the \(n\)th term:
   75 \[
   76 a_n=$an
   77 \]
   78 END_SOLUTION
   79 
   80 #$a=1
   81 #$b=5
   82 
   83 $p = random(1,3,1);
   84 $q = $p + random(1,3,1);
   85 
   86 $bnumer=Formula("n+$p");
   87 $bdenom=Formula("n+$q");
   88 
   89 $bn=Formula("($bnumer)/($bdenom)");
   90 $bn->{test_points} = [[1],[2],[3],[4],[5],[6]];
   91 
   92 foreach $j (1..3) {
   93 $bnum[$j]=$bnumer->eval(n=>$j);
   94 $bden[$j]=$bdenom->eval(n=>$j);
   95 $b[$j]=Formula("$bnum[$j]/$bden[$j]");
   96 }
   97 
   98 BEGIN_TEXT
   99 $PAR
  100 (b) \($b[1],$b[2],$b[3],\ldots\)
  101 $PAR
  102 \( b_n = \) \{ans_rule()\}$BR
  103 $BITALIC Assume a starting index of \(n=1\). $EITALIC
  104 
  105 END_TEXT
  106 Context()->normalStrings;
  107 
  108 ANS($bn->cmp);
  109 
  110 Context()->texStrings;
  111 $diff = $q-$p;
  112 
  113 SOLUTION(EV3(<<'END_SOLUTION'));
  114 $PAR
  115 $SOL
  116 Assuming a starting index of \(n=1\), we see that each numerator is $p more than the index and the denominator is $diff more than the numerator.  Thus, the general term \(b_n\) is:
  117 \[
  118 b_n=$bn
  119 \]
  120 END_SOLUTION
  121 
  122 
  123 ENDDOCUMENT();