View of /trunk/NationalProblemLibrary/WHFreeman/Rogawski_Calculus_Early_Transcendentals_Second_Edition/10_Infinite_Series/10.2_Summing_an_Infinite_Series/10.2.42.pg

    1 ## DBsubject('Calculus')
    2 ## DBchapter('Limits and Derivatives')
    3 ## DBsection('Definition of the Derivative')
    4 ## KEYWORDS('calculus', 'derivatives', 'slope')
    5 ## TitleText1('Calculus: Early Transcendentals')
    6 ## EditionText1('2')
    7 ## AuthorText1('Rogawski')
    8 ## Section1('10.2')
    9 ## Problem1('42')
   10 ## Author('Keith Thompson')
   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 loadMacros("PGauxiliaryFunctions.pl");
   18 loadMacros("PGgraphmacros.pl");
   19 
   20 #$showPartialCorrectAnswers=1;
   21 Context()->variables->add(n=>'Real');
   22 $base=random(2,9,1);
   23 $num=random(2,9,1);
   24 
   25 $ans1=$base-$num/100;
   26 $ans2=$num*(1/9-1/256);
   27 $ans3=$num/4-$num/9;
   28 $ans4=Formula("$num*(2*n-1)/((n*(n-1))**2)");
   29 $ans5=$base;
   30 Context()->texStrings;
   31 BEGIN_TEXT
   32 \{ beginproblem() \}
   33 \{ textbook_ref_exact("Rogawski ET 2e", "10.2","42") \}
   34 $PAR
   35 Let \(S=\sum\limits_{n=1}^\infty a_n\) be an infinite series such that \(S_N=$base-\frac{$num}{N^2}\).
   36 $PAR
   37 $BBOLD (a) $EBOLD What are the values of \(\sum\limits_{n=1}^{10} a_n\) and \(\sum\limits_{n=4}^{16} a_n\)?
   38 $BR  \(\sum\limits_{n=1}^{10} a_n=\) \{ans_rule()\}
   39 $BR \(\sum\limits_{n=4}^{16} a_n=\)  \{ans_rule()\}
   40 $PAR
   41 $BBOLD (b) $EBOLD What is the value of \(a_3\)?
   42 $BR \(a_3=\)  \{ans_rule()\}
   43 $PAR
   44 $BBOLD (c) $EBOLD Find a general formula for \(a_n\).
   45 $BR \(a_n=\)  \{ans_rule()\}
   46 $PAR
   47 $BBOLD (d) $EBOLD Find the sum \(\sum\limits_{n=1}^\infty a_n\).
   48 $BR \(\sum\limits_{n=1}^\infty a_n=\)  \{ans_rule()\}
   49 END_TEXT
   50 
   51 Context()->normalStrings;
   52 
   53 ANS(Real($ans1)->cmp);
   54 ANS(Real($ans2)->cmp);
   55 ANS(Real($ans3)->cmp);
   56 ANS($ans4->cmp);
   57 ANS(Real($ans5)->cmp);
   58 Context()->texStrings;
   59 SOLUTION(EV3(<<'END_SOLUTION'));
   60 $PAR
   61 $SOL
   62 $BBOLD (a) $EBOLD
   63 \[\sum_{n=1}^{10} a_n=S_{10}=$base-\frac{$num}{10^2}=$ans1;\]
   64 \[\sum_{n=4}^{16} a_n=(a_1+\ldots+a_{16})-(a_1+a_2+a_3)=S_{16}-S_3\]
   65 \[=\left($base-\frac{$num}{16^2}\right)-\left($base-\frac{$num}{3^2}\right)=\frac{$num}{9}-\frac{$num}{256}\approx $ans2.\]
   66 
   67 $BBOLD (b) $EBOLD
   68 \[a_3=(a_1+a_2+a_3)-(a_1+a_2)=S_3-S_2=\]
   69 \[\left($base-\frac{$num}{3^2}\right)-\left($base-\frac{$num}{2^2}\right)=\frac{$num}{4}-\frac{$num}{9}\approx $ans3.\]
   70 
   71 $BBOLD (c) $EBOLD Since \(a_n=S_n-S_{n-1}\), we have:
   72 \[a_n=S_n-S_{n-1}=\left($base-\frac{$num}{n^2}\right)-\left($base-\frac{$num}{(n-1)^2}\right)=\frac{$num}{(n-1)^2}-\frac{$num}{n^2}\]
   73 \[=\frac{$num(n^2-(n-1)^2)}{(n(n-1))^2}=\frac{$num(n^2-n^2+2n-1)}{(n(n-1))^2}=\frac{$num(2n-1)}{(n(n-1))^2}.\]
   74 
   75 $BBOLD (d) $EBOLD The sum \(\sum\limits_{n=1}^\infty a_n\) is the limit of the partial sums \(\{S_N\}\). Hence;
   76 \[\sum_{n=1}^\infty a_n = \lim_{N\rightarrow \infty} S_N = \lim_{N\rightarrow \infty} \left($base - \frac{$num}{N^2}\right) = $base.\]
   77 
   78 END_SOLUTION
   79 ENDDOCUMENT();