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Revision 2584 - (download) (annotate)
Tue Nov 8 15:17:41 2011 UTC (2 years, 5 months ago) by aubreyja
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Rogawski problems contributed by publisher WHFreeman. These are a subset of the problems available to instructors who use the Rogawski textbook. The remainder can be obtained from the publisher.

    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('31')
   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',t=>'Real',M=>'Real');
   21 
   22 #$a=0
   23 #$b=1
   24 
   25 $a = random(0,3,1);
   26 $b = $a + random(1,3,1);
   27 
   28 $an=Formula("(n+$a)/(n+$b)")->reduce;
   29 
   30 $ans1=Real(999*$b-1000*$a);
   31 $ans2=Real(99999*$b-100000*$a);
   32 
   33 $an1=Formula("(n+$a-(n+$b))/(n+$b)");
   34 $ab=$a-$b;
   35 $an2=Formula("$ab/(n+$b)");
   36 $ab2=-$ab;
   37 $an3=Formula("$ab2/(n+$b)");
   38 $an4=$an3->substitute(n=>M);
   39 
   40 $ans3=Formula("$ab2/t-$b");
   41 
   42 BEGIN_TEXT
   43 \{ beginproblem() \}
   44 \{ textbook_ref_exact("Rogawski ET 2e", "10.1","31") \}
   45 $PAR
   46 Let \(a_n=$an\).  Find the smallest number \(M\) such that:
   47 $PAR
   48 $BBOLD (a) $EBOLD \(|a_n-1|\le 0.001\) for \(n\ge M\)$PAR
   49 \(M=\)\{ans_rule(10)\}
   50 END_TEXT
   51 SOLUTION(EV3(<<'END_SOLUTION'));
   52 $PAR
   53 $SOL
   54 We have
   55 \[\left|a_n-1\right|=\left|$an-1\right|=\left|$an1\right|=\left|$an2\right|=$an3\]
   56 Therefore \(|a_n-1|\le 0.001\) provided \($an3\le 0.001\), that is, \(n\ge $ans1\).  It follows that we can take \(M=$ans1\).
   57 END_SOLUTION
   58 
   59 
   60 BEGIN_TEXT
   61 $PAR
   62 $BBOLD (b) $EBOLD \(|a_n-1|\le 0.00001\) for \(n\ge M\)$PAR
   63 \(M=\)\{ans_rule(10)\}
   64 END_TEXT
   65 
   66 SOLUTION(EV3(<<'END_SOLUTION'));
   67 $PAR
   68 $SOL
   69 By part (a), \(|a_n-1|\le 0.00001\) provided \($an3\le 0.00001\), that is, \(n\ge $ans2\).  It follows that we can take \(M=$ans2\).
   70 END_SOLUTION
   71 
   72 BEGIN_TEXT
   73 $PAR
   74 $BBOLD (c) $EBOLD Now use the limit definition to prove that \(\displaystyle\lim_{n\to\infty}a_n=1\).  That is, find the smallest value of \(M\) (in terms of \(t\)) such that \(\left|a_n-1\right|<t\) for all \(n>M\).$BR
   75 (Note that we are using \(t\) instead of \(\epsilon\) in the definition in order to allow you to enter your answer more easily).$BR
   76 \(M=\)\{ans_rule(20)\} (Enter your answer as a function of \(t\))
   77 END_TEXT
   78 
   79 SOLUTION(EV3(<<'END_SOLUTION'));
   80 $PAR
   81 $SOL
   82 Using part (a), we know that \[|a_n-1|=$an3<t,\] provided \(n>$ans3\).  Thus to complete the proof, let \(t>0\) and take \(M=$ans3\).  Then, for \(n>M\), we have
   83 \[|a_n-1|=$an3<\{$an4\}=t.\]
   84 END_SOLUTION
   85 
   86 
   87 ANS($ans1->cmp);
   88 ANS($ans2->cmp);
   89 ANS($ans3->cmp);
   90 
   91 
   92 Context()->texStrings;
   93 
   94 ENDDOCUMENT();

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