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Tue Nov 8 15:17:41 2011 UTC (19 months, 1 week 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('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();
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();