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Tue Nov 8 15:17:41 2011 UTC (2 years, 5 months ago) by aubreyja
File size: 4082 byte(s)
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('')
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();
17
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();