View of /trunk/NationalProblemLibrary/Michigan/Chap9Sec5/Q23.pg

    1 # DESCRIPTION
    2 # Problem from Calculus, single variable, Hughes-Hallett et al., 4th ed.
    3 # WeBWorK problem written by Gavin LaRose, <glarose@umich.edu>
    4 # ENDDESCRIPTION
    5 
    6 ## KEYWORDS('calculus', 'integral', 'series', 'power series', 'interval of convergence', 'radius of convergence')
    7 ## Tagged by glr 02/08/09
    8 
    9 ## DBsubject('Calculus')
   10 ## DBchapter('Infinite Sequences and Series')
   11 ## DBsection('Power Series')
   12 ## Date('')
   13 ## Author('Gavin LaRose')
   14 ## Institution('University of Michigan')
   15 ## TitleText1('Calculus')
   16 ## TitleText2('Calculus')
   17 ## EditionText1('4')
   18 ## EditionText2('5')
   19 ## AuthorText1('Hughes-Hallett')
   20 ## AuthorText2('Hughes-Hallett')
   21 ## Section1('9.5')
   22 ## Section2('9.5')
   23 ## Problem1('23')
   24 ## Problem2('23')
   25 
   26 ## Textbook tags
   27 ## HHChapter1('Sequences and Series')
   28 ## HHChapter2('Sequences and Series')
   29 ## HHSection1('Power Series and Interval of Convergence')
   30 ## HHSection2('Power Series and Interval of Convergence')
   31 
   32 
   33 
   34 DOCUMENT();
   35 
   36 loadMacros(
   37 "PGstandard.pl",
   38 "MathObjects.pl",
   39 "PGchoicemacros.pl",
   40 # "parserNumberWithUnits.pl",
   41 # "parserFormulaWithUnits.pl",
   42 # "parserFormulaUpToConstant.pl",
   43 # "PGcourse.pl",
   44 );
   45 
   46 Context("Interval");
   47 $showPartialCorrectAnswers = 1;
   48 
   49 $a = random(2,9,1);
   50 
   51 TEXT(beginproblem());
   52 Context()->texStrings;
   53 BEGIN_TEXT
   54 
   55 Consider the series
   56 \[ \sum_{n=1}^{\infty} \frac{($a x)^n}{n}. \]
   57 
   58 Find the interval of convergence of this power series by first
   59 using the ratio test to find its radius of convergence and then
   60 testing the series' behavior at the endpoints of the interval
   61 specified by the radius of convergence.
   62 $BR
   63 interval of convergence = \{ ans_rule(15) \}
   64 $BR
   65 ${BITALIC}(Enter your answer as an interval: thus, if the interval
   66 of convergence were \(-3 < x\le 5\), you would enter ${BBOLD}(-3,5]$EBOLD.
   67 Use ${BBOLD}Inf$EBOLD for any endpoint at infinity.)$EITALIC
   68 
   69 END_TEXT
   70 # fix emacs' hiliting
   71 Context()->normalStrings;
   72 
   73 ANS(Compute("[-1/$a,1/$a)")->cmp() );
   74 
   75 Context()->texStrings;
   76 SOLUTION(EV3(<<'END_SOLUTION'));
   77 $PAR SOLUTION $PAR
   78 
   79 Let \(C_n=$a^n/n\).  Then replacing \(n\) by \(n+1\) gives
   80 \(C_{n+1}=$a^{n+1}/(n+1)\). Using the ratio test, we have
   81 \[
   82 \frac{|a_{n+1}|}{|a_n|}=
   83   |x|\frac{|C_{n+1}|}{|C_n|}=
   84   |x|\frac{$a^{n+1}/(n+1)}{$a^n/n}=
   85   |x|\frac{$a^{n+1}}{n+1} \cdot \frac{n}{$a^n}=
   86   $a |x|\left(\frac{n}{n+1}\right).
   87 \]
   88 Thus
   89 \[
   90 \lim_{n \rightarrow \infty}  \frac{|a_{n+1}|}{|a_n|}=$a|x|.
   91 \]
   92 The radius of convergence is \(R=1/$a\).
   93 
   94 $PAR
   95 For \(x=1/$a\) the series becomes the harmonic series
   96 \[
   97 \sum_{n=1}^{\infty} \frac{1}{n}
   98 \]
   99 which diverges.
  100 
  101 $PAR
  102 For \(x=-1/$a\) the series becomes the alternating series
  103 \[
  104 \sum_{n=1}^{\infty} \frac{(-1)^n}{n}
  105 \]
  106 which converges.
  107 
  108 $PAR
  109 Thus the interval of convergence is \( [-\frac 1 $a, \frac1 $a) \).
  110 
  111 END_SOLUTION
  112 Context()->normalStrings;
  113 
  114 
  115 COMMENT('MathObject version');
  116 ENDDOCUMENT();