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Power series

  1. Dec 20, 2007 #1

    tony873004

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    This is from the teacher's notes
    1. The problem statement, all variables and given/known data
    [tex]f(x) = \frac{x}{{9 + x^2 }} = \frac{x}{9} \cdot \frac{1}{{1 - \left( { - \frac{{x^2 }}{9}} \right)}} = \frac{x}{9}\sum\limits_{n = 1}^\infty {\left( { - \frac{{x^2 }}{9}} \right)^n } = \sum\limits_{n = 1}^\infty {\frac{{( - 1)^n x^{2n + 1} }}{{9^{n + 1} }}} [/tex]

    I can see distributing the n inside the parenthesis, to -1, x^2, and 9. But what's the justification for chaning it to n+1 for x^2 and for 9?

    The next step is
    This converges for [tex]\left| { - \frac{{x^2 }}{9}} \right| < 1\,\,\,\,\,\,or\,\,\,\,x^2 < 9\,\,\,\,\,\,\,\,\,\,\, - 3 < x < 3[/tex]
    (−3, 3)

    So what was the point in doing the last step in my first tex, if she just resorted to the 2nd to last step to determine the interval of convergence?
    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 20, 2007 #2
    she'll probably want you to give it's power series representation as well. my teacher made me give the power series in order to receive credit for the interval of convergence.
     
    Last edited: Dec 20, 2007
  4. Dec 20, 2007 #3

    HallsofIvy

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    You have x/9 outside the sum. Taking it inside the sum, and multiplying each term contributes one factor of x in the numerator and one factor of 9 in the denominator.

    What was the point of doing the whole problem? Yes, a geometric series, with common ratio r, converges for |r|< 1. You don't need the "x/9" for that. But putting the "x/9" inside the series makes it simpler.
     
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