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Deriving a geometric series

  1. Oct 23, 2006 #1
    I could use some help with this question:

    Derive the geometric series representation of 1/(1-x) by finding a0, a1,
    a2,... such that
    (1-x)(a0+a1x+a2x^2+a3X^3+...)=1

    Thank you.
     
  2. jcsd
  3. Oct 23, 2006 #2

    AKG

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    [tex](1 - x)(a_0 + a_1x + a_2x^2 + a_3x^3 + \dots) = ? = \dots = ? = a_0 + (a_1-a_0)x + (a_2-a_1)x^2 + (a_3-a_2)x^3 + \dots[/tex]
     
  4. Oct 23, 2006 #3

    Office_Shredder

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    I recommend multiplying through by 1, then multiplying through by -x, then rearranging the terms
     
  5. Oct 23, 2006 #4

    mathman

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    How about (1-x)/(1-x)=1. Expand the denominator into a power series (binomial expansion of 1/(1-x)).
     
  6. Oct 27, 2006 #5
    If |x| <1, (1-x)*(1+x+x^2+x^3+x^4+...) = 1.
    So your coefficients a_i = 1, for all i.
     
  7. Oct 27, 2006 #6

    Hurkyl

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    (1) You (presumably) know how to multiply.
    (2) You (presumably) know how to tell when two power series are equal.

    There's no "trick" to this one -- you just do exactly what the equation suggests.
     
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