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Trying to verify function as a powerseries

  1. May 5, 2010 #1
    1. The problem statement, all variables and given/known data

    Let [tex]\frac{1}{1-z+z^2} = \sum_{n=0}^{\infty} a_n \cdot z^n[/tex]

    where [tex]a_{1} = a_0 = 1 \Leftrightarrow a_{1}-a_{0} = 0[/tex] and [tex]a_2=0[/tex] and [tex]a_{n+3} + a_n = 0[/tex]

    3. The attempt at a solution

    By the theorem of Identity

    [tex]1 = 1 - z + z^2 \sum a_n \cdot z^n = a_0 + (a_{1}-a_{0})z + a_{2} \cdot z^2 + \sum_{n=3}^{\infty} a_n \cdot z^{n} - \sum_{n=1}^{\infty} a_{n} \cdot z^{n+1} + \sum_{n=0}^{\infty} a_n z^{n+2}[/tex]

    If I write out the first sum I get [tex]\sum_{n=3}^{\infty} a_n \cdot z^{n} = a_3 z^3 + a_{4}z^4 + \cdots [/tex]

    I write out a couple terms of the second sum [tex]- \sum_{n=1}^{\infty} a_{n} \cdot z^{n+1} = - a_1 \cdot z^{2} - a_2 \cdot z^{3} - \cdots [/tex]

    If I make the commen index n = 0

    it looks like the index increase at n+3 and at a power of n+2 such that

    [tex]a_0 + (a_{1}-a_{0})z + a_{2} \cdot z^2 +\sum_{n=3}^{\infty} + a_n \cdot z^{n} - \sum_{n=1}^{\infty} a_{n} \cdot z^{n+1} + \sum_{n=0}^{\infty} a_n z^{n+2} = \sum_{n=0}^{\infty} (a_{n+3} +a_n)z^{n+2}[/tex]

    How is this?

    Sincerely
    Susanne
     
    Last edited: May 5, 2010
  2. jcsd
  3. May 5, 2010 #2

    tiny-tim

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    Hi Susanne! :smile:

    Sorry, but this is really messy. :redface:

    Write out the total coefficient of zn+2 for n≥0. to get a general equation for an an-1 and an-2, and then deal with z0 and z1 separately. :wink:
     
  4. May 5, 2010 #3
    Just be clear tiny-tim,

    Is my formulation correct until here

    Let [tex]\frac{1}{1-z+z^2} = \sum_{n=0}^{\infty} a_n \cdot z^n[/tex]

    where [tex]a_{1} = a_0 = 1 \Leftrightarrow a_{1}-a_{0} = 0[/tex] and [tex]a_2=0[/tex] and [tex]a_{n+3} + a_n = 0[/tex]

    3. The attempt at a solution

    [tex]1 = 1 - z + z^2 \sum a_n \cdot z^n = a_0 + (a_{1}-a_{0})z + a_{2} \cdot z^2 + \sum_{n=3}^{\infty} a_n \cdot z^{n} - \sum_{n=1}^{\infty} a_{n} \cdot z^{n+1} + \sum_{n=0}^{\infty} a_n z^{n+2}[/tex]

    My attempt to expand the three sums (using the original n = 0)
    1) [tex]\sum_{n=0}^{\infty} a_n \cdot z^{n} =a_{0} + a_{1}z + a_{2}z^2+ a_{3}z^3 + a_4 z^4 + a_5 z^5 + \cdots + [/tex]

    2) [tex]\sum_{n=0}^{\infty} a_{n} \cdot z^{n+1} = a_0 z + a_1 z^{2} + a_2 z^3 + \cdots [/tex]

    3) [tex]\sum_{n=0}^{\infty} a_n z^{n+2} = a_{0}z^2 + a_{1}z^3 + a_2 z^4[/tex]

    I can see that if I subtract the second sum from the first I get

    [tex]\sum a_{n-1}z^{n+1}[/tex]

    But whats my next step? to add

    [tex]\sum_{n=1}^{\infty} a_{n-1}z^{n+1}+ \sum_{n=0}^{\infty} a_{n}z^{n+2}[/tex]

    Can that be right tiny-tim?? Cause I keep getting something else that [tex]a_{n+3}+a_n[/tex] inside the sum??

    If I do I end up with [tex]\sum_{n=0}^{\infty} a_{n}z^{n+1}[/tex] as the joined sum of the three. But can't be right?
     
    Last edited: May 5, 2010
  5. May 5, 2010 #4

    tiny-tim

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    Hi Susanne! :smile:
    This is a bad idea …

    it might turn out ok, but it's so easy to make a mistake if you do it that way. :redface:

    Instead, keep all the ∑s so far as you can, but rewrite them so that they all say zn, and they all start from n = 2 (this makes it really easy to add the ∑s :wink:)

    For example, with ∑n=0 anzn+1,

    rewrite it as ∑n=2 an-1zn + a0z.

    Try again. :smile:
     
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