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Derivation problem

  1. Apr 4, 2009 #1

    I have [tex]\sum_{j=\kappa}^{\inf} \varphi^{2j-\kappa}\sigma^2[/tex]

    I know the answer is [tex] = \frac{\sigma^2}{1-\sigma^2} \varphi^\kappa [/tex]

    Can someone explain the mathematics involved in this derivation?

  2. jcsd
  3. Apr 4, 2009 #2
    Does [itex]\sigma[/itex] have an exponent related to j? Otherwise, the sum is a straightforward geometric series, once you reparameterize the index of summation. Note that
    [tex]\sum_{j=\kappa}^\infty \varphi^{2j-\kappa} = \varphi^\kappa + \varphi^{\kappa + 2} + \varphi^{\kappa + 6} + \cdots = \varphi^{\kappa}\sum_{n=0}^\infty \varphi^{2n} = \varphi^{\kappa}\sum_{n=0}^\infty (\varphi^2)^n[/tex]
    Do you know the sum of a geometric series?
    Last edited: Apr 4, 2009
  4. Apr 5, 2009 #3
    Thanks for that. I have a question on your reparameterization. Isn't [tex]\varphi^{\kappa}[/tex] to the negative index, and therefore shouldn't it be [tex] \frac{1}{\varphi^{\kappa}}[/tex] when you write out the series? Or did I miss something?
  5. Apr 6, 2009 #4
    [itex]\kappa[/itex] is the same as that in the original sum. When j starts at [itex]\kappa[/itex], we get 2K - K = K, 2(K + 1) - K = K + 2, 2(K + 2) - K = K + 4, and so on.
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