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Simple question regarding formulae for power series

  1. Sep 4, 2006 #1
    Sorry if this is in the wrong forum, I think it's right but I'm not totally sure.

    I'm (slowly) working my way through Roger Penrose's brilliant book Road to Reality. Recently I finished chapter four (Magical Complex Numbers) and one thing had me confused. How does he go about getting those formulae for the power series? For example, the series

    1 + x^2 + x^4 + x^6 + x^8 ....

    He gives the formula

    (1 - x^2)^-1

    How is this formula derived from the power series?

    Thanks in advance :)
     
  2. jcsd
  3. Sep 4, 2006 #2
    Welcome to PF :smile:
    [tex] \begin{gathered}
    \sum\limits_{k = 0}^n {x^{2k} } = 1 + x^2 + x^4 + \ldots + x^{2n} \Rightarrow \hfill \\
    \left( {1 - x^2 } \right)\sum\limits_{k = 0}^n {x^2 ^k } = \left( {1 + x^2 + x^4 + \ldots + x^{2n} } \right) - \left( {x^2 + x^4 + x^6 + \ldots + x^{2\left( {n + 1} \right)} } \right) = 1 - x^{2\left( {n + 1} \right)} \Rightarrow \hfill \\
    \sum\limits_{k = 0}^n {x^2 ^k } = \frac{{1 - x^{2\left( {n + 1} \right)} }}{{1 - x^2 }} \hfill \\ \end{gathered} [/tex]

    Since you have an infinite series (1+x2+x4+ ...),
    [tex]\sum\limits_{k = 0}^\infty {x^{2k} } = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 0}^n {x^{2k} } = \mathop {\lim }\limits_{n \to \infty } \frac{{1 + x^{2\left( {n + 1} \right)} }}{{1 - x^2 }} = \frac{1}{{1 - x^2 }}\,{\text{ IFF}}\;\left| x \right| < 1. [/tex]

    (If |x|≥1, the series obviously diverges. When |x|>1, [itex]\mathop {\lim }\limits_{n \to \infty } x^{2n} = \infty [/itex],
    and when x=±1, your series 1+x2+x4+... = 1+1+1+... = ∞).
     
    Last edited: Sep 4, 2006
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