# Simple question regarding formulae for power series

1. Sep 4, 2006

### iamaelephant

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?

2. Sep 4, 2006

### bomba923

Welcome to PF
$$\begin{gathered} \sum\limits_{k = 0}^n {x^{2k} } = 1 + x^2 + x^4 + \ldots + x^{2n} \Rightarrow \hfill \\ \left( {1 - x^2 } \right)\sum\limits_{k = 0}^n {x^2 ^k } = \left( {1 + x^2 + x^4 + \ldots + x^{2n} } \right) - \left( {x^2 + x^4 + x^6 + \ldots + x^{2\left( {n + 1} \right)} } \right) = 1 - x^{2\left( {n + 1} \right)} \Rightarrow \hfill \\ \sum\limits_{k = 0}^n {x^2 ^k } = \frac{{1 - x^{2\left( {n + 1} \right)} }}{{1 - x^2 }} \hfill \\ \end{gathered}$$

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

(If |x|≥1, the series obviously diverges. When |x|>1, $\mathop {\lim }\limits_{n \to \infty } x^{2n} = \infty$,
and when x=±1, your series 1+x2+x4+... = 1+1+1+... = ∞).

Last edited: Sep 4, 2006