My teacher has done this: $\frac{1}{z^3(1-z^2/3+O(z^4))} = \frac{1+z^2/3+O(z^4)}{z^3}$ How does that work? I don't understand why he can claim this.
(easy) rearranging of power series denominator
3
$\begingroup$
sequences-and-series
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0It's 4. I edited the expression. – 2011-05-30
1 Answers
6
Isn't it simply because $\frac{1}{1-u} = 1+u+O(u^2)$ ?
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2@A. Top: Don't! This happens to all of us :) Next time you'll remember. – 2011-05-30