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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0What is $n$ here? – 2011-05-30
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0It's 4. I edited the expression. – 2011-05-30
1 Answers
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Isn't it simply because $\frac{1}{1-u} = 1+u+O(u^2)$ ?
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4In case OP finds this a little opaque, the idea is to let $u=z^2/3-O(x^4)$ and notice that $u^2=O(z^4)$. – 2011-05-30
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0@Gerry, thanks for fleshing it out. – 2011-05-30
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0Thanks all! I feel so stupid. – 2011-05-30
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2@A. Top: Don't! This happens to all of us :) Next time you'll remember. – 2011-05-30