3
$\begingroup$

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.

  • 0
    What is $n$ here?2011-05-30
  • 0
    It's 4. I edited the expression.2011-05-30

1 Answers 1

6

Isn't it simply because $\frac{1}{1-u} = 1+u+O(u^2)$ ?

  • 4
    In 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
  • 0
    @Gerry, thanks for fleshing it out.2011-05-30
  • 0
    Thanks all! I feel so stupid.2011-05-30
  • 2
    @A. Top: Don't! This happens to all of us :) Next time you'll remember.2011-05-30