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How can I derive this $$(1+\xi) \left(1+\frac{1}{2}\xi-\frac{1}{12}\xi^{2}\right)+O(\xi^3)$$ from $$\frac{1+\xi}{1-\frac{1}{2}\xi+\frac{1}{3}\xi^{2}}+O(\xi^3)$$ ?

The whole formula is below. This is from "A first course in the numerical analysis of differential equations by Arieh Iserles" $$\frac{\rho(w)}{\ln w}=\frac{\xi+\xi^{2}}{\xi-\frac{1}{2}\xi^2+\frac{1}{3}\xi^{3}+O(\xi^4)}= \frac{1+\xi}{1-\frac{1}{2}\xi+\frac{1}{3}\xi^{2}}+O(\xi^3) $$

$$=(1+\xi) \left(1+\frac{1}{2}\xi-\frac{1}{12}\xi^{2}\right)+O(\xi^3)=1+\frac{3}{2}\xi+\frac{5}{12}\xi^{2}+(\xi^3)$$

user26767
  • 299

2 Answers2

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Use the series $\frac1{1-x}=1+x+x^2+O(x^3)$ to get $$ \begin{align} \frac1{1-\frac12\xi+\frac13\xi^2} &=1+\left(\frac12\xi-\frac13\xi^2\right)+\left(\frac12\xi-\frac13\xi^2\right)^2+O(\xi^3)\\ &=1+\frac12\xi-\frac13\xi^2+\frac14\xi^2+O(\xi^3)\\ &=1+\frac12\xi-\frac1{12}\xi^2+O(\xi^3) \end{align} $$

robjohn
  • 345,667
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Take the Taylor expansion of $$\frac{1}{1-\frac{1}{2}\xi+\frac{1}{3}\xi^2}$$ around $0$ and insert. All the terms of order greater than $2$ are absorbed in the $O(\xi^3)$ part.

Daniel Robert-Nicoud
  • 29,795