0
$\begingroup$

I have an equation: $(1-\epsilon)x^2 -2x +1=0$ (regularly perturbed problem, we anticipate all roots to remain bounded when $\epsilon$ goes to $0$)

I substitute $x=\displaystyle \sum_{n\geq0}C_n \epsilon^n$ and have to work out the equations for $C_0,C_1,C_2$.

So I get: $C_0^2-2C_0+1=0$ so $C_0=1$

and there a problem starts: as the equation for $C_1$ is: $2C_0C_1-C_0^2-2C_1=0$ so we get $1=0$. What is the reason this approach fails and how to devise a way to fix it?

  • 0
    C_0 has one possible value, not two.2011-02-21
  • 0
    Well the equation is quadratical, so theres 2 roots2011-02-21
  • 0
    You will know the reason when you solve for the roots in terms of $\epsilon$. You will get a term with $\sqrt{\epsilon}$ and you try to write it as a power series about origin which is not possible since the derivatives at the origin are not bounded. A better way would be to replace $\epsilon$ by $\epsilon^2$ (if you know $\epsilon$ is positive) and then go about2011-02-21

1 Answers 1