I have been given the question: "Find the first three terms in the asymptotic expansions as $x^3 + x - \epsilon = 0$ of the roots of $\epsilon \rightarrow 0$ using the iterative method."
The method of iteration is stated as follows:
- You find the first term of the asymptotic expansion by finding the roots when $\epsilon = 0$.
For this example: You find $x^3 + x= 0$ which has roots $x = 0$ and $x = \pm i$
- Rearrange the original equation to find an iterative process:
For this equation I have rearranged the equation into: $x_n = \epsilon -x_{n-1}^3$.
Using this equation and the starting point x = 0, we get $x = \epsilon + - \epsilon^3 + 3\epsilon^5 + ...$ This seems to work for x = 0.
However when I put in $x = \pm i$, the iterative method doesn't come out with constant coefficients.
Does anybody know if this method works for complex numbers or if my method is incorrect?
You want the expansion to be for small values of the parameter.
For the root, say, $x=i$, you want to consider $i+c$ for small $c$.
So, if $x_n = i+c_n$, then
$\begin{array}\\ i+c_{n+1} &=\epsilon-(i+c_n)^3\\ &=\epsilon-(i^3+3i^2c_n+3ic_n^2+c_n^3)\\ &=\epsilon-(-i-3c_n+3ic_n^2+c_n^3)\\ &=\epsilon+i+3c_n-3ic_n^2-c_n^3\\ \text{so}\\ c_{n+1} &=\epsilon+3c_n-3ic_n^2-c_n^3\\ \end{array} $
Starting with $c_0 = 0$, $c_1 = \epsilon$, $c_2 =\epsilon+3\epsilon-3i\epsilon^2-\epsilon^3 =4\epsilon-3i\epsilon^2-\epsilon^3 $, and so on.
(added later)
Looking at this, I am perturbed by the fact that the coefficient of $\epsilon$ changes from $c_1$ to $c_2$. I don't see anything wrong, so I'll leave it at this. If I made a mistake, maybe someone will find something.