[solution]
**Step 1: Identify the unperturbed roots and their simplicity.**
At $\epsilon = 0$, the equation reduces to $x^2 - 2x + 1 = (x - 1)^2 = 0$, which has a double root at $x_0 = 1$. We compute $\partial_x f(x, \epsilon) = 3\epsilon x^2 + 2x - 2$, so $\partial_x f(1, 0) = 2 - 2 = 0$. Since the derivative vanishes, the root is **non-simple** with multiplicity $m = 2$, and regular perturbation does not apply. We also note that $\partial_x^2 f(1, 0) = 6\epsilon \cdot 1 + 2 \big|_{\epsilon = 0} = 2 \neq 0$, confirming the multiplicity is exactly $2$.
Since the perturbed equation is cubic ($\epsilon \neq 0$) but the unperturbed equation is quadratic, one root is "missing" at $\epsilon = 0$ — it has escaped to infinity.
**Step 2: Fractional power expansion for the double root at $x = 1$.**
We substitute $x = 1 + a\epsilon^{1/2} + b\epsilon + c\epsilon^{3/2} + \cdots$ into $f(x, \epsilon) = \epsilon x^3 + x^2 - 2x + 1 - 2\epsilon$. Writing $\eta = x - 1$:
\begin{align*}
\epsilon(1 + \eta)^3 + (1 + \eta)^2 - 2(1 + \eta) + 1 - 2\epsilon.
\end{align*}
Expanding:
\begin{align*}
&\epsilon(1 + 3\eta + 3\eta^2 + \eta^3) + 1 + 2\eta + \eta^2 - 2 - 2\eta + 1 - 2\epsilon \\
&= \eta^2 + \epsilon(1 + 3\eta + 3\eta^2 + \eta^3) - 2\epsilon \\
&= \eta^2 - \epsilon + 3\epsilon\eta + 3\epsilon\eta^2 + \epsilon\eta^3.
\end{align*}
Now substituting $\eta = a\epsilon^{1/2} + b\epsilon + \cdots$:
\begin{align*}
\eta^2 &= a^2\epsilon + 2ab\epsilon^{3/2} + \cdots, \\
3\epsilon\eta &= 3a\epsilon^{3/2} + \cdots.
\end{align*}
Collecting by powers of $\epsilon$:
**Order $\epsilon$:** The equation reads $(a^2 - 1)\epsilon = 0$, giving $a^2 = 1$, so $a = \pm 1$.
**Order $\epsilon^{3/2}$:** The contributions are $2ab$ (from $\eta^2$) and $3a$ (from $3\epsilon\eta$). The equation reads
\begin{align*}
2ab + 3a = a(2b + 3) = 0.
\end{align*}
Since $a \neq 0$, we obtain $b = -3/2$.
The two root branches from the double root are therefore
\begin{align*}
x_\pm(\epsilon) = 1 \pm \epsilon^{1/2} - \frac{3}{2}\epsilon + O(\epsilon^{3/2}).
\end{align*}
These arise from **fractional power expansion** (irregular perturbation) because the root is non-simple.
**Step 3: Rescaling to find the root at infinity.**
We set $x = \epsilon^{-\beta} X$ and substitute into $f$:
\begin{align*}
\epsilon^{1-3\beta} X^3 + \epsilon^{-2\beta} X^2 - 2\epsilon^{-\beta} X + 1 - 2\epsilon = 0.
\end{align*}
The exponents of $\epsilon$ on the four terms are $1 - 3\beta$, $-2\beta$, $-\beta$, and $0$. For the first two terms to balance: $1 - 3\beta = -2\beta$, giving $\beta = 1$. With this choice, multiply through by $\epsilon^2$:
\begin{align*}
X^3 + X^2 - 2\epsilon X + \epsilon^2 - 2\epsilon^3 = 0.
\end{align*}
At leading order ($\epsilon = 0$): $X^2(X + 1) = 0$. The solution $X_0 = -1$ corresponds to the root at infinity (the solution $X_0 = 0$ reproduces the two finite roots from the rescaled perspective and is not relevant here).
To find the correction, write $X = -1 + d_1\epsilon + \cdots$ and substitute:
\begin{align*}
(-1 + d_1\epsilon)^3 + (-1 + d_1\epsilon)^2 - 2\epsilon(-1 + d_1\epsilon) + \epsilon^2 - 2\epsilon^3 &= 0.
\end{align*}
Expanding each term and collecting at order $\epsilon$:
\begin{align*}
(-1 + 3d_1\epsilon + \cdots) + (1 - 2d_1\epsilon + \cdots) + (2\epsilon - 2d_1\epsilon^2 + \cdots) + \epsilon^2 - 2\epsilon^3 &= 0, \\
(3d_1 - 2d_1 + 2)\epsilon + O(\epsilon^2) &= 0, \\
(d_1 + 2)\epsilon &= 0.
\end{align*}
Thus $d_1 = -2$, and the root at infinity is
\begin{align*}
x = \epsilon^{-1}(-1 - 2\epsilon + O(\epsilon^2)) = -\epsilon^{-1} - 2 + O(\epsilon).
\end{align*}
This root arises from **rescaling and dominant balance** (singular perturbation).
**Step 4: Summary.**
The three roots of $\epsilon x^3 + x^2 - 2x + 1 - 2\epsilon = 0$ for small $\epsilon > 0$ are:
\begin{align*}
x_+(\epsilon) &= 1 + \epsilon^{1/2} - \frac{3}{2}\epsilon + O(\epsilon^{3/2}) && \text{(fractional power, from double root)}, \\
x_-(\epsilon) &= 1 - \epsilon^{1/2} - \frac{3}{2}\epsilon + O(\epsilon^{3/2}) && \text{(fractional power, from double root)}, \\
x_\infty(\epsilon) &= -\epsilon^{-1} - 2 + O(\epsilon) && \text{(rescaling, from root at infinity)}.
\end{align*}
The analysis required three of the four techniques from this article: the [Inverse Function Theorem](/theorems/51) to classify the double root as non-simple, fractional power expansion to track the two branches emanating from it, and dominant balance with rescaling to capture the root that escapes to $-\infty$ as $\epsilon \to 0$.
[/solution]
