Perturbation theory for algebraic equations I'm trying to find expansion (up to the 2nd non zero term) for the roots of:
$x^5-x^2+\epsilon=0$ as $\epsilon\rightarrow0$
So I've assumed the solution may be written as a power series $$x(\epsilon)=\sum x^n\epsilon^n=x_0+\epsilon x_1+\epsilon^2 x_2+...$$
Substituting this in and equating the orders of $\epsilon$ I have:
$O(\epsilon^0)\Rightarrow x_0=0, x_0^3=1$
$O(\epsilon^1)\Rightarrow x_1=?, x_1=-\epsilon/3$
$O(\epsilon^2)\Rightarrow x_1=0, x_2=-\epsilon^2/3$
Is this along the right lines? My problem is that this is a quintic equation yet I only seem to have 2 solutions.
 A: Alright I'll expand on my comment a bit more and make it more of an answer.
First let's try to expand it as a solution varying analytically in $\epsilon$, we'll see why this isn't enough.  If we are looking for a solution of the form $x(\epsilon)=x_0 + x_1\epsilon + x_2\epsilon^2 + ...$ to this equation, solving for the lead term $x_0$ amounts to solving $x^5 - x^2 = 0$ which has a double root at $0$ and then single roots at the three complex third roots on unity.
For each of these roots we will have to do separate calculations, as a warm up let's do the easiest case when $x_0 = 1$. We plug into our equation $x = x(\epsilon)=1 + x_1\epsilon + x_2\epsilon^2 + ...$ and set the coefficient of $\epsilon$ to $0$, in this case this amounts to $5x_1 - 2x_1 + 1 =0$ so we get $x_1 = -\frac{1}{3}$. We can then use this value to solve for $x_2$ and so on, at each step from here on the next coefficient will be completely determined. The same thing will happen if we start with either of the other third roots of unity, the answers will indeed depend analytically on $\epsilon$ and this method will work.
Now let's try to do the same thing for $x_0 = 0$. Let's try plugging in $x(\epsilon)=x_0 + x_1\epsilon + x_2\epsilon^2 + ...$ into the equation and setting the coefficient of epsilon to $0$.  Well if we expand it out we get $5\cdot0 \cdot x_1 - 2\cdot0\cdot x_1+1=0 \implies 1=0$. We are in big trouble already, and have no way to continue.  The problem is that in this case your assumption that we can write the root as a power series in $\epsilon$ is actually wrong.  The problem is that near $x=0$ this polynomial looks very much like $x^2 + \epsilon$, and we know that the function $f(\epsilon)=\sqrt \epsilon$ does not have an analytic expansion at $0$.
We can however save ourselves by introducing a square root of $\epsilon$, I'll stick with the variable $t$ that I used in my comment although formally writing $\epsilon^{\frac{1}{2}}$ would also be reasonable notation if you are careful. Now the polynomial we are trying to solve is $x^5-x^2+t^2$ and we will try to do it with a power series $x(t) = a_0 + a_1t + a_2t^2 + ...$ with $a_0 =0$ as that was the root giving us trouble before.
Now we can proceed as before by plugging in $x(t)$ into our polynomial and setting the coefficient of $t$ to zero.  This time something funny happens, we get: $5\cdot0 \cdot a_1 - 2\cdot0\cdot a_1=0 \implies 0a_1 = 0$. Note that we don't have the $+1$ term anymore since we have $t^2$ not $t$ in the polynomial we are trying to solve. We don't have a contradiction, which is a good sign, but we still can't solve for $a_1$, so what do we do?
Well let's not jump ship just yet, we still have to set coefficients of higher powers of $t$ to zero. Let's leave $a_1$ as a variable for now and set the coefficient of $t^2$ to $0$. We get: $10\cdot 0 \cdot a_1^2 + 5 \cdot 0 \cdot a_2 - a_1^2 - 2 \cdot 0 \cdot a_2 + 1 =0 \implies a_1^2 = 1$.  This gives us two solutions with $a_1 = i$ and $a_1 = -i$.  As a reality check we can use our heuristic that our polynomial near zero is very close to $x^2 + \epsilon$ which we expect to have two roots at $\pm i t$.  Now we can pick one of these values and keep going, setting higher and higher coefficients of $t^n$ to $0$ and solving for the $a_m$ successively.
Want some general theory? Well too bad, you are getting some anyway.  Suppose we want to consider a general analytic perturbation of a (monic for simplicity) polynomial. Something of the form $x^n + a_{n-1}(\epsilon)x^{n-1} + ...+ a_0(\epsilon) = 0$ where all the $a_i(\epsilon)$ are analytic at $\epsilon = 0$.  Turns out that in general the roots will not be analytic in $\epsilon$ at $\epsilon = 0$.  They will however be analytic in some variable $t$ with $t^k = \epsilon$, in other words you might need to allow yourself to take roots of $\epsilon$ in order to get something analytic but you don't need anything else. Moreover, it turns out that the value of $k$ can just be taken to be the $\gcd$ of the multiplicities of the roots of the polynomial at $\epsilon = 0$, or if you do each root separately with different variables you just need $k$ to be the multiplicity of that root.
