Solving $x^2+bx^{1+\varepsilon}+c =0$

Let $x \in \mathbb{R}$. Is it possible to find the roots of $x^2+bx^{1+\varepsilon}+c =0$ where $b,c \in \mathbb{R}$ and $\varepsilon$ is small. I am guessing that an explicit expression might not be possible but is it possible to write the roots as an expansion?

Solution 1:

Write $x^{\epsilon} = e^{\epsilon \log{x}} = 1 + \epsilon \log{x} + O(\epsilon^2)$. Thus, to zeroth order,

$$x_0^2+b x_0+c=0 \implies x_0 = \frac{-b \pm \sqrt{b^2-4 c}}{2}$$

Write the solution as $x = x_0 + \epsilon x_1 + \epsilon^2 x_2 + \cdots$. Let's find $x_1$ by equating coefficients of $\epsilon$:

$$(x_0+\epsilon x_1)^2 + b (x_0+\epsilon x_1) (1+\epsilon \log{(x_0 + \epsilon x_1)}) + c=0$$

Now $\log{(x_0+\epsilon x_1)} = \log{x_0} + O(\epsilon)$. Thus, to $O(\epsilon)$, we have

$$x_0^2+b x_0+c + \epsilon (2 x_0 x_1+ b x_1+b x_0 \log{x_0})=0$$

The quadratic is zero by definition. The quantity in the parentheses, we set to zero and solve for $x_1$:

$$x_1 = -\frac{b \,x_0 \log{x_0}}{b+2 x_0}$$

We thus have $x=x_0+\epsilon x_1 + O(\epsilon^2)$. Thus procedure may be carried out to arbitrary order.

Note: for practical reasons, you should be careful if one or both of the solutions $x_0$ are negative or complex. I will not go into detail here, so the above assumes $x_0 \gt 0$.

EDIT

@Michael correctly points out that the above expansion is not valid in the case of a double root where $2 x_0+ b=0$. In this case, a different expansion is needed. That is, assume instead that

$$x=x_0+\epsilon^a x_1 + \cdots$$

for some real value of $a$. By doing out the expansion, we find that $a=1/2$. To find $x_1$, we write

$$(x_0+\epsilon^{1/2} x_1 + \epsilon x_2)^2 + b (x+0 + \epsilon^{1/2} x_1 + \epsilon x_2) (1+\epsilon \log{x_0}) + c=0$$

Collecting orders of $\epsilon$, I get

$$x_0^2+b x_0+c + \epsilon^{1/2} (2 x_0 x_1 + b x_1) + \epsilon (x_1^2 + b x_0 \log{x_0} + 2 x_0 x_2 + b x_2)=0$$

Using the fact that $2 x_0+b=0$ for the double root, we find that

$$x_1^2 = -b x_0 \log{x_0}$$

which imposes constraints on $x_0$ for a real solution.