Roots of modified polynomials

Question

Consider the following two polynomials:

$$ g=x^3 - x^2 - (c + 2)x + c $$

and

$$ h=x^3 - x^2 - cx + c $$

The roots of $h$ are $1$ and $\pm \sqrt{c}$. I am interested in obtaining the roots of $g$, using the fact that $g=h-2x$.

Numerical calculation indicates that the roots of $g$ are indeed not very far from those of $h$. For example, the largest root of $g$ is $\sqrt{c}+\Theta(\frac{1}{\sqrt{c}})$.

In this example, I can derive such formulas using elementary methods but I wonder if there is a principled way to derive them.

Answer 1

I guess, Chapter 2, §1 of "Perturbation Theory of Linear Operators" by Kato will answer your question.

Answer 2

Ten years too late and too long for a comment.

It is a bit more convenient to let $c=d^2$

Let $x_i$ the roots of $g=0$ with $(x_1>x_2>x_ 3)$

We have $$x_1=d+\frac{1}{d} \left(1+\frac{1}{d}+\frac{1}{2 d^2}-\frac{1}{d^3}-\frac{3}{d^4}+O\left(\frac {1}{d^5}\right)\right)$$ $$x_2=1-\frac 2 {d^2} \left(1-\frac{1}{d^2}-\frac{3}{d^4} +O\left(\frac{1}{d^{6}}\right)\right)$$ $$x_3=-d -\frac{1}{d} \left(1-\frac{1}{d}+\frac{1}{2 d^2}+\frac{1}{d^3}-\frac{3}{d^4}+O\left(\frac {1}{d^5}\right)\right)$$

Trying for $c=12.345$, this would give $$x_1=3.87853 \quad \quad x_2=0.85430 \quad\quad x_3=-3.72964$$ while the exact solutions are $$x_1=3.87813\quad \quad x_2=0.85313 \quad\quad x_3=-3.73126 $$

For $g$, using the above truncated series, the errors are $\left\{\frac{6}{d^4},-\frac{30}{d^6},\frac{6}{d^4}\right\}$.