Regularize I

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We will now present (without justification) a method for regularizing a singular problem involving a polynomial. We want to find the roots of :

$\displaystyle P(x) = x^6 - x^4 - x^3 + 8$

We decide to perturb the above polynomial as follows:

$\displaystyle P_{\epsilon}(x) = \epsilon^2 x^6 - \epsilon x^4 - x^3 + 8$

For $\epsilon = 0$ we have:

$\displaystyle P_{\epsilon= 0}(x) = -x^3 + 8$

This polynomial has three roots. We have a singular problem since the original polynomial has six roots.

We can try to fix the problem using a change of variable:

$\displaystyle x := \epsilon^p w$

$\displaystyle P_{\epsilon}(\epsilon^p w) = \epsilon^2 (\epsilon^p w)^6 - \epsilon (\epsilon^p w)^4 - (\epsilon^p w)^3 + 8$

$\displaystyle = \epsilon^2 (\epsilon^{6p} w^6) - \epsilon (\epsilon^{4p} w^4) - (\epsilon^{3p} w^3) + 8$

$\displaystyle = \epsilon^{6p+2} w^6 - \epsilon^{4p+1} w^4 - \epsilon^{3p} w^3 + 8$

Now we collect the exponents of $\epsilon$ : ( $6p+2, 4p+1$ , $3p, 0$ ) and plot them in a graph as a function of p (see figure below). If we think of the lines on the graph as delimiting a figure in the plane, we can select the vertex of this figure with the largest p value and the smallest f(p) value. In our case, the point corresponding to this criterion is the point of intersection of the lines $6p+2$ and $3p$ . We thus obtain the point of intersection with coordinates ( $-2/3, -2$ ).