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Month: July 2026

Regularization III

In this post, we justify the regularization procedure introduced in previous posts for singular polynomial problems. Consider:

$\displaystyle P(x)=a_0+a_1x+\dots+a_nx^n$

$\displaystyle P(x)=1+A_1x+\dots+A_nx^n$

$\displaystyle P_{\epsilon}(x)=f_0(\epsilon)+f_1(\epsilon)\epsilon^{\alpha_1}A_1x+\dots+f_n(\epsilon)\epsilon^{\alpha_n}A_nx^n$

$\displaystyle P_{\epsilon}(x)=(1+b_0\epsilon+c_0\epsilon^2+\dots)+(1+b_1\epsilon+c_1\epsilon^2+\dots)\epsilon^{\alpha_1}A_1x+\dots+(1+b_n\epsilon+c_n\epsilon^2+\dots)\epsilon^{\alpha_n}A_nx^n$

where $f_0(\epsilon),f_1(\epsilon),\dots,f_n(\epsilon)$ admit power-series expansions in $\epsilon$ , $\alpha_i$ ‘s are rational numbers and $b_i$ ‘s, $c_i$ ‘s are constants.

The last expression is therefore a generalized version of a perturbed polynomial

. We can rearrange this expression as follows:

$\displaystyle P_{\epsilon}(x)=(1+b_0\epsilon+c_0\epsilon^2+\dots)$

$\displaystyle +(1+b_1\epsilon+c_1\epsilon^2+\dots)\epsilon^{\alpha_1}A_1x$

$\displaystyle +\dots$

$\displaystyle +(1+b_n\epsilon+c_n\epsilon^2+\dots)\epsilon^{\alpha_n}A_nx^n$

Collecting terms according to their powers of $\epsilon$ , we obtain:

$\displaystyle P_{\epsilon}(x)=1+\epsilon^{\alpha_1}A_1x+\dots+\epsilon^{\alpha_n}A_nx^n$

$\displaystyle +b_0\epsilon+b_1\epsilon^{\alpha_1+1}A_1x+\dots+b_n\epsilon^{\alpha_n+1}A_nx^n$

$\displaystyle +\dots$

Theorem (Newton–Puiseux). Each root of the polynomial above is of the form:

$\displaystyle x(\epsilon)=\epsilon^pw(\epsilon),\quad w(0)\neq0$

where $w(\epsilon)$ is a continuous function near zero. For convenience, we simply write $x=\epsilon^pw$ . If the theorem holds, the polynomial above becomes:

$\displaystyle P_{\epsilon}(\epsilon^pw)=1+\epsilon^{\alpha_1}A_1\epsilon^pw+\dots+\epsilon^{\alpha_n}A_n(\epsilon^pw)^n$

$\displaystyle +b_0\epsilon+b_1\epsilon^{\alpha_1+1}A_1\epsilon^pw+\dots+b_n\epsilon^{\alpha_n+1}A_n(\epsilon^pw)^n$

$\displaystyle +\dots$

We define $Q_{\epsilon}(w):=1+\epsilon^{\alpha_1+p}A_1w+\dots+\epsilon^{\alpha_n+np}A_nw^n$ and have:

$\displaystyle P_{\epsilon}(\epsilon^pw)=Q_{\epsilon}(w)+\epsilon(b_0+b_1\epsilon^{\alpha_1+p}A_1w+\dots+b_n\epsilon^{\alpha_n+np}A_nw^n)+\dots$

If $\epsilon^pw$ is a root of $P_{\epsilon}(x)$ , we must satisfy:

$\displaystyle \lim_{\epsilon\to0}P_{\epsilon}(\epsilon^pw)=\lim_{\epsilon\to0}Q_{\epsilon}(w)=0$

Since $w(\epsilon)$ is a continuous function near zero, we inspect the structure of $Q_{\epsilon}(w(0))$ to determine the leading-order behavior:

$\displaystyle Q_{\epsilon}(w(0))=1+\epsilon^{\alpha_1+p}A_1w(0)+\dots+\epsilon^{\alpha_n+np}A_nw(0)^n$

Considering the set of its exponents of $\epsilon$ :

$\displaystyle E=\{0,\alpha_1+p,\dots,\alpha_k+kp,\dots,\alpha_n+np\}$

In order for the condition $\lim_{\epsilon\to0}Q_{\epsilon}(w)=0$ to hold, the minimal value of the set $E$ must be exactly $0$ , and this minimum must be shared by at least two distinct exponents. These identical minimal exponents define the dominant terms of $Q_{\epsilon}(w(0))$ which can then compensate each other to balance the equation at leading order.

July 2, 2026