Roots of polynomials with logarithmic functions in coefficients (Newton Puiseux with log coefficients)

134 Views Asked by At

I am interested in the extension of Newton-Puiseux's theorem to polynomials with log terms in the coefficients.

The specific polynomial I am considering is: $f(\lambda,k)=\log(k)\sum_{i=0}^{N-D} b_i(k)\lambda^i+\sum_{i=N-D+1}^{N}a_i(k)\lambda^i$

Suppose $\{n_1,\dots n_{D-1}\}$ are strictly increasing negative integers.

For $0\leq i\leq N-D$, $b_i(k)=b'_i k^{\sum_{l=1}^{D-1}n_l}+$higher order terms in k, $b'_{N-D}\neq 0$.

For $N-D+1\leq i\leq N$, $a_i(k)=a'_ik^{\sum_{l=1}^{N-i}n_l}$+higher order terms in $k$, $a'_i\neq 0$.

I believe the roots of $f(\lambda,k)=0$ satisfy the following properties, but I do not know how to prove them,

  • $N-D$ roots of $f(\lambda,k)=0$ labelled $\lambda_{i}$ for $i=1,\dots N-D$ approach the roots of $\sum_{i=0}^{N-D} b'_i \lambda^i=0$ when $k\rightarrow 0$;

  • Amongst the rest of the roots of $f(\lambda,k)=0$, $D-1$ of them labelled by $\lambda_i(k)$ for $i=N-D+2,\dots N$ satisfies $\lambda_i(k)=-\frac{a'_{i-1}}{a'_{i}}k^{n_i}+$ higher orders in $k$ ;

  • The last remaining root labelled $\lambda_{N-D+1}(k)$ satisfies $\lambda_{N-D+1}(k)=-\frac{b'_{N-D}}{a'_{N-D+1}}\log(k)+$higher orders in $k$.


Some observations

Note that if there is no log(k) term in $f(\lambda,k)$, one can easily obtain the leading terms of the roots of $f(\lambda,k)=0$. Define $g(\lambda,k)= \sum_{i=0}^{N-D} b_i(k)\lambda^i+\sum_{i=N-D+1}^{N}a_i(k)\lambda^i$. The Newton polygon for $g(\lambda,k)$ is illustrated in the figure, where the slopes are strictly increasing for $i\geq N-D+1$ because $n_i$ are strictly increasing negative integers.

Using Newton-Puiseux's theorem,

  • $N-D$ roots of $g(\lambda,k)=0$ labelled $\lambda_{i}$ for $i=1,\dots N-D$ approach the roots of $\sum_{i=0}^{N-D} b'_i \lambda^i=0$ when $k\rightarrow 0$;

  • Amongst the rest of the roots of $f(\lambda,k)=0$, $D-1$ of them labelled by $\lambda_i(k)$ for $i=N-D+2,\dots N$ satisfies $\lambda_i(k)=-\frac{a'_{i-1}}{a'_{i}}k^{n_i}+$ higher orders in $k$;

  • The last remaining root labelled $\lambda_{N-D+1}(k)$ satisfies $\lambda_{N-D+1}(k)=-\frac{b'_{N-D}}{a'_{N-D+1}}+$higher orders in $k$.

I believe the only difference between the behaviors of the roots of $f(\lambda,k)$ and $g(\lambda,k)$ is in $\lambda_{N-D+1}(k)$.