finding roots of polynomial equation with Taylor series I have the following equation which is obtained by the characteristic equation of a matrix:
$$
(-1-\lambda)^N +  c \sum_{i=1}^N (-1)^{i+1} (-1-\lambda)^{N-i} \beta_i = 0
$$
$N$ is the size of the matrix. And $\beta_i$ comes from an exponential kernel which I can choose another type of kernel if I want. let's simplify the equation by changing the variable $(-1-\lambda = x)$.
$$
x^N + c \sum_{i=1}^N (-1)^{i+1} x^{N-i} \beta_i.
$$
I know this equation is very complicated and most probably I should use numerical methods to find the roots which are my eigenvalues.
But out of curiosity, say we can choose a kernel $\beta_i$ such that we can create some sort of coefficients that lead to an expanded Taylor series of a know function. Then is the problem approachable? I wonder what you think of this and if you have seen something similar?
Or an equivalent question is whether there's a way to write the above equation in the form of:
$$
x^N + c\sum_{i=1}^N (-i)^{i+1} \frac{x^{N-i}}{\beta_i^\prime !} =0
$$
update:
I solved this equation numerically and surprisingly found out that for an exponential kernel, the roots lie on a circle of radius 1 in the complex domain. The result is invariant to the change of N and $\tau$. There's one negative eigenvalue close to -3 which is changing if I change the coefficient $\eta$ in the exponential kernel formula:
$$
\beta = \eta e^{-t/\tau}
$$
Figure right is an exponential kernel but bouncing between $-\beta$ and $\beta$ because of the $(-1)^{i+1}$ term in my equations. and on the left is the roots of the second equation.

 A: With $\,\beta_i = \eta e^{−i/\tau}\,$ per OP's comment, let $\,n = N\,$, $\,\mu = c \eta\,$, $\,\alpha = -e^{-1/\tau}\,$, $\,z = \dfrac{x}{\alpha}\,$, then $\,\beta_i = (-1)^i \eta \alpha^i\,$ and the equation can be rewritten as:
$$
\begin{align}
x^N + c \sum_{i=1}^N (-1)^{i+1} x^{N-i} \beta_i &= x^n + c \sum_{i=1}^n (-1)^{i+1} x^{n-i} \cdot  (-1)^i\eta \alpha^i
\\ &= x^n - c \eta \sum_{i=1}^n x^{n-i} \alpha^i
\\ &= x^n - \mu \alpha \sum_{i=0}^{n-1} x^{n-i-1} \alpha^i
\end{align}
$$
If $\,x = \alpha\,$ the latter reduces to:
$$
\alpha^n - \mu \alpha \sum_{i=0}^{n-1} \alpha^{n-i-1} \alpha^i = \left(1 - n \mu\right)\alpha^n
$$
Therefore $\,x = \alpha\,$ is not a solution unless $\,n \mu = 1\,$, otherwise if $\,x \ne \alpha\,$:
$$
\begin{align}
x^n - \mu \alpha \sum_{i=0}^{n-1} x^{n-i-1} \alpha^i ​&= x^n - \mu\alpha \,\frac{x^n - \alpha^n}{x - \alpha}
\\ &= \frac{x^{n+1} - \alpha x^n -\mu\alpha x^n + \mu \alpha^{n+1}}{x - \alpha}
\\ &= \frac{x^{n+1} - \alpha(\mu + 1) x^n + \mu \alpha^{n+1}}{x - \alpha}
\\ &= \alpha^{n}\,\frac{z^{n+1} - (\mu + 1) z^n + \mu}{z - 1}
\end{align}
$$
It follows that the roots of the original equation are $\,x = \alpha z\,$ where $\,z \ne 1\,$ and:
$$
z^{n+1} - (\mu + 1) z^n + \mu = 0
$$

*

*The "geometry" of the roots is dictated by the polynomial in $\,z\,$ i.e. by $\,\mu=c\eta\,$ alone. The "scale" of the roots is dictated by the $\,\alpha\,$ factor in $\,x = \alpha z\,$.


*When $\,n\,$ is large and $\,|z| \gt 1\,$ the non-constant terms dominate, so it is expected that a root exists near $\,z^{n+1} - (\mu + 1) z^n \simeq 0\,$ $\,\iff z \simeq \mu+1\,$ $\,\iff x \simeq -(c \eta+1)e^{-1/\tau}\,$.


*When $\,\mu\,$ is large the last two terms dominate, so it is expected that roots exist near $\,-(\mu + 1) z^n + \mu \simeq 0\,$ i.e. near the $\,n^{th}\,$ roots $\,z \simeq \left(\dfrac{\mu}{\mu+1}\right)^{1/n}\,$. In this case, the roots in $\,x\,$ will lie near the circle $\,|x| \simeq |\alpha|\left(\dfrac{\mu}{\mu+1}\right)^{1/n} = e^{-1/\tau} \left(\dfrac{c \eta}{c \eta + 1}\right)^{1/n}\,$.


*The equation can be solved numerically, but there is no closed form solution for $\,n \ge 5\,$.
Incidentally, the same equation was discussed in Annual interest rate in amortized loan, though the focus there was on the real roots.
