implies that the term of degree $n_K$ must be nonzero. Thus, in the product $\phi_{L{^\prime}}{\left(x\right)}^{n_K}\phi_K{\left(\frac{1}{\alpha}\frac{\phi_{L}{\left(x\right)}}{\phi_{L{^\prime}}{\left(x\right)}}\right)}$, there must be a term which is *indivisible* by the polynomial $\phi_{L{^\prime}}{\left(x\right)}$. Therefore, the zero of the right-hand side cannot be a root of the polynomial $\phi_{L{^\prime}}$.

We are seeking values of $x$ such that the polynomial $\phi_K{\left(\frac{1}{\alpha}\frac{\phi_{L}{\left(x\right)}}{\phi_{L{^\prime}}{\left(x\right)}}\right)}$ evaluates to zero. In other words, we are looking for $x$ such that the term $\frac{1}{\alpha}\frac{\phi_{L}{\left(x\right)}}{\phi_{L{^\prime}}{\left(x\right)}}$ is a root of $\phi_K$. Therefore, we solve Eq.  for $x$, for all roots $\kappa_i$ of $K$.

If the forms of $\phi_L$ and $\phi_{L{^\prime}}$ are known (and if each have sufficiently low degree), then computing the roots of $\phi_{{\mathbin{\Pi}}}$ becomes tractable, even if $K$ is a large matrix. This suggests a recursive procedure for computing the spectrum of a $k$-level hierarchy, by writing it as a product of the ${\left(k-1\right)}$-level hierarchy with the $k$-th base graph. We now frame this as our main result of this section:

\[th:specMain\] Suppose we have a $k$-level hierarchy $L^{{\mathbin{\Pi_{\vec{\alpha}}}}k}$ described by base graph Laplacians $L_1,L_2,\ldots,L_k$ and weights ${{\vec{\alpha}}}={\left(1,\alpha_2,\ldots,\alpha_{k}\right)}$ as follows, $$L^{{\mathbin{\Pi_{\vec{\alpha}}}}k}=L_{k}{\mathbin{\Pi}}_{\alpha_{k}}L_{k-1}{\mathbin{\Pi}}_{\alpha_{k-1}}\cdots {\mathbin{\Pi}}_{\alpha_3}L_{2}{\mathbin{\Pi}}_{\alpha_{2}}L_1 .$$ Define a new set of weights ${{\vec{\beta}}}={\left(1,\beta_2,\ldots,\beta_{k}\right)}$ with $\beta_i = \alpha_i/\alpha_{i-1}$, and a new set of Laplacians $M_{k},M_{k-1},\ldots, M_{1}$ recursively as $$\begin{aligned}
