|b_k| \\
                                 & \leq & 2,\end{aligned}$$ by our assumption on the $b_k$. Hence there is a choice of $a_j$, either $+1$ or $-1$, such that $u_j = a_j - \sum_{k=0}^{n-1} b_k u_{j+k-n} \in [-1, 1]$.

We claim that this sequence of $a_j$ has the desired properties. To see this, we first consider the base case (we put $b_n = 1$ for ease of notation). Observe that: $$\begin{aligned}
\sum_{j=0}^\infty a_j y^j
    & = & \sum_{j=0}^\infty \left( \sum_{k=0}^n b_k u_{j+k-n} \right) y^j \\
    & = & \sum_{k=0}^n b_k y^{-k} \sum_{j=0}^\infty u_{j+k-n}  y^{j+k} \\
    & = & \sum_{k=0}^n b_k y^{-k} \left(\sum_{t=k}^{n-1} u_{t-n}  y^{t} +
                                          \sum_{t=n}^\infty u_{t-n}  y^{t} \right) \\
    & = & \sum_{k=0}^n \sum_{t=k}^{n-1} b_k y^{t-k} u_{t-n} +
          P(y^{-1}) \sum_{t=n}^\infty u_{t-n}  y^{t}.\end{aligned}$$

Evaluating at $y = {\lambda}_i$ and observing that $P({\lambda}_i^{-1}) = 0$, this simplifies to $$\sum_{t=0}^{n-1} u_{t-n} B_t({\lambda}_i).
\label{eq:base}$$

We further see that $$\frac{1}{s!} \frac{d^s}{d y^s} \left( \sum_{j=0}^\infty a_j y^j \right)
    = \frac{1}{s!} \frac{d^s}{d y^s} \left( \sum_{k=0}^n \sum_{t=k}^{n-1} b_k y^{t-k} u_{t-n} +
          P(y^{-1}) \sum_{t=n}^\infty u_{t-n}  y^{t} \right).$$ Taking derivatives and evaluating at ${\lambda}_i$, this simplifies to $$\sum_{t=0}^{n-1} u_{t-n} B_t^{(s)}({\lambda}_i),
\label{eq:induct}$$

Combining equations , with Lemma \[lem:jordan-exp\] gives $$\begin{aligned}
\pi(a_0 a_1 a_2 \dots) & =
        \begin{pmatrix}
        \frac{1}{(k_1-1)!}\frac{d^{k_1-1}}{d {\lambda}_1^{k_1-1}} \sum_{i=0}^\infty a_i {\lambda}_1^i \\
        \frac{1}{(k_1-2)!}\frac{d^{k_1-2}}{d 