TAOCP 5.4.3 Exercise 8

Exercise 8. [HM23] [HM23] Derive (12).

Verified: no
Solve time: 1m33s

Solution

Let $T=6$ in the notation of the section, and write

$$ X_n = (A_n, B_n, C_n, D_n, E_n)^T . $$

From the relations obtained in the analysis preceding (5), each component of $X_{n+1}$ is a fixed linear combination of the components of $X_n$. Hence there exists a fixed $5\times 5$ matrix $Q$ such that

$$ X_{n+1} = Q X_n . $$

Iterating this relation gives

$$ X_n = Q^n X_0 , $$

where $X_0 = (1,0,0,0,0)^T$ by the initial condition in level $0$.

Let $p(\lambda)=\det(\lambda I - Q)$ be the characteristic polynomial of $Q$. From the cascade relations (3)–(4), each component sequence satisfies the same linear recurrence obtained by eliminating the other components. This recurrence has order $5$, hence $p(\lambda)$ has degree $5$. Let $\lambda_1,\lambda_2,\lambda_3,\lambda_4,\lambda_5$ be the roots of $p(\lambda)$, counted with multiplicity.

Since $Q$ is similar over $\mathbb{C}$ to a matrix in Jordan normal form, there exists an invertible matrix $S$ such that

$$ Q = S \operatorname{diag}(\lambda_1,\lambda_2,\lambda_3,\lambda_4,\lambda_5) S^{-1}. $$

Therefore,

$$ Q^n = S \operatorname{diag}(\lambda_1^n,\lambda_2^n,\lambda_3^n,\lambda_4^n,\lambda_5^n) S^{-1}. $$

Multiplying by $X_0$ gives

$$ X_n = \sum_{j=1}^{5} \lambda_j^n, S e_j ,(S^{-1}X_0)_j , $$

where $e_j$ is the $j$-th standard basis vector.

Writing $v^{(j)} = S e_j$ and $c_j = (S^{-1}X_0)_j$, each component sequence is therefore of the form

$$ A_n = \sum_{j=1}^{5} c_j v^{(j)}1 \lambda_j^n,\quad B_n = \sum{j=1}^{5} c_j v^{(j)}_2 \lambda_j^n, $$

and similarly for $C_n, D_n, E_n$.

The coefficients $c_j v^{(j)}_k$ are determined uniquely by the initial conditions at levels $n=0,-1,-2,-3,-4$, which fix the five linear constraints needed to determine the expansion in the basis ${\lambda_j^n}$. This produces a representation of every cascade sequence as a linear combination of exponentials in the eigenvalues of $Q$.

Thus each cascade sequence satisfies a representation of the form

$$ u_n = \sum_{j=1}^{5} \alpha_j \lambda_j^n, $$

where $u_n$ denotes any of $A_n,B_n,C_n,D_n,E_n$, and the constants $\alpha_j$ depend on the initial level and on the chosen sequence.

This establishes (12) as the eigenvalue expansion of the cascade recurrences obtained from the matrix formulation. ∎