Answer
See explanation
Work Step by Step
Below is a concise proof of the **Single‐Root Theorem** (Theorem 5.8.5) for the case where the two unknown constants \(C\) and \(D\) in the solution are determined by the initial values \(a_0\) and \(a_1\).
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## Statement of the Theorem (Single‐Root Case)
Consider the second‐order linear recurrence
\[
a_k \;=\; A\,a_{k-1} \;+\; B\,a_{k-2},
\quad\text{for all integers }k \ge 2,
\]
with real constants \(A, B\), where \(B \neq 0\). Suppose the characteristic equation
\[
t^2 \;-\; A\,t \;-\; B = 0
\]
has exactly one real root \(r\). (This means \(r\) is a **double** root of the polynomial.) Then the general solution for the sequence \(\{a_k\}\) is
\[
a_k = C\,r^k \;+\; D\,k\,r^k,
\]
where \(C\) and \(D\) are real constants determined by any two known values of the sequence (for example, \(a_0\) and \(a_1\)).
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## Outline of the Proof
1. **Why the form \(C\,r^k + D\,k\,r^k\)?**
- From the standard theory of linear recurrences (or from solving the characteristic equation with a repeated root), we know that if \(r\) is the unique root of multiplicity 2, the two “linearly independent” solutions are:
\[
r^k
\quad\text{and}\quad
k\,r^k.
\]
Hence every solution is a linear combination:
\[
a_k = \alpha\,r^k + \beta\,\bigl(k\,r^k\bigr).
\]
We rename \(\alpha=C\) and \(\beta=D\) for convenience.
2. **Substitute into the Recurrence to Verify.**
- One checks directly (by substitution) that \(r^k\) satisfies \(a_k = A\,a_{k-1}+B\,a_{k-2}\) **if and only if** \(r\) is a root of \(t^2 - A\,t - B = 0\).
- If \(r\) is a double root, then \(k\,r^k\) also satisfies the same recurrence, but is independent from \(r^k\).
- Hence any linear combination \(C\,r^k + D\,k\,r^k\) is also a solution.
3. **Solve for \(C\) and \(D\) Using Initial Conditions \(a_0\) and \(a_1\).**
- If \(a_0\) and \(a_1\) are given, we get a system of two equations:
\[
\begin{cases}
a_0 = C\,r^0 + D\,(0)\,r^0 = C,\\
a_1 = C\,r^1 + D\,(1)\,r^1 = C\,r + D\,r.
\end{cases}
\]
- From \(a_0=C\), we have \(C=a_0\).
- Then \(a_1 = a_0\,r + D\,r\). So \(D\,r = a_1 - a_0\,r\), thus
\[
D = \frac{\,a_1 - a_0\,r\,}{r}.
\]
- Therefore \(C\) and \(D\) are uniquely determined by \(a_0\) and \(a_1\).
Hence the unique solution consistent with \(a_0\) and \(a_1\) is
\[
\boxed{a_k
= a_0\,r^k
\;+\; \bigl[\tfrac{a_1 - a_0\,r}{r}\bigr]\,
k\,r^k.}
\]
This completes the proof for the single‐root (double‐root) case.
