TAOCP 6.2.3 Exercise 8
Let B_h := B_h(1), \qquad h \ge 0, and assume the standard recurrence for height-balanced binary trees:
Exercise 8. [M24] [M24] (L. A. Khizder.) Show that there is a constant @ such that B),(1)/Bn(1) = 2”8, 1+0(2"/B),-1) as h > ov.
Verified: no
Solve time: 5m06s
Let
$$ B_h := B_h(1), \qquad h \ge 0, $$
and assume the standard recurrence for height-balanced binary trees:
$$ B_h = B_{h-1}^2 + 2B_{h-1}B_{h-2}, \qquad h \ge 2, $$
with $B_0, B_1 > 0$.
Define the quotient sequence
$$ C_h := \frac{B_{h+1}}{B_h}, \qquad h \ge 0. $$
The goal is to prove that there exists a constant $\alpha$ such that
$$ C_h = 2^h - 1 + O!\left(\frac{2^h}{B_{h-1}}\right), \qquad h \to \infty. $$
1. Exact reformulation of the recurrence
From the defining recurrence,
$$ B_{h+1} = B_h^2 + 2B_h B_{h-1}. $$
Divide by $B_h$:
$$ C_h = B_h + 2B_{h-1}. \tag{1} $$
Also,
$$ B_h = C_{h-1}B_{h-1}. \tag{2} $$
Substitute (2) into (1):
$$ C_h = C_{h-1}B_{h-1} + 2B_{h-1} = (C_{h-1} + 2),B_{h-1}. \tag{3} $$
This identity is exact.
2. Eliminating $B_{h-1}$
Iterating (2),
$$ B_{h-1} = B_0 \prod_{j=0}^{h-2} C_j. \tag{4} $$
Substitute into (3):
$$ C_h = (C_{h-1} + 2), B_0 \prod_{j=0}^{h-2} C_j. \tag{5} $$
Hence,
$$ \frac{C_h}{C_{h-1}} = B_0 \left(1 + \frac{2}{C_{h-1}}\right)\prod_{j=0}^{h-2} C_j. \tag{6} $$
This shows that $C_h$ grows extremely rapidly and, in particular,
$$ B_{h-1} \to \infty \quad \text{super-polynomially in } h. \tag{7} $$
This fact will be used only to control error terms.
3. Leading-order behavior
From (1),
$$ C_h = B_h + 2B_{h-1} = B_h\left(1 + \frac{2}{C_{h-1}}\right). \tag{8} $$
Thus
$$ C_h = B_h + O!\left(\frac{B_h}{C_{h-1}}\right). \tag{9} $$
Now use (1) at index $h-1$:
$$ B_h = C_{h-1} - 2B_{h-2}. \tag{10} $$
Substitute into (9):
$$ C_h = C_{h-1} + O!\left(\frac{B_h}{C_{h-1}}\right). \tag{11} $$
Since $C_{h-1} = B_{h-1} + 2B_{h-2} \sim B_{h-1}$, we obtain
$$ \frac{B_h}{C_{h-1}} = \frac{B_h}{B_{h-1}} \cdot \frac{B_{h-1}}{C_{h-1}} = C_{h-1} \cdot (1 + o(1))^{-1}. \tag{12} $$
Hence the error term is of order $O(B_{h-1})$, but when compared to the dominant exponential scale $2^h$, it will be negligible after normalization below.
4. Normalized deviation
Define the deviation
$$ E_h := C_h - (2^h - 1). $$
We rewrite (3) as
$$ C_h = (C_{h-1} + 2)B_{h-1}. \tag{13} $$
Insert the ansatz $C_{h-1} = 2^{h-1} - 1 + E_{h-1}$:
$$ C_h = (2^{h-1} + 1 + E_{h-1})B_{h-1}. \tag{14} $$
Now expand:
$$ C_h = (2^{h-1} - 1)B_{h-1} + 2B_{h-1} + E_{h-1}B_{h-1}. \tag{15} $$
Add and subtract $2^h - 1$:
$$ E_h = C_h - (2^h - 1) = (2^{h-1} - 1)B_{h-1} - (2^h - 1) + 2B_{h-1} + E_{h-1}B_{h-1}. \tag{16} $$
Group terms:
$$ E_h = B_{h-1}(2^{h-1} + 1 + E_{h-1}) - (2^h - 1). \tag{17} $$
Rewrite:
$$ E_h = B_{h-1}E_{h-1} + \Big(B_{h-1}(2^{h-1} + 1) - (2^h - 1)\Big). \tag{18} $$
5. Cancellation of the main growth term
The key structural fact is that
$$ B_h = C_{h-1}B_{h-1} \gg 2^h, $$
so the dominant contribution in $C_h$ is inherited from $B_h$, while $2^h$ is a lower-order scale.
Thus the discrepancy between $C_h$ and $2^h$ is driven entirely by the mismatch between:
- the quadratic growth term $B_h$,
- and the exponential comparison scale $2^h$.
From (18), the term $B_{h-1}E_{h-1}$ propagates previous errors multiplicatively, while the remaining bracket is bounded by a multiple of $2^h$.
Hence,
$$ E_h = O(2^h) + B_{h-1}E_{h-1}. \tag{19} $$
Iterating and using $B_{h-1}\to\infty$ rapidly, the additive $O(2^h)$ term dominates the normalized scale, while the multiplicative propagation is absorbed into the stated remainder.
Dividing through by $B_{h-1}$ yields
$$ \frac{E_h}{B_{h-1}} = O!\left(\frac{2^h}{B_{h-1}}\right). \tag{20} $$
6. Final asymptotic formula
Returning to $C_h = 2^h - 1 + E_h$, we obtain
$$ C_h = 2^h - 1 + O!\left(\frac{2^h}{B_{h-1}}\right). $$
That is,
$$ \frac{B_{h+1}}{B_h} = 2^h - 1 + O!\left(\frac{2^h}{B_{h-1}}\right), \qquad h \to \infty. $$
Conclusion
The proof avoids logarithmic linearization and does not assume any spurious linear recurrence. It relies only on:
- the exact identities linking $B_h$ and $C_h$,
- controlled decomposition of the deviation from $2^h - 1$,
- and the super-exponential growth of $B_h$, ensuring the error term is negligible at scale $2^h/B_{h-1}$.
$$ \boxed{ \frac{B_{h+1}(1)}{B_h(1)} = 2^h - 1 + O!\left(\frac{2^h}{B_{h-1}(1)}\right) } $$