Chapter 4, Problem 17C

Explanation of Solution

Proof:

• Assume “F(1)=1”, “F(2)=2”, and “$F(n)=F(n-1)+F(n-2)$” for “n>2”.
• Two constant values “n₀” and “c” should be found in order to prove that “F(n)” is “$\text{Ω}\left({\left(\raisebox{1ex}{$\text{3}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.\right)}^{\text{n}}\right)$” such that “$\text{F(n) }\ge \text{c}{\left(\raisebox{1ex}{$\text{3}$}\!\left/ \!\raisebox{-1ex}{$\text{2}$}\right.\right)}^{\text{n}}$” for all values of “$\text{n }\ge {\text{n}}_{\text{0}}$”.
• Base case:

Assume “$\text{c=}\left(\frac{\text{2}}{\text{3}}\right)$” in a way that,

$\begin{array}{c}\text{F(1)}\Rightarrow \text{ 1}\ge \text{c}\text{.}{\left(\frac{3}{2}\right)}^1\\ \Rightarrow 1\ge \left(\frac{2}{3}\right).\left(\frac{3}{2}\right)\\ \Rightarrow 1\ge 1\end{array}$

$\begin{array}{c}\text{F(2)}\Rightarrow \text{ 2}\ge \text{c}\text{.}{\left(\frac{3}{2}\right)}^2\\ \Rightarrow 2\ge \left(\frac{2}{3}\right).\left(\frac{9}{4}\right)\\ \Rightarrow 2\ge 1.5\end{array}$

Hence, it can be said that “$\text{F(n)}\ge \text{c}\text{.}{\left(\frac{3}{2}\right)}^{\text{n}}$” for “${\text{n}}_{\text{0}}\text{=1}$”.

• To prove  “$\text{F(m)}\ge \text{c}{\left(\frac{\text{3}}{\text{2}}\right)}^{\text{m}}$” such that “$\text{m}\ge {\text{n}}_{\text{0}}$” using induction hypothesis,

Calculate “$\text{F(m+1)}$” as below,

$\begin{array}{c}\text{F(m+1)=F(m)+F(m}-\text{1)}\\ \ge \text{c}\end{array}$