SICP Exercise 1.13

The following is my solution to Exercise 1.13 in Structure and Interpretation of Computer Programs

Problem statement

$\mathit{Fib}\left(n\right)=\left\{\begin{array}{ll}0& \text{if}n=0\\ 1& \text{if}n=1\\ \mathit{Fib}\left(n-1\right)+\mathit{Fib}\left(n-2\right)& \text{otherwise}\end{array}\right.$

Prove that $\mathit{Fib}\left(n\right)$ is the closest integer to $\frac{{\phi }^n}{\sqrt{5}}$, where $\phi =\frac{1+\sqrt{5}}{2}$. Hint: Let $\psi =\frac{1-\sqrt{5}}{2}$. Use induction and the definition of the Fibonacci numbers (above) to prove that $\mathit{Fib}\left(n\right)=\frac{{\phi }^n-{\psi }^n}{\sqrt{5}}$.

Proof

Let's restate the problem more formally. For $\mathit{Fib}\left(n\right)$ to be the closest integer to $\frac{{\phi }^n}{\sqrt{5}}$, the distance between the two must be less than or equal to $\frac{1}{2}$. In other words, we must prove:

$\left|\mathit{Fib}\left(n\right)-\frac{{\phi }^n}{\sqrt{5}}\right|\le \frac{1}{2}$

For now, let's assume the hint in the problem statement is true (we'll prove it later).

$\mathit{Fib}\left(n\right)=\frac{{\phi }^n-{\psi }^n}{\sqrt{5}}$, or $\mathit{Fib}\left(n\right)-\frac{{\phi }^n}{\sqrt{5}}=-\frac{{\psi }^n}{\sqrt{5}}$

We can substitue this into our problem statement, simplifying it remarkably:

$\left|\left(-\frac{{\psi }^n}{\sqrt{5}}\right)\right|\le \frac{1}{2}$, or $\left|{\psi }^n\right|\le \frac{\sqrt{5}}{2}$

Now, everybody knows that for $\left|x\right|\le 1$ and $n\ge 0$

$\left|x^n\right|\le 1$

Therefore, if we can show that $\left|\psi \right|\le 1$ and $\frac{\sqrt{5}}{2}>1$, we'll have our proof.

We can see immediately that $\frac{\sqrt{5}}{2}>1$, as $\frac{\sqrt{5}}{2}$ is obviously greater than $\frac{\sqrt{4}}{2}$, which is equal to $1$.

Let's show that $\left|\psi \right|\le 1$. That is. let's show:

$\left|\frac{1-\sqrt{5}}{2}\right|\le 1$, or $\left|\frac{\sqrt{5}-1}{2}\right|\le 1$

It is easy to see when we express the value of $1$ differently, say

$1=\left|\frac{\sqrt{9}-1}{2}\right|$

It would be even easier to see by punching the expression into a calculator, but how fun would that be?

Anyways, it should be obvious that

$\left|\frac{\sqrt{5}-1}{2}\right|<\left|\frac{\sqrt{9}-1}{2}\right|$

So there's our proof. Unfortunately, it rests on an unproven assumption: the hint that was given in the problem statement. Let's prove it.

Recall, the hint:

$\mathit{Fib}\left(n\right)=\frac{{\phi }^n-{\psi }^n}{\sqrt{5}}$

We'll approach this proof inductively as the problem statement recommends. Let's start by establishing some base cases. From the definition of $\mathit{Fib}\left(n\right)$,

For $n=0$,

$\mathit{Fib}\left(0\right)=0$

For $n=1$,

$\mathit{Fib}\left(1\right)=1$

For $n=2$,

$\begin{array}{ll}\hfill \mathit{Fib}\left(2\right)& =\mathit{Fib}\left(1\right)+\mathit{Fib}\left(0\right)\\ & =1+0\\ & =1\end{array}$

Now let's look at the right hand side.

For $n=0$,

$\frac{{\phi }^0-{\psi }^0}{\sqrt{5}}=\frac{1-1}{\sqrt{5}}=0$

For $n=1$,

$\begin{array}{ll}\hfill \frac{{\phi }^1-{\psi }^1}{\sqrt{5}}& =\frac{\left(\frac{1+\sqrt{5}}{2}\right)-\left(\frac{1-\sqrt{5}}{2}\right)}{\sqrt{5}}\\ & =\frac{\frac{1}{2}+\frac{\sqrt{5}}{2}-\frac{1}{2}+\frac{\sqrt{5}}{2}}{\sqrt{5}}\\ & =\frac{\sqrt{5}}{\sqrt{5}}\\ & =1\end{array}$

For $n=2$,

$\begin{array}{ll}\hfill \frac{{\phi }^2-{\psi }^2}{\sqrt{5}}& =\frac{{\left(\frac{1+\sqrt{5}}{2}\right)}^2-{\left(\frac{1-\sqrt{5}}{2}\right)}^2}{\sqrt{5}}\\ & =\frac{\left(1+2\sqrt{5}+2\right)-\left(1-2\sqrt{5}+2\right)}{4\sqrt{5}}\\ & =\frac{4\sqrt{5}}{4\sqrt{5}}\\ & =1\end{array}$

You can see that the for these base cases, the left-hand side equals the right.

Let's move on to the inductive case. If we assume $\mathit{Fib}\left(n\right)=\frac{{\phi }^n-{\psi }^n}{\sqrt{5}}$ and $\mathit{Fib}\left(n-1\right)=\frac{{\phi }^{n-1}-{\psi }^{n-1}}{\sqrt{5}}$, then we should expect

$\mathit{Fib}\left(n+1\right)=\frac{{\phi }^{n+1}-{\psi }^{n+1}}{\sqrt{5}}$

Does it?

$\begin{array}{ll}\hfill \mathit{Fib}\left(n+1\right)& =\mathit{Fib}\left(n\right)+\mathit{Fib}\left(n-1\right)\\ & =\left(\frac{{\phi }^n-{\psi }^n}{\sqrt{5}}\right)+\left(\frac{{\phi }^{n-1}-{\psi }^{n-1}}{\sqrt{5}}\right)\\ & =\frac{\left({\phi }^n+{\phi }^{n-1}\right)-\left({\psi }^n+{\psi }^{n-1}\right)}{\sqrt{5}}\\ & =\frac{{\phi }^{n+1}\left({\phi }^{-1}+{\phi }^{-2}\right)-{\psi }^{n+1}\left({\psi }^{-1}+{\psi }^{-2}\right)}{\sqrt{5}}\\ & =\frac{{\phi }^{n+1}{\phi }^{-1}\left(1+{\phi }^{-1}\right)-{\psi }^{n+1}{\psi }^{-1}\left(1+{\psi }^{-1}\right)}{\sqrt{5}}\\ & =\frac{{\phi }^{n+1}\left(\frac{2}{1+\sqrt{5}}\right)\left(1+\frac{2}{1+\sqrt{5}}\right)-{\psi }^{n+1}\left(\frac{2}{1-\sqrt{5}}\right)\left(1+\frac{2}{1-\sqrt{5}}\right)}{\sqrt{5}}\\ & =\frac{{\phi }^{n+1}\left(\frac{2}{1+\sqrt{5}}\right)\left(\frac{1+\sqrt{5}+2}{1+\sqrt{5}}\right)-{\psi }^{n+1}\left(\frac{2}{1-\sqrt{5}}\right)\left(\frac{1-\sqrt{5}+2}{1-\sqrt{5}}\right)}{\sqrt{5}}\\ & =\frac{{\phi }^{n+1}\left(\frac{2}{1+\sqrt{5}}\right)\left(\frac{3+\sqrt{5}}{1+\sqrt{5}}\right)-{\psi }^{n+1}\left(\frac{2}{1-\sqrt{5}}\right)\left(\frac{3-\sqrt{5}}{1-\sqrt{5}}\right)}{\sqrt{5}}\\ & =\frac{{\phi }^{n+1}\left(\frac{6+2\sqrt{5}}{1+2\sqrt{5}+5}\right)-{\psi }^{n+1}\left(\frac{6-2\sqrt{5}}{1-2\sqrt{5}+5}\right)}{\sqrt{5}}\\ & =\frac{{\phi }^{n+1}\left(\frac{6+2\sqrt{5}}{6+2\sqrt{5}}\right)-{\psi }^{n+1}\left(\frac{6-2\sqrt{5}}{6-2\sqrt{5}}\right)}{\sqrt{5}}\\ & =\frac{{\phi }^{n+1}\left(1\right)-{\psi }^{n+1}\left(1\right)}{\sqrt{5}}\\ & =\frac{{\phi }^{n+1}-{\psi }^{n+1}}{\sqrt{5}}\end{array}$

QED