# Exercise 1.13 of SICP

Exercise 1.13: Prove that Fib(n) is the closest integer to n/5, where $$\phi = \frac{1 + \sqrt{5}}{2}$$. Hint: Let $$\psi = 1-\frac{\sqrt{5}}{2}$$ Use induction and the definition of the Fibonacci numbers (see section 1.2.2) to prove that $$Fib(n) = \frac{\phi^n-\psi^n}{\sqrt5}$$.

$$F_n = F_{n-1} + F_{n-2}$$

$$F_0 = 0$$

$$F_1 = 1$$

Fn is a linear second-order recurrence with constant coefficients. The characteristic equation for it is: $$r^2-r-1=0$$

$$r = \phi , \psi$$

This means that the closed form solution is of the form $$F_n = \alpha\phi^n-\beta\psi^n$$ The constants $$\alpha$$ and $$\beta$$ are determined by the initial conditions bellow.

$$F_0 = \alpha + \beta = 0$$

$$\alpha = -\beta$$

$$F_1 = \alpha\phi + \beta\psi = 1$$

$$\beta(\psi-\phi)=1$$

$$\beta=\frac{1}{\psi-\phi}$$

$$\alpha=\frac{-1}{\psi-\phi}$$

The closed expression has the following form:

$$F_n = \frac{-\phi^n}{\psi-\phi}-\frac{\psi^n}{\psi-\phi}=\frac{\phi^n-\psi^n}{\sqrt5}$$

Now that the derivation of the closed form for Fn is done here’s the proof by induction.

$$\phi = \frac{1 + \sqrt{5}}{2}$$

$$\psi = \frac{1-\sqrt{5}}{2}$$

$$F_n = \frac{\phi^n-\psi^n}{\sqrt5}$$

$$F_0 = \frac{\phi^0-\psi^0}{\sqrt5}=0$$

$$F_1 = \frac{\phi^1-\psi^1}{\sqrt5}=1$$

Perform the induction step:

Assume: $$F_n = \frac{\phi^n-\psi^n}{\sqrt5}$$

$$F_{n+1} = F_{n}+F_{n-1} = \frac{\phi^n-\psi^n}{\sqrt5} + \frac{\phi^{n-1}-\psi^{n-1}}{\sqrt5}=\frac{\phi^{n+1}-\psi^{n+1}}{\sqrt5}$$

$$F_{n+1} = \frac{\phi^{n-1}(\phi+1)-\psi^{n-1}(\psi+1)}{\sqrt5}$$

Because $$\phi^2 = \phi+1$$

$$F_{n+1} = \frac{\phi^{n-1}(\phi+1)-\psi^{n-1}(\psi+1)}{\sqrt5}=\frac{\phi^{n+1}-\psi^{n+1}}{\sqrt5}$$

The proof that $$F_n = \frac{\phi^n}{\sqrt5}$$ is straightforward.

$$\psi = \frac{1-\sqrt{5}}{2} = -0.618$$

$$\lim_{n \to \infty}\psi^n = 0$$ for example $$\psi^{30} \approx 0.0000005374904998555718$$ and $$\phi^{30} \approx 1860498$$

$$\epsilon = 1-\frac{\phi^{30}-\psi^{30}}{\phi^{30}} \approx 2.8899105330992825 \cdot 10^{-13}$$

Clearly for n>30 $$F_n = \frac{\phi^n}{\sqrt5}$$ is a good approximation.

 n approx exact epsilon 0 0.4472135955 0.894427191 0.4472135955 1 0.72360679775 0.4472135955 0.27639320225 2 1.17082039325 1.3416407865 0.17082039325 3 1.894427191 1.788854382 0.105572809 4 3.06524758425 3.1304951685 0.0652475842499 5 4.95967477525 4.9193495505 0.0403252247502 6 8.0249223595 8.049844719 0.0249223594996 7 12.9845971347 12.9691942695 0.0154028652506 8 21.0095194942 21.0190389885 0.00951949424901 9 33.994116629 33.988233258 0.00588337100159