FT10: More on the de Moivre Analogue

In part 9, we looked at an analogue to de Moivre's theorem. Slightly rephrased, we saw that:

• L(n)/2 + √5 F(n)/2 = φⁿ

so that:

• L(kn)/2 + √5 F(kn)/2 = (L(n)/2 + √5 F(n)/2)^k.

However, we would like to eliminate the factors of ½, since, when we derive other formulas from these, the ½'s end up obscuring which numbers are integers.

We'll actually do this in two ways, one using the Fibonacci numbers and one using the Lucas numbers.

Taking the formula φⁿ = L(n)/2 + √5 F(n)/2, rewriting √5 as 2φ − 1, and using the fact that L(n) = 2F(n−1) + F(n) yields:

• φⁿ = F(n−1) + φ F(n)

Similarly, but using the fact that F(n) = (2L(n−1) + L(n))/5, we get:

• φⁿ = (L(n−1) + φL(n))/√5

Alternatively, one can prove these two formulas for φⁿ by mathematical induction.

Plugging these expressions into the formula (φⁿ)^k = φ^nk results in two more forms of our analogue to de Moivre's theorem:

• F(kn−1) + φ F(kn) = (F(n−1) + φ F(n))^k
• (L(kn−1) + φL(kn))/√5 = (L(n−1) + φL(n))^k/√5^k

Next we'll look at some applications of these formulas.