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I've always enjoyed that formula for the Fibonacci numbers, primarily because it's so unintuitive that the right hand side would actually even be an integer. It
by tgb 10y ago
I've always enjoyed that formula for the Fibonacci numbers, primarily because it's so unintuitive that the right hand side would actually even be an integer. It's a fun exercise to prove that (without resorting to the fact that the Fibonacci numbers are obviously integers). Hint: start with -1/phi = 1-phi and use phi=(sqrt(5)+1)/2.
- jacobolus 10y ago> start with -1/phi = 1-phi and use phi=(sqrt(5)+1)/2. This is the same fact twice. (Well, the second part is just one of the two roots of the polynomial equation x^2 = x + 1.)
- btilly 10y agoThat is a special case of an even more amazing result. Given a polynomial P(x), any expression that can be written as a symmetric polynomial in the roots of P, can be written as a polynomial in the coefficients of P. See https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial#The_fundamental_theorem_of_symmetric_polynomials https://en.wikipedia.org/wiki/Elementary_symmetric_polynomia... for more. This applies to arbitrary rings. So if P(x) and Q(x) are both polynomials, then any polynomial which is symmetric in the roots of P(x) and also symmetric in the roots of Q(x) can be written as polynomials in the coefficients of Q(x) and P(x). Therefore, for example, (x-sqrt(2)-sqrt(3))(x-sqrt(2)+sqrt(3))(x+sqrt(2)-sqrt(3))(x+sqrt(2)+sqrt(3)) will work out to be a polynomial in the coefficients of x^2-2 and x^2-3. It will therefore be an integer polynomial with those 4 roots. This is actually the original way in which people proved that the algebraic integers formed a ring.