Golden powers are nearly integers

Re:Golden powers are nearly integers by jacobolus on Wednesday March 22, 2017 @06:08PM (#13935463)

Another way to get at this:

  φ^1 = φ
  φ^2 = φ + 1 [this is the definition of φ]
  φ^3 = φ(φ + 1) = 2φ + 1
  φ^4 = φ(2φ + 1) = 3φ + 2
  φ^5 = φ(3φ + 2) = 5φ + 3
  φ^6 = φ(5φ + 3) = 8φ + 5
  ...
  φ^n = F(n)φ + F(n-1)

Note that F(n+1)/F(n) → φ as n → ∞. Or flipped around, F(n)φ → F(n+1).

So φ^n = F(n)φ + F(n-1) → F(n+1) + F(n-1) = L(n) as n → ∞. This gets close to being an integer because F(n+1) and F(n-1) are both integers.

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Re:Golden powers are nearly integers by adrianratnapala on Thursday March 23, 2017 @05:08AM (#13938268)

I'm almost with you, but you need a bit more.

To show what we want, we must prove the stronger condition that F(n)φ - F(n+1) → 0. Merely knowing that F(n+1)/F(n) → φ only shows that the difference grows more slowly than F(n).

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Re:Golden powers are nearly integers by winstonsmith on Thursday March 23, 2017 @09:23AM (#13939470)

>To show what we want, we must prove the stronger condition that F(n)φ - F(n+1) → 0.

Noting that {F(n+1)/F(n)} is the sequence of convergents of the continued fraction expansion of φ, your stronger condition follows from Theorem 5 at https://en.wikipedia.org/wiki/Continued_fraction .

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