4 ms·
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 +
by jacobolus 10y ago
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.
- deleted 10y ago[deleted]
- adrianratnapala 10y agoI'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).
- winstonsmith 10y ago>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 https://en.wikipedia.org/wiki/Continued_fraction .
- deleted 10y ago[deleted]