How Close Do Fibonacci Ratios Get to φ?

Open with the ratios 1/1, 2/1, 3/2, 5/3, 8/5, 13/8, 21/13 displayed in a small table, and pose the driving question directly beneath them. Frame the tension: do these numbers settle, and if so, where and how quickly?

• Display the first six ratios F(n+1)/F(n) side by side
• State the driving question: how close to φ, and how fast?
• Hint that the answer involves both a limit and a rate

Ask the learner to predict how the gap |F(n+1)/F(n) − φ| behaves as n grows, choosing among linear halving, geometric shrinking by a fixed factor, slowing to zero very slowly, or staying roughly constant.

• One focused commitment to a convergence behavior
• Tests intuition about rate vs. amount of shrinking

A simulation where the learner increases n and sees two traces update in real time: the ratio F(n+1)/F(n) approaching the horizontal line φ, and the absolute gap |F(n+1)/F(n) − φ| plotted on a log scale. Sliders control n and the log-scale toggle.

• Step n from 1 to 20 and watch the ratio hug the φ line
• Toggle a log scale to see whether the gap looks linear or curved there
• Record the gap at n = 5, 10, 15, 20 to compare rates

A static table of the actual numerical gaps |F(n+1)/F(n) − φ| at n = 1, 2, 5, 10, 15, 20, together with the ratio of successive gaps. Set up the observation that successive ratios of gaps are approaching a constant near 0.382.

• Tabulate gaps to at least 8 decimal places
• Show the ratio gap(n)/gap(n−1) settling near 0.382
• Highlight that 0.382 is approximately 1/φ²

Show that if the ratio is r_n = F(n+1)/F(n), then r_{n+1} = 1 + 1/r_n. A short geometric argument using the fixed-point equation r = 1 + 1/r gives r_n − φ ≈ (r_{n−1} − φ) / φ², so the error is multiplied by 1/φ² ≈ 0.382 each step.

• Recurrence for ratios: r_{n+1} = 1 + 1/r_n
• Fixed point of r = 1 + 1/r is exactly φ
• Linearization gives error multiplied by 1/φ²

An interactive simulation that lets the learner change the initial ratio r_1 and watch the iteration r_{n+1} = 1 + 1/r_n. They see that any positive starting ratio still converges to φ, and the gap still multiplies by roughly 1/φ² once it is small. Includes a brief comparison to a slower map (e.g., r_{n+1} = (r_n + 2/r_n)/2 toward √2) to anchor what 'geometric rate' means.

• Try r_1 = 1, 1.5, 2, 3 — all still converge to φ
• Confirm the error factor near φ is still ≈ 1/φ²
• Contrast with Newton's method on √2, whose error squares each step

Note explicitly what this investigation does not claim: it does not derive the closed-form Binet formula, does not extend to negative indices, and does not generalize to every linear recurrence. It also marks that 'geometric with ratio 1/φ²' is the asymptotic statement, not the exact finite-step identity.

• Result is asymptotic; finite gaps differ from C · (1/φ²)ⁿ by a tiny correction
• Scope is positive-index Fibonacci ratios only
• Other recurrences have their own characteristic rates

Resolve the driving question in two parts: the ratios converge to φ, and the error shrinks geometrically by a factor of 1/φ² per step. Show a closing visual: the ratio curve flattening onto φ, the gap curve diving on a log scale, and a final line giving the approximate size of the gap at n = 20.

• F(n+1)/F(n) → φ as n → ∞
• |F(n+1)/F(n) − φ| is roughly C · (1/φ²)ⁿ
• By n = 20 the gap is already on the order of 10⁻⁵