A challenge! (1 Viewer)
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Proving this would make a good (hard) 4u question.Xayma said:The golden ratio is a cool number, φ<sup>2</sup>=φ+1, 1/φ=φ-1, as fibbonachi numbers get higher, F<sub>n</sub>/F<sub>n-1</sub> approaches φ, and lots of cool other things
Xayma
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No, Ive known that formula for a few years so finding out how it works wasnt really a top pirority when I first saw it (when I was about 12). Hmm would be interested to know why it does work though. I would never put that I was using it in an exam, I would just put it in the calculator to speed it up.CM_Tutor said:
Got down to the fractional equivalent of φ, got no idea how to prove that it is equal to (1+√5)/2CM_Tutor said:
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Here is a 4u induction question:
With the Fibonacci Numbers defined in the usual way (ie. F<sub>1</sub> = 1, F<sub>2</sub> = 1 and F<sub>n</sub> = F<sub>n-1</sub> + F<sub>n-2</sub> for n ≥ 3),
(a) Prove by induction that F<sub>n</sub> = (1 / √5) * {[(1 + √5) / 2]<sup>n</sup> - [(1 - √5) / 2]<sup>n</sup>} for integers n ≥ 1
(b) Hence, or otherwise, show that as n --> ∞, F<sub>n+1</sub> / F<sub>n</sub> ---> φ, where φ = (1 + √5) / 2
Xayma - you should be able to explain your approximation method using part (a) as well.
Now, here is a question with a more formal proof that F<sub>n+1</sub> / F<sub>n</sub> ---> φ:
Let φ = (1 + √5) / 2, and define the Fibonacci numbers as above, and define R<sub>n</sub> as F<sub>n+1</sub> / F<sub>n</sub>
(a) Show that if φ < R<sub>n</sub> < φ + ε, where ε ≤ 1 / 2, then φ - ε / 2 < R<sub>n+1</sub> < φ
(b) Show that if φ - ε < R<sub>n</sub> < φ, where ε ≤ 1 / 2, then φ < R<sub>n+1</sub> < φ + ε / 2
(c) Hence, show that |R<sub>n</sub> - φ| < 1 / 2<sup>n</sup>
(d) Explain why F<sub>n+1</sub> / F<sub>n</sub> ---> φ as n --> ∞
With the Fibonacci Numbers defined in the usual way (ie. F<sub>1</sub> = 1, F<sub>2</sub> = 1 and F<sub>n</sub> = F<sub>n-1</sub> + F<sub>n-2</sub> for n ≥ 3),
(a) Prove by induction that F<sub>n</sub> = (1 / √5) * {[(1 + √5) / 2]<sup>n</sup> - [(1 - √5) / 2]<sup>n</sup>} for integers n ≥ 1
(b) Hence, or otherwise, show that as n --> ∞, F<sub>n+1</sub> / F<sub>n</sub> ---> φ, where φ = (1 + √5) / 2
Xayma - you should be able to explain your approximation method using part (a) as well.
Now, here is a question with a more formal proof that F<sub>n+1</sub> / F<sub>n</sub> ---> φ:
Let φ = (1 + √5) / 2, and define the Fibonacci numbers as above, and define R<sub>n</sub> as F<sub>n+1</sub> / F<sub>n</sub>
(a) Show that if φ < R<sub>n</sub> < φ + ε, where ε ≤ 1 / 2, then φ - ε / 2 < R<sub>n+1</sub> < φ
(b) Show that if φ - ε < R<sub>n</sub> < φ, where ε ≤ 1 / 2, then φ < R<sub>n+1</sub> < φ + ε / 2
(c) Hence, show that |R<sub>n</sub> - φ| < 1 / 2<sup>n</sup>
(d) Explain why F<sub>n+1</sub> / F<sub>n</sub> ---> φ as n --> ∞
Xayma
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Bah phooey to those, I prefer lim (n-->∞.) F<sub>n+1</sub>/F<sub>n</sub> --> (via continually expanding and dividing) 1+1/(1+1/[1+1/{1+1/(........ which is = to φ (although I dont know how to prove that last bit)
Anyway to part a) I understand that it becomes (φ<sup>n</sup>/√5)*-[(1-√5)/2]<sup>n</sup>/√5 but I dont know how *-[(1-√5)/2]<sup>n</sup>/√5 transforms it into the nint function.
Anyway to part a) I understand that it becomes (φ<sup>n</sup>/√5)*-[(1-√5)/2]<sup>n</sup>/√5 but I dont know how *-[(1-√5)/2]<sup>n</sup>/√5 transforms it into the nint function.
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