Fibonacci Complex

Algebra Level 5

Let a 0 = 0 a_0 = 0 , a 1 = 1 a_1 = 1 , and for n ≥ 1 n \geq 1 , a n + 1 = − a n − 1 + i a n . a_{n+1} = -a_{n-1} + i\: a_n.

Let S n = ∑ k = 1 4 n a k \displaystyle S_n = \sum_{k=1}^{4n} a_k . Determine lim ⁡ n → ∞ arg S n \displaystyle \lim_{n\to\infty} \text{arg}\ S_n .

(Give your answer in radians, between − π -\pi and + π +\pi .)

Clarification : i = − 1 i = \sqrt{-1} .


The answer is -2.12437.

Arjen Vreugdenhil by Arjen Vreugdenhil , 44, USA

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1 solution

Note that the sequence a n a_n are the Fibonacci numbers , but each next value is rotated 90 degrees in the complex plane: a n = i n − 1 F n . a^n = i^{n-1}\:F_n.

Thus we have tan ⁡ arg S n = Im S n Re S n = F 2 − F 4 + F 6 + − ⋯ − F 4 n F 1 − F 3 + F 5 + − ⋯ − F 4 n − 1 . \tan \text{arg}\ S_n = \frac {\text{Im}\ S_n} {\text{Re}\ S_n} = \frac{F_2 - F_4 + F_6 +- \cdots - F_{4n}}{F_1 - F_3 + F_5 +- \cdots - F_{4n-1}}. Now as n → ∞ n\to\infty , F n + 1 F n → ϕ = 1 2 ( 1 + 5 ) . \frac{F_{n+1}}{F_n} \to \phi = \tfrac12(1+\sqrt 5). Therefore N D → ϕ , \frac N D \to \phi, and the desired argument is arg S n → inv tan ⁡ ϕ − π = − 2.12437 . \text{arg}\ S_n \to \text{inv}\:\tan\phi - \pi = \boxed{-2.12437}. We subtract π \pi because clearly S n S_n lies in the third quadrant, requiring the argument to be between − π -\pi and π \pi .

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Some issues with your solution:

  1. The first line is false since it doesn't hold for n = 1 n = 1 . You want a n = i n − 1 F n a_n = i ^ { n-1} F_n .
  2. The third line needs to be justified further, and you need the fact that lim ⁡ F N = ∞ \lim F_N = \infty .

Calvin Lin Staff - 5 years, 3 months ago

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As you wish... :) I left this out because the properties of FIbonacci numbers are well known.

It can be shown that F n = ( 1 2 + 1 2 5 ) n − ( 1 2 − 1 2 5 ) n 5 . F_n = \frac{(\tfrac12+\tfrac12\sqrt 5)^n - (\tfrac12-\tfrac12\sqrt 5)^n}{\sqrt 5}. (It is a simple exercise that this generates the correct values for F 1 F_1 and F 2 F_2 , and that the recursion relation holds true. It is more work to derive this formula, but that is not my business here...)

The first term is an exponential function with base ϕ = 1 2 + 1 2 5 > 1 \phi = \tfrac12 + \tfrac12\sqrt 5 > 1 , and therefore tends to infinity. The second term is exponential with base 1 − ϕ 1-\phi which lies between 0 and 1; therefore its absolute value tends to zero. Thus we have lim ⁡ n → ∞ F n ϕ n = 1 5 , \lim_{n\to\infty} \frac{F_n}{\phi^n} = \frac 1{\sqrt 5}, from which the third line follows easily.

Arjen Vreugdenhil - 5 years, 3 months ago

Thanks for the correction.

Arjen Vreugdenhil - 5 years, 3 months ago

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