Recursion Cursion Cursion

Calculus Level pending

Let $F(x)$ represent the x-th term of a sequence defined by $F(x+1) = F(x) + F(x-1)$ , where $F(1) =1, F(2) = 2$ .

Evaluate $\displaystyle \lim_{x \rightarrow \infty} \dfrac{F(x+10)}{F(x)+F(10)}$ to the nearest integer.

The answer is 123.

1 solution

Guilherme Dela Corte
Dec 29, 2013

$r = \dfrac{F(x+a+1)}{F(x+a)}, \; a \in \mathbb{Z}$

$\dfrac{F(x+1)}{F(x)} \times \dfrac{F(x+2)}{F(x+1)} \times (\cdots) \times \dfrac{F(x+9)}{F(x+8)} \times \dfrac{F(x+10)}{F(x+9)} \Rightarrow \boxed{\dfrac{F(x+10)}{F(x)} = r^{10}}$

$r = \dfrac{F(x+a)+F(x+a-1)}{F(x+a)} \Rightarrow r = 1 + \dfrac{F(x+a-1)}{F(x+a)} \Rightarrow \boxed{r = 1 + \dfrac{1}{r}}$

$r^2 = r + 1\Rightarrow r^4 = 3r+2 \Rightarrow r^8 = 21r+13 \Rightarrow \boxed{r^{10} = 55r +34}$ .

$r^2 = r + 1 \Rightarrow r = \dfrac{1 + \sqrt{5}}{2} \Rightarrow r^{10} = 55 \left (\dfrac{1 + \sqrt{5}}{2} \right ) + 34 \Rightarrow \boxed{r^{10} \approx 123}$ .

$x \rightarrow \infty, F(x) \rightarrow \infty \Rightarrow \displaystyle \lim_{x \rightarrow \infty} \dfrac{F(x+10)}{F(x)+F(10)} = \dfrac{F(x+10)}{F(x)}$

$\Rightarrow \displaystyle \lim_{x \rightarrow \infty} \dfrac{F(x+10)}{F(x)+F(10)} = \boxed{123.}$

You could have just said that r= $\phi$ and then use the golden ratio's properties.That could have saved you a lot of Latex-ing.

Bogdan Simeonov - 7 years, 5 months ago

1 - Not at all. We define $\varphi$ = $\frac{1+\sqrt{5}}{2}$ , and clearly $\varphi = \varphi^{-1} + 1$ has more than this one solution.

2 - To use golden ratio's properties we must prove them, or, at least, explain them.

3 - I only kept the words "golden ratio", "Fibonacci" and "phi" out of the problem because, if people find out right from the start that $F(x)$ is the Fibonacci Sequence, all the magic in finding the limit is lost.

4 - What if I changed the sequence a little bit? What if we have now $G(x)$ , such that

$G(x+1) + G(x) = G(x-1), G(0) = 499$ ?

What will our new limit be?

Guilherme Dela Corte - 7 years, 5 months ago

For this we would get r= $\frac{1}{r}-1$ and r=0,618...Then,the limit $r^{10}$ would be equal to 34-55r(again Fibonacci-like) and by carefully using the calculator , we derive that it is going to be approximately 0.01?

Bogdan Simeonov - 7 years, 5 months ago

@Bogdan Simeonov – Bogdan, you only forgot to round it to the nearest integer... you could have even stopped at $r \approx 0.618$ , because already $[r^2] = 0$ .

Guilherme Dela Corte - 7 years, 5 months ago

@Guilherme Dela Corte – I was close enough :D

Bogdan Simeonov - 7 years, 5 months ago

Recursion Cursion Cursion

The answer is 123.

1 solution

0 pending reports