Page 48 - Ramanujan Yatra

We already know that the Fibonacci sequence is given by the recurrence relation,F(n+1)=Fn+F(n-1) (1)
Also from the sequence of ratios we observe that after some value of n, we have
F(n+1) = Fn Fn F(n-1)
Let this ratio be equal to x. Using (1) and (2) we get
(2)
Fn+ F(n-1) = Fn or 1 + F(n-1) = Fn (3) Fn Fn-1) Fn F(n-1)
In terms of x (3) becomes 1+ 1 =x which simplifies to the quadratic equation x
x2 – x – 1 = 0 with roots 1±√5 . 2
We ignore the value 1-√5 as x (being a ratio) cannot be negative. Thus 2
x= 1+√5 (which is approximately equal to 1.618 up to three decimal places). 2
This is also known as the Golden Ratio.
Students were excited by this exploration as they had heard about the Golden Ratio but did not know about its relationship to the Fibonacci sequence. They had understood that recurrence relation F(n+1)=Fn+F(n-1) can be used to generate new Fibonacci numbers but wondered if there was a formula for the nth Fibonacci number? After all, if we wanted the 100th or the 1000th Fibonacci number, wouldn’t it be far more convenient to have a formula in which we could insert the value of n (say n = 100 for the 100th Fibonacci number) and obtain the required number. With such explorations the student’s mathematical thinking took a leap and they wanted to find a more general or explicit formula to obtain Fibonacci numbers.
Ramanujan
YATRA
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