Page 49 - Ramanujan Yatra

Ramanujan
YATRA
40
A general formula for Fibonacci numbers
To find the required formula, the students had to be acquainted with a simple but useful tool called the left shift operator. Given a sequence of numbers x1, x2, x3,..xn......we define an operator L as follows
L(xn )=x(n+1) (4)
Where xn and xn+1 are the nth and (n+1)th terms of the sequence x1, x2, x3,..xn
If L is made to operate on this sequence, we obtain
L(x1), L(x2), L(x3),....
which is x2, x3, x4,.... i.e., the original sequence shifted to the left by one term (x1 has been deleted).
Operating L on the Fibonacci sequence we get F(n+1) = L(Fn).
Also, F(n+2) = L(F(n+1)) = L(L(Fn ) = L2 (Fn) (5) Replacing F(n+2) with F(n+1) + Fn we get
F(n+2) - F(n+1) - Fn = O (6) Rewriting this in terms of L and using (6) we get
L2 (Fn ) - L(Fn ) - Fn = O (7)
This can be further rewritten as (L2-L-1) Fn = O
This interesting property of the left shift operator L allows us to treat it like a variable to manipulate it algebraically. We already know that the roots of the
quadratic equation x2 – x – 1 = O are 1±√5 Let θ=1+√5 and φ=1-√5 2
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We can solve the equation (7) to obtain
((L – θ) (L – φ )) Fn = O