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In other words, we need to solve (L – θ) F = O and (L – φ) F = O separately and combine the results linearly.
Now (L-θ) Fn = O implies L(Fn ) = θ Fn = F(n+1).
Further F(n+1) = θFn implies that Fn = Aθn for some constant A. This is an
example of a geometric sequence.
Similarly, solving (L-φ) Fn = 0 we get Fn = Bφn for some constant B. Finally we obtain the general solution of (L2 - L - 1) Fn = O as
Fn= Aθn + Bφn (8)
This relation will help us find the formula for the nth Fibonacci number.
Substituting n = O and n = 1 respectively in (8) and recalling that F0 = O
and F1 = 1, we get A=1/√5 and B = (-1)/√5. Finally the formula for the nth Fibonacci number is
Fn = 1 [(1 + √5)n-(1 - √5 )n ]
√522 (9)
For the interested reader, a more detailed treatment of the derivation is available in [2].
The students inserted different values of n in the general formula given by (9) and verified that the formula actually works. They thoroughly enjoyed the derivation even though the concept of the use of operators is beyond the scope of school mathematics. While engaging in this exploration they observed patterns, made and verified their conjectures numerically as well as graphically (on the spreadsheet) and finally proved their conjecture using their knowledge of the theory of equations. Most importantly, all students participated in the investigations. The overall positive feedback encouraged the author to plan similar explorations for students of grades X, XI and XII.
Nature is replete with instances of the occurrence of the Fibonacci numbers. They appear in the patterns of the florets of a flower, the bracts of a pinecone and on the scales of a pineapple [3]. There are also numerous examples of the golden ratio in nature, which may be explored by the students. A spreadsheet such as MS Excel or LibreOffice Calc can prove handy in exploring the patterns in the sequence. Such explorations can enhance students’ mathematical thinking, give them access to higher level mathematical concepts and provide opportunities to
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