Well actually we have only defined the nth Fibonacci number in terms of the two before it: the n-th Fibonacci number is the sum of the (n-1)th and the (n-2)th. So to calculate the 100th Fibonacci number, for instance, we need to compute all the 99 values before it first - quite a task, even with a calculator! A natural question to ask therefore is: Can we find a formula for F(n) which involves only n and does not need any other (earlier) Fibonacci values?
Yes! It involves our golden section number Phi and its reciprocal phi: Here it is: Fib(n) = Phin – (–Phi)–n = Phin – (–phi)n sqrt5 sqrt5 where Phi = 1·61803 39887 49894 84820 45868 34365 63811 77203 09179 80576 ... . The next version uses just one of the golden section values: Phi, and all the powers are positive: Fib(n) = Phin – (–1)n Phin sqrt5 Since phi is the name we use for 1/Phi on these pages, then we can remove the fraction in the numerator here and make it simpler, giving the second form of the formula at the start of this section.
We can also write this in terms of sqrt5 since Phi = 1 + sqrt5 and –phi = 1 – sqrt5 : 2 2 Fib(n) = Phin – (–phi)n = Phin–(–phi)n = 1 (( 1 +?5 ) n – ( 1 –?5 ) n ) -- -- -- -- -- Phi – (–phi)?5?5 2 2 If you prefer values in your formulae, then here is another form:- Fib(n) = 1.6180339..n – (–0.6180339..)n 2.236067977.. This is a surprising formula since it involves square roots and powers of Phi (an irrational number) but it always gives an integer for all (integer) values of n!
I cant really gove you an answer,but what I can give you is a way to a solution, that is you have to find the anglde that you relate to or peaks your interest. A good paper is one that people get drawn into because it reaches them ln some way.As for me WW11 to me, I think of the holocaust and the effect it had on the survivors, their families and those who stood by and did nothing until it was too late.