IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

devesh_l2k007

Find a function such that

f(x)=f(x-1)+f(x-2)

req great brains

Think of a step series

titun

The Fibonacci series is the required answer.

In mathematics, the Fibonacci numbers form a sequence defined by the following recurrence relation:

$F(n):= \begin{cases} 0 & \mbox{if } n = 0; \\ 1 & \mbox{if } n = 1; \\ F(n-1)+F(n-2) & \mbox{if } n > 1. \\ \end{cases}$

The nth number Fib(n) in the Fibonacci series, where ( n > = 1 ), is given by the following formula:

Fib(n) = 1 /

5 x [ { (1+

5)/2 }ⁿ - { (1-

5)/2 }ⁿ ]

If you prefer values in your formulae, then here is another form:-

Well the Fibonacci series is the following:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987

i.e any number is the sum of the preceeding two numbers.

Cheers !!!

devesh_l2k007

what i asked was xcan u define a plynomial ffunction whose roots give

devesh_l2k007

anyone ???\