1. ## Fibonacci Numbers

F(n) = F(n-1) + F(n-2):

Can someone explain to me how to use this formula, and maybe show an example. I'm uncertain of what I'm supposed to be plugging in for 'n'.

2. Hi

For instance for n=2 you get F(2) = F(1) + F(0)

Given F(0) and F(1) you can compute F(2)

Then for n=3 you get F(3) = F(2) + F(1)

Knowing F(1) and F(2) calculated just before you can get F(3)

And so on ...

3. ## Two things

1) F has to be a natural number greater than 2

2) What do you know about Fibonacci numbers? Why are you interested in
them?

4. Pick a first number - say, 3 - and call it F(1), meaning the first number in the sequence. Then pick a second number - perhaps 5 - and assign this to F(2). To find the third number F(3) you use F(3)= F(2) + F(1) = 5 + 3 = 8. Then to find the 4th you use F(4) = F(3) + F(2) = 8 + 5 = 13. In general, to find the nth number in the sequence you add F(n-1) and F(n-2) together. Typically people start the Fibonacci series off with F(1) =1 and F(2)=1, but you don't have to. This leads to:

F(1) = 1
F(2) = 1
F(3) = 2
F(4) = 3
F(5) = 5
F(6) = 8
F(7) = 13
F(8) = 21
etc.

Hope this helps!

5. Well, it was part of a problem I had to do for school. When I was given a number I was supposed to correspond with a Fibonacci number. I guess like:

1. 1
2. 2
3. 3
4. 5
5. 8
6.13

So, if I wanted to find the Fibonacci number for 2 (which is two), I thought I'd let n be 2 and plug it in...

F(2)=F(2-1)+F(2-2)
F(2)=F(1)+F(0)
but F(1)+F(0)=F(1)?

but the fibonacci number for 2 isn't 1 is it?

I'm a little confused here

6. The Fibonacci series needs to be "seeded" with the first two numbers F(1) and F(2). In other words - the first two numbers must be given to you, so that you can then determine the 3rd, and so on.

7. @spacegirl777 you pluged-in well.

Here is what you need to remember.

F(0)= 1
F(1)= 1
F(2)= 2
F(3)= 3
F(4)= 5
F(5)= 8

..... etc ....
so F(2)=F(2-1)+F(2-2)=F(1)+F(0)=1+1=2 which is correct.

8. A more suggestive and general approach is to consider the Fibonacci sequence as solution of the difference equation...

$x_{n+2} - x_{n-1} - x_{n}=0$ (1)

... with 'initial consitions' $x_{0}=0$ and $x_{1}=1$. If u and v are solution of the second order equation...

$z^{2} - z -1 =0$ (2)

... then the solution of (1) is of the form...

$x_{n}= c_{1} u^{n} + c_{2} v^{n}$ (3)

In this case is...

$u= \frac{1 - \sqrt{5}}{2}$

$v= \frac{1 + \sqrt{5}}{2}$ (4)

... and the constants $c_{1}$ and $c_{2}$ are found imposing the initial conditions and solving the system...

$c_{1} + c_{2} =0$

$c_{1} \frac{1 - \sqrt{5}}{2} + c_{2} \frac{1 + \sqrt{5}}{2}= 1$ (5)

Kind regards

$\chi$ $\sigma$

9. Originally Posted by p0oint
@spacegirl777 you pluged-in well.

Here is what you need to remember.

F(0)= 1
F(1)= 1
F(2)= 2
F(3)= 3
F(4)= 5
F(5)= 8

..... etc ....
so F(2)=F(2-1)+F(2-2)=F(1)+F(0)=1+1=2 which is correct.
Okay, I understand it now Thank you very much!