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Fibonacci formula?

  1. Aug 31, 2008 #1
    im just curious. is there a formula for the fibonacci formula in terms of..well terms. like the nth term =..?
    iv been trying to figure it out for a couple of days now but am not that smart.
  2. jcsd
  3. Aug 31, 2008 #2
  4. Aug 31, 2008 #3


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    You could derive it, if you know enough elementary linear algebra and in particular diagonalisation of matrices. It's not that difficult. You start off with recursive definition of the n+1 and nth term and n-1 term, put them all into a matrix and show that it is diagonalisable, then write out the matrix equation.
  5. Sep 2, 2008 #4
    done the algebra but have only learnt +-x/ matrices.
    how does the n+1 thing work
    like i said am not that smart.
  6. Sep 2, 2008 #5


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    There are probably other ways to derive it, but I'm only familiar with the one with matrices. There's a current thread on this here:
  7. Sep 2, 2008 #6


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    The terms in a Fibonacci sequence obey the recursive rule Fn+2= Fn+1+ Fn. One common way of solving such equations is to try a solution of the form Fn= an. Then Fn+1= an+1 and Fn+2= an+2 so the equation becomes an+2= an+1+ an. Dividing by an gives a2= a+ 1 or a2- a- 1= 0. Solving that by the quadratic formula,
    [tex]a= \frac{1\pm\sqrt{5}}{2}[/tex]
    In other words,
    [tex]F_n= \left(\frac{1+\sqrt{5}}{2}\right)^n[/tex]
    [tex]F_n= \left(\frac{1-\sqrt{5}}{2}\right)^n[/tex]
    both satisfy Fn+2= Fn+1+ Fn.

    Since that is a linear equation, any solution of that equation can be written
    [tex]A\left(\frac{1+\sqrt{5}}{2}\right)^n+ B\left(\frac{1-\sqrt{5}}{2}\right)^n[/tex]

    Now, looking at the first two terms of the Fibonacci sequence
    [tex]F_0= A+ B= 1[/tex]
    [tex]F_1=A\left(\frac{1+\sqrt{5}}{2}\right)+ B\left(\frac{1-\sqrt{5}}{2}\right)= 1[/tex]
    gives two equations to solve for A and B.
  8. Aug 6, 2010 #7


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    Sorry for the bump, but could you show me how you would solve for A and B?

    I'm not able to solve simultaneous equations in this form;

    A + B = 1
    Ax + By = 1

  9. Aug 6, 2010 #8


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    Simply let B=1-A and then substitute this into the second equation, solve for A there and then substitute back into the first to find B.
  10. Aug 7, 2010 #9


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    Of course! Solving by substitution. Thanks, I forgot about doing that.
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