1. ## Fibonacci Simultaneous Equations

Something I have come across recently ...

Take the Fibonacci sequence ( 1,1,2,3,5,8,13,21,...) and us it as coeff. for linear simultaneous equations:

For example

x+y =2
3x+5y=8

OR

2x+3y=5
8x+13y=21

These simultaneous equations and any others formed in the same way, have the same solution

I know its possible to see that x=1, y=1 and perhaps quite obvious..

Are there other sequences we can use in the same that might give the same solutions every time?

2. ## Re: Fibonacci Simultaneous Equations

Yes, the Fibonacci numbers have the defining property that $a_{n+2}= a_{n}+ a_{n+1}$ so it should be no surprise that $a_nx+ a_{n+1}y= a_{n+2}$ is satisfied by x= y= 1. Construct another sequence using some other rule connecting $a_n$, $a_{n+1}$, and $a_{n+2}$ and you will get a similar property.

For example, suppose we define $a_1= 1$, $a_2= 1$, and $a_{n+2}= 2a_n+ 3a_{n+1}$, giving the sequence 1, 1, 5, 17, 44, .... Then we will have every equation of the form $2a_nx+ 3a_{n+1}y= a_{n+2}$, where $a_n$, $a_{n+1}$ and $a_{n+2}$, is satisfied by x= y= 1.

3. ## Re: Fibonacci Simultaneous Equations

So if we use arithmetic series, for example

2x+5y=8
11x+14y=17

(2,5,8,11,14, 17) they seem to always intersect at (-2,1).
So regardless of first term and common difference.

I dont see how the recurrence definition helps here? I tend to think of the closed form as a+(n-1)d

how do i use your thinking on a_nx+ a_n+1y = a_n+2 where a_n+1 = d+a_n ?

4. ## Re: Fibonacci Simultaneous Equations

But the nth Fibonacci term cannot be calculated by a single formula.

5. ## Re: Fibonacci Simultaneous Equations

Originally Posted by DenisB
But the nth Fibonacci term cannot be calculated by a single formula.

I am now asking about general arithmetic series and it appears if we use these terms as coefficients as above, they solve to the same values.
And i am wondering how to prove this using the recurrence defintion of a general arithmetic series.

6. ## Re: Fibonacci Simultaneous Equations

Originally Posted by rodders
So if we use arithmetic series, for example

2x+5y=8
11x+14y=17

(2,5,8,11,14, 17) they seem to always intersect at (-2,1).
So regardless of first term and common difference.

I dont see how the recurrence definition helps here? I tend to think of the closed form as a+(n-1)d

how do i use your thinking on a_nx+ a_n+1y = a_n+2 where a_n+1 = d+a_n ?
$f_n = \dfrac{1}{\sqrt{5}}\left(\phi^{n+1}-\left(1-\phi\right)^{n+1}\right)$

Where $\phi = \dfrac{1+\sqrt{5}}{2}$ is the Golden Ratio.

The proof that this equation generates the Fibonacci sequence involves slightly more complicated combinatorics. Since you are posting in a Pre-University forum, I will not delve into the proof that this is the correct formula. But, check it out for the first few numbers. This sequence starts at $f_0 = 1$, $f_1 = 1$. If you want it to start at 1 instead, you would use the following:

$g_n = \dfrac{1}{\sqrt{5}}\left( \phi^n-\left(1-\phi\right)^n \right)$