Composite functions and fibonacci

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apple2357

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Can anyone help me see why this particular function seems to generate coefficients of the Fibonacci sequence when constructed in the following way:

f(x)= (x+1)/(x+2) , then ff(x) = (2x+3)/(3x+5) , fff(x) = (5x+8)/(8x+13), ffff(x) = (13x+21)/(21x+34) etc.

What is it about the original function that makes this happen? I tried explored f(x) = (x+a)/(x+b) and it started getting messy and led to an equation b^2= a+ab+1, but i am not sure that is the right approach...
 
I have put this in the wrong section! Sorry, can it be moved to algebra please!
I dont know how i did this!
 
Can anyone help me see why this particular function seems to generate coefficients of the Fibonacci sequence when constructed in the following way:

f(x)= (x+1)/(x+2) , then ff(x) = (2x+3)/(3x+5) , fff(x) = (5x+8)/(8x+13), ffff(x) = (13x+21)/(21x+34) etc.

What is it about the original function that makes this happen? I tried explored f(x) = (x+a)/(x+b) and it started getting messy and led to an equation b^2= a+ab+1, but i am not sure that is the right approach...

And is this a special type of function?
 
Very interesting. What have you tried? Where are you stuck? It is hard to help you if you show us no work for you to build on. Please post back showing us your work.

I would look closely at what happens to f(x) = (x+a)/(x+b)
 
I think your use of parameters is a good approach. So let's take a look at what appears to be your general form:

[MATH]f(x)=\frac{F_{n}x+F_{n+1}}{F_{n+1}x+F_{n+2}}[/MATH]
So then, we find:

[MATH]f(f(x))=\frac{F_{n}\dfrac{F_{n}x+F_{n+1}}{F_{n+1}x+F_{n+2}}+F_{n+1}}{F_{n+1}\dfrac{F_{n}x+F_{n+1}}{F_{n+1}x+F_{n+2}}+F_{n+2}}=\frac{F_{n}(F_{n}x+F_{n+1})+F_{n+1}(F_{n+1}x+F_{n+2})}{F_{n+1}(F_{n}x+F_{n+1})+F_{n+2}(F_{n+1}x+F_{n+2})}=\frac{(F_n^2+F_{n+1}^2)x+(F_{n}F_{n+1}+F_{n+1}F_{n+2})}{(F_{n}F_{n+1}+F_{n+1}F_{n+2})x+(F_{n+1}^2+F_{n+2}^2)}[/MATH]
It can be shown that:

[MATH]F_{n+m}=F_{n-1}F_{m}+F_{n}F_{m+1}[/MATH]
And so, with \(m=n-1\) this becomes:

[MATH]F_{2n-1}=F_{n-1}^2+F_{n}^2[/MATH]
And with \(m=n\) this becomes:

[MATH]F_{2n}=F_{n-1}F_{n}+F_{n}F_{n+1}[/MATH]
And so we may write:

[MATH]f(f(x))=\frac{F_{2(n+1)-1}x+F_{2(n+1)}}{F_{2(n+1)}x+F_{2(n+1)+1}}[/MATH]
If we let [MATH]m=2n+1[/MATH] we then have:

[MATH]f(f(x))=\frac{F_{m}x+F_{m+1}}{F_{m+1}x+F_{m+2}}[/MATH]
 
apple2357 said:
This presumably explains why this result falls out? Is this a basic property of Fibonacci sequences?
Click to expand...

It is an identity for Fibonacci numbers. I was expecting to get in my final result (based on the pattern you presented):

[MATH]m=n+2[/MATH]
That is, that I'd find:

[MATH]f(f(x))=\frac{F_{n+2}x+F_{n+3}}{F_{n+3}x+F_{n+4}}[/MATH]
 
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