# Fibonacci function

Dear Professors and Mathematcians,

Now, I am introducing Fibonacci sequence and function. Consider, $F(x)$ is a Fibonacci function and $f_n$ is Fibonacci sequence. For fixing the initial values by definition, like $f_0= 0$, $f_1=1=f_2$. I made the following true proposition which is true by trial and error method for $n\geq2$ and $n$ is some integer. For all real value of $x$, the following proposition is true. But, we can state as a theorem if one can produce a proof. Of course, I failed to prove and seeking helps to prove the statement. $$F(x +n) = f_nF(x+1)+f_{n-1}F(x).$$ Thanks in advance

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What is a Fibonacci function? – Andrea Mori Jul 30 '12 at 10:03
If you don't give the definition of a "Fibonacci function", we won't be able to help you... – J. M. Jul 30 '12 at 10:13
Do you mean the generating function of the Fibonacci sequence? – Dennis Gulko Jul 30 '12 at 10:18

I assume that you define the Fibonacci function as the natural continuation of the Fibonacci sequence, $$F(x) = \frac{\alpha^x-\beta^x}{\sqrt5}$$ where $\alpha = (1+\sqrt5)/2, \beta=(1-\sqrt5)/2$. If not, I'll just delete this answer and will have wasted my time.

For later use, note that $\alpha\beta+1=0,\alpha+1/\alpha=\sqrt5\text{, and }\beta+1/\beta=-\sqrt5.$

Now notice that for any integer $n$ we have $F(n)=f_n,$ so you wish to prove $$F(x+n)=F(n)F(x+1)+F(n-1)F(x)$$ Using the supposed definition of $F(x)$ the right side of your equality, $F(n)F(x+1)+F(n-1)F(x)$, becomes \begin{align*} &= \frac{1}{5}[(\alpha^n-\beta^n)(\alpha^{x+1}-\beta^{x+1})+(\alpha^{n-1}-\beta^{n-1})(\alpha^x-\beta^x)]\\ &= \frac{1}{5}[\alpha^{x+n+1}-\alpha^n\beta^{x+1}-\alpha^{x+1}\beta^n+\beta^{x+n+1}+\alpha^{x+n-1}-\alpha^{n-1}\beta^x-\alpha^x\beta^{n-1}+\beta^{x+n-1}]\\ &=\frac{1}{5}[\alpha^{x+n}(\alpha+1/\alpha)+\beta^{x+n}(\beta+1/\beta)-\alpha^{n-1}\beta^x(\alpha\beta+1)-\alpha^x\beta^{n-1}(\alpha\beta+1)]\\ &=\frac{1}{5}[\alpha^{x+n}(\alpha+1/\alpha)+\beta^{x+n}(\beta+1/\beta)]\quad\text{(using our first observation above)}\\ &=\frac{1}{5}[\alpha^{x+n}\sqrt5-\beta^{x+n}\sqrt5]\quad\text{(using our second and third observations)}\\ &=\frac{\alpha^{x+n}-\beta^{x+n}}{\sqrt5}\\ &=F(x+n) \end{align*}

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! I am really very happy with your solution, which I really make me mad and big fan of you. Thank you so much. I have a habit to do some rough work in my free time on fib series. This one is one of my rough work problem, which is ends with your great help. I also posted the similar one. Anyhow, thank you so much to this site and my special thanks to you. I love your solution and I love you. – BMSA Jul 31 '12 at 5:51
The formal manipulations work out fine. However, there are some difficulties in assigning a meaning to $\beta^x$ for general real $x$, since $\beta$ is negative. – André Nicolas Jul 31 '12 at 6:00

It also can be proven by induction on $n$ :

• for $n = 1$, we have: $$F(x + 1) = f_1 F(x + 1) + f_0 F(x) = F(x + 1)$$
• assuming the equality holds for $\forall i \leq n$, we'll prove that it's true for $n + 1$. Using the equality $F(x + 2) = F(x) + F(x + 1)$, it follows: \begin{align*} &F(x+n+1) = F(x+n) + F(x+n-1)\\ &= f_n F(x+1) + f_{n-1} F(x) + f_{n-1} F(x+1) + f_{n-2} F(x)\\ &= (f_n + f_{n-1}) F(x+1) + (f_{n-1} + f_{n-2}) F(x)\\ &= f_{n+1} F(x+1) + f_n F(x)\\ \end{align*}
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! I am so happy for your short solution and one of the alternative best solution. I am happy. how much happy I am, I cant say in my words. Thank a lot. – BMSA Jul 31 '12 at 5:52