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

  • $\begingroup$ What is a Fibonacci function? $\endgroup$ – Andrea Mori Jul 30 '12 at 10:03
  • $\begingroup$ If you don't give the definition of a "Fibonacci function", we won't be able to help you... $\endgroup$ – J. M. is a poor mathematician Jul 30 '12 at 10:13
  • $\begingroup$ Do you mean the generating function of the Fibonacci sequence? $\endgroup$ – 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 $$ \begin{align*} &= \frac{1}{5}[(\alpha^n-\beta^n)(\alpha^{x+1}-\beta^{x+1})+(\alpha^{n-1}-\beta^{n-1})(\alpha^x-\beta^x)]\\ &= \frac{1}{5}[\alpha^{x+n+1}-\alpha^n\beta^{x+1}-\alpha^{x+1}\beta^n+\beta^{x+n+1}+\alpha^{x+n-1}-\alpha^{n-1}\beta^x-\alpha^x\beta^{n-1}+\beta^{x+n-1}]\\ &=\frac{1}{5}[\alpha^{x+n}(\alpha+1/\alpha)+\beta^{x+n}(\beta+1/\beta)-\alpha^{n-1}\beta^x(\alpha\beta+1)-\alpha^x\beta^{n-1}(\alpha\beta+1)]\\ &=\frac{1}{5}[\alpha^{x+n}(\alpha+1/\alpha)+\beta^{x+n}(\beta+1/\beta)]\quad\text{(using our first observation above)}\\ &=\frac{1}{5}[\alpha^{x+n}\sqrt5-\beta^{x+n}\sqrt5]\quad\text{(using our second and third observations)}\\ &=\frac{\alpha^{x+n}-\beta^{x+n}}{\sqrt5}\\ &=F(x+n) \end{align*} $$

  • $\begingroup$ ! 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. $\endgroup$ – BMSA Jul 31 '12 at 5:51
  • 1
    $\begingroup$ 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. $\endgroup$ – 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: $$ \begin{align*} &F(x+n+1) = F(x+n) + F(x+n-1)\\ &= f_n F(x+1) + f_{n-1} F(x) + f_{n-1} F(x+1) + f_{n-2} F(x)\\ &= (f_n + f_{n-1}) F(x+1) + (f_{n-1} + f_{n-2}) F(x)\\ &= f_{n+1} F(x+1) + f_n F(x)\\ \end{align*} $$
  • $\begingroup$ ! 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. $\endgroup$ – BMSA Jul 31 '12 at 5:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.