Fibonacci sequence and ODE

We know that the Fibonacci sequence is given by $F_{n+2}=F_{n+1}+F_n,F_0=1,F_1=1$$F_{n+2}=F_{n+1}+F_n,F_0=1,F_1=1$

But how to get the explicit formula?

Observe an interesting way to approach it by considering the Taylor Series and ODE.

Consider representing a sequence to a Taylor Series,

${\displaystyle (a{)}_{n=1}^{\mathrm{\infty }}\mapsto f_{a_n}=a_0+a_{1}x+a_2\frac{x^2}{2!}+a_3\frac{x^3}{3!}...}$$\displaystyle(a)_{n=1}^{\infty}\mapsto f_{a_n}=a_0+a_1x+a_2\frac{x^2}{2!}+a_3\frac{x^3}{3!}...$

And ${\displaystyle \frac{d}{dx}}$$\displaystyle\frac{d}{dx}$ act on $f_{a_n}$$f_{a_{n}}$ gives ${\displaystyle f_{a_{n+1}}=a_1+a_{2}x+a_3\frac{x^2}{2!}+...}$$\displaystyle f_{a_{n+1}}=a_1+a_2x+a_3\frac{x^2}{2!}+...$

For convenience, we will denote $f(x)$$f(x)$ for the Fibonacci sequence

So we can rewrite $F_{n+2}=F_{n+1}+F_n$$F_{n+2}=F_{n+1}+F_n$ as ${\displaystyle ({\left(\frac{d}{dx}\right)}^2-\frac{d}{dx}-1)f(x)=0}$$\displaystyle\left(\left(\frac{d}{dx}\right)^2-\frac{d}{dx}-1\right) f(x)=0$

And we can factor the polynomial, ${\displaystyle (\frac{d}{dx}-\frac{1+\sqrt{5}}{2})(\frac{d}{dx}-\frac{1-\sqrt{5}}{2})}$$\displaystyle\left(\frac{d}{dx}-\frac{1+\sqrt{5}}{2}\right)\left(\frac{d}{dx}-\frac{1-\sqrt{5}}{2}\right)$

So the solution for the ODE is ${\displaystyle C_1{e}^{\frac{1+\sqrt{5}}{2}x}+C_2{e}^{\frac{1-\sqrt{5}}{2}x}}$$\displaystyle C_1e^{\frac{1+\sqrt{5}}{2}x}+C_2 e^{\frac{1-\sqrt{5}}{2}x}$

And consider the Taylor Series, we get $F_n=C_1(\frac{1+\sqrt{5}}{2}{)}^n+C_2(\frac{1+\sqrt{5}}{2}{)}^n$$F_n=C_1(\frac{1+\sqrt{5}}{2})^n+C_2(\frac{1+\sqrt{5}}{2})^n$

And $F_0=1,F_1=1$$F_0=1,F_1=1$, So $C_1+C_2=1,C_1(\frac{1+\sqrt{5}}{2})+C_2\frac{1-\sqrt{5}}{2})=1$$C_1+C_2=1,C_1(\frac{1+\sqrt{5}}{2})+C_2\frac{1-\sqrt{5}}{2})=1$

So $C_1={\displaystyle \frac{1+\sqrt{5}}{2\sqrt{5}},C_2=\frac{1-\sqrt{5}}{2\sqrt{5}}}$$C_1=\displaystyle\frac{1+\sqrt 5}{2\sqrt 5},C_2=\frac{1-\sqrt 5}{2\sqrt 5}$

So $F_n={\displaystyle \frac{1}{\sqrt{5}}({\left(\frac{1+\sqrt{5}}{2}\right)}^{n+1}+{\left(\frac{1-\sqrt{5}}{2}\right)}^{n+1})}$$F_n=\displaystyle\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}+\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}\right)$