
> 
> The operator $E$ is defined as $Eu\_n=u\_{n+1}$.
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I encountered a strange equality. when I tried out
Let $u\_n$ represent a series such that 


$$u\_{n+2}=u\_{n+1}+u\_n. \tag{$\star$}$$


Or
$E^2{u\_n}=Eu\_{n+1}+Eu\_n$


$\left(E^2-E-1\right)u\_n=0$



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> $E\equiv \frac{1\pm \sqrt{5}}{2} $
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***What interpretation should be made on such an equality ?.*** 


This equality seems interesting to me as two well know sequences that obeys$\huge\star$ Fibonacci sequence($u\_1=u\_2=1$) and Lucas sequence($u\_1=1,u\_2=3$) has it's $n$ th term represented by 


$\Large {\frac{\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n}{\sqrt{5}}}$ and $\Large {\frac{\left(\frac{1+\sqrt{5}}{2}\right)^n+\left(\frac{1-\sqrt{5}}{2}\right)^n}{2}}$ respectively 


If I am to follow that $E$ and $\frac{1\pm \sqrt{5}}{2} $. Then in case of the above two particular case of the recursion . The formula for the $n$ the term is not true(Of course I can't determine which sign($\pm$) should be taken and hence such an argument is bad).But interestingly , when we assume the same value for $E$, $\huge \star$ is valid with both the signs(the $\pm$).


I am really interested with such an equivalence, as I think it's interpretation can be exploited for a proof of those formulas. And also possibly a general solution. As 



> 
> $E^n{u\_n}=u\_{n+1}$
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A strange equality of the operator E ($Eu_n=u_{n+1}$)
$E$ acts on the finite-dimensional vector space consisting of sequences satisfying that recurrence relation, and those are its eigenvalues on that space. More generally, $E$ acts on the nullspace of $p(E)$ for any polynomial $p$ with eigenvalues the roots of $p$, and you can use this together with the theory of Jordan normal form to write down general solutions to linear recurrence relations. 


