Consider $F_n$ the $n^{th}$ Fibonacci number.
$$g(x) = \sum^\infty_{n=0}F_n \frac{x^n}{n!}$$
Prove the following:
$$g''(x)=g'(x)+g(x)$$
I've never dealt with derivatives in the above form so I am not exactly sure where to start with this problem. Would love some help!
Hint. One has $$ g'(x) = \left(\sum^\infty_{n=0}F_n \frac{x^n}{n!}\right)'=\sum^\infty_{n=1}n \times F_n \frac{x^{n-1}}{n!}=\sum^\infty_{n=1}F_n \frac{x^{n-1}}{(n-1)!}=\sum^\infty_{n=0}F_{n+1} \frac{x^n}{n!} $$ Can you do the same with $g''(x)$?
Then use $$ F_{n+2}=F_{n+1}+F_n, $$ to get the announced identity.