# Isn't it still 2016?

**Calculus**Level 3

Consider the functional equation, \(\dfrac { { d }^{ 2017 }f(x) }{ d{ x }^{ 2017 } } =f(x)\) for a function \(f\) which is defined for all real.

Let the solution to this equation be of the form of: \(\displaystyle \sum _{ i=1 }^{ 2017 }{ { c }_{ i }{ e }^{ { r }_{ i }x } } \) where \({ c }_{ i }\) are arbitrary constants independent of \(x\).

Now, given that \({ r }_{ 2017 }=1\) and \(\left| \displaystyle \sum _{ i=1 }^{ 2016 }{ { r }_{ i } } \right| =S\).

Also, \(\displaystyle \lim _{ x\rightarrow 2017 }{ \frac { { e }^{ 2016S-x }-{ e }^{ -1 } }{ 2017-x } } 2017=A{ e }^{ -a }\).

Then calculate \(A-a\).