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simplifying expressions - How to simplify exponents of the form $expleft(ifrac{2pi}{N}(N+1)right)$ inside a function


I'm really annoyed with Mathematica and I need your help. I defined a discrete Fourier-Transformation(I use II instead of N, because Mathematica wont let me-.-):


Subscript[Ï•, i_] = Sum[Exp[I (2 Pi)/II i k] f[k], {k, II}];

And I want to check that



Subscript[Ï•, 1] == Subscript[Ï•, II + 1]

This gives me $$\sum _k^{\text{II}} \left(f(k) e^{\frac{2 i \pi k}{\text{II}}}\right)=\sum _k^{\text{II}} \left(f(k) e^{\frac{2 i \pi (\text{II}+1) k}{\text{II}}}\right)$$


And by simply splitting the exponents on the right side of the equations this can be seen to be true.($\exp(i2\pi k)=1\forall k\in\mathbb N$ ) I really would have thought that Mathematica would be able to simplify that on its own, considering that $k$ is the summation index and is therefore an integer, but it does not. So I told Mathematica:


Simplify[Subscript[Ï•, 1] == Subscript[Ï•, II + 1],     Assumptions -> k ∈ Integers]

Didn't help, so I tried:


Simplify[Subscript[Ï•, 1] == Subscript[Ï•, II + 1] /. 
        Exp[a_] :> Exp[Expand[a]], Assumptions -> k ∈ Integers]


Still does not work! Please help me. Thank you



Answer



Apparently, simplifying a Sum does not result in simplifying each term in the Sum. Try


Subscript[Ï•, 1] == Subscript[Ï•, II + 1] /. 
    Exp[a_] :> Simplify[Exp[a], Assumptions -> k ∈ Integers]
(* True *)

Undoubtedly, there are more elegant approaches.


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