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I am obtaining in Mathematica:

Cos[2*Pi*FractionalPart[1/2*(i + j + k)]]

We know that for $i, j, k$ being positive integers this expression after simplification should give:

$$ (-1)^{i+j+k} $$

I would like to know why the one of the solutions proposed for this similar question:

 Refine[Cos[2*Pi*FractionalPart[1/2*(i + j + k)]], 
    Assumptions -> {Element[{i, j, k}, Integers], i > 0, j > 0, k > 0}]

did not work for this case?

share|improve this question
@Kuba sorry, that was a copy paste issue from the real code.. I fixed now... – Saullo Castro Feb 3 '14 at 8:15
@Artes this is not a duplicate because the question is asking why the Refine is not working in this case, while the other question was asking how to simplify the expression... – Saullo Castro Feb 3 '14 at 14:13
@Artes I think that it is not a duplicate since Refine/FullSimplify works bad with FractionalPart. – ybeltukov Feb 3 '14 at 15:32
@ybeltukov I would retract my close vote if you provided at least a bit more optimal approach e.g. Simplify[Cos[ 2 Pi (# - Floor@#)&[(i + j + k)/2]], (i | j | k) ∈ Integers] – Artes Feb 3 '14 at 17:00
up vote 6 down vote accepted

One can use x-Floor[x] instead of FractionalPart[x] for positive x

FullSimplify[Cos[2 Pi ((i + j + k)/2 - Floor[(i + j + k)/2])], 
   Assumptions -> (i | j | k) ∈ Integers]

(-1)^(i + j + k)

share|improve this answer
ybeltukov, you haven't been active on this site recently. I want to let you know that your contributions are missed. – Mr.Wizard Jun 3 '14 at 21:26
@Mr.Wizard I'm sorry that I have no spare time to participate on this site now. I hope I'll be back in September or earlier. I was glad to talk with Leonid at the recent Wolfram conference in St. Petersburg. – ybeltukov Jun 4 '14 at 13:46

Using the solution proposed in this answer also works:

FullSimplify[Cos[2 Pi FractionalPart[1/2 (i + j + k)]], 
             Assumptions -> {Element[i + j + k, Integers], i > 0, j > 0, k > 0},
             ComplexityFunction -> LeafCount]


(-1)^(i + j + k)

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