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Let me give the background for this question. I'm using WebAssign to assign homework problems, and WebAssign has a built-in mechanism which can grade problems using Mathematica commands. For instance, if the solution to a problem is $e^{x+y}$ then it is possible to use the Simplify command to ensure that $e^x e^y$ is marked as correct.

I'm currently assigning problems on Fourier series, and expressions like $\cos(n \pi)$ occur frequently. It is fairly standard practice to rewrite $\cos(n \pi)$ as $(-1)^n$, and I want these two expressions to be treated as equivalent (as well as similar expressions for $\cos(n \pi /2)$, etc.) The problem is that these expressions are only equal on the integers, and so Mathematica has no reason to believe they are the same. I suppose I could just write some code which plugs in the first hundred integers and checks to see if the values are the same, but I'm wondering if there's a more elegant solution.

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closed as off-topic by Louis, MarcoB, Edmund, Karsten 7., ubpdqn May 26 at 8:33

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "The question is out of scope for this site. The answer to this question requires either advice from Wolfram support or the services of a professional consultant." – Louis, Edmund, Karsten 7., ubpdqn
If this question can be reworded to fit the rules in the help center, please edit the question.

If you denote in Simplify that n is an integer it will simplify it to the form $(-1)^n$. – Spawn1701D Apr 24 '13 at 13:05
Silly answer: did you contact the WebAssign math editor, or other WebAssign tech support, for help on this? – murray Apr 24 '13 at 15:04
Table[Simplify[f[Pi n], Element[n, Integers]], {f, {Cos, Sin}}]
(* {(-1)^n, 0} *)

Simplify[Cos[Pi n/2], Element[n, Integers]]
(* Cos[(n Pi)/2] *)

You could do something about this differently I guess ... what, as @J. M. suggests, you could append an assumption that n is odd eg.

Simplify[Cos[Pi n/2],
 Element[n, Integers],
 Assumptions -> Mod[n, 2] == 1]
(* 0 *)
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Of course, one could also use Assuming[]. – J. M. Apr 24 '13 at 13:43
Huh, I had something like Simplify[Cos[π n/2], n ∈ Integers && Mod[n, 2] == 1] in mind... – J. M. Apr 25 '13 at 12:43
You said "append", but this is not correct: Note that using Simplify[Sqrt[x^2], Assumptions -> x>0 ] will use only that one assumption in the Simplify command. If you previously set global assumptions in $Assumptions, these will (temporarily) be ignored. Instead, you can use Simplify[Sqrt[x^2], x>0 ] - here, the new assumption is truly appended. – Martin J.H. Aug 20 '14 at 14:45

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