Hello Neil,

You are complete right, this is the point which I did not understand in the beginning. The expression "1 - Ceiling ... + Floor" encodes not for one condition, but for many. This way the sum of KroneckerDeltas (each of these represents one condition) can be absorbed into one expression.

Thank you very much for explanations and enlighten me,

All the best,

Michael

Hello Neil, thank you for your response. I agree to your explanation, but there is still a problem: let us assume for simpliciity that the condition t = 7 tau will hold. In this case I would expect that the KroneckerDelta[t, 7 tau] will be equal to one. But I don't see how this can be achieve with the expression given above, since there is no "n" left in the argument of Ceiling and Floor.

Therefore I expect to have a sum of Kroneckerterms, of which just one will be equal to one and all others are equal to zero (under the premise that t is a multiple of tau). Since t is not fixed in the expression above yet, there has to be the sum to be returned which then after t gets an assignment collapses to just one term being unequal to zero. Do you share my view?

Regards, Michael

Hello Neil,

Thank you for pointing this out, this was definitely a typo. The expression look smuch better now. I changed the condition slightly ( tau >0 instead of tau >= 0).

Assuming[{N > 0 && t >= 0 && \[Tau] > 0 && {n, N} \[Element] 
            Integers && {x , t, \[Tau]} \[Element] Reals}, 
  FullSimplify[ te[x] + v[x] \[Tau]   \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(N\)]\(KroneckerDelta[t, 
      n\ \[Tau]]\)\)]]

As result I get now:

But something still puzzles me: I would expect to get n-times this expression. For example, put N=3, I would expect that the 4 KroneckerDeltas will add up to 4 times this expression with corresponding arguments for Ceiling, Floor, respectively, t/tau = { t/0, t/tau, t/( 2tau), t/(3 tau). Where do I go wrong here?