Simplify sum expression containing Kronecker delta?

[mhelmle (Michael Helmle)]

Hello Community,

I tried to formulate an expression that contains a sum over KroneckerDeltafunctions.

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

Using Simplify and FullSimplify with the appropiate conditions on the Integer and Real variables I get the following result:

t1 = FullSimplify[ te[x] + v[x] \[Tau]   \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(N\)]\(KroneckerDelta[t, 
      n\ \[Tau]]\)\), 
  N > 0 && t >= 0 & \[Tau] >= 0 && {n, N} \[Element] 
    Integers && {x , t, \[Tau]} \[Element] Reals]

Apparently Mathematica has not considered all of the constraints like N >0 etc. Any ideas on how to improve the output of this calculation?

[neils (Neil Singer)]

Michael,

You have a typo (I think). Your second & should be &&

N > 0 && t >= 0 && \[Tau] >= 0 && {n, N} \[Element] 
    Integers && {x , t, \[Tau]} \[Element] Reals

Now it can simplify. You can also do:

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]]\)\)]]

Regards,

Neil

[mhelmle (Michael Helmle)]

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?

[neils (Neil Singer)]

Michael,

KroneckerDelta[a,b] gives 0 unless a==b in which case it gives 1. If you are summing a bunch of KroneckerDelta[a, nb] as n goes from 0 to N, a ==bn will happen only once, if ever. If a is a multiple of b then you will eventually get one “1” in the sum (assuming a<=N*b). If a is not a multiple of b, then you will never get a 1. I hope this helps.

Regards,

Neil

[mhelmle (Michael Helmle)]

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

[neils (Neil Singer)]

Michael,

I think Mathematica is correct. If you have KroneckerDelta[t, n tau], if you sum to a number N equal to or larger than 7, you will have:

 KroneckerDelta[t, 0 tau] + KroneckerDelta[t, 1 tau]+... + KroneckerDelta[t, 7 tau]+  KroneckerDelta[t, 8 tau]+...

since only KroneckerDelta[t, 7 tau] is 1 and the rest are zero, your sum is 1. If t is not an integer multiple of tau then you will never get a 1.

Regards,

Neil

[mhelmle (Michael Helmle)]

But if you look at the expression above which I repost here:

for all values of t <= N tau you get only t/tau as argument in the Ceiling and Floor function. How would get the 1 for t = 7 N then? You need have a remaining (1 - Ceiling[ t / (7 tau)] + Floor[ t /(7 tau)]). But n is not contained in the expression anymore. This is what is not obvious to me.

Thank you for your patience,
regards
Michael

[neils (Neil Singer)]

Michael,

The only thing that matters is that t/tau is an integer. It does not matter to the answer if t=7 tau or if t = 5 tau. The answer is the same as long as N is larger than 7 (in this case). This is precisely what MMA is saying: if t<=N tau it means that the summation must include the situation where t = n*tau during the summation. The best way to convince yourself it is correct is try some numerical values.

Grid[Prepend[
  Table[{t, t/\[Tau], t1} /. N -> 10  /. \[Tau] -> 1, {t, 1, 15, 
    1}], {"t", "t/\[Tau]", "t1"}], Background -> {None, {LightGray}}, 
 Frame -> All]

Now change numbers, for example:

Grid[Prepend[
  Table[{t, t/\[Tau], t1} /. N -> 10  /. \[Tau] -> 1.1, {t, 1, 
    15}], {"t", "t/\[Tau]", "t1"}], Background -> {None, {LightGray}},
  Frame -> All]

to get

I hope this is clear.

Regards

Neil

[mhelmle (Michael Helmle)]

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