1
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What else is needed to make Mathematica to simplify the following expression to $z[j]$?

enter image description here

Code:

Assuming[
 {n \[Element] Integers, p \[Element] Integers,
  i \[Element] Integers, j \[Element] Integers,
  r \[Element] Integers, s \[Element] Integers,
  n >= 1, p >= 1,
  i >= 1, i <= n,
  j >= 1, j <= p,
  r >= 1, r <= p,
  s >= 1, s <= n,
  },
 FullSimplify[\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(r = 1\), \(p\)]\(z[r]*\(
\*UnderoverscriptBox[\(\[Sum]\), \(s = 1\), \(n\)]\*
TemplateBox[{RowBox[{"i", ",", "s"}]},
"KroneckerDeltaSeq"] \*
TemplateBox[{RowBox[{"j", ",", "r"}]},
"KroneckerDeltaSeq"]\)\)\)]
 ]
$\endgroup$
5
  • $\begingroup$ 1) Use KroneckerDelta instead of "KroneckerDeltaSeq" (no idea what's that supposed to mean); 2) what is it that you really want to achieve? As shown, you're multiplying a sum of zs by a sum of $\delta$s - why should it even be equal to z[j]? $\endgroup$
    – corey979
    Jan 20, 2018 at 8:46
  • $\begingroup$ @corey979 1) "KroneckerDeltaSeq" is produced by copy and paste, replacing it with "KroneckerDelta" doesn't make a difference. 2) I want to simplify the expression to it's simplest form, which IS $z[j]$: The first delta in the second sum is not zero only when $s=i$, thus the second sum resolves to $\delta _{j,r}$, so the second sum is one only when $s=i, r=j$, which makes the entire expression $z[j]$ $\endgroup$ Jan 20, 2018 at 9:03
  • $\begingroup$ @corey979 oh, I think you misunderstood the expression, it's $\Sigma (z \Sigma \_)$, not $(\Sigma z)(\Sigma \_)$ $\endgroup$ Jan 20, 2018 at 9:08
  • $\begingroup$ So you want FullSimplify[ Sum[z[r]* Sum[KroneckerDelta[i, s]*KroneckerDelta[j, r], {s, 1, n}], {r, 1, p}]]. $\endgroup$
    – corey979
    Jan 20, 2018 at 9:29
  • $\begingroup$ Yes, but just using the above equation (FullSimplify...) wouldn't yield the desired result. $\endgroup$ Jan 20, 2018 at 12:27

1 Answer 1

1
$\begingroup$

Try this:

    FullSimplify[
 Sum[z[r]*Sum[
    KroneckerDelta[i, s] KroneckerDelta[j, r], {s, 1, n}], {r, 1, 
   p}], {{i, j, n, p, s, r} \[Element] Integers, n >= 1, p >= 1, 
  i >= 1, j >= 1, r >= 1, i <= n, j <= p, s <= n, r <= p}]

yielding the following:

enter image description here

Have fun!

$\endgroup$
2
  • $\begingroup$ The code in the question also produces this output, the problem is, it is not the simplest form. $\endgroup$ Jan 20, 2018 at 14:30
  • $\begingroup$ Here's an observation: DiscreteDelta[j - r] // PiecewiseExpand gives the bracketed expression in the answer above. Is there an inverse to PiecewiseExpand? $\endgroup$
    – bill s
    Jan 20, 2018 at 15:09

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