# Form of Kronecker Delta Not Recognized in Mathematica

1. Apr 18, 2012

### DocZaius

kron[m_,n_]:=\!$$\*SubsuperscriptBox[\(\[Integral]$$, $$0$$, $$2$$]$$\(Sin[ \*FractionBox[\(n\ \[Pi]\ x$$, $$2$$]]\ Sin[
\*FractionBox[$$m\ \[Pi]\ x$$, $$2$$]]\) \[DifferentialD]x\)\)

This is the integral over x of sin(n pi x / 2) times sin(m pi x / 2) from 0 to 2. This is one way to define the Kronecker Delta function. Given m and n as integers and not both zero, if m = n then the output is 1, if m ≠ n the output is 0.

Why then is it that when I type:

Simplify[kron[m, n],
Element[m, Integers] && Element[n, Integers]]

that I get zero?

If I do kron[1, 1] I get 1! How can Mathematica say a function is 0 over the integers when I just put in two integers and got a non-zero ouput?

It is certainly not identical to zero right?

#### Attached Files:

File size:
1.1 KB
Views:
139
• ###### output.png
File size:
3.8 KB
Views:
151
Last edited: Apr 18, 2012
2. Apr 21, 2012

### djelovin

When you force Mathematica to simplify expression it is assumed that m and n are not equal. This:

Code (Text):
Assuming[{Element[m, Integers], Element[n, Integers], m == n}, kron[m, n]]
Assuming[{Element[m, Integers], Element[n, Integers], m != n}, kron[m, n]]
gives 1 and 0 respectively!

3. Apr 21, 2012

### DocZaius

Thank you for looking into it! I didn't know Mathematica assumed m and n are not equal. Why is Mathematica making that assumption? Given two integers m and n, the possibility that they are equal is always there.

Rather worrisome Mathematica assumption that I will keep in mind from now on! I will have to handhold Mathematica by separately asking it the = and ≠ cases.

4. Apr 21, 2012

### djelovin

When you call Simplify[expression] Mathematica assumes that all symbols in expression have unique values, otherwise not!