Need help with evaluating:
$\displaystyle \delta _{i,j}\delta _{i,j}$
$\displaystyle \delta _{i,j}\delta _{j,k}$
Thanks
My attempt at expanding $\displaystyle \delta _{i,j}\delta _{i,j}$:
When i=j
$\displaystyle \delta _{i,i}\delta _{i,i}$
$\displaystyle (\delta _{1,1}+\delta _{2,2}+\delta _{3,3})$ *$\displaystyle (\delta _{1,1}+\delta _{2,2}+\delta _{3,3})$
(1+1+1)x(1+1+1)
9
Have you changed the question?
Where did $\displaystyle (\delta _{1,1}+\delta _{2,2}+\delta _{3,3})$ come from?
That was not part of the OP!
The OP asked for $\displaystyle \delta _{i,j}\delta _{i,j}=\left\{ {\begin{array}{lr} {1,} & {i = j} \\ {0,} & {i \ne j} \\\end{array}} \right.$
The second part of the OP asked $\displaystyle \delta _{ij} \delta _{jk} = \left\{ {\begin{array}{lr} {1,} & {i = j = k} \\ {0,} & {else} \\\end{array}}\right.$
So why did you change it?
I did post the question word for word as it is given to me and I did ask for help on evaluating them. I guess dropping the summation notation is what is throwing you off. We were told that the index notation implies that the quantities are being summed. Let me rewrite the problem although this is not how the question is given to us.
$\displaystyle \sum_{i,j,k=1}^{i,j,k=3}\delta _{i,j}\delta _{i,j}$
Similarly the second problem is being summed too.
$\displaystyle \sum_{i,j,k=1}^{i,j,k=3}\delta _{i,j}\delta _{i,j}=3$
There are twenty-seven triples $\displaystyle (i,j,k)$ where $\displaystyle i=1,2,3~j=1,2,3~k=1,2,3$.
Only three of them have $\displaystyle i=j=k$ (see reply #4) so you three 1's and twenty-four 0's in that sum.