How Does the Kronecker Delta Function as a Substitution Operator?

[bossman007]

Homework Statement

[PLAIN]http://postimage.org/image/s7m1kohst/ [/PLAIN]

Homework Equations

The Kronecker Delta = 1 ; if i=j

The Kronecker Delta = 0 ; i (not equal) j

The Attempt at a Solution

I have no idea what to do from here, or even if I did this first step right?
[PLAIN]http://postimage.org/image/t3d3u9qg7/ [/PLAIN]

 

[bossman007]

Still no luck on my own :(

 

[STEMucator]

Still no luck on my own :(

Okay so expanding your sum :

\sum_{i=1}^{3}a_iδ_{ij} = a_1δ_{1j} + a_2δ_{2j} + a_3δ_{3j}

Note that the only term which survives is the term a_jδ_{jj} where i=j, but δ_{jj} = 1 as per the Delta Kronecker. So a_jδ_{jj} = a_j

Are you sure that's the question? I feel as if you're missing something. Some important intervals are not mentioned here.

 

[bossman007]

Thats the entire problem...im still confused from ur response, but I feel u gave me a good nudge in the right direction. thx

 

[bossman007]

That's the entire problem...hmmm

 

[bossman007]

I solved it for good, I understand it now. many thanks. The Kronecker Delta has the role of a substitution operator, basically replacing a repeated indice