# Einstein Summation Convention

You tell me, $\delta_{ij}x_j$ simplifies to $x_i$...what about $(x_k x_k)^{n/2-1}$?...that looks like a power of $r$ to me

What does that give you for $\mathbf{\nabla}r^n$?
2xieir^n-1?

gabbagabbahey
Homework Helper
Gold Member
2xieir^n-1?
Why $r^{n-1}$? Remember, $r^2=(x_kx_k)$....what exponent do you have to raise each side of this equation to in order for the LHS to be $(x_kx_k)^{n/2-1}$?

Also, what is $x_i\textbf{e}_i$?