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Einstein Summation Convention

  • Thread starter joe:)
  • Start date
  • #26
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You tell me, [itex]\delta_{ij}x_j[/itex] simplifies to [itex]x_i[/itex]...what about [itex](x_k x_k)^{n/2-1}[/itex]?...that looks like a power of [itex]r[/itex] to me:wink:

What does that give you for [itex]\mathbf{\nabla}r^n[/itex]?
2xieir^n-1?
 
  • #27
gabbagabbahey
Homework Helper
Gold Member
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2xieir^n-1?
Why [itex]r^{n-1}[/itex]? Remember, [itex]r^2=(x_kx_k)[/itex]....what exponent do you have to raise each side of this equation to in order for the LHS to be [itex](x_kx_k)^{n/2-1}[/itex]?

Also, what is [itex]x_i\textbf{e}_i[/itex]?
 

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