Understanding Einstein Tensor Conventions for Tensor Summation

[DeShark]

Homework Statement

Write out $c_{j}x_{j}+c_{k}y_{k}$ in full, for n=4.

Homework Equations

The Attempt at a Solution

So I figure we have to sum over both j and k. So the answer I obtained is:
$(c_1x_1+c_1y_1)+(c_1x_1+c_2y_2)+(c_1x_1+c_3y_3)+(c_1x_1+c_4y_4)+$
$(c_2x_2+c_1y_1)+(c_2x_2+c_2y_2)+(c_2x_2+c_3y_3)+(c_2x_2+c_4y_4)+$
$(c_3x_3+c_1y_1)+(c_3x_3+c_2y_2)+(c_3x_3+c_3y_3)+(c_3x_3+c_4y_4)+$
$(c_4x_4+c_1y_1)+(c_4x_4+c_2y_2)+(c_4x_4+c_3y_3)+(c_4x_4+c_4y_4)$

i.e. $4(c_1x_1+c_2x_2+c_3x_3+c_4x_4+c_1y_1+c_2y_2+c_3y_3+c_4y_4)$

but the book I'm working from just gives the answer:
$c_1x_1+c_2x_2+c_3x_3+c_4x_4+c_1y_1+c_2y_2+c_3y_3+c_4y_4$

so I'm a factor of 4 out. Am I doing it wrong or is the book.

Surely the answer the book gave can be written

$c_ix_i+c_iy_i$

Apologies for the noobiness of the question, but I'm trying to self-teach tensor calculus and I want to nail the basics before I progress much further.

 

[dextercioby]

there are two different summations, the first with the dummy index j will give 4 possible terms, while the second with the dummy index k will give other 4. So the whole sum will have 4+4 terms.

 

[DeShark]

Ah... I guess that makes sense if the indices are over different ranges, e.g. j=1,2,3 k=1,2,3,4. It confused me in this case because why would you use two separate indices when one is perfectly adequate. It seems simpler, more obvious and more elegant to just use the one index, given that n=4 for both. Thank you.

 

[HallsofIvy]

Without the summation convention, this would be $\sum_{j=0}^4 x_jc_j+ \sum_{k=0}^4 y_kc_k= x_1c_1+ x_2c_2+ x_3c_3+ x_4c_4+ y_1c_1+ y_2c_2+ y_3c_4+ y_4c_4$ which has, as dextercioby said.

 

[Ray Vickson]

Ah... I guess that makes sense if the indices are over different ranges, e.g. j=1,2,3 k=1,2,3,4. It confused me in this case because why would you use two separate indices when one is perfectly adequate. It seems simpler, more obvious and more elegant to just use the one index, given that n=4 for both. Thank you.

Yes, using one is simpler, but maybe the point of the exercise is to get you to understand the conventions better, and I think it has now succeeded.