A question on a summation

  • Thread starter julypraise
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Homework Statement


The Spivak's Calculus Answer Book (3ed) states that, on page 17,

[itex]\sum_{i \neq j} (x_{i}^{2}y_{j}^2 - x_{i}y_{i}x_{j}y_{j}) = 2\sum_{i < j}(x_{i}^{2}y_{j}^2 + x_{j}^{2}y_{i}^2 - x_{i}y_{i}x_{j}y_{j})[/itex]

But as I speculate, I've got the following:

[itex]\sum_{i \neq j} (x_{i}^{2}y_{j}^2 - x_{i}y_{i}x_{j}y_{j}) = \sum_{i < j}(x_{i}^{2}y_{j}^2 + x_{j}^{2}y_{i}^2 - 2x_{i}y_{i}x_{j}y_{j})[/itex]

Could you check which is right?

Thanks.


Homework Equations





The Attempt at a Solution



[itex]\sum_{i \neq j} (x_{i}^{2}y_{j}^2 - x_{i}y_{i}x_{j}y_{j}) = \sum_{i < j}(x_{i}^{2}y_{j}^2 + x_{j}^{2}y_{i}^2 - 2x_{i}y_{i}x_{j}y_{j})[/itex]
 

Answers and Replies

  • #2
CompuChip
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I suspect that the answer in the book is wrong.
You can easily write it out manually for i, j running through {1, 2}.

It's also possible to prove using
[tex]\sum_{i \neq j} a_{ij} = \sum_{i < j} a_{ij} + \sum_{i > j} a_{ij}[/tex]
where
[tex]\sum_{i > j} a_{ij} = \sum_{j > i} a_{ji} = \sum_{i < j} a_{ji}[/tex]
 

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