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Revision difference : Lagrange's identity |
| Version 5 |
Version 4 |
| Let $R$ be a commutative ring, and let |
Let $R$ be a commutative ring, and let |
| $a_1, \ldots, a_n, b_1, \ldots, b_n$ be arbitrary elements in R. Then |
$a_1, \ldots, a_n, b_1, \ldots, b_n$ be arbitrary elements in R. Then |
| $$(\sum_{k=1}^n a_kb_k)^2 =(\sum_{k=1}^n a_k^2)(\sum_{k=1}^n b_k^2) |
$$(\sum_{k=1}^n a_kb_k)^2 =(\sum_{k=1}^n a_k^2)(\sum_{k=1}^n b_k^2) |
| - \sum_{1 \le k < i \le n} (a_kb_i -a_ib_k)^2\mbox{.}$$ |
- \sum_{1 \le k < i \le n} (a_kb_i -a_ib_k)^2\mbox{.}$$ |
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