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Identitatea lui Lagrange

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Joseph-lagrange.jpg

Joseph-Louis Lagrange

\left  (\sum_{i=1}^n a^2_i \right ) \cdot \left  (\sum_{i=1}^n b^2_i \right )  =\left  (\sum_{i=1}^n a_i b_i \right )^2 + \sum_{1\le i < j \le n} (a_i b_j - a_j b_i )^2

unde a_i \in \mathbb C, \; \forall i \in \overline {1, n} şi b_i \in \mathbb C, \; \forall i \in \overline {1, n}.


Relaţia este un caz particular al identităţii Binet–Cauchy.

Caz particular Edit

Considerăm vectorii \vec a (a_x, a_y, a_z), \vec b (b_x, b_y, b_z). \!

Atunci:

(a_yb_z - a_zb_y)^2+ (a_zb_x-a_xb_z)^2+ (a_xb_y-a_yb_x)^2 + \!
+ (a_xb_x+ a_yb_y+a_zb_z)^2= (a_x^2+a_y^2+a_z^2)(b^2_x+b^2_y+b^2_z). \!


Identităţi similare Edit

\sum_{1 \le i<j \le n} (a_i - a_j) (b_i - b_j) = n \sum_{i=1}^n a_i b_i - \left ( \sum_{i=1}^n a_i \right ) \cdot \left ( \sum_{i=1}^n b_i \right ) .


\sum_{1 \le i<j \le n} (a_i + a_j) (b_i + b_j) = (n-2) \sum_{i=1}^n a_i b_i + \left ( \sum_{i=1}^n a_i \right ) \cdot \left ( \sum_{i=1}^n b_i \right ) .


Vezi şi Edit

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