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Ishaan94

  • 3 years ago

If \(\large a_1, a_2,a_3,\ldots,a_n\) are real numbers, show that \[\large\sum_{i=1}^{n}\sum_{j=1}^{n}i\cdot j\cdot \cos \left(a_i - a_j\right)\ge0\] @mukushla

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  1. mukushla
    • 3 years ago
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    \[\sum_{i=1}^{n} \sum_{j=1}^{n} ij \cos (a_{i}-a_{j}) = \sum_{i=1}^{n} \sum_{j=1}^{n} ij ( \cos a_{i} \cos a_{j} +\sin a_{i} \sin a_{j} )\\ =\sum_{i=1}^{n} i \cos a_{i} \sum_{j=1}^{n} j \cos a_{j}+\sum_{i=1}^{n} i \sin a_{i} \sum_{j=1}^{n} j \sin a_{j}\\=(\sum_{k=1}^{n} k \cos a_{k})^2+(\sum_{k=1}^{n} k \sin a_{k})^2\ge0\]

  2. Ishaan94
    • 3 years ago
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    you're really good mukushla. thanks for helping me

  3. mukushla
    • 3 years ago
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    welcome :D

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