最后, 根据 jensen 不等式: [引理 3] 对于任意的正定矩阵 r>0 ,以及所有连续可微的向量函数 \omega: 考虑jensen不等式 \left (\sum_ {i=1}^k\frac {a_i} {\sum_ {j=1}^k a_j} x_i\right)^n\le\sum_ {i=1}^k \frac {a_i} {\sum_ {j=1}^k a_j} x_i^n\\ i.e.
Jensen Ackles to return for 'Tracker' season 2
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