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Machine Learning: Difference between revisions

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&= \sum_k \sum_i  v_{i} \phi_k(\mathbf{x}^{(i)}) \sum_j \phi_k(\mathbf{x}^{(j)})v_{j}\\
&= \sum_k \sum_i  v_{i} \phi_k(\mathbf{x}^{(i)}) \sum_j \phi_k(\mathbf{x}^{(j)})v_{j}\\
&= \sum_k (\sum_i  v_{i} \phi_k(\mathbf{x}^{(i)}))^2\\
&= \sum_k (\sum_i  v_{i} \phi_k(\mathbf{x}^{(i)}))^2\\
&\geq 0
&\geq 0 \quad
\end{aligned}
\end{aligned}
</math>
</math>