Machine Learning: Difference between revisions

Machine Learning (view source)

55 bytes removed , 11 November 2019

5,304

edits

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 Then <br>
 <math>
-\begin{alignat}{2}
+\mathbf{v}^T \mathbf{K} \mathbf{v}= \mathbf{v}^T [\sum_j K_{ij}v_j]
-\mathbf{v}^T \mathbf{K} \mathbf{v}&= \mathbf{v}^T [\sum_j K_{ij}v_j]\\
+= \sum_i \sum_j v_{i}K_{ij}v_{j}
-&= \sum_i \sum_j v_{i}K_{ij}v_{j}\\
+= \sum_i \sum_j v_{i}\phi(\mathbf{x}^{(i)})^T\phi(\mathbf{x}^{(j)})v_{j}
-&= \sum_i \sum_j v_{i}\phi(\mathbf{x}^{(i)})^T\phi(\mathbf{x}^{(j)})v_{j}\\
+= \sum_i \sum_j v_{i} \sum_k \phi_k(\mathbf{x}^{(i)}) \phi_k(\mathbf{x}^{(j)})v_{j}
-&= \sum_i \sum_j v_{i} \sum_k \phi_k(\mathbf{x}^{(i)}) \phi_k(\mathbf{x}^{(j)})v_{j}\\
+= \sum_k \sum_i \sum_j v_{i} \phi_k(\mathbf{x}^{(i)}) \phi_k(\mathbf{x}^{(j)})v_{j}
-&= \sum_k \sum_i \sum_j v_{i} \phi_k(\mathbf{x}^{(i)}) \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)}) \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
-\end{alignat}
 </math>
 }}