Unsupervised Learning: Difference between revisions

Unsupervised Learning (view source)

72 bytes added , 12 November 2019

5,321

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 Taking the gradient and setting it to 0 we get:<br>
 <math>
-kmo
+\nabla L(\mu, \mathbf{z}) &= \nabla \sum_{i} \Vert x^{(i)} - \mu_{z^{(i)}} \Vert ^2
-\begin{align}
+</math><br>
-\nabla L(\mu, \mathbf{z}) &= \nabla \sum_{i} \Vert x^{(i)} - \mu_{z^{(i)}} \Vert ^2\\
+<math>
-&= \nabla \sum_{j=1}{k} \sum_{i\mid z(i)=j} \Vert x^{(i)} - \mu_{z^{(i)}} \Vert ^2\\
+&= \nabla \sum_{j=1}{k} \sum_{i\mid z(i)=j} \Vert x^{(i)} - \mu_{z^{(i)}} \Vert ^2
-&= \nabla \sum_{j=1}{k} \sum_{i\mid z(i)=j} \Vert x^{(i)} - \mu_{j} \Vert ^2\\
+</math><br>
-&= \sum_{j=1}{k} \sum_{i\mid z(i)=j}  \nabla  \Vert x^{(i)} - \mu_{j} \Vert ^2\\
+<math>
-&=  \sum_{j=1}{k} \sum_{i\mid z(i)=j} \nabla \Vert x^{(i)} - \mu_{j} \Vert ^2\\
+&= \nabla \sum_{j=1}{k} \sum_{i\mid z(i)=j} \Vert x^{(i)} - \mu_{j} \Vert ^2
-&=  \sum_{j=1}{k} \sum_{i\mid z(i)=j} 2(x^{(i)} - \mu_{j})\\
+</math><br>
+<math>
+&= \sum_{j=1}{k} \sum_{i\mid z(i)=j}  \nabla  \Vert x^{(i)} - \mu_{j} \Vert ^2
+</math><br>
+<math>
+&=  \sum_{j=1}{k} \sum_{i\mid z(i)=j} \nabla \Vert x^{(i)} - \mu_{j} \Vert ^2
+</math><br>
+<math>
+&=  \sum_{j=1}{k} \sum_{i\mid z(i)=j} 2(x^{(i)} - \mu_{j})
+</math><br>
+<math>
 \implies \mu_{j} &= (\sum_{i\mid z(i)=j} x^{(i)})/(\sum_{i\mid z(i)=j} 1) \quad \forall j
-\end{align}
 </math>