Unsupervised Learning: Difference between revisions

Unsupervised Learning (view source)

30 bytes added , 12 November 2019

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