Unsupervised Learning: Difference between revisions

Unsupervised Learning (view source)

7 bytes removed , 12 November 2019

5,321

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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 L(\mu, \mathbf{z}) = \nabla \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 \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 \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  \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 \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})
+=  \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
+\implies \mu_{j} = (\sum_{i\mid z(i)=j} x^{(i)})/(\sum_{i\mid z(i)=j} 1) \quad \forall j
 </math>