Unsupervised Learning: Difference between revisions

Unsupervised Learning (view source)

385 bytes added , 14 November 2019

5,321

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 Taking the derivative with respect to <math>\beta</math>, we get:<br>
 <math>
-(\frac{1}{(-1/\beta)(\sum Q)})(-\sum Q)(-\beta^{-2}) - \frac{1}{k} = (\beta)(-\beta^{-2}) - \frac{1}{k} = \frac{-1}{\beta} - \frac{1}{k} = 0
+\sum_{j=1}^{k} [(\frac{1}{(-1/\beta)(\sum Q)})(-\sum Q)(-\beta^{-2})(\sum Q) - \frac{1}{k}]
 </math><br>
 <math>
-\implies \beta = -k
+=\sum_{j=1}^{k} [ (\beta)(-\beta^{-2})(\sum Q) - \frac{1}{k}]
 </math><br>
-Plugging in <math>\beta = -k</math> into our equation for <math>\phi_j</math> we get <math>\phi_j = \frac{1}{k}\sum_{i=1}^{m}Q^{(i)}_{(j)}</math>
+<math>
+=  \sum_{j=1}^{k} [\frac{-1}{\beta}(\sum_{i=1}^{m} Q) - \frac{1}{k}]
+</math><br>
+<math>
+=[\sum_{i=1}^{m} \frac{-1}{\beta} \sum_{j=1}^{k}P(z^{(i)} = j | x^{(i)}) - \sum_{j=1}^{k}\frac{1}{k}]
+</math><br>
+<math>
+=  [\frac{-1}{\beta}\sum_{i=1}^{m}1 - 1]
+</math><br>
+<math>
+= \frac{-1}{\beta} (m-1) = 0 \implies \frac{-m}{\beta} - 1 = 0
+</math><br>
+<math>
+\implies \beta = -m
+</math><br>
+Plugging in <math>\beta = -m</math> into our equation for <math>\phi_j</math> we get <math>\phi_j = \frac{1}{m}\sum_{i=1}^{m}Q^{(i)}_{(j)}</math>
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