Github user tillrohrmann commented on a diff in the pull request:
https://github.com/apache/flink/pull/1397#discussion_r50383721
--- Diff: docs/libs/ml/optimization.md ---
@@ -256,6 +271,79 @@ The full list of supported prediction functions can be
found [here](#prediction-
</tbody>
</table>
+#### Effective Learning Rate ##
+
+Where:
+- $j$ is the iteration number
+- $\eta_j$ is the step size on step $j$
+- $\eta_0$ is the initial step size
+- $\lambda$ is the regularization constant
+- $k$ is the decay constant
+
+<table class="table table-bordered">
+ <thead>
+ <tr>
+ <th class="text-left" style="width: 20%">Function Name</th>
+ <th class="text-center">Description</th>
+ <th class="text-center">Function</th>
+ <th class="text-center">Called As</th>
+ </tr>
+ </thead>
+ <tbody>
+ <tr>
+ <td><strong>Default</strong></td>
+ <td>
+ <p>
+ The function default method used for determining the step
size. This is equivalent to the inverse scaling method for $\tau$=0.5. This
special case is kept as the default to maintain backwards compatibility.
+ </p>
+ </td>
+ <td class="text-center">$\eta_j = \eta_0/\sqrt{j}$</td>
+ <td class="text-center">`default`</td>
+ </tr>
+ <tr>
+ <td><strong>Constant</strong></td>
+ <td>
+ <p>
+ The step size is constant throughout the learning task.
+ </p>
+ </td>
+ <td class="text-center">$\eta_j = \eta_0$</td>
+ <td class="text-center">`constant`</td>
+ </tr>
+ <tr>
+ <td><strong>Leon Bottou's Method</strong></td>
+ <td>
+ <p>
+ This is the `'optimal'` method of sklearn. Chooses optimal
initial $t_0 = \lambda \cdot eta_0$, based on Leon Bottou's [Learning with
Large Data Sets ](http://leon.bottou.org/slides/largescale/lstut.pdf)
+ </p>
+ </td>
+ <td class="text-center">\eta_j = \frac{1}{\lambda \cdot
(\frac{1}{\lambda \cdot eta_0 } +j -1) }</td>
+ <td class="text-center">`bottou`</td>
+ </tr>
+ <tr>
+ <td><strong>Inverse Scaling</strong></td>
+ <td>
+ <p>
+ A very common method for determining the step size.
+ </p>
+ </td>
+ <td class="text-center">$\eta_j = \frac{\lambda}{j^{\tau}}$</td>
--- End diff --
Maybe don't use `frac` here but instead `\lambda/ j^\tau`, because the
exponent of `j` is rendered really small with `frac`.
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