Question 21: Given the update rule for Adagrad \begin{aligned} v_t &=…

Given the update rule for Adagrad

$\begin{aligned} v_t &= v_{t-1} + (\nabla w_t)^2 \\ w_{t+1} &= w_t - \frac{\eta}{\sqrt{v_t + \epsilon}} \nabla w_t \end{aligned}$

where $v_{-1}$ and $\epsilon$ are set to $0$. The gradient for the function of single variable is $1$ at any point. The effective learning rate at timestamp $t$ is $\frac{\eta}{\sqrt{v_t}}$.

$\beta$ lies between $0$ and $1$.

Note: $w_0 \to w_1$ is the first update, $w_1 \to w_2$ is the second update and so on.

What is the effective learning rate for the parameter while using Adagrad at the ninth update step (to get $w_9$) if the initial learning rate is $1$? Round your answer to 2 decimal places.