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ReLUs don't suffer from vanishing gradients, but they have their own issues. The main issue for ReLU is the dead-ReLU phenomenon, i.e. a neuron whose activation becomes zero for all samples due to ReLU. This neuron is irrecoverable, because gradients through it are always zero. Leaky ReLU and other activation functions were created to deal with this issue.


According to the update equation (below the figure), there is vector subtraction. To perform it, just draw a line from the point to the tip of $\theta$. Your new $\theta$ would be this vector, so move it to the origin as others, and draw a normal as your hyperplane. This would be the case for $\alpha=1$, i.e. $\theta-x$. Assuming a smaller $\alpha$, e.g. $0....


No because $a^2 = [(-1)(-a)]^2 = [(-1)^2(-a)^2] = (-a)^2$. For the gradient, you'd have $2(-a)\left[\frac{\partial}{\partial \theta}(-a) \right] = 2a \left[\frac{\partial}{\partial \theta} a \right]$ due to the chain rule.


Because it's better to keep the parameters close to 0, then the closer to 0 the smaller the gradient should be. See this question: Three reasons to keep the parameters small(Source: Probabilistic Deep Learning: with Python, Keras and Tensorflow Probability): Experience shows that trained NNs often have small ...


Hope the following text from Wikipedia may dissipate your concerns: A number of methods have been proposed to accelerate the sometimes slow convergence of the EM algorithm, such as those using conjugate gradient and modified Newton's methods (Newton–Raphson).[25] Also, EM can be used with constrained estimation methods. Parameter-expanded expectation ...


Another presentation, with matrix notation. Preparation: $\sigma(t)=\frac{1}{1+e^{-t}}$ has $\frac{d \ln \sigma(t)}{dt}=\sigma(-t)=1-\sigma(t)$ hence $\frac{d \sigma}{dt}=\sigma(1-\sigma)$ and hence $\frac{d \ln (1- \sigma)}{dt}=\sigma$. We use the convention in which all vectors are column vectors. Let $X$ be the data matrix whose rows are the data points $...

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