# Tag Info

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. $0.... 4 No because$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. 0 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: https://stackoverflow.com/q/34569903/3552975 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 ... 1 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 ... 0 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$...