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Assuming that the error function is $$J(w)$$ with one parameter $$w$$, to minimize the error, we can update the weight $$w$$ as follows:

$$w = w - \alpha \cdot \frac{dJ(w)}{dw}$$

where $$\alpha$$ is the learning rate, and $$\frac{dJ(w)}{dw}$$ is the derivative of $$J(w)$$ with respect to $$w$$. If the error function has two or more parameters, for example, a weight $$w$$ and a bias $$b$$, we can update them one by one:

$$w = w - \alpha \cdot \frac{\partial J(w,b)}{\partial w}$$

$$b = b - \alpha \cdot \frac{\partial J(w,b)}{\partial b}$$

where $$\partial$$ denotes the partial derivative.

Batch Gradient Descent Update Formula

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Machine Learning Yearning (Deeplearning.ai)

The gradient $\nabla_x f(x)$ of a scalar function $f(x_1, x_2, x_3, ..., x_n)$ is defined as the unique vector field whose dot product with any vector $v$ at each point $x$ is the directional derivative of $f$ along $v$. That is,
$ \nabla_x f(x) \cdot  v = \nabla_v f(x) $

The directional derivative in direction $v$ (a unit vector) is the slope of the function $f$ in direction $v$, namely the rate of increase of $f$ per unit of distance moved in the direction given by $v$. 

To minimize $f$, we would like to ﬁnd the direction in which $f$ decreases the fastest. We can do this using the directional derivative:
$\min_{v, v^Tv = 1} \nabla_x f(x) \cdot  v= \min_{v, v^Tv = 1} ||\nabla_x f(x)||_2 ||v||_2 \cos \theta$
where θ is the angle between $v$ and the gradient. Substituting in $||v||_2= 1$ and ignoring factors that do not depend on $v$, this simpliﬁes to $\min_{v}cos θ$.

This is minimized when $v$ points in the opposite direction as the gradient. In otherwords, the gradient points directly uphill, and the negative gradient points directly down hill. We can decrease $f$ by moving in the direction of the negative gradient.

Hence we have $x' = x-\alpha \frac{df(x)}{dx}$ where $\alpha$ is the learning rate, a positive scalar determining the size of the step.

Derivation of the Gradient Descent Formula

Batch gradient descent computes each update using the full training set, so one update can be expensive when the dataset is large. Mini-batch gradient descent reduces that cost by splitting the data into small random groups and using one group at a time to estimate the gradient. The batch size is usually larger than 1 but smaller than the full dataset. This gives more frequent updates and often makes training much faster and easier to scale.

Mini-Batch Gradient Descent

Epoch is every iteration of gradient descent through the entire training set.

Epoch in Gradient Descent

The basic idea of gradient descent with momentum is to compute an exponentially weighted average of your gradients, and then use that gradient to update your weights instead. It almost always works faster than the standard gradient descent algorithm.


Gradient Descent with Momentum

Does adding polynomial features (e.g., instead using $h\theta(x)=g(\theta0+\theta1x1+\theta2x2+\theta3x21+\theta4x1x2+\theta5x22) )$ could increase how well we can fit the training data?

Suppose you have the following training set, and fit a logistic regression classifier $h\theta(x)=g(\theta0+\theta1x1+\theta2x2)$.

Backpropagation is a systematic computational procedure for applying the chain rule to calculate gradients automatically. It operates by traversing a computational graph in a backwards direction—from the output loss back to the input parameters—multiplying matrices of partial derivatives at each step to determine how parameters affect the final output.

Backpropagation

If we only have two features, $$x_1$$ and $$x_2$$, in order to minimize the loss function, we can apply gradient descent to update $$w_1$$, $$w_2$$, and $$b$$. To compute the derivatives of $$\mathcal{L}(a, y)$$ with respect to $$w_1$$, $$w_2$$, and $$b$$, we need to compute the derivatives of $$\mathcal{L}(a, y)$$ with respect to $$a$$ and $$z$$ first. $$\mathcal{L}(a, y) = -(y \log(a) + (1 - y) \log(1 - a)) \Rightarrow$$ $$\frac{d\mathcal{L}(a, y)}{da} = -\frac{y}{a}+\frac{1-y}{1-a}$$ $$a = \sigma(z) = \frac{1}{1 + e^{-z}} \Rightarrow \frac{da}{dz} = a(1-a) \Rightarrow$$ $$\begin{aligned} \frac{d\mathcal{L}(a, y)}{dz} & = \frac{d\mathcal{L}(a, y)}{da}\frac{da}{dz} \\ & = \left(-\frac{y}{a}+\frac{1-y}{1-a}\right)(a(1-a)) = a-y \end{aligned}$$ $$\begin{aligned} \frac{d\mathcal{L}(a, y)}{dw_1} & = \frac{d\mathcal{L}(a, y)}{dz}\frac{dz}{dw_1} = (a-y)x_1 \end{aligned}$$ $$\begin{aligned} \frac{d\mathcal{L}(a, y)}{dw_2} & = \frac{d\mathcal{L}(a, y)}{dz}\frac{dz}{dw_2} = (a-y)x_2 \end{aligned}$$ $$\begin{aligned} \frac{d\mathcal{L}(a, y)}{db} & = \frac{d\mathcal{L}(a, y)}{dz}\frac{dz}{db} = (a-y) \cdot 1 = a-y \end{aligned}$$

Logistic Regression Gradient Descent Derivation

For logistic regression, the gradient is given by ∂∂θjJ(θ)=1m∑mi=1(hθ(x(i))−y(i))x(i)j. Which of these is a correct gradient descent update for logistic regression with a learning rate of α?

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