(Solved): The first 3 picture for question 1 and the last one of it is like a hint same like the second questi ...

Please complete the compute_cost() function below to compute the \( \operatorname{cost} J(w, b) \). Exercise 1 Complete the compute_cost below to: - Iterate over the training examples, and for each example, compute: - The prediction of the model for that example - The cost for that example \( f_{\omega b}\left(x^{(i)}\right)=w x^{(i)}+b \) \[ \operatorname{cost}^{(i)}=\left(f_{w b}-y^{(i)}\right)^{2} \] - Return the total cost over all examples \[ J(\mathbf{w}, b)=\frac{1}{2 m} \sum_{i=0}^{m-1} \operatorname{cost}^{(i)} \] - Here, \( m \) is the number of training examples and \( \sum \) is the summation operator If you get stuck, you can check out the hints presented after the cell below to help you with the implementa ]: H # UNe Cl # GRADED FUNCTION: compute_cost def compute_cost \( (x, y, w, b) \) : Computes the cost function for linear regression. Args: x (ndarray): Shape \( (m, \), Input to the model (Population of cities) y (ndarray): Shape \( (\mathrm{m}, \), Label (Actual profits for the cities) # GRADED FUNCTION: compute_cost def compute_cost \( (x, y, w, b) \) : Computes the cost function for linear regression. Args: \( x \) (ndarray): Shape \( (m, \), Input to the model (Population of cities) y (ndarray): Shape \( (m, \), Label (Actual profits for the cities) w, b (scalar): Parameters of the model. Returns total_cost (float): The cost of using \( w, b \) as the parameters for linear regression to fit the data points in \( x \) and \( y \) * number of training examples. \( m=x . \) shape \( [\theta] \) # You need to return this variable correctly total_cost \( =\theta \) #\# START CODE HERE ## H \#A\# END CODE HERE \#\#\# return tota1_cost Click for hints You can check if your implementation was correct by running the following test code: You can check if your implementation was correct by running the following test code: Please complete the compute_gradient function to: - Iterate over the training examples, and for each example, compute: - The prediction of the model for that example \[ f_{a b}\left(x^{(i)}\right)=w x^{(i)}+b \] - The gradient for the parameters \( w, b \) from that example \[ \begin{aligned} \frac{\partial J(w, b)^{(i)}}{\partial b} &=\left(f_{w, b}\left(x^{(i)}\right)-y^{(i)}\right) \\ \frac{\partial J(w, b)^{(i)}}{\partial w} &=\left(f_{w, b}\left(x^{(i)}\right)-y^{(i)}\right) x^{(i)} \end{aligned} \] - Return the total gradient update from all the examples \[ \frac{\partial J(w, b)}{\partial w}=\frac{1}{m} \sum_{i=0}^{m-1} \frac{\partial J(w, b)^{(i)}}{\partial w} \] - Here, \( m \) is the number of training examples and \( \sum \) is the summation operator If you get stuck, you can check out the hints presented after the cell below to help you with the implementation. ]: M # UNQ C2 # GRADED FUNCTION: compute_gradient def compute_gradient \( (x, y, w, b) \) : "wn Computes the gradient for linear regression Args: If you get stuck, you can check out the hints presented after the cell below to help you with the implementation. Click for hints Run the cells below to check your implementation of the compute_gradient function with two different initializations of Run the cells below to check your implementation of the compute_gradient function with two different initializations of the parameters \( w \), \( b \). H "compute and display gradient with w initialized to zeroes. initial_w = 0 initial_ \( b=0 \) tmp_dj_dw, tmp_dj_db = compute_gradient(x_train, y_train, initial_w, initial_b) print('Gradient at initial w, b (zeros):', tap_dj_dw, tmp_dj_db) coepute_gradient_test(compute_gradient) Now let's run the gradient descent algorithm implemented above on our dataset. Expected Output: Gradient at initial , b (zeros) \( -65.32884975-5.83913505154639 \) ]: M #Compute and display cost and gradient with non-zero w test_\( H=0.2 \) test_b \( =0.2 \) tmp_dj_dw, tmp_dj_db = compute_gradient (x_train, y_train, test_w, test_b) print('Gradient at test \( w, b i^{*} \), tmp_dj_dw, tmp_dj_db)