The theorem from question $1.($e$)$ provides an upper bound on the number of steps of the Perceptron algorithm and implies that it indeed converges. In this question, we will show that the result still holds even when e is not initialized to $0.$ In other words: Given a set of training examples that are linearly separable through the origin, show that the initialization of does not impact the perceptron algorithm's ability to eventually converge. To derive the bounds for convergence, we assume the following inequalities holds: There exists such that y $(0*20)>$ y for all i $=1,...,$ n and some $7>0||04||$ All the examples are bounded $||20||<$ R$,$ i $=1,...,$ n If @ is initialized to $0,$ we can show by induction that: old$).$ A $||-||>$ ky For instance, g$(2+1).0||0||0=($g$($x$)+$ y$($x$()).||0||>($k$+1)7$ If we initialize $0$ to a general $($not necessarily $0)6(0),$ then: ol$).$ zatky $||0$o$||$ Determine the formulation of a in terms of $6$ and $0(0)$; Important: Please enter $*$ as theta" $($star$)$ and $(0)$ as theta" to$),$ and use norn$(...)$ for the vector norm $||...||.-$