only e5.1 needed ,A simple hand writing answer will be so nice,thanks a lot

P5.1 Consider the single-neuron perceptron network shown in Figure P5.1. Recall from Chapter 3 (see Eq. (3.6)) that the decision bound ary for this network is given by wp + b = 0 . Show that the decision boundary is a vector space if b 0. Inputs Sym. Hard Limit Layer 1XR /2 1x1 a-hardlims (Wp+b) Figure P5.1 Single-Neuron Perceptron To be a vector space the boundary must satisfy the ten conditions given at the beginning of this chapter. Condition 1 requires that when we add two vectors together the sum remains in the vector space. Let p, and p2 be two vectors on the decision boundary. To be on the boundary they must satisfy WP2 = 0 If we add these two equations together we find W(pi + p) = 0 Therefore the sum is also on the decision boundary Conditions 2 and 3 are clearly satisfied. Condition 4 requires that the zero vector be on the boundary. Since W0 = 0 , the zero vector is on the decision boundary. Condition 5 implies that if p is on the boundary, then -p must also be on the boundary. If p is on the boundary, then wp = 0 If we multiply both sides of this equation by -1 we find Therefore condition 5 is satisfied. 5-17 5 Signal and Weight Vector Spaces Condition 6 will be satisfied if for any p on the boundary ap is also on the boundary. This can be shown in the same way as condition 5. Just multiply both sides of the equation by a instead of by 1. Conditions 7 through 10 are clearly satisfied. Therefore the perceptron de- cision boundary is a vector space. P5.1 Consider the single-neuron perceptron network shown in Figure P5.1. Recall from Chapter 3 (see Eq. (3.6)) that the decision bound ary for this network is given by wp + b = 0 . Show that the decision boundary is a vector space if b 0. Inputs Sym. Hard Limit Layer 1XR /2 1x1 a-hardlims (Wp+b) Figure P5.1 Single-Neuron Perceptron To be a vector space the boundary must satisfy the ten conditions given at the beginning of this chapter. Condition 1 requires that when we add two vectors together the sum remains in the vector space. Let p, and p2 be two vectors on the decision boundary. To be on the boundary they must satisfy WP2 = 0 If we add these two equations together we find W(pi + p) = 0 Therefore the sum is also on the decision boundary Conditions 2 and 3 are clearly satisfied. Condition 4 requires that the zero vector be on the boundary. Since W0 = 0 , the zero vector is on the decision boundary. Condition 5 implies that if p is on the boundary, then -p must also be on the boundary. If p is on the boundary, then wp = 0 If we multiply both sides of this equation by -1 we find Therefore condition 5 is satisfied. 5-17 5 Signal and Weight Vector Spaces Condition 6 will be satisfied if for any p on the boundary ap is also on the boundary. This can be shown in the same way as condition 5. Just multiply both sides of the equation by a instead of by 1. Conditions 7 through 10 are clearly satisfied. Therefore the perceptron de- cision boundary is a vector space