The figure below shows a Recurrent Neural Network (RNN) with one input unit z. o

Question

The figure below shows a Recurrent Neural Network (RNN) with one input unit z. one logistic hidden unit h. and one linear output unit y. The RNN is unrolled in time for T=0, 1, and 2. The network parameters are: W_zh= -0.1. W_hh = 0.5 W_hg = 0.25 h_ = 0.4 and y_ = 0.0. If the input x takes the values 18, 9, - 8 at time steps 0, 1, 2 respectively, the hidden unit values will be 0.2, 0.4, 0.8 and the output unit values will be 0.05, 0.1, 0.2 (you can check these values as an exercise). A variable z is defined as the total input to the hidden unit before the logistic nonlinearity. If we are using the squared loss, with targets t_0, t_1, t_2, then the sequence of calculations required to compute the total error E is as follows: If the target output values are t_0 =0.1, t_1 = -0. 1, t_2 = -0.2 and the squared error loss is used, what is the value of the error derivative just before the hidden unit nonlinearity at T = 1 (i.e.)? Write your answer up to at least the fourth decimal place.

Explanation / Answer

A tree is by definition a connected (hence non empty) graph GG while not cycles. Here may be a proof by induction of the quantity m m of edges that each such (finite) tree has m+1m+1 nodes. (Then in fact each tree with nn nodes has n1n1 edges). If m=0m=0 (no edges) one cannot have over eleven nodes (connectivity) and no less either (non-emptyness). within the case m>0m>0 select any edge (a,b)(a,b). If there have been a path from aa to pellet while not mistreatment that edge, then one would have a cycle in GG in conjunction with that edge. thus removing the sting (a,b)(a,b) disconnects aa from pellet within the remaining graph GG. conjointly each node is connected in GG either to aa or to pellet (every path in GG passing neither through aa nor pellet remains a path in GG). thus GG may be a disjoint union of 2 trees; say they need p,qp,q edges severally. Then m=p+q+1m=p+q+1, and by induction GG (hence GG) has (p+1)+(q+1)=p+q+2=m+1(p+1)+(q+1)=p+q+2=m+1 nodes.
  
A tree is by definition a connected (hence non empty) graph GG while not cycles. Here may be a proof by induction of the quantity m m of edges that each such (finite) tree has m+1m+1 nodes. (Then in fact each tree with nn nodes has n1n1 edges). If m=0m=0 (no edges) one cannot have over eleven nodes (connectivity) and no less either (non-emptyness). within the case m>0m>0 select any edge (a,b)(a,b). If there have been a path from aa to pellet while not mistreatment that edge, then one would have a cycle in GG in conjunction with that edge. thus removing the sting (a,b)(a,b) disconnects aa from pellet within the remaining graph GG. conjointly each node is connected in GG either to aa or to pellet (every path in GG passing neither through aa nor pellet remains a path in GG). thus GG may be a disjoint union of 2 trees; say they need p,qp,q edges severally. Then m=p+q+1m=p+q+1, and by induction GG (hence GG) has (p+1)+(q+1)=p+q+2=m+1(p+1)+(q+1)=p+q+2=m+1 nodes.

For any finite graph the total of the degrees of the nodes is double the quantity of edges. For a tree with nn nodes and n1n1 edges it follows that each one nodes cannot have degree22. in reality either there area unit a minimum of 2 nodes of degree one one (leaves) or one among degree zero 0; the latter solely happens for n=1n=1. within the proof on top of I needed to avoid mistreatment this truth en passant, as its direct proof is a couple of troublesome because the on top of proof itself.

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